Heat Transfer in Hydronic Systems

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8 1 y 2 0 j u l 23

JOURNAL OF DESIGN INNOVATION FOR HYDRONIC PROFESSIONALS 8 1 y 2 0 j u l 23 Caleffi North America, Inc. Caleffi North America, Inc. 38839850 SouthW. Milwaukee 54th Street Rd JOURNAL OF DESIGN INNOVATION FOR HYDRONIC PROFESSIONALS Milwaukee,Franklin, WI 53132Wisconsin 53208 T:T: 414.238.2360414.421.1000 F: F: 414.421.2878 414.238.2366

Dear Plumbing and Hydronic Professional, Dear Hydronic Professional, Is an underperforming heat emitter the result of insufficient flow nd rate,Welcome low fluid to the temperature, 2 edition of idronics undersizing, – Caleffi’s improper semi-annual placement, design journal for hydronic professionals. trapped air, or sludge deposits? It could be any or all of these. The 1st edition of idronics was released in January 2007 and distributed to over 80,000 people in North America. It focused on the topic hydraulic separation. From Anthe understanding feedback received, of it’show evi dentheat we moves attained and our goalwhat of factorsexplaining enhance the benefits (orand inhibit) proper that application movement of this modernis vitally design important technique to for those hydronic who systems. A Technical Journal design, troubleshoot, or maintain hydronic systems. Conduction, If you haven’t yet received a copy of idronics #1, you can do so by sending in the from convection,attached reader and response radiation card, all or play by registering rolls, as online do material at www.caleffi.us selections. The Caleffi Hydronic Solutions andpublication the placement will be mailed of those to you materialsfree of charge. within You can a system.The also download ability the to completeestimate journal heat as transfer a PDF file rates from ourusing Web basic site. formulas is also importantThis second when edition designing addresses systems.air and dirt in hydronic systems. Though not a new CALEFFI NORTH AMERICA, INC topic to our industry, the use of modern high-efficiency equipment demands a 3883 W. Milwaukee Rd thorough understanding of the harmful effects of air and dirt, as well as knowledge This issue of idronics discusses the fundamentals of heat transfer Milwaukee, Wisconsin 53208 USA on how to eliminate them. Doing so helps ensure the systems you design will asoperate they apply at peak to efficiency hydronic and heatingprovide long and trouble-free cooling systems.service. Tel: 414-238-2360 It describes how the rate of heat movement in these systems We trust you will find this issue of idronics a useful educational tool and a handy FAX: 414-238-2366 is referenceinfluenced for yourby flow future rate hydronic and systemtemperature designs. Wedifferences. also encourage It also you to send describesus feedback practical on this issue details of idronics that enhance using the attached heat transfer reader response in a wide card or by E-mail: [email protected] varietye-mailing of systemus at [email protected] applications. . Website: www.caleffi.us To receive future idronics issues WeSincerely, hope you enjoy this issue of idronics and encourage you to FREE, register online www.caleffi.us send us any feedback by e-mailing us at [email protected].

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INDEX

1. INTRODUCTION 2. HEAT TRANSFER FUNDAMENTALS 3. HOW FLOW AND TEMPERATURE CHANGE INFLUENCE HEAT EXCHANGE 4. OPERATING TEMPERATURE OF HYDRONIC HEATING SYSTEMS 5. HEAT EXCHANGER FUNDAMENTALS APPENDIX A: PIPING SYMBOL LEGEND APPENDIX B: THERMAL MODELS FOR FINNED-TUBE BASEBOARD APPENDIX C: THERMAL MODEL FOR A RADIANT PANEL CIRCUIT

Disclaimer: Caleffi makes no warranty that the information presented in idronics meets the mechanical, electrical or other code requirements applicable within a 10% given jurisdiction. The diagrams presented in idronics are conceptual, and do not represent complete schematics for any specific installation. Local codes may require differences in design, or safety devices relative to those shown in idronics. It is the responsibility of those adapting any information presented in idronics to Cert no. XXX-XXX-XXXX verify that such adaptations meet or exceed local code requirements. Heat Transfer in Hydronic Systems

1. INTRODUCTION: 2. HEAT TRANSFER FUNDAMENTALS

All hydronic heating and cooling systems move thermal Two fundamental concepts apply to all situations involving energy (e.g., heat) from one location to another. In heat transfer: heating applications, this movement takes place across many components, starting at a heat source and exiting 1. Heat always moves from a material at some temperature the hydronic system at the heat emitter(s). Between these to another material at a lower temperature. “end points” of the system, several intermediate heat transfer processes must occur. Examples include heat 2. The rate of heat transfer depends on the temperature moving from combustion gases to water flowing through difference between the two materials. The greater this a boiler, heat passing from one fluid to a different fluid difference, the higher the rate of heat transfer. through a heat exchanger, and heat passing from water inside a copper tube to aluminum fins attached to the To gain a better detailed understanding of how a hydronic outer surface of that tube and finally into room air. system works, heat transfer needs to be classified. A well-established scientific principal is that all heat It’s important for those who design or troubleshoot transfer processes occur through one or more of the hydronic systems to understand the physical processes following modes: by which heat moves. That knowledge can help them quickly identify factors that may constrict heat • Conduction movement. Examples of such constrictions include scale • Convection formation on a boiler’s heat exchanger, heat emitters • Thermal radiation with insufficient surface area, or a thick rug placed over a heated floor slab. CONDUCTION: Conduction heat transfer takes place within solid This issue of idronics discusses the fundamentals of heat materials. It’s the result of atomic vibrations within those transfer as they apply to hydronic heating and cooling materials. The atoms in all materials having temperatures systems. It also shows how the rate of heat movement above absolute zero (-458ºF) vibrate to some extent. The in hydronic systems is influenced by flow rate and higher the material’s temperature, the more vigorous the temperature change of water or water-based fluids. atomic vibrations.

Another concept called thermal equilibrium, which When heat is added to a solid material, the atomic applies to all hydronic heating and cooling systems, vibrations become more energetic. These vibrations will be described. The relationship between heat emitter spread out across trillions of atoms that are bonded surface area and required water temperature for a given together, moving towards material at a lower temperature. rate of heat transfer is also be examined. The property that determines how well a material transfers Heat transfer can be a complex subject from a theoretical heat by conduction is called its thermal conductivity. The standpoint. Mathematical models that predict the higher a material’s thermal conductivity value, the faster heat transfer performance of various heat emitters heat can pass through it, with all other conditions being have been developed, and some are presented in the equal. The thermal conductivity of a material is usually appendices. However, the primary objective of this issue determined by testing. The thermal conductivity of many is not to dive deeply into the mathematical theory of heat materials is listed in references such as the ASHRAE transfer. Rather, it’s to provide a solid understanding of Handbook of Fundamentals. fundamentals, along with many practical applications of those fundamentals. Such an understanding can help root Heat moving from the inner surface of a pipe to its outer out potential performance problems at the design stage, surface is an example of conduction heat transfer. Heat long before they reveal themselves as costly mistakes passing from tubing embedded in a concrete slab to within installed systems. It can also help technicians the surface of that slab is another example. The warmth quickly identify performance problems caused by heat felt on hands wrapped around a cup of hot coffee is the transfer “bottlenecks,” rather than faulty equipment. result of heat conduction through the cup.

3 Where: Figure 2-1 Q = rate of heat transfer through the material (Btu/hr) k = thermal conductivity of the material (Btu/°F•hr•ft) ∆x = thickness of the material in the direction of heat flow (ft) ∆T = temperature difference across the material (°F) A = area heat flows across (ft2)

Although the value of thermal conductivity for a given material is usually treated as a constant, it does vary slightly based on factors such as the material’s temperature, age and moisture content. The thermal conductivity of certain foam insulations increases slightly as the foam ages. This happens because the gas used to inflate the cells in the DRAFT (4-16-18) foam is slowly replaced with air. The thermal conductivity of freshly sawn wood decreases as its moisture content decreases. Similarly,The mathematical the thermal term conductivity (k/∆x) in formula of dry 2.1 soil is sometimes called a heat transfer coefficient, and can be substantiallyrepresented lower by than the symbolwet soil. (U). The The latter reciprocal is an of the heat transfer coefficient U (e.g., 1/U) is important consideration when designing earth loopsDRAFT for (4-16-18) geothermal heatcalled pump “thermal systems resistance,” that dissipate and represented significant by (R). amounts of heat into the soil, causing it to dry. The mathematical term (k/∆x) in formula 2.1 is sometimes called a heat transfer coefficient, and Formula 2-1 can therefor be modified into formula 2-2:

Source: http://stillstandingmag.com/2014/02/epiphany/ Source: The mathematicalrepresented term (k/ by∆x) the in symbolFormula (U). 2-1 The is sometimes reciprocal of the heat transfer coefficient U (e.g., 1/U) is called a heat transfer coefficient, and is represented The rate of heat transfer by conduction is directly called “thermal resistance,” and represented by (R). by the symbolFormula (U). The 2-2 reciprocal of the heat transfer proportional to both the temperature difference across coefficient U (e.g., 1/U) is called “thermal resistance,” and the material and its thermal conductivity. It is inversely is representedFormula by (R). 2-1 can therefor be modified into formula 2-2: proportional to the thickness of a material. Thus, if A one were to double the thickness of a material while Q = (∆ T ) Formula 2-1 can thereforeR be modified into Formula 2-2: maintaining the same temperature difference between its Formula 2-2 sides, the rate of heat transfer through the material would Formula 2-2: be cut in half. Where: A Q = (∆ T ) Materials such as copper, aluminum, and steel have R DRAFT (4-16-18) relatively high values of thermal conductivity, and thus, Where: Q = rate of heat transfer through the material (Btu/hr) allow high rates of heat conduction. Copper and aluminum R = thermal resistance ( or “R-value) of a material (ºF•hr•ft2/Btu) Where: in particular areare commonly commonly used used inin applications where where the the ratio Qof =the rate heat of transfer heat transfer rate by throughconduction, the per material (Btu/hr) ratio of the heat transfer rate by conduction, per pound ∆T = temperature difference across the material (°F) pound of metal used, needs to be high Other materialsR = thermal such as resistancewood, concrete, (or “R-value) and of a material2 of metal used, needs to be high. Other materials, such as (ºF•hr•ft2/Btu) A = area heat flows across (ft ) wood, concretepolyethylene and polyethylene, have lower have values lower of thermal values conductivity of relative to Qmetals. = rate Materialsof heat transfer such as through the material (Btu/hr) ∆T = temperature difference across the material (°F) R = thermal resistance2 ( or “R-value) of a material (ºF•hr•ft2/Btu) thermal conductivityexpanded relative polystyrene, to metals. and polyurethaneMaterials such foam as have A very= area low heat thermal flows conductivities, across (ft ) and are expanded polystyrene and polyurethane foam have very The R-value of a material can be determined based on its thermal conductivity (k), and its commonly used as thermal insulation to minimize conduction heat∆ transfer.T = temperature difference across the material (°F) low thermal conductivities, and are commonly used as thickness (∆X). The relationship is given as Formula 2-3. The R-value of a material can be determined2 based on thermal insulation to minimize conduction heat transfer. A = area heat flows across (ft ) its thermal conductivity (k), and its thickness (∆X). The The rate of conduction heat transfer depends on the thermal conductivity of the material, its relationship is givenFormula as 2-3: Formula 2-3. The rate of conduction heat transfer depends on the The R-value of a material can be determined based on its thermal conductivity (k), and its thickness, the area across which heat is passing, and the temperature difference across it. The thermal conductivity of the material, its thickness, the Formula 2-3: area across whichrelationship heat is between passing, these and quantities the temperature is given as Formula 2-1. thickness (∆X). The relationship is given as Formula 2-3. ∆ x difference across it. The relationship between these R = quantities is given as Formula 2-1. k Formula 2-1: Formula 2-3: Where: Formula 2-1: k = thermal conductivityWhere: of the material (Btu/°F•hr•ft) ⎛ ⎞ x = thickness of ∆thex material in the direction of heat k ∆ Rk == thermal conductivity of the material (Btu/°F•hr•ft) "Q = A⎜ ⎟(∆T ) k ⎝ ∆ x ⎠ flow (ft) ∆x = thickness of the material in the direction of heat flow (ft)

Where: Where: For example: The thermal conductivity of a specific polyurethane foam is listed as 4 k = thermal conductivity of the material (Btu/°F•hr•ft) Q = rate of heat transfer through the material (Btu/hr) k = 0.015 (Btu/°F•hr•ft). Determine the R-value of a 2 inch thick slab of this material. ∆x = thickness of the material in the direction of heat flow (ft) k = thermal conductivity of the material (Btu/°F • hr • ft) ∆x = thickness of the material in the direction of heat flow (ft) For example: The thermal conductivity of a specific polyurethane foam is listed as ∆T = temperature difference across the material (°F) Page 6" of 128" k = 0.015 (Btu/°F•hr•ft). Determine the R-value of a 2 inch thick slab of this material. A = area heat flows across (ft2)

Although the value of thermal conductivity for a given material is usually treated as a constant, it Page 6" of 128" does vary slightly based on factors such as the material’s temperature, age, and moisture content. The thermal conductivity of certain foam insulations increases slightly as the foam ages. This happens because the gas used to inﬂate the cells in the foam is slowly replaced with air. The thermal conductivity of freshly sawn wood decreases as its moisture content decreases. Similarly, the thermal conductivity of dry soil can be substantially lower than wet soil. The latter is an important consideration when designing earth loops for geothermal heat pump systems that dissipate signiﬁcant amounts of heat into the soil, causing it to dry.

Page 5" of 128" DRAFT (4-16-18)

For example: The thermal conductivity of a specific DRAFT (4-16-18) polyurethaneSolution: foam is listed as Figure 2-3 k = 0.015 (Btu/°F•hr•ft). Determine the R-value of a 2-inch-thickSolution: slab of this material. 2" extruded polystyrene (R=10.8ºF•hr•ft2/Btu) ⎛ 1ft ⎞ 8" concrete (R=0.8 ºF•hr•ft2/Btu) Solution: (2in)⎜ ⎟ 2 ∆ x ⎝ 12in⎠ ⎛ ft ⋅º F ⋅hr ⎞ 1/2" plywood (R=0.61 ºF•hr•ft2/Btu) R = = ⎛ 1ft ⎞= 11.1⎜ ⎟ k ⎛ (2iBn)tu⎜ ⎞ ⎟ ⎝ B2tu ⎠ ∆0.x015⎜ ⎝ 12i⎟n⎠ ⎛ ft ⋅º F ⋅hr ⎞ R = = ⎝ ⋅ ⋅ ⎠ = 11.1⎜ ⎟ k ft º⎛F hBrtu ⎞ ⎝ Btu ⎠ 0.015⎜ ⎟ ⎝ ft⋅º F ⋅hr ⎠ Note that it was necessary to convert the thickness of the material from inches to feet so that Note that it was necessary to convert the thickness of 2 the resultingmaterialNote that R-valuefrom it was inches necessaryis expressed to feet to convert inso traditional that the thethickness Northresulting of American the material units from of inches(ºF•hr•ft to feet/Btu). so that R-valuethe is resulting expressed R-value in traditional is expressed North in traditional American North units American units of (ºF•hr•ft2/Btu). of (ºF•hr•ft2/Btu). The R-value of many common materials can be found in references such as the ASHRAE The R-value of many common materials can be found in references such as the ASHRAE TheHandbook R-value of ofFundamentals many common. The materials R-value canof insulation be found materials is usually listed on their in referencesHandbook suchof Fundamentals as the ASHRAE. The R-value Handbook of insulation of materials is usually listed on their packaging. The table in figure 2-2 lists the R-value of some common materials on a per inch of Fundamentalspackaging.. The The R-valuetable in figure of insulation 2-2 lists the materials R-value of is some common materials on a per inch of usuallythickness listed basis. on their packaging. The table in Figure 2-2 lists thethickness R-value basis. of some common materials on a per inch of thickness basis. [insert [insertfigure figure2-2] 2-2] Figure 2-2

Total R-value of assembly = MaterialMaterial R-valueR-value per per inch inch of of 2 0.61+0.8+10.8 = 12.21 (ºF•hr•ft2/Btu) thicknessthickness (ºF•hr•ft (ºF•hr•ft2/Btu)/Btu)

Many situations in which heat is transferred by ﬁberglass insulation 3.1 ﬁberglass insulation 3.1 conduction involve multiple materials with different thermal conductivities. Heat passes from one material concrete 0.1 concrete 0.1 to another, depending on how those materials are extruded polystyrene 5.4 arranged. Materials with high thermal conductivity extruded polystyrene 5.4 create very little temperature drop as heat passes plywood 1.24 through them at a given rate (e.g., Btu/hr). Materials with plywood 1.24 lower thermal conductivity create higher temperature copper 0.0051 drops while passing heat at the same rate. The overall copper 0.0051 heat transfer rate of an assembly of materials is typically limited by the material with lower thermal conductivity. When the R-value of a one inch thick layer of a material is known the R-value of other When thicknessesthe R-value can of abe one-inch-thick found by multiplying layer the of R-valuea material per inchOne of thickness,example byof thickness. this is aFor concrete floor slab with isWhen known, the theR-value R-value of a of one other inch thicknesses thick layer canof a bematerial found is knownembedded the R-value heating of tubing. other It’s logical to assume that example, the R-value of a 2 inch thick layer of extruded polystyrene would be 2 x 5.4 = 10.4 bythicknesses multiplying can the be foundR-value by permultiplying inch of the thickness, R-value perby inchusing of thickness, a tubing bymaterial thickness. with highFor thermal conductivity thickness.(ºF•hr•ft For2/Btu), example, and the the R-value R-value of a 1/2”of a thick 2-inch-thick layer of plywood would would increase be 0.5 xthe 1.24 slab’s = 0.61 upward heat output. But by example, the R-value of a 2 inch thick layer of extruded polystyrene would be 2 x 5.4 = 10.4 layer of extruded2 polystyrene would be 2 x 5.4 = 10.8 how much? (ºF•hr•ft2 /Btu). (ºF•hr•ft2/Btu),/Btu), andand the the R-value R-value of ofa 1/2”a ½” thick thick layer layer of plywoodof would be 0.5 x 1.24 = 0.61 plywood would be 0.5 x 1.24 = 0.62 (ºF•hr•ft2/Btu). Consider a 4-inch-thick concrete slab with tubing (ºF•hr•ft2/Btu). embedded at 12-inch spacing. The slab has a layer of Page 7" of 128" When several materials are combined in layers, the total 3/8” hardwood flooring fully bonded to its surface. The R-value of the assembly is found by adding the R-values tubing options are copper or PEX, both in ½” nominal of those materials. An example is shown Pagein Figure 7" of 2-3. 128" tube size.

5 Figure 2-4 87.8 ºF

3/8" oak ﬂooring bonded to slab 70 ºF room air temperature 74.5 ºF

4" concrete slab nominal 1/2" tubing

65 ºF constant underside temperature assumed 1" extruded polystyrene unside insulation isotherms (lines of constant temperature)

Figure 2-5 35 copper tubing

25 PEX tubing

Heat output (Btu/hr/sq. ft.) 5

0 (water temp. - room air temp) (ºF)

6 The thermal conductivity of the tubing options is: Figure 2-6

Copper: k= 223 (Btu/hr•ft•ºF) PEX: k= 0.237 (Btu/hr•ft•ºF)

The thermal conductivity of the copper is 941 times greater than that of the PEX tubing. Does this imply that the heat output of the floor slab would be 941 times greater if copper tubing were used instead of PEX?

It’s a reasonable question, given the large difference in thermal conductivity. To find the answer one could build two otherwise identical slabs, one with the copper tubing and the other with PEX tubing, and perform accurate testing. An alternative is to simulate the same scenario using finite element analysis software.

Figure 2-4 shows the result of a finite element analysis of Figure 2-7 the slab described above.

The section of the floor shown spans 6 inches to the left and right of the tubing centerline. The 4-inch concrete slab (k= 0.833 Btu/ft•F•hr) rests on 1” thick extruded polystyrene (k= 0.0154 Btu/ft•F•hr). The wood flooring (k = 0.0833 Btu/ft•F•hr) is assumed to be fully bonded to the top of the slab.

The contour lines show in Figure 2-4 are called isotherms. Each one represents the locations within the material that Courtesy of Harvey Youker are at the same temperature. The farther an isotherm the slab using the PEX tubing. This example demonstrates is from the tube, (e.g., the source of heat input), the that even when some materials within an assembly have lower the temperature it represents. Heat flow is always high thermal conductivity relative to the other materials, perpendicular to any isotherm at any location. In this it’s the assembly of all materials that dictates overall heat case, the isotherms can be thought of as “waves” of heat transfer rates. flowing outward from the tube. They show the combined effect of all the materials, including their shape, relative Another example that demonstrates conduction heat position and thermal conductivity. transfer is the use of formed aluminum heat transfer plates in combination with PEX or PEX-AL-PEX tubing The red curve above the floor section is a surface in an “underside” tube & plate radiant floor construction. temperature profile. It’s a plot of the surface temperature versus position across the floor section. The floor is Figure 2-6 shows an example of ½” PEX tubing stabled warmest directly above the tube and coolest half way to the bottom of a plywood subfloor. The tubing is to the adjacent tube (assuming that any adjacent tubes spaced approximately 8 inches apart. No heat transfer contain water at the same temperature). plates were installed.

By analyzing the surface temperature profile, it’s possible Figure 2-7 shows a similar installation, but one where to estimate the rate of upward heat output from the floor. 6-inch-wide by 0.024-inch-thick pre-formed aluminum Figure 2-5 shows this upward heat output versus the plates were installed to help conduct heat away from the difference between the average water temperature in the tubing and spread it across the floor. The tube spacing is tubing and the room air temperature. The type of tubing the same nominal 8 inches. used is the only variable. Figure 2-8 shows a thermal model of these two As expected, the slab using the copper tubing has the installations produced using finite element analysis. The higher heat output, but only about 16 percent higher than

7 Figure 2-8 140 ºF water (w/ plates) 8” o.c. tubing 140 ºF water (no plates)

Floor surface temperature proﬁles 100 ºF water (w/ plates) 90 90 100 ºF water (no plates) 86 86

82 82

78 78

74 74

70 70

3/8" ceramic tile

3/4" plywood 4" ﬁberglass

70 ºF air temperature

3/8" ceramic tile

3/4" plywood 4" ﬁberglass

70 ºF air temperature only variable between the two models is the presence of Figure 2-9 6-inch-wide by 0.024-in.-thick pre-formed aluminum heat transfer plates in the upper graphic, and the absence of any plates in the lower graphic.

Analysis of the surface temperature profiles of both systems shows that, for the water temperatures simulated, the system using aluminum plates has almost 3 times more heat output than the “plateless” system, all other factors being the same.

Figure 2-9 is a thermograph that also shows the importance of aluminum heat transfer plates in conducting heat away from tubing in a radiant ceiling heating application. This image is based on thermal radiation released from the lower surface of ½” drywall installed over the tubing and plates.

8 The location of the 6-inch-wide aluminum plates, placed on tubing Figure 2-10 spaced 8 inches apart, is easily tube wall seen as the bright yellow color, boundary layer of slow moving fluid limits convection even though the plates and tubing velocity profile of fluid are covered with ½” drywall. That Note low velocity near tube wall fact that the drywall surface is significantly warmer under the plates compared to the gap between the plates indicates that the plates are effectively spreading heat, by conduction, away from "core" of flow stream the tubing. Contrast this with the “striping” seen where the tubing is present without the plates in the lower part of the image. As was the case with floor heating, the heat flows from water, high thermal conductivity of the through boundary layer, tube aluminum plates greatly enhances wall, into surrounding material lateral heat diffusion across the surface of the radiant ceiling panel. The plates make the ceiling panel a Most people have experienced the increased “wind chill” more efficient heat emitter, allowing it to deliver heat to effect as cold outside air blows past them, compared the room at a relatively low water temperature. to how they would feel if standing in still air at the same temperature. The faster the air blows past them the CONVECTION: greater the rate of convective heat transfer between their Convection heat transfer occurs as the result of fluid skin or clothing surfaces and the air stream. Although movement. The fluid can be a liquid or a gas. When the person may feel as if the moving air is colder, it isn’t. heated, fluids expand. This lowers their density relative Instead, they’re experiencing an increased rate of heat to surrounding cooler fluid. Lowered density increases loss due to enhanced convective heat transfer. Achieving buoyancy, which causes the warmer fluid to rise upward. the same cooling sensation from calm air would require a Examples of the latter include warm air rising toward much lower air temperature. the ceiling in a room, and heated water rising to the upper portion of tank-type water heater. Both processes Convective heat transfer increases with increasing fluid occur without circulators or blowers. As such, they are speed. This happens because a layer of fluid called the examples of “natural” convection. “boundary layer,” which clings to surfaces, gets thinner as the fluid’s velocity increases (see Figure 2-10). Convection heat transfer is also responsible for moving heat between a fluid and a solid. The thinner the boundary layer, the lower the thermal resistance between the bulk of the fluid stream and the For example, consider water at 100ºF flowing along a surface. Less thermal resistance allows for higher rates solid surface that has a temperature of 120ºF. The cooler of heat transfer between the fluid molecules in the bulk water molecules contacting the warmer surface absorb of the stream and the tubing wall. heat from that surface. These molecules are churned about as the water moves along. Molecules that have The relationship between convective heat transfer and absorbed heat from the surface are constantly being fluid speed past a surface can be seen in many hydronic swept away from that surface into the bulk of the water systems. For example, the faster water flows through a stream and replaced by cooler molecules. One can finned-tube baseboard heat emitter, the higher it’s heat envision this form of convective heat transfer as heat output, with all other conditions being the same. This being “scrubbed” off the surface by the flowing water. effect is evident in the heat output rating data listed by manufacturers of finned-tube baseboard. An example of The speed of the fluid moving over the surface significantly such ratings, using numbers from a specific make and affects the rate of convective heat transfer. model of baseboard, is shown in Figure 2-11.

9 DRAFT (4-16-18) Figure 2-11

ﬂow rate heat output (Btu/hr/ft) at stated water temperature (ºF) (gpm) ﬂow rate 110 ºF 120 ºF 130 ºF 140 ºF 150 ºF 160 ºF 170 ºF 180 ºF (gpm) 1 gpm 150 200 260 310 370 430 490 550

4 gpm 160 210 270 330 390 450 520 580

At a given water temperature, the heat output at a flow heat transfer is determined by the multiplication of flow rate of 4 gallons per minute (gpm) will always be slightly rate times temperature drop. This is why the rate of higher than at a flow rate of 1 gpm. This increase is due heat transfer in Figure 2-12 increases as the flow rate toAt a athinner given boundary water temperature, layer between the bulkheat of output the fluid at aincreases, flow rate even of 4 though gallons the per temperature minute (gpm)drop across will the stream and the inner tube wall of the finned-tube element. heat emitter is decreasing. This will be discussed more always be slightly higher than at a flow rate of 1 gpm.in section This 3. increase is due to a thinner One way to rationalize the increase in heat output with increasingboundary flow layer rate between is to consider the bulkwhat ofhappens the fluid to thestream From and the the standpoint inner tube of heat wall transfer of the only, finned-tube there is no such averageelement. water temperature inside the heat emitter with thing as the water moving “too fast” through a finned- changes in flow rate. Figure 2-12 shows an example. tube element. This is also true for other heat emitters, such as the coil of an air handler, or a tubing circuit in a As flow rate increases, the temperature drop along floor heating system. The faster the flow rate, the higher theOne finned-tube way to rationalize element decreases. the increase Since in heatthe inletoutput the with rate increa of heatsing transfer. flow rate is to consider what temperature remains constant, decreasing temperature drophappens implies to a thehigher average average water water temperaturetemperature in insidethe Another the heat example emitter of withincreasing changes convective in flow heat rate. transfer finnedFigure tube. 2-12 Higher shows average an example. water temperature, in any with increasing flow rate can be found in heat output heat emitter, results in higher heat output. ratings for fan-coils. Figure 2-13 shows an example of how heat output increases with increasing water Never “judge” the rate of heat transfer from a heat flow rate through the coil of the air handler, while the emitter[insert based figure solely 2-12] on temperature drop. The rate of inlet water temperature and incoming air temperature remain constant.

Figure 2-12 The heat output rate Btu increases rapidly Btu = 488(gpm)(∆ T ) = 488(1)(34.8) = 16,980 hr = 488(gpm)(∆ T ) = 488(1)(34.8) = 16,980 at low flow rates. It Tin = 180 ºF hr Tout = 145.2 ºF Tin = 180 ºF Tout = 145.2 ºF continues to increase flow rate = 1 gpm flow rate = 1 gpm as flow rate increases, flow rate = 1 gpm average water temperature = 162.6 ºF flow rate = 1 gpm average water temperature = 162.6 ºF but the rate of increase is much slower at higher flows. Btu Btu = 488(gpm)(∆ T ) = 488(4)(11.3) = 22,060 hr = 488(gpm)(∆ T ) = 488(4)(11.3) = 22,060 Tin = 180 ºF hr Tout = 168.7 ºF This demonstrates Tin = 180 ºF Tout = 168.7 ºF flow rate = 4 gpm flow rate = 4 gpm that there are practical flow rate = 4 gpm average water temperature = 174.4 ºF flow rate = 4 gpm limits on how much average water temperature = 174.4 ºF the heat output of a hydronic heat emitter Btu = 488(gpm)(∆ T ) = 488(8)(5.9) = 23,030 can be increased Bhtru based on increasing Tin = 180 ºF = 488(gpm)(∆ T ) = 488(8)(5.9) = 23,030 Tout = 174.2 ºF T = 180 ºF hr T = 174.2 ºF flow rates. flowin rate = 8 gpm flowout rate = 8 gpm average water temperature = 177.1 ºF flow rate = 8 gpm flow rate = 8 gpm average water temperature = 177.1 ºF

Page "18 of "128 Heat output also increases as the air flow rate through the Figure 2-13 air handler’s coil increases. This happens because faster 4500 air flow reduces the resistance of the boundary layer between the bulk air stream and the surface of the coil. 4000 The “non-linear” relationship between heat output and 3500 flow rate seen in Figure 2-13 is a characteristic of 3000 all hydronic heat emitters. Figure 2-14 shows the relationship for the upward heat output of a 250-foot- 2500 long circuit of 1/2-inch PEX tubing embedded at 12-inch 2000 spacing in a 4-inch bare concrete slab. The supply water temperature to this circuit is constant at 110ºF. The only 1500 thing being varied is flow rate. Heat output (Btu/hr)

1000 As was true with the fan-coil, the gains in heat output are 500 much more noticeable at lower flow rates. At 0.2 gallons per minute, which is only 10% of the maximum flow rate 0 shown on the graph, the circuit releases about 44 percent 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 of the maximum heat output. Increasing flow from 1 to 2 water ﬂow rate (gpm) gpm only increases heat output about 11 percent.

The head loss created by fluid moving through the heat NATURAL VS. FORCED CONVECTION emitter, in theory, increases with the cube of flow rate (e.g., When fluid motion is caused by a circulator, a blower or if the flow rate is doubled, the head loss increases 23 = (8) any other powered device, the resulting convective heat times). Thus, attempting marginal gains in heat transfer by transfer is called “forced convection.” When the fluid selecting circulators that can maintain high flow rates will motion is strictly the result of buoyancy differences within significantly increase the installation and operating cost the fluid, the resulting convective heat transfer is called of the circulator. Velocity noise and erosion corrosion also “natural convection.” become concerns at high flow rates. To avoid erosion corrosion, the flow velocity in smaller copper tubes should The room side heat output of hydronic heat emitters not exceed 5 feet per second. such as finned-tube baseboard and wall cabinet heaters with internal coils (but no blower) is largely the result of natural convection. Such heat emitters are appropriately Figure 2-14 called “convectors.”

However, when a circulator is used to move water 8000 through these heat emitters, the convection occurring between the water and the inner surfaces of the heat 7000 emitter is forced convection. 6000 Heat emitters that use fans or blowers to force air 5000 through a heat exchanger are usually called fan-coils or air handlers. The convective heat transfer that occurs 4000 on the room (air) side of these heat emitters is another example of forced convection. 3000 Another situation in which both natural and forced 2000 Upward heat output (Btu/hr) Upward convection are present is a heat exchanger coil immersed in a thermal storage tank, as shown in Figure 2-15. 1000

0 Natural convection is usually a “weaker” form of heat 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 transfer in comparison to forced convection. It’s the result of slower fluid motion created by buoyancy Circuit flow rate (gpm)

11 speed of the fluid, and the physical properties of the Figure 2-15 fluid (e.g. thermal conductivity, specific heat, density and viscosity). In many cases, the value of (h) needs to be determined experimentally. Heat transfer textbooks circulator sometimes list values of h, or methods for determining (h) for very specific and often simplifiedDRAFT situations. (4-16-18) The table in Figure 2-16 shows how wide the range of (h) values heat can be for specific situations. [Figure 2-16] source FigureDRAFT 2-16 (4-16-18) situation range of convection coefficient (h) (Btu/hr•ft2•ºF) Natural convection is usually a “weaker” form of heat transfer in comparison to forced natural convection 1-5 forcedconvection. convection It’s the result of slower fluid motioninvolving created air by buoyancy differences versus much (inside surface of coil) faster motion created by a circulator. The slower fluid motion increases the thickness of the natural convection forced convection 2-100 involving air (ousideboundary surface of layer, coil) which creates greater resistance to heat transfer between the bulk of the fluid and the surface. convection involving 20-3000 differences versus much faster motion created by a water circulator. The slower fluid motion increases the thickness convection involving 500-5000 of the boundary layer,Forced which convection creates greater versus resistance natural toconvection boilingexplains water why a small wall-mounted fan coil that's heat transfer betweenonly the 18 bulk inches of the wide fluid canand providethe surface. the equivalent heat output of 10+ feet of finned-tube baseboard

Forced convectionwhen versus both natural are operating convection at theexplains same waterTHERMAL supply RADIATION: temperature and flow rate. The rate of heat why a small wall-mounted fan coil that’s only 18 inches Thermal radiation is probably the least understood form THERMAL RADIATION: wide can provide thetransfer equivalent from theheat finned-tube output of 10+ element feet inof theheat baseboard transfer. To is many, limite dthe by word natural “radiation” convection denotes heat of finned-tube baseboard when both are operating at an undesirable or harmful effect associated with nuclear transfer between its outer surfaces andradiation. the surrounding However thermalair. radiation and nuclear radiation the same water supply temperature and flow rate. The Thermal radiation is probably the least understood form of heat transfer. To many, the word are very different. Human skin releases thermal radiation rate of heat transfer from the finned-tube element in the “radiation” denotes an undesirable or harmful effect associated with nuclear radiation. However baseboard is limited by natural convection heat transfer similar to that emitted by a low-temperature hydronic The rate of convective heat transfer can be estimated using formula 2-4. between its outer surfaces and the surrounding air. radiantthermal panel.radiation A andlightly nuclear clothed radiant human are very body different. often releases Human skin releases thermal radiation over half of its metabolic heat production as thermal similar to that emitted by a low temperature hydronic radiant panel. A lightly clothed human radiation. It’s an entirely natural and harmless process. The rate of convective heat transfer can be estimated body releases over half of its metabolic heat production as thermal radiation. It’s an entirely using Formula 2-4.Formula 2-4: Thinknatural of and thermal harmless radiation process. as light. More specifically, the Formula 2-4: thermal radiation emitted by components such as a panel radiator in a hydronic heating system is infrared light. Q = hA(∆ T ) Think of thermal radiation as light. More specifically, the thermal radiation emitted by Humancomponents eyes such cannot as a panel see radiatorinfrared in alight. hydronic However, heating systemjust is infrared light. Where: like visible light, infrared light (a.k.a. thermal radiation) Where: travels outward from its source in straight lines at the Q = rate of heat transfer by convection (Btu/hr) Human eye’s cannot see infrared light. However, just like visible light, infrared light (a.k.a. h = convection coefficient (Btu/hr•ft2•ºF) speed of light (186,000 miles per second). Although it thermal radiation) travels outward from its source in straight lines at the speed of light (186,000 A = area over which fluid contacts a surface with which it cannot “bend” around corners, thermal radiant can be Q = rate of heat transfer by convection (Btu/hr) exchanges heat (ft2) reflectedmiles per second).by some Although surfaces. it cannot It also “bend” travels around equally corners, well thermalin radiant can be reflected any direction, from some surface that emits it to another ∆T = temperature hdifference = convection between coefficient bulk fluid (Btu/hr•ft stream2 •ºF)by some surfaces. It also travels equally well in any direction, from some surface that emits it to and surface (ºF) surface that absorbs it. This is why a heated ceiling can A = area over which fluid contacts a surfacewarmanother thewith surface objects which that absorbsandit exchanges floor it. inThis the isheat whyroom (ft a 2heatedbelow.) ceiling Just canas warm the objects and floor in a ceiling-mounted lighting fixture can shine visible light Although Formula ∆2-4T = istemperature relatively simple, difference determining between bulk fluid stream and surface (ºF) the value of the convection coefficient (h) is often onto objects below, thermal radiation can shine down complex. The value of (h) can vary widely depending on from a warm ceiling and be absorbed by objects below. the “geometry” of the surface relative to the fluid, the Page "24 of "128 Although formula 2-4 is relatively simple, determining the value of the convection coefficient (h) is often complex. The value of (h) can vary widely depending on the “geometry” of the surface relative to the fluid, the speed of the fluid, and the physical properties of the fluid (e.g. thermal 12 conductivity, specific heat, density, and viscosity). In many cases the value of (h) needs to be determined experimentally. Heat transfer textbooks sometimes list values of h, or methods for determining (h) for very specific and often simplified situations. The table in figure 2-16 shows how wide the range of (h) values can be for specific situations.

Page "23 of "128 The instant thermal radiation is absorbed by the surface Consider a person standing a few feet away from a of an object, it becomes heat that warms that object. campfire on a cold day. If pointed toward the fire, their face probably feels warm, even though the air Consider the crucible of molten metal in a dark room surrounding them is cold. This sensation is the result of shown in Figure 2-17. thermal radiation emitted by the fire traveling through the cold air and being absorbed by their exposed skin. Figure 2-17 The air between the fire and the person absorbs very little of the energy being transferred from the flames to the person’s face. Likewise, thermal radiation emitted from the warm surface of a heat emitter such as a panel radiator can pass through the air in a room without first heating that air. When the thermal radiation strikes another surface in the room, most of it is absorbed. At that instant, the energy carried by the thermal radiation becomes heat.

Source: matrix.edu.au Source: Every surface continually emits thermal radiation to any cooler surface within sight of it. The surface of a heat The molten metal is likely in the temperature range of emitter that is warmer than our skin or clothing surfaces 2,500ºF. At that temperature, it emits both visible light (since transfers heat to us by thermal radiation. Likewise, our it’s obviously visible in an otherwise dark room) and thermal skin and clothing continually give off thermal radiation to radiation. The latter would be immediately detected by any any surrounding surfaces at lower temperatures. A lightly skin surface within a few feet of the crucible. clothed person standing next to a large cold window surface emits significant thermal radiation to that cold If the metal were allowed to cool, the color of visible surface. This eventually leads to discomfort, even when light it emits would change from the bright yellowish/ the air temperature surrounding the person is in the white seen in Figure 2-17 to deeper and deeper shades normal comfort range of 68-72ºF. of red. The intensity of the visible light (e.g., its perceived brightness) would also decrease. Figure 2-18

When the surface of the metal dropped below approximately 970ºF, it would no longer emit visible light. If the surrounding space was dark, the human eye could no longer detect the presence of the metal by sight. However, any exposed skin within a few feet of the cooling metal would confirm that it’s still emitting profuse amounts of heat. That sensation is caused by thermal radiation emitted by the surface of metal being absorbed by the surface of the exposed skin.

The initial situation where the molten metal was emitting both visible light and thermal radiation changed to one where only thermal radiation was emitted. The change was caused by a shift in the wavelengths of the radiation When thermal radiation strikes an opaque surface, part emitted from the surface of the metal. of it is absorbed as heat and part is reflected away from the surface. The percentage of incoming radiation that Unlike conduction or convection, thermal radiation needs is absorbed or reflected is determined by the optical no material (e.g., a fluid or solid) to transfer heat from characteristics of the surface and the wavelength of one location to another. Thermal radiation cannot pass the radiation. Most interior building surfaces absorb the through a solid (as radiation). It can, to differing extents, majority of thermal radiation that strikes them. The small pass through gases with minimal warming effect on those percentage that is reflected typically strikes another gases. Intense sunlight passing through very cold air in surface within the room where most of it will be absorbed, the upper layers of the earth’s atmosphere is an example and so on. Very little, if any, thermal radiation emitted by of the latter. warm surfaces in a room escapes from the room.

13 Although the human eye cannot see thermal radiation, Figure 2-20 there are devices that can detect it, and display an image that uses colors to represent different surface temperatures. Such an image is called an infrared thermograph. Figure 2-19 shows an example of such an image for a hot wood stove.

Figure 2-19

The rate thermal radiation transfers heat between two surfaces depends upon their temperatures, an optical property of each surface called emissivity, as well as the angle and distance between the surfaces. DRAFT (4-16-18) The rate of heat exchange by radiation between two flat surfaces having the same area can be estimated using The thermal radiation coming from the wood stove is Formula 2-5. captured by the imaging system in a special camera and converted to visible colors. The correspondence between Formula 2-5: these colors and the surface temperatures is shown on 4 4 the scale at the right of the image. This image shows that sAF12 (T1 − T2 ) Q = the sides and top of the stove are hot (around 300ºF), ⎡ 1 1 ⎤ ⎢ + −1⎥ while some of the materials behind the stove are at a e e normal room temperature of about 70ºF. ⎣ 1 2 ⎦ Where: Objects don’t necessarily have to be “hot” to emit Q = rate of heat transfer from hotter to cooler surface by thermal radiation. The image in Figure 2-20 is an infrared thermal radiationWhere: (Btu/hr) thermograph of a small heated building on a cold winter s = Stefan Boltzmann constant = 0.1714x10-8 Btu/hr•ft2•ºR4 night. The temperature scale shows that the hottest Q = rate of heat transfer from hotter to cooler surface by thermal radiation (Btu/hr) F12 = shape factor between the two surfaces (unitless) surface temperature detected is about 20ºF. 2 A = area of eithers surface= Stefan (ft Boltzmann) constant = 0.1714x10-8 Btu/hr•ft2•ºR4 T1 = absolute temperature of the hotter surface (ºR) A close look at Figure 2-20 shows the location of the F12 = shape factor between the two surfaces (unitless) T2 = absolute temperature of the cooler surface (ºR) wooden studs in the exterior walls. They show up because 2 e1 = emissivity ofA the = area hotter of eithersurface surface (unitless) (ft ) the exterior surface at the stud locations is slightly e2 = emissivity of the cooler surface (unitless) warmer than the exterior surface between the studs. The T1 = absolute temperature of the hotter surface (ºR) wooden studs have a higher thermal conductivity than This formula is moreT2 =complex absolute that temperature those used of for the estimating cooler surface (ºR) the insulation material fitted between the studs, and thus conduction and convection heat transfer. create higher temperatures at the exterior surface of the e1 = emissivity of the hotter surface (unitless) building. The window areas show the highest exterior 2 The value of the e“shape = emissivity factor” of(F 12the) is cooler a number surface between (unitless) surface temperatures because they have lower thermal 0 and 1.0. It’s determined based on the relative angle and resistance than the insulated walls. The relatively dark distance between the two surfaces exchanging radiant heat. color of the roof indicates a relatively cool surface. This Heat transfer textbooksThis formula give specificis more complexmethods thatfor thosefinding used for estimating conduction and convection heat suggests that the roof structure is well insulated, and thus values of the shapetransfer. factor (F12) for different surfaces and there is minimal heat transfer from the interior space to orientations. For two parallel planes having infinite width and the roof surface. depth, the value of the shape factor (F12) is 1.0. The value of the “shape factor” (F12) is a number between 0 and 1.0. It’s determined based on the relative angle and distance between the two surfaces exchanging radiant heat. Heat transfer

14 textbooks give specific methods for finding values of the shape factor (F12) for different surfaces and orientations. For two parallel planes having infinite width and depth the value of the shape

factor (F12) is 1.0.

e1 and e2 are the emissivities of surfaces 1 and 2. Emissivity is a surface property determined experimentally based on how well the surface emits thermal radiation. It must be a number between 0 and 1. A high value indicates that the surface is a good emitter, and vice versa. Emissivity values for various surfaces can be found in references such as heat transfer handbooks. Interestingly, the emissivity of a surface is not necessarily correlated with its color. A rough metal surface coated with white enamel paint has an emissivity of 0.91, and a ﬂat black painted surface has an emissivity of 0.97. Freshly fallen snow can have an emissivity over 0.90.

Page "30 of "128 DRAFT (4-16-18)

surface combined radiative / DRAFTconvective (4-16-18) coefficient (m) (Btu/hr/ft2/ºF) e1 and e2 are the emissivities of surfaces 1 and 2. Emissivity Figure 2-21 is a surface property determined experimentally based on radiantDRAFT floor (4-16-18) 2.0 how well the surface emits thermal radiation. It must be surface combined radiative / a number between 0 and 1. A high value indicates that radiantconvective wall coefficient (m) 1.8 the surface is a good emitter, and vice versa. Emissivity surface combined radiative / convective coefficient(Btu/hr/ft (m) 2/ºF) values for various surfaces can be found in references radiant(Btu/hr/ft2 /ºF)ceiling 1.6 such as heat transfer handbooks. Interestingly, the radiantradiant floor floor 2.0 2.0 emissivity of a surface is not necessarily correlated with its color. A rough metal surface coated with white enamel radiantradiant wall wall 1.8 1.8 paint has an emissivity of 0.91, and a flat black painted radiant ceiling 1.6 surface has an emissivity of 0.97. Freshly fallen snow can radiant ceiling 1.6 have an emissivity over 0.90. The emissivity of a polished These combined heat transfer coefficients greatly simply estimates of heat transfer in common copper surface is 0.023, while a heavily oxidized copper These combined heat(but transfer limited) coefficients situations. greatly These simplify coefficients can be used in formula 2-5. surface has an emissivity of 0.78. Most highly polished theThese estimating combined heat of transferheat transfercoefficients in greatly common simply estimat(but limited)es of heat transfer in common metal surfaces have low emissivities, and thus would situations.(but limited) situations. These coefficients These coefficients can canbe beused used in in Formula formula 2-5. 2-6. not be good choices for the surface of a hydronic heat These combined heatFormula transfer 2-5: coefficients greatly simply estimates of heat transfer in common emitter that’s expected to radiate heat into a room. FormulaFormula(but limited) 2-5: 2-6: situations. These coefficients can be used in formula 2-5. It’s also important to understand that the temperatures Q = mA(∆ T ) Q = mA(∆ T ) (T1) and (T2) in Formula 2-5 must be absolute temperatures. Temperatures in ºF can be converted to Where:Formula 2-5: absolute temperatures in degrees Rankine (ºR) by adding QWhere: = rate of heat transfer from both thermal radiation and Where: 458 degrees. Thus, 32ºF becomes 32 + 458 = 490ºR. convectionQ = rate of heat (Btu/hr) transfer from both thermal radiation and convection (Btu/hr) Q = mA(∆ T ) 2 mm == combined combined heat heattransferQ transfer =coefficient rate coefficientof (Btu/hr/ft heat 2transfer/ºF) (Btu/hr/ft from/ºF) both thermal radiation and convection (Btu/hr) The mathematical result of the calculation (T 4-T 4) 2 1 2 AA == area area of ofthe thepanel panel releasing releasing heat to the heat room to (ft the2) room (ft ) changes much more than the simple ∆T term used in the m = combined heat transfer coefficient (Btu/hr/ft2/ºF) ∆∆T == difference difference between between the average the temperatureaverage temperature of the surface releasing of the heat and room air formulas for conduction and convection. For example, surfaceWhere: releasing heat and room air temperature (ºF). consider two surfaces exchanging radiant heat with temperature ( ºF). A = area of the panel releasing heat to the room (ft2) Q = rate of heat transfer from both thermal radiation and convection (Btu/hr) temperatures of 100ºF and 80ºF. These temperatures For example: If the∆ Taverage = difference surface between temperature the average of a temperature of the surface releasing heat and room air would convert to 558ºR and 538ºR. The difference between 2 heatedForm =example: combined floor If wasthe averageheat 80ºF, transfer surfacethe floor temperature coefficient area was of a 100 (Btu/hr/ftheated ft floor, and2 was/ºF) the 80 ºF, the floor area was these temperatures would be only 20ºR, the same as room100 ft2 ,air and temperature the room air temperature was 70ºF, was the 70( ºF).estimatedºF. The estimated heat heat output output from the floor to the the difference between 100ºF and 80ºF. However, when 2 fromroomA = area wouldthe floor be:of the to panelthe room releasing would be:heat to the room (ft ) these Rankine temperatures are used in Formula 2-5 the 4 4 4 ∆T = difference between the average temperature of the surface releasing heat and room air resulting number for the term (T1 -T2 ) is 131,700,000ºR . For example: If the average surface temperature of a heated floor was 80 ºF, the floor area was temperature ⎛( ºF). Btu ⎞ Btu = = 2 2 − = Due to mathematical complexities, as well as variability Q mA(∆ T ) ⎜ 2.0 1002 ft ⎟, (and100 ft the)(80 ºroomF 70 ºairF) temperature2,000 was 70 ºF. The estimated heat output from the floor to the ⎝ (hr ⋅ ft ⋅º F)⎠ hr or uncertainty in properties such as surface emissivities, theoretical calculations of radiant heat transfer are often room would be: For example: If the average surface temperature of a heated floor was 80 ºF, the floor area was limited to relatively simple situations. Keep in mind that this is an estimate based on a typical 2 rangeKeep100 inft mindof, and operating that the this roomis an conditions estimate air temperature based and on a assumedtypical was range 70 commonofºF. operating The estimatedconditions and heat output from the floor to the Other than situations where heat is exchanged in a buildingassumedroom would common materials. be: building It materials.is also basedIt is also basedon⎛ average on averageB tsurfaceu surface⎞ temperature rather Btu vacuum, radiant heat transfer occurs in combination temperature rather thanQ =them waterA(∆ T temperature) = 2.0 circulating 100 ft 2 (80º F − 70º F) = 2,000 with convective heat transfer, since air is a fluid and is ⎜ 2 ⎟ ( ) through tubing in the radiant surface.Page "32 The ⎝of " 128average(hr ⋅ f tsurface⋅º F)⎠ hr present between the objects exchanging heat. To simplify temperature of the panel, as well as the relationship estimating overall heat transfer rates, some references between it and the⎛ water temperatureBtu ⎞ in the tubing, are Btu provide hybrid heat transfer coefficients that combine the highly= influenced= by the material and “geometry”2 of how− = Q mA(∆ T ) ⎜ 2.0 2 ⎟ (100 ft )(80º F 70º F) 2,000 effects of radiant and convective heat transfer for typical the radiant panel ⎝is constructed.(hr ⋅ ft ⋅º F Other)⎠ references can hr situations where the variation in conditions is limited. be used to estimate the room side heat output of radiant Keep in mind that this is an estimate based on a typical range of operating conditions and An example is a hybrid heat transfer coefficient (m) for floor, wall and ceiling panels based on specific panel a heated floor, wall or ceiling panel that embodies the construction and theassumed average common water temperature building materials. within It is also based on average surface temperature rather effect of both radiant and convective heat output. Figure the embedded tubing. 2-21 lists some values of hybrid heat transfer coefficients that embody both convective and radiant effects for Keep in mind that this is an estimate based on a typical rangePage of operating"32 of "128 conditions and commonly constructed interior building surfaces. assumed common building materials. It is also based on average surface temperature rather

Page "32 of "128 15 DRAFT (4-16-18)

3. HOW FLOW AND TEMPERATURE CHANGE INFLUENCE HEAT EXCHANGE

All hydronic systems transport heat using streams of water. The water absorbs heat at a heat source, carries it through piping, fittings, and other components, and eventually releases it to the space to be heated through one or more heat emitters. The flow rate of the water as well as the 3. HOW FLOW AND TEMPERATURE CHANGE Formula 3-2 is technically only valid for cold water INFLUENCE HEATtemperature EXCHANGE change it undergoes duringbecause the heat the “delivery”factor 500 pro iscess based affect on thethe ratedensity of heat and specific heat of water at 60ºF. However, because the All hydronic systemstransfer. transport This heat section using describes streams theof relationshipfactor 500 isbetween easy to these remember, factors, Formula presents 3-2 methods is often for water. The water absorbsestimating heat rates at a ofheat heat source, transfer, carries and cautionsused for quick against mental certain calculations assumptions of the thatrate areof sensible sometimes it through piping, fittings and other components, and heat transfer involving water. While Formula 3-2 is fine for eventually releases madeit to the about space the to fundamental be heated through relationship initial between estimates, flow, Formula temperature 3-1 is more change, accurate, and especially rates of heat one or more heat emitters. The flow rate of the water, as for higher water temperatures, because it accounts for transfer. well as the temperature change it undergoes during the variations in both the density and specific heat of the heat “delivery” process, affect the rate of heat transfer. This fluid. Formula 3-1 can also be used for liquids (such as section describes the relationship between these factors, antifreeze solutions) that have different densities and presents methods forSENSIBLE estimating HEATrates of RATE heat transfer, EQUATION and specific heats relative to water. cautions against certain assumptions that are sometimes made about the fundamental relationship between flow, Example: Water flows into a radiant panel circuit at 110°F temperature changeHydronic and rates system of heat designerstransfer. often needand to leavesknow theat 92°F. rate The of heat flow transferrate is measured to or from as 1.5a fluid gpm, flowing as shown in Figure 3-1. Calculate the rate of heat transfer SENSIBLE HEAT RATEthrough EQUATION a device such as a heat sourcefrom or theheat water emitter. to the This heat can emitter be determined using Formula using 3-1: the Hydronic system designerssensible often heat need rate toformula, know the given rate ofas formula 3-1 heat transfer to or from a fluid flowing through a device such as a heat source or heat emitter. This can be determined Figure 3-1 using the sensible heat rate formula, given as Formula 3-1. Formula 3-1: 92 ºF Formula 3-1: 110 ºF 1.5 gpm q = (8.01Dc) f (∆ T ) where: Q = rate of heat transfer into or out of the water stream Solution: To use Formula 3-1, the density of water at its (Btu/hr) where: averageDRAFT temperature (4-16-18) of 101ºF must first be estimated 8.01 = a constant based on the units used using Figure 3-2: D = density of the fluid (lb/ft3) c = specific heat ofWhenQ the = ratefluid using of(Btu/lb/ºF) heat Formula transfer 3-1, into the ordensity out of and the specificwater stream heat of (Btu/hr) the wat er (or water-based anti-freeze Figure 3-2 f = flow rate of fluidsolution) through shouldthe device be based(gpm) on the average temperature of the liquid during the process by which ∆T = temperature change8.01 =of a the constant fluid through based the on device the units(°F) used the fluid is gaining or losing heat. D = density of the fluid (lb/ft3) When using Formula 3-1, the density and specific heat of the water (or water-basedc = specific antifreeze heat ofsolution) the fluid should (Btu/lb/ºF) be based on the averageFor temperature cold water onlyof the, Formulaliquid during 3-1 thesimplifies to Formula 3-2: process by which thef = fluidflow israte gaining of fluid or losingthrough heat. the device (gpm) ∆T = temperature change of the fluid through the device (°F) For cold water onlyFormula, Formula 3-23-1 simplifies to Formula 3-2: Formula 3-2 q = 500 f (∆ T ) Page "34 of "128 where: Q = rate of heat transferwhere: into or out of the water stream (Btu/hr) f = flow rate of water through the device (gpm) 500 = constant rounded off from 8.33 x 60 ∆T = temperature changeQ = rate of ofthe heat water transfer through into the ordevice out of the water stream (Btu/hr) (°F) f = flow rate of water through the device (gpm) 500 = constant rounded off from 8.33 x 60 ∆T = temperature change of the water through the device (°F) 16

Formula 3-2 is technically only valid for cold water because the factor 500 is based on the density and specific heat of water at 60 ºF. However, because the factor 500 is easy to remember, formula 3-2 is often used for quick mental calculations of the rate of sensible heat transfer involving water. While formula 3-2 fine for initial estimates, formula 3-1 is more accurate, especially for higher water temperatures because it accounts for variations in both the density and specific heat of the fluid. Formula 3-1 can also be used for liquids (such anti-freeze solutions) that have different densities and specific heats relative to water.

Example: Water ﬂows into a radiant panel circuit at 110 °F, and leaves at 92 °F. The ﬂow rate is measured as 1.5 gpm, as shown in Figure 3-1. Calculate the rate of heat transfer from the water to the heat emitter using formula 3-1:

[insert ﬁgure 3-1]

Page "35 of "128 DRAFT (4-16-18)

corroded), and the condition of the external surfaces of the heat emitters (e.g., are they bent or covered with dust, pet hair, etc.).

Formula 3-1 should be used as a “what is required” rather than a “what will be” design tool.

For example, what would the temperature drop of a water stream ﬂowing 0.5 gallons per minute have to be to deliver 50,000 Btu/hr to a load using water at an average temperature of 101 ºF?

DRAFTDRAFT (4-16-18) (4-16-18) Mathematically this is simple. Just put the given numbers into formula 3-1 and calculate. 3 D = 61.96 3lb/ft3 D D= =61.96 61.96 lb/ft lb/ft Btu q 50,000 ∆ T = = hr = 201.5º F The specific heat of water can be assumed to remain 1.0 ⎛ 3 ⋅ ⎞ ⎛ ⎞ TheThe speciﬁc speciﬁc heat heat of of water water can can be be assumed assumed to to remain remain 1.0 1.0 Btu/lb/º Btu/lb/ºF.F. 8.01cd( f ) ft min ⎛ Btu ⎞ 61.96lb gal ⎜ 8.01 ⎟ ⎜1 ⎟ ⎜ 3 ⎟ (0.5 ) Btu/lb/ºF. ⎝ gal ⋅hr ⎠ ⎝ lb⋅º F ⎠ ⎝ ft ⎠ min

PuttingPutting these these numbers numbers into into Formula Formula 3-1 3-1yields:yields: Putting these numbers into Formula 3-1 yields: If the 4.0 gpmIf the flow 4.0 rate gpm used inflow the previousrate used example in the was previous reduced to 0.5example gpm, would was the temperaturereduced drop of the to circuit 0.5 “automatically”gpm, would change the totemperature 201.5 ºF to deliver drop the intendedof the 50,000 q q==(8(.80.10D1Dc)cf)(f∆(∆TT) =) =(8(.80.10×1×616.19.696×1×.10.00)0×)1×.15.5××(1(11010−−929)2=) =131,34,0400B0tButu/ h/rhr Btu/hr heatcircuit output? “automatically” change to 201.5ºF to deliver the intended 50,000 Btu/hr heat output? Formula 3-1 can be rearranged to determine the Formula 3-1 can be rearranged to determine the temperature drop or flowAbsolutely rate that not! would be temperatureFormula 3-1 can bedrop rearranged or flow to determinerate that the would temperature be requireddrop or flow forrate thatAbsolutely would be not! arequired requiredspecific for for a rateaspecific specific of rate heatrate of of heat transfer.heat transfer. transfer. While it isWhile possible it to is operate possible very specialized to operate industrial very hyd specializedronic systems withindustrial a 201.5 ºF ForFor example, example, what what iswhat isthe the temperature temperatureis the temperature drop drop required required to to dropdeliver deliver hrequired eatheat at at a aratetemperature rate to of of 50,000 50,000hydronic drop, Btu/hr, Btu/hr, under systems specific conditions, with a thosetemperature systems are dropvery different in the from range typical deliverusingusing a adistribution distributionheat at systema system rate operating of operating 50,000 with with Btu/hr,water water at at a using aflow flow rate ratea ofdistribution of 4 4gpm? gpm?hydronic Assume Assume spaceof the the average200ºF, heatingaverage systems. under Therespecific is no way conditions, that a practical orthose cost effective systems residential or systemwaterwater temperature temperature operating is is101 101 withºF ºF water at a flow rate of 4 gpm?light commercial are very hydronic different heating system from could typical operate hydronic with this calculated space-heating 201.5 ºF drop Assume the average water temperature is 101ºF between thesystems. supply and There return sideis no of waythe distribution that a practicalsystem. or cost-effective

Solution:Solution: The The density density of of water water at at 101 101 ºF ºF was was previously previously determined determined to to be be 61.96 61.96 lb/ft residentiallb/ft3. 3After. After or light commercial hydronic heating system Solution:rearranging The formula density 3-1the numbers of water are enteredat 101ºF and calculated was previously. Formula 3-1could provided operate a “what withis required” this valuecalculated based solely 201.5ºF on a mathematical drop between relationship of rearranging formula 3-1the numbers are3 entered and calculated. determined to be 61.96 lb/ft . After rearranging Formulathe variables. the Itsupply does not and assure return that the side resulting of thenumbers distribution accurately describesystem. a “what will be” 3-1, the numbers are entered and calculated. situation in a real system. When “what is required” values determined by calculation seem very Formula 3-1 provided a “what is required” value based out of the ordinary designers should carefully check their assumptions and their calculations. BtButu q 505,00,00000 solely on a mathematical relationship of the variables. It = q = hrhr = It’s likely that a situation that seems too good to be true will be just that. ∆∆TT= = 3 = 252.52.º2Fº F 8.80.10c1dc(df()f ) ⎛ ⎛ ft 3ft⋅ m⋅ mini⎞n⎛⎞ ⎛ BtButu⎞ ⎛⎞ 6⎛16.19.69l6bl⎞b⎞ gaglal does not assure that the resulting numbers accurately ⎜ 8⎜.80.101 ⎟ ⎜⎟1⎜1 ⎟ ⎜⎟ ⎜ 3 3 ⎟ (⎟4(4 ) ) ⎝ ⎝ gagla⋅lh⋅rhr⎠ ⎝⎠ ⎝lbl⋅bº ⋅Fº F⎠ ⎝⎠ ⎝ ftft ⎠ ⎠ mminin describe a “what will be” situation in a real system. When “what is required” values determined by calculation seem It’s important to realize that the 25.2ºF temperature very out of the ordinary, designers should carefully check dropIt’sIt’s important important calculated to to realize realize inthat that this the the 25.2 example25.2 ºF ºF temperature temperature is what drop drop wouldcalculated calculated have in in this this toexample example their is iswhat whatassumptions andPage their "38 calculations. of "128 It’s likely that a occurwouldwould have havefor to ato occur wateroccur for for astream awater water stream stream flowing flowing flowing at at at 44 4gallonsgallons gallons per per per minute minute minute to to deliver deliver 5situation 0,00050,000 Btu/hr. Btu/hr. that seems too good to be true will be just that. to deliver 50,000 Btu/hr. Using Formula 3-1 does not UsingUsing formula formula 3-1 3-1 does does not not “guarantee” “guarantee” that that a awater water stream stream flowi flowingng at at 4 4gallons gallons per per minute, minute, with with “guarantee” that a water stream flowing at 4 gallons per When all but one of the variables in Formula 3-1 are an average water temperature of 101 ºF, will drop 25.2 ºF as it passes through a distribution minutean average with water an temperature average of 101water ºF, will temperature drop 25.2 ºF as of it passes101ºF through will a distributionknown based on accurate physical measurements, the systemsystem that that was was intended intended to to deliver deliver 50,000 50,000 Btu/hr. Btu/hr. Whether Whether it itdoes does or or doesn’t doesn’t deliver deliver this this rate rate of of drop 25.2ºF as it passes through a distribution system remaining unknown can be calculated with confidence. heat transfer depends on system design, the actual (not assumed) room air temperature, the thatheat transfer was dependsintended on systemto deliver design, the50,000 actual (notBtu/hr. assumed Whether) room air ittemperature, the doesheatheat loss lossor of doesn’tof the the supply supply deliver and and return return this piping piping rate that that of connect connectheat thetransfer the heat heat emitter(s) emitter(s) depends to to the the balance balanceFor example,of of the the if a streamDRAFT of (4-16-18) water with an accurately onsystem,system, system the the condition condition design, of of the the theinternal internal actual surfaces surfaces (not of of the theassumed) heat heat emit emittersters room(e.g., (e.g., are areair they they scaledmeasured scaled or or flow rate of 4.00 gpm passed through a temperature, the heat loss of the supply and return hydronicWhen all but one distribution of the variables in formula system, 3-1 are known and based onif accuratethe physicalaccurately measurements, the remaining unknown can be calculated with confidence. piping that connect the heat emitter(s) to the balance of measured temperature drop from the supply to return of PagePage "37 "37 of of "128 "128 the system, the condition of the internal surfaces of the thatFor example, distribution if a stream of system water with an was accurately 25.19ºF, measured andflow rate if of the 4.00 gpmaccurately passed heat emitters (e.g., are they scaled or corroded), and the measuredthrough a hydronic average distribution system,water and temperatureif the accurately measured in the temperature system drop from was condition of the external surfaces of the heat emitters 101ºFthe supply (e.g.,to return ofso that distributionthat the system density was 25.19 ºF,is and known if the accurately to measuredbe 61.96 (e.g., are they bent or covered with dust, pet hair, etc.). lb/ftaverage3), water then temperature the inactual the system ratewas 101 ofºF (e.g., heat so th atdelivery the density is knowncould to be be 3 calculated61.96 lb/ft ), then usingthe actual Formularate of heat delivery 3-1. could be calculated using formula 3-1. Formula 3-1 should be used as a “what is required” rather ⎛ ft 3 ⋅ min⎞ ⎛ Btu ⎞ ⎛ 61.96lb⎞ ⎛ gal ⎞ Btu Btu than a “what will be” design tool. = = = ≈ q 8.01cd( f )(∆ T ) ⎜ 8.01 ⎟ ⎜1 ⎟ ⎜ 3 ⎟ ⎜ 4.00 ⎟ (25.19º F) 50,007 50,000 ⎝ gal ⋅hr ⎠ ⎝ lb⋅º F ⎠ ⎝ ft ⎠ ⎝ min⎠ hr hr For example, what would the temperature drop of a water stream flowing 0.5 gallons per minute have to be to UsingUsing formula Formula 3-1 to calculate 3-1 a rateto of heatcalculate transfer in ana operating rate systemof heat requires transfer deliver 50,000 Btu/hr to a load using water at an average ininstrumentation an operating for measuring flowsystem rate and temperaturerequires change. instrumentation Flow rate can be measured for using a flow meter or a balancing valve equipped with an accurate flow indicator, such as shown temperature of 101ºF? measuringin figure 3-3. flow rate and temperature change. Flow rate can be measured using a flow meter or a balancing valve Mathematically this is simple. Just put the given numbers equipped[insert figure 3-3] with an accurate flow indicator, such as shown into Formula 3-1 and calculate. in Figure 3-3.

Page "39 of "128 Figure 3-3 Figure 3-4 processing unit load to/from to/from heat source

flow meter

Temperature drop can be measured with any accurate LAMINAR VS. TURBULENT FLOW temperature indicator. Devices that simultaneously read two It’s helpful to imagine liquids flowing through pipes as sensors and give a direct reading of ∆T are also available. a stream containing millions of tiny “water particles.” The path that any given water particle takes as it moves The instantaneous rate of heat transfer, as well as the through the pipe is called a streamline. total heat passing a given location in the system can also be measured using a heat meter, such as shown in In some situations, the streamlines of all the water Figure 3-4. particles are parallel to each other, like multiple lanes of traffic on a major highway. The water particles near the Heat meters combine an accurate flow meter with two center of the pipe move faster than those near the inner calibrated temperature sensors. The latter are mounted pipe wall, but they don’t “change lanes.” This type of fluid in special fittings that allow a very precise measurement movement, represented by Figure 3-5 is called laminar of ∆T. flow. It can be pictured 3-dimensionally as multiple concentric “layers” of liquid, each moving at a different The sensors and flow meter send data to a processing speed, sliding past each other, as shown in Figure 3-6. unit that calculates the instantaneous rate of heat transfer, based on Formula 3-1. The processing unit also Laminar flow can be desirable or undesirable depending integrates the heat transfer rate over time to determine on where it occurs in hydronic systems, and the design the total amount of heat that has passed through the goals for the system. If the goal is to move fluid long meter. The latter can be used to verify operation of the distances with minimal pumping power, laminar flow system, or as the basis of billing a customer for the total is desirable. However, if the goal is to maximize heat amount of heat used. Caleffi CONTECA heat meters are transfer from a flowing fluid to a tube wall or other solid in conformance with the newly released ASTM E3137/ surface, laminar flow is very undesirable. E3137M-17 heat metering standard.

18 The erratically shaped streamlines present during turbulent Figure 3-5 flow create good mixing. A small quantity of fluid that’s imaginary "fluid particles" close to the tube wall one instant could be swept into the bulk of the fluid stream the next instant. This decreases streamlines the boundary layer thickness, and significantly improves convective heat transfer.

The convection coefficient associated with a relatively viscous liquid at low temperatures, such as a 25% solution of propylene glycol antifreeze at 30ºF, under laminar flow, can be only 7% of the convection coefficient of the same fluid under turbulent flow. This implies a very significant Laminar ﬂow streamlines drop in heat transfer between a fluid and the inner wall of a tube if the flow is laminar versus turbulent. Such a change could have profound effect on the ability of a hydronic circuit, such as a ground loop heat exchanger supplying a geothermal heat pump, to transfer heat at Figure 3-6 the required rate.

Although turbulent flow improves convective heat transfer, it also increases head loss, and thus requires more pumping power to maintain a given flow rate relative to the power required for laminar flow. However, the increased heat transfer capabilities of turbulent flow are often more important than the penalty associated with the increased pumping power.

REYNOLDS NUMBER It’s possible to predict if flow through a pipe will be laminar or turbulent. It’s based on calculating a dimensionless quantity called the Reynolds number of Because the water particles move in parallel streamlines the fluid (abbreviated as Re#). If the calculated Re# is during laminar flow, there is very little mixing between over 4000, the flow is turbulent. If it is below 4,000, the them. This increases the thickness of the boundary layer flow may be either laminar or turbulent. If the Re# is between the bulk of the fluid stream and the inner wall of below 2,300, the flow will be laminar. the pipe. Thicker boundary layers decrease convective heat transfer. The Reynolds number for flow in a pipe can be calculated DRAFT (4-16-18) using Formula 3-3: Turbulent flow causes the streamlines of imaginary fluid particles to bend and twist as the fluid moves down the Formula 3-3: pipe, as shown in Figure 3-7. vdD " Re# = Figure 3-7 µ imaginary "fluid particles" where: streamlines where: v = average flow velocity of the fluid (ft/sec) d = internal diameterv = ofaverage pipe (ft) flow velocity of the fluid (ft/sec) D = fluid’s density (lb/ft3) µ = fluid’s dynamicd viscosity= internal (lb/ft/sec) diameter of pipe (ft) D = fluid’s density (lb/ft3) For example, determine the Reynolds number of water at 140°F flowing µat = 5 fluid’s gpm throughdynamic a viscosity 3/4-inch (lb/ft/sec)type M copper tube. Is this flow laminar or turbulent? turbulent flow streamlines

For example, determine the Reynolds number of water at 140 °F ﬂowing at 5 gpm through a 3/4-inch type M copper tube. Is this19 ﬂow laminar or turbulent?

Solution: To calculate the Re# the density and dynamic viscosity of the water must be determined. The density of water can be found from ﬁgure 3-2. The dynamic viscosity of water can be found from ﬁgure 3-8.

[insert ﬁgure 3-8]

Page "44 of "128 DRAFT (4-16-18)DRAFT (4-16-18) Solution: To calculate the Re#, the density and dynamic The Reynolds number can now be calculated. viscosity of the water must be determined. The density 3 (3.106 ft/s3.106ec)(0.06758 ft/sec ft)0.06758(61.35 lb/ fftt )61.35 lb/ft3 of water can be found from Figure 3-2. The dynamic " Re# = ( )( )( = 40,242 ) " Re# = 0.00032 lb/ft/sec = 40,242 viscosity of water can be found from Figure 3-8. 0.00032 lb/ft/sec This value is well above the threshold of 4,000, and Figure 3-8 Thistherefore value is thewell flowabove is the turbulent. threshold of 4,000, and therefore the flow is turbulent. This value is well above the threshold of 4,000, and therefore the flow is turbulent. 0.0009 Notice that the units on the quantities used in the Re# Notice that the units on the quantities used in the Re# formula cancel out completely. This 0.0008 formula cancel out completely. This serves as a check servesthat the Noticeas a proper check that that theunits the units properare on being theunits quantities areused being in usedFormulaused in in Formula the 3-3. Re# 3-3. formula cancel out completely. This 0.0007 serves as a check that the proper units are being used in Formula 3-3. TEMPERATURE DROP IN A HYDRONIC SYSTEM 0.0006 TEMPERATURE DROP IN A HYDRONIC SYSTEM Formula 3-1 indicates that, with all other factors and DRAFT (4-16-18) 0.0005 conditionsTEMPERATURE being equal, DROP the INrate A HYDRONICof heat delivery SYSTEM from Formulaa stream 3-1 indicatesof water that, (or with solution all other factorsof antifreeze) and conditions is directly being equal, the rate of heat 0.0004 deliveryproportional from a stream to ofthe water temperature (or solution of anti-freeze)drop that is directly stream proportional to the undergoesFormula as 3-1 it indicatespasses through that, with a all hydronic other factors distribution and conditions being equal, the rate of heat 0.0003 temperature drop that stream undergoes as it passes through a hydronic distribution system. DRAFTsystem. (4-16-18) At 140 ºF the density of water is 61.35 lb/ft3. From figure 3-8 thedelivery dynamic from viscosity a stream of ofwater water 140 (or solution of anti-freeze) is directly proportional to the 0.0002

dynamic viscosity (lb/ft/sec) The ∆T at which a hydronic circuit operates is always determined, at any time, by the ºF is 0.00032 lb/ft/sec. DRAFT (4-16-18)The ∆Ttemperature at which drop a hydronic that stream circuit undergoes operates as it ispasses always through a hydronic distribution system. 0.0001 circuit’sdetermined, ability to at release any heat. time, by the circuit’s ability to release heat. 0 The ∆T at which a hydronic circuit operates is always determined, at any time, by the The inside diameter of a 3/4-inch type M copper tube 3is 0.811 inches. This must be converted to 50 At100 140 ºF the150 density of200 water is 61.35250 lb/ft .Consider From figure a hydronic 3-8 the heating dynamic distribution viscosity system of water that’s 140 releasing heat at a rate of 100,000 Btu/hr Considercircuit’s a hydronic ability to heatingrelease heat.distribution system that’s feet to matchºFwater isthe 0.00032 stated temperature unitslb/ft/sec. for Formula(ºF) 3-3: into a space at 70 ºF air temperature when supplied with 180 ºF water at 9 gpm. The At 140 ºF the density of water is 61.35 lb/ft3. From figurereleasing 3-8 the heat dynamic at a rateviscosity of 100,000 of water 140Btu/hr into a space at temperature70ºF air temperature drop of the circuit when under supplied these conditions with 180ºF is found water using formulaat 3-1: ºF is 0.00032 lb/ft/sec. 3 Consider a hydronic heating distribution system that’s releasing heat at a rate of 100,000 Btu/hr At 140ºF, the density of water⎛ 1 ft is⎞ 61.35 lb/ft . From Figure 3-8, 9 gpm. The temperature drop of the circuit under these " d = (0.811 iThen ⎜ inside⎟ = diameter0.06758 foft a 3/4-inch type M copper tube is 0.811 inches. This must be converted to the dynamic viscosity )of⎝12 water in.⎠ 140ºF is 0.00032 lb/ft/sec. conditionsinto a spaceis found at 70using ºF air Formula temperature 3-1: when supplied with 180 ºF water at 9 gpm. The feet to match the stated units for Formula 3-3: temperature drop of the circuit under these conditions is found using formula 3-1: The inside diameterThe inside of a diameter 3/4-inch of typea 3/4-inch M copper type M tubecopper is tube is 0.811 inches. This must be convertedBtu to The average flow velocity corresponding to a flow rate of 5 gpm canq be found using formula100,00 03-4. 0.811 inches. Thisfeet mustto match be theconverted stated units to feetfor Formula to match 3-3: the ∆ T = = hr = 22.85º F ⎛ 3 ⋅ ⎞ ⎛ ⎞ ⎛ 1 ft ⎞ 8.01cd( f ) ft min ⎛ Btu ⎞ 60.7lb gal stated units for Formulad = 3-3:0.811 in = 0.06758 ft ⎜ 8.01 ⎟ ⎜1 ⎟ ⎜ 3 ⎟ (9 ) " ( )⎝⎜ ⎠⎟ ⎝ gal ⋅hr ⎠ ⎝ lb⋅º F ⎠ ⎝ ft ⎠ min Formula 3-4: 12 in. ⎛ 1 ft ⎞ Btu " d = (0.811 in)⎜ ⎟ = 0.06758 ft The average⎝ 12 in.⎠ flow velocity corresponding to abased flow rate on of this 5 gpm qcalculation, can be found the using return formula10 water0,0 03-4.0 temperature ∆ T = = hr = 22.85º F from the distribution system⎛ would3 ⋅ be:⎞ ⎛ ⎞ ⎛ 0.408⎞ based on this calculation,8.01cd( f ) the return fwatert m itemperaturen ⎛ Btu from⎞ 6 the0.7 ldistributionb gal system would be: The averagev flow= The velocity averagef correspondingflow velocity corresponding to a flow rateto a flowof rate180 of - 5 22.85 gpm can = 157.15ºF be found⎜ using8.01 formula ⎟3-4.⎜1 ⎟ ⎜ 3 ⎟ (9 ) ⎝⎜ 2 ⎠⎟ ⎝ gal ⋅hr ⎠ ⎝ lb⋅º F ⎠ ⎝ ft ⎠ min 5 gpm can be foundd Formulausing Formula 3-4: 3-4. 180-22.85 = 157.15 ºF How would the output of this distribution system change if: Formula 3-4: Formula 3-4: Where: How would the output of this distribution system change if: ⎛ 0.408⎞ 1. The room temperature was reduced from 70ºF to 65ºF = v = averagev fluid⎜ velocity2 ⎟ f (ft/sec) ⎝ d ⎠ with allbased other on conditions this calculation, being the thePage return same? "46 water of "128 temperature from the distribution system would be: d - exact⎛ 0inside.408⎞ diameter of pipe (inches) 180-22.85 = 157.15 ºF v = ⎜ ⎟ f Where: ⎝ d 2 ⎠ 2. The supply water temperature was changed to 140ºF v = averagef =fluid flow velocity rateWhere: through (ft/sec) pipe (gpm) with all other conditions being the same. d = exact inside diameterv = average of pipe fluid (inches) velocity (ft/sec) How would the output of this distribution system change if: f = flow rate throughWhere: pipe (gpm) Both changes can be estimated based on a fundamental For 3/4” Typed - exactM copper inside tubing diameter operating of pipe at (inches)a flow rate oftenant 5 gpm of the hydronic average heating,flow velocity specifically: is: v = average fluid velocity (ft/sec) Page "46 of "128 For ¾” Type M copperf = flowtubing rate operating through pipe at (gpm)a flow rate of 5 gpm, the averaged - exact flow inside velocity diameter is: of pipe (inches) The heat output of any hydronic distribution system f = flow rate through pipe (gpm) will be approximately proportional to the difference ⎛ 0.408⎞ ⎛ 0.408 ⎞ ft = For =3/4” Type M copper= tubing operating at abetween flow rate ofthe 5 gpm supply the average water flowtemperature velocity is: and the room v ⎜ 2 ⎟ f ⎜ 2 ⎟ 5 3.1 ⎝ d ⎠ ⎝ (0.811) ⎠ sec air temperature. For 3/4” Type M copper tubing operating at a flow rate of 5 gpm the average flow velocity is:

The Reynolds number⎛ 0.408⎞ can now⎛ 0 .be40 8calculated.⎞ ft = = = 20 v ⎜ 2 ⎟ f ⎜ 2 ⎟ 5 3.1 ⎝ d ⎠ ⎝ (0.811) ⎠ sec ⎛ 0.408⎞ ⎛ 0.408 ⎞ ft = = = v ⎜ 2 ⎟ f ⎜ 2 ⎟ 5 3.1 ⎝ d ⎠ ⎝ (0.811) ⎠ sec The Reynolds number can nowPage be calculated."45 of "128

The Reynolds number can now be calculated.

Page "45 of "128

Page "45 of "128 DRAFT (4-16-18)

For the distribution system being considered it follows that the heat output formula is as follows:

q = c × (Tsupply − Tair ) = 909 × (Tsupply − Tair )

This formula can be treated as the “heat output formula” for this speciﬁc distribution system.

DRAFT (4-16-18) If the room air temperature was reduced from 70 ºF to 65 ºF, the estimated heat output of the DRAFT (4-16-18) distribution system would be approximately: DRAFT (4-16-18) 1. The room temperature was reduced from 70 ºF to 65 ºF with all other conditions being the Btu q = 909 × (180 − 65) ≅ 104,500 1. Thesame? room temperature was reduced from 70 ºF to 65 ºF with all other conditions being the hr same?1. The room temperature was reduced from 70 ºF to 65 ºF with all other conditions being the same? 2. The supply water temperature was changed toThe 140 circuit’s ºF with heat all otheroutput conditions increased because being the the ∆T between its supply water temperature and 2. The supply water temperature was changed to 140 ºF with all other conditions being the same. the room air temperature increased. The new ∆T for the system would be: same.2. The supply water temperature was changed to 140 ºF with all other conditions being the same. Both ofBoth these of changes these changes can be estimated can be based estimated on a fundamental based on tenanta fundamental of hydronic tenant heating, of hydronic heating, Btu specifically:Bothspecifically: of these changes can be estimated based on a fundamental tenantq of hydronic heating, 104,500 ∆ T = = hr = 23.88º F specifically: 8.01cd( f ) ⎛ ft 3 ⋅ min⎞ ⎛ Btu ⎞ ⎛ 60.7lb⎞ gal ⎜ 8.01 ⎟ ⎜1 ⎟ ⎜ 3 ⎟ (9 ) The heat output and any hydronic distribution system will be approximately proportional⎝ gal ⋅ hr ⎠ ⎝ lb⋅º F ⎠ ⎝ ft ⎠ min The heat output and any hydronic distribution system will be approximately proportional to theThe difference heat output between and any the hydronic supply water distribution temperature system and will the be room approximately air temperature. proportional toto the the difference difference between between the supply the supplywater temperature water temperature and the room and air thetemperature. room air temperature. This implies that the following mathematical relationship can be written:Note that the ∆T of the distribution system changed based solely on a small change in the room ThisThis implies implies that thatthe following the following mathematical mathematical relationship relationship can airbe temperature.written: can be written: Formula 3-5 This implies thatFormula the following 3-5 mathematical relationship If the supplyIf thewater supply temperature water temperature was reduced was fromreduced 180ºF from 180 ºF to 140 ºF, and all other conditions can be written: Formula 3-5 to 140ºF, and all other conditions remained the same, the remained the same, the heat output of the distribution system would be: q = c × (Tsupply − Tair ) heat output of the distribution system would be: Formula 3-5 q = c × (Tsupply − Tair ) q = c × (T − T ) Btu supply air q = 909 × (140 − 70) ≅ 63,630 Where: hr Where: q = rateWhere: of heat output from the circuit (Btu/hr) q = rate of heat output from the circuit (Btu/hr) c = aq constant= rate of ofheat proportionality output from thatthe circuitcan be (Btu/hr) determined for for each system c = a constant ofWhere: proportionality that can be determined HOW SUPPLY TEMPERATURE AFFECTS ∆T c = a constant of proportionality that can be determined for for each system for each systemTsupply = supply water temperature to the circuit (ºF) Because heat output is approximately proportionalPage to"48 of "128 T = supply waterq = rate temperature of heat output to the fromcircuit the (ºF) circuit (Btu/hr)the difference between the supply water temperature supply Tair =T supplyroom = air supply temperature water temperature (where heat to is the being circuit released) (ºF) (ºF) Tair = room air temperaturec = a constant (where of heat proportionality is being released) that can and be thedetermined room air fortemperature, for each systemand when the circuit flow (ºF) Tair = room air temperature (where heat is being released)rate is constant (ºF) , the ∆T between the supply and return of If the circuit’sTsupply =heat supply output water is known temperature at a specific to set the ofany operatingcircuit hydronic (ºF) conditions, system includingmust decrease a flow rate, as the supply water If the circuit’s heat output is known at a specific set of temperature decreases. supplyIf Tthe waterair circuit’s= room temperature, heat air outputtemperature and is room known air (where at temperature, a specific heat set isthe ofbeing value operating released)of (c) conditions, in formula (ºF) including3-5 can be a flow rate, operating conditions, including a flow rate, supply water temperature,found. andsupply Substitutingroom water air temperature,temperature, the previously and the stated room value airconditions oftemperature, (c) intoFigure form the 3-9 ulavalue 3-5shows of yields: (c) inthe formula relationship 3-5 can betweenbe the supply in Formula 3-5 found.can be Substituting found. Substituting the previously the previouslystated conditions and into return form waterula 3-5 temperatureyields: for a specific distribution stated conditionsIf into the Formulacircuit’s 3-5heat yields: output is known at a specificsystem that set suppliesof operating a design conditions, output 100,000 including Btu/hr a flow into rate, = × − 100,00supply0 c (water180 7 temperature,0) andDRAFT room (4-16-18)air atemperature, space at 70ºF, the using value a of supply (c) in formulawater temperature 3-5 can be of 100,000 = c × (180 − 70) DRAFT (4-16-18)180ºF and a flow rate of 9 gpm. found. SubstitutingDRAFT (4-16-18) the previously stated conditions into formula 3-5 yields: 100,000Btu Btu c = = 909 Figure 3-9 (180º F10−07,00º0FB)tu hr⋅º FBtu Forc =the distribution= × system= 90 −being9 consideredPage "47it follows of "128 that the heat output formula is as follows: 10(108,0º0F0− 7c0º F()180 7h0r⋅)º F Supply & return water temperature based on constant flow rate For the distributionFor thesystem distribution being considered system it follows being that considered the heat output Pageit follows formula "47 ofthat is "128as the follows: heat output formula is as follows: For the distribution system being considered, it follows design load = 100,000 Btu/hr that the heat output formula is as follows: flow rate = 9 gpm (water) q = c × (Tsupply − Tair ) = 909 × (Tsupply − Tair ) supply water temp. @ design load = 180 ºF = × − = × − room air temperature = 70 ºF (constant) q c (Tsupply Ta=ir ) ×909 (Tsu−1ppl0y 0T,a0ir=)00Btu× − Btu q c (cT=supply Tair ) 909 (Tsu=ppl9y 09Tair ) design load ∆T = 22.9 ºF (180º F − 70º F) hr⋅º F 190 Page "47 of "128 ThisThis formula formula can becanThis treated beformula astreated the can “heat asbe output thetreated “heatformula” as outputthe for th“heatis specificformula” output distribution formula” system. for this specific distribution system. for this specific distribution system. 170 ∆T =22.9 ºF This formula can be treated as the “heat output formula” for this specific distribution system. (@ design load) IfIf the the room room air temperature airIf the temperature room was reducedair temperature was from 70reduced ºF was to 65 reduced ºF,from the estimated70ºF from 70to heat ºF outputto 65150 ºF,of the the estimated heat output of the distribution system would be approximately: 65ºF, the estimatedIf thedistribution room heat air temperaturesystem output would of the was be distribution approximately:reduced from system 70 ºF to 65 ºF, the estimated heat output of the would be approximately: 130 distribution system would be approximately: Btu = × − ≅ q 909 (180 65) 104,500 Btu 110 q = 909 × (1h8r0 − 65) ≅ 104,500

hr water temperature (ºF) Btu Supply water q = 909 × (180 − 65) ≅ 104,500 90 The circuit’s heat output increased because the ∆T between its supply water temperature and Return water The circuit’s heat output increased becausehr the ∆T between the room air temperature increased. The new ∆T for the system would be: its supply waterThe temperature circuit’s heat and output the increased room air becausetemperature the ∆ T between70 its supply water temperature and 0 10 20 30 40 50 60 70 80 90 100 increased. Thethe new room ∆T air for temperature the system increased. would be: The new ∆T for the system would be: The circuit’s heat output increased because the ∆T between its supply% water of design temperature heating load and

the room air temperatureBtu increased. The new ∆T for the systemno would be: design q 104,500 ∆ T = = hr = 23.88º F load load 8.01cd( f ) ⎛ ft 3 ⋅ min⎞ ⎛ Btu ⎞ ⎛ 60.7lb⎞ gal 8.01 ⎜1 ⎟ (9 ) Btu ⎝⎜ ⋅ ⎠⎟ ⎝ ⋅ ⎠ ⎝⎜ 3 ⎠⎟ gal hqr lb º F ft mi1n04,500 ∆ T = = hr = 23.88º F ⎛ 3 ⋅ ⎞ Btu⎛ Although⎞ the value of the supply water minus return water 8.01cd( f ) ft mi1n04⎛,50B0tu ⎞ 60.7lb gal q ⎜ 8.01 ⎟ ⎜1 ⎟ ⎜ 3 ⎟ (9 ) Note that the∆ T ∆=T of the distribution= ⎝ systemgal ⋅h rchanged⎠ ⎝ lb⋅º F hbasedr⎠ ⎝ ft ∆⎠T willmin be= 2 3different.88º F for different hydronic systems using Notesolely that onthe ∆aT small of the8 distribution.change01cd( f ) insystem the⎛ changedroomft 3 ⋅ mair basedin temperature.⎞ ⎛ solelyBtu on⎞ a⎛ small60.7 lchangeb⎞different gina thel roomheat emitters, the relationship between supply air temperature. ⎜ 8.01 ⎟ ⎜1 ⎟ ⎜ 3 ⎟ (9 ) ⎝ gal ⋅hr ⎠ ⎝ lb⋅º F ⎠ ⎝ ft ⎠ min

If the supply water Notetemperature that the was ∆ Treduced of the from distribution 180 ºF to system140 ºF, and changed all other based conditions solely on a small change in the room remained the same,air the temperature. heat output of the distribution system would be: Note that the ∆T of the distribution system changed based solely on a small change in the room 21 Btu q = 909 × (140air− 7 0temperature.) ≅ 63,630 If the supplyhr water temperature was reduced from 180 ºF to 140 ºF, and all other conditions remained the same, the heat output of the distribution system would be: If the supply water temperature was reduced from 180 ºF to 140 ºF, and all other conditions remained the same, the heat outputBtu of the distribution system would be: q = 909 × (140 − 7Page0) ≅ 6 "483,6 of3 0"128 hr Btu q = 909 × (140 − 70) ≅ 63,630 hr

Page "48 of "128

Page "48 of "128 water temperature and return water temperature will be When design load is no longer present in one zone, similar to that shown in Figure 3-9 (e.g., the ∆T will shrink and it turns off, less heat would be removed from the from a maximum value at design load to 0 at no load). distribution system. This change would reveal itself as an increase in return water temperature (assuming that the CONSTRAINED ∆T supply water temperature is fixed). The ∆T of the circuit It is possible to vary the flow rate through a hydronic would decrease. A controller that senses this decrease heating distribution system so that the temperature drop could respond by reducing the speed of a circulator so from the supply to the return remains constant. This that the design load ∆T was reestablished for the zones method of control is called constrained ∆T. It’s appropriate that remain active. This process would repeat when in systems that have all the following attributes: another zone turned off. This method of control reduces circulator energy use during partial load conditions. 1. Multiple heating zones controlled with valves. When a zone turned on, and the supply water temperature 2. Low thermal mass distribution systems. remains fixed at the design load value, the return water temperature decreases because more heat is being 3. Maintain a constant supply water temperature, at the removed from the distribution system. A controller could design load value, whenever any zone is calling for heat. sense this increase in ∆T and respond by increasing circulator speed to reestablish the design load ∆T for all Consider an accurately designed low thermal mass the active zones. hydronic heating system that supplies design load heat output when all zones are active and the supply water The requirement that the distribution system have low temperature remains constant at the design load value. thermal mass implies that the temperature changes on An example of such a system is shown in Figure 3-10. the return side of the system would occur quickly as zones turn on and off. A high thermal mass distribution If the heat source was not oversized for the design load, system, such as a heated concrete floor slab, could all zones would, in theory, remain on until the design load significantly delay these temperature changes due to heat condition subsided (or other factors such as internal gains being absorbed into or released from the thermal mass. or intentional thermostat setbacks began influencing the zone loads). This method of control forces the active portion of the system to operate at design load conditions based on supply and Figure 3-10 return water temperatures. When a constant ∆T controller zone doesn’t require design load heat variable speed circulator input, the thermostat in that zone would have to completely stop flow. supply temperature sensor This type of system, in effect, directs “pulses” of heat into each zone zone whenever the associated zone valves thermostat calls for heat and opens the corresponding zone valve. The

VENT low thermal mass rate of heat delivery during each distribution system pulse remains at the design load rate. The duration of each pulse is the time that the zone valve is open. The heat transfer rate multiplied by the time duration of the heat input pulse determines the total heat added to return temperature sensor the space during that time. This “pulsed” method of heat delivery has been used in millions of North constant temperature heat source, set American hydronic systems. It is for design load supply water temperature generally acceptable if the thermostat differential is reasonable.

22 However, not all hydronic systems meet the three Consider a 50-foot-long finned-tube baseboard, as previously stated constraints. Many modern systems now shown in Figure 3-11. use outdoor reset control to vary the water temperature supplied to the distribution system. When this method of Imagine an experiment where the inlet water temperature water temperature control is combined with a controller to the baseboard will be varied from 180ºF down to that varies flow rate to maintain a fixed ∆T, the heat output 120ºF in increments of 20ºF. The flow rate at each inlet from the distribution system decreases faster than it temperature will be adjusted to produce a nominal 20ºF should based on outdoor reset control theory. temperature drop along the baseboard. The corresponding heat output of the baseboard will then be calculated. The This can be illustrated by determining the flow rates that results are shown in Figure 3-12. would be necessary to maintain a constant ∆T as the supply water temperature to the system decreases. As the supply water temperature decreases, with the ∆T constrained to 20.1ºF, the heat output of the baseboard, shown by the orange line, drops rapidly. For comparison, Figure 3-11 the red line shows how the baseboard’s heat output 50 ft x 3/4" baseboard would normally decrease in proportion to the difference between the supply water temperature and the room air (rated @ 600 Btu/hr/ft @200 ºF water) temperature. This proportional relationship is the basis of operation for many outdoor reset controllers (standalone or integrated into boilers). At a supply water temperature of 120ºF, and with the ∆T constrained to 20.1ºF, the heat output of the baseboard is only about 55% of the output, ∆T constrained to 20.1 ºF based on the previously mentioned proportionality. This suggests that heat output under the combined effects of 1) reduced supply water temperature based on outdoor Figure 3-12 reset, and 2) constrained T, could result heat output based on holding ∆ ∆T constrained to 20.1 ºF in inadequate comfort under partial load conditions. heat output being approx. proportional to (Ts-Tr) assuming room temperature ﬁxed at 70 ºF WHY 20? flow rate One misconception that has long been 20000 = 2.01 gpm entrenched in the North American hydronics 18000 market is that all hydronic systems should 16000 operate at a temperature drop of 20ºF. The origin of this number in the North 14000 flow rate American hydronics market is unknown. = 1.45 gpm 12000 Perhaps it came from the concept that 10000 if a system using water operates with a 20ºF temperature drop, the rate of heat 8000 flow rate = 0.95 gpm delivery would be approximately 10,000 6000 heat output (Btu/hr) Btu/hr per gpm of flow rate. While this is 4000 flow rate a close approximation to the rate of heat = 0.50 gpm ºF 20.1 flow rates necessary to hold ∆ T = 2000 delivery, it all hinges on the word “if.” More specifically, if the system operates with a 0 20ºF temperature drop, then each gpm 60 80 100 120 140 160 180 of water flow rate would transport about Supply water temperature (ºF) 10,000 Btu/hr. 50 ft x 3/4" finned-tube baseboard There is nothing sacrosanct about operating a hydronic system with a 20ºF temperature drop. Systems can operate ∆T very well using much lower and much higher ∆Ts.

23 For example, if a very high flow rate was maintained would be 7.3ºF. The heat output of the circuit increases through a radiant floor tubing circuit, that circuit might with increasing flow rate because the average water operate at a ∆T as low as perhaps 4ºF. That would temperature in the circuit is increasing. However, the make the average water temperature in the circuit gains become smaller and smaller at higher flow rates. only 2ºF lower than the supply water temperature. It would also make the floor area served by the circuit It’s also possible to operate hydronic circuits at relatively uniform in surface temperature. So, what’s temperature drops much higher than 20ºF. In Europe, wrong with operating a circuit under this 4ºF ∆T? panel radiator systems are sometimes designed for a From the standpoint of heat transfer, and achieving 20ºC temperature drop under design load conditions. relatively uniform floor surface temperatures, there’s 20ºC converts to 36ºF. Using the higher ∆T allows the nothing wrong. However, maintaining the high flow flow rates to each panel radiator to be reduced. If the rate necessary for the low ∆T would require a larger ∆T was increased from 20 to 36ºF, the required flow rate circulator that would use significantly more electrical would decrease to (20/36) = 56 percent of the flow rate energy compared to that required to operate the system required at 20ºF. This may allow smaller tubing to be at a higher ∆T. used to supply the radiator. It may also allow a smaller circulator with a lower power motor to be used. Figure 3-13 shows the ∆Ts that would occur in a radiant floor heating circuit based on holding the inlet Figure 3-14 shows a thermograph of a panel radiator temperature constant at 110ºF, and varying the flow rate operating with an estimated 40ºF ∆T between its supply from 0.52 to 2.65 gpm. and return water temperature. The ∆T was estimated based on the temperature scale and colors on the supply At the 0.52 gpm flow rate, the circuit’s ∆T is almost and return piping. 26ºF. At the 2.65 gpm flow rate, the circuits flow rate

Figure 3-13

300 ft x 1/2" PEX tubing embedded in 4" bare slab 12" tube spacing

Tin = 110 ºF Tout = 84.4 ºF 10000 30 flow = 0.52 gpm 9000 25 8000 ∆T = 25.6 ºF 7000 20 6000 Tin = 110 ºF Tout = 97.9 ºF 5000 15

flow = 1.48 gpm 4000 10 3000 heat output (Btu/hr) 2000 temperature drop (ºF) ∆T = 12.1 ºF 5 1000

0 0 0 0.5 1 1.5 2 2.5 3 Tin = 110 ºF Tout = 102.7 ºF ﬂow rate (gpm) flow = 2.65 gpm

∆T = 7.3 ºF

24 Figure 3-14

Notice the relatively even temperature gradient from the There are also applications in which very high ∆Ts are top to the bottom of the panel radiator. This indicates possible in portions of a hydronic heating system. They even flow distribution among the panels’ vertical flow typically involve the combination of a higher temperature heat channels. source, such as a conventional boiler, and low-temperature heat emitters, such as radiant floor panels. Figure 3-15 shows an example. Figure 3-15 Under design load conditions, this low temperature heat emitters system supplies water at 180ºF to 105 ºF 90 ºF three Caleffi distribution stations. At

distribution distribution distribution station #1, this hot water station station is mixed with return water from the #1 #2 radiant panel circuits at 88ºF, to achieve a supply water temperature of 105ºF. The radiant panel circuits served by 180 ºF water ∆T = 92 ºF PEX or the distribution station operate at a T 88 ºF water ∆ PEX-AL-PEX of 105 - 88 = 17ºF under design load tubing 110 ºF conditions. The portion of the 88ºF water not used for mixing returns from master the mixing manifold station. mainfold station Although the radiant panel circuits 180 ºF distribution station operate at a typical design load #3 temperature drop, the ∆T for the tubing set carrying water between the boiler manifold and distribution station #1 this 3-way thermostatic valve protects boiler operate at a ∆T of 180 - 88 = 92ºF. from sustained flue gas condensation At this ∆T, each gallon per minute of water flow transports approximately 130 ºF 45,000 Btu/hr from the boiler manifold

VENT to the distribution station. This allows small diameter flexible tubing to carry a substantial rate of heat transfer. Low conventional boiler flow rates and small flexible tubing help reduce installation cost.

25 4: OPERATING TEMPERATURE OF Precise analysis of energy flows in high thermal mass HYDRONIC HEATING SYSTEMS hydronic systems must address transient system response. The word transient implies that temperatures The first law of thermodynamics can be expressed as and possibly flow rates within the system are changing follows: significantly with time. This further implies that the system’s thermal mass is absorbing or releasing heat. Energy input = Energy Output + Energy Stored Transient system analysis is mathematically complex, and as such, is usually done through computer simulation. This principle of energy conservation is a widely used “starting point” when analyzing physical systems involving A simplified analysis examines a system that is at or very energy flow. It cannot be circumvented. Any analysis, close to steady state conditions. Steady state means that method or product that claims to have “side-stepped” the the heat stored in the system’s thermal mass is neither first law of thermodynamics is flawed. increasing or decreasing. Under steady state conditions, the water temperatures and flow rates in various parts of When this law is applied to hydronic heating systems, the system are not changing with time. Steady state analysis the energy flows can be categorized as thermal energy produces “snapshots” of system performance, rather than and electrical energy. Thermal energy (e.g., heat) is showing how operating conditions change over time. the dominant form of energy in hydronic systems. The electrical energy used in hydronic systems (other One of the most common “snapshots” of steady state than those with electrically operated heat sources) is performance are when the system is operating at its for operating the distribution system. In well-designed maximum heat conveyance rate. This is called “design hydronic systems, electrical energy use is very small in load conditions.” Designers use this condition to size comparison to thermal energy use. It is often treated equipment that can maintain specified comfort conditions as insignificant, and as such, is not included in the when the building is losing heat at a maximum rate based thermodynamic analysis of the system. In such cases, on the statistically lowest outdoor temperature, and the first law can be limited to thermal energy flows in the assuming all zones in the system are operating. system and stated as follows: Because design load conditions assume steady state, the Heat generated by the heat source = heat released thermal mass of the system is not included in the analysis. The first law of thermodynamics now simplifies to: from the system + heat stored within the system.

The ability of hydronic heating systems to store heat Rate of heat input by the heat source = rate of heat varies significantly. Systems with high thermal mass release by the distribution system components, such as a concrete floor slab with embedded tubing, or a large thermal storage tank holding several The condition described by this relationship is called hundred gallons of water, can store large quantities of thermal equilibrium. It’s a condition that all hydronic heat. Systems with low metal and water volume store systems inherently try to establish and maintain. Assuming minimal amounts of heat. that controls in the system do not intervene, and flow rates remain steady, thermal equilibrium will eventually occur at The way in which a hydronic heating system is operated any rate of heat input by the heat source. When it does, the also affects how the heat it stores behaves. For example, temperatures in all parts of the system remain constant. in a high thermal mass system operated with constant circulation and outdoor reset control of supply water When a hydronic heating system begins operation from a temperature, the influence of the stored heat is minimal cold start, the heat source is injecting heat to the system’s under normal conditions. This type of system makes very fluid at a rate much higher than the rate of heat dissipation gradual changes in its rate of heat output. However, if from the relatively cool distribution system. The fluid the water flow through that system is turned on and off, temperature in all portions of the system increases under and if the supply water temperature always remains close this condition. Assuming there is no interference by system to the temperature required at design load conditions, controls, this increase in temperature continues until the the influence of the thermal mass in determining room system “finds” it thermal equilibrium condition, where temperature will be much greater. Wider swings in room the rate of heat input exactly matches the rate of heat temperature are likely. dissipation. If there are no changes to the heat input rate, or the ability of the distribution system to release heat, the system remains at thermal equilibrium indefinitely.

26 If the rate of heat output at the heat source changes, or Assume that there are no controllers in this system, no there’s a change in the distribution system, such as a zone pressure relief valve, and that all components used have circuit turning on or off, thermal equilibrium is temporarily very high temperature and pressure ratings. Also, assume disrupted. However, the system immediately seeks to that the room air temperature is 70ºF. reestablish thermal equilibrium under the new operating conditions. If the system’s thermal mass is small, thermal What average water temperature would be necessary in equilibrium will be reestablished within a few minutes. the baseboard at thermal equilibrium? Systems with high thermal mass take longer, in some cases several hours, to reestablish thermal equilibrium Based on a detailed performance model for the baseboard, conditions. the required average water temperature in the 20-foot- long finned-tube baseboard would be 420ºF! It is possible When the rate of heat input to the system increases, so to maintain water in a liquid state at this temperature, do the water temperatures within the distribution system. but it requires a gauge pressure of 294 psi! This water These temperature increases allow the distribution system temperature and its requisite pressure to maintain a liquid to dissipate heat at a higher rate. When the rate of heat state is much higher than practical in any residential or input decreases, so do the water temperatures within the light commercial hydronic system. Still, it represents a system. The reduced temperature causes the distribution thermodynamic condition that the system is striving for. system to dissipate heat at a lower rate. Fortunately, if this were a real system, one or more safety devices, such as a boiler high limit aquastat or pressure It’s important to understand that EVERY hydronic system relief valve, would interrupt the process long before it always seeks to operate at thermal equilibrium. The reached such a dangerous condition. conditions necessary for this to occur may not produce the desired heat delivery. They may not allow the system Many hydronic systems operate in a similar manner. Each to operate efficiently. They may not even allow the system time the heat source is turned on, the system’s water to operate safely. Still, unless controllers within the temperature begins climbing toward thermal equilibrium, system intervene, the system always strives to achieve but the operating controls on the system interrupt that thermal equilibrium. progression when a specific temperature or pressure condition is attained at which the designer wants the heat Consider a hypothetical system with a 50,000 Btu/hr heat source to stop producing heat. A common example is source and 20 feet of finned-tube baseboard, as shown the high limit controller on a boiler turning off the burner in Figure 4-1. when the water temperature in the boiler reaches a set “high limit” value, such as 190ºF. The system’s circulator continues to operate, and thus heat continues to flow Figure 4-1 from the thermal mass of the boiler to the distribution system. After a few minutes, the water temperature 20 feet finned-tube 50,000 Btu/hr inside the boiler has dropped to perhaps 170ºF due to baseboard convector heat disipaton heat dissipation, and the burner is turned back on. Each boiler cycle represents an attempt by the system to reach thermal equilibrium that is subsequently interrupted by the high limit controller, so that operating conditions remain safe and reasonably efficient. Taverage = 420 ºF (at thermal equilibrium Here’s another example: Imagine a hydronic floor heating system that has eight parallel 350-foot circuits of ½” PEX tubing embedded in a bare concrete slab. The system is directly piped to a 50,000 Btu/hr boiler, as shown in Figure 4-2. The boiler’s temperature limiting controller has been set by the installer for 140ºF because that’s the temperature the installer thinks the boiler should supply to the distribution system. Also, assume that the load on the system remains constant at 50,000 Btu/hr.

50,000 Btu/hr When the system turns on, the boiler’s outlet temperature heat input climbs over a period of several hours and eventually

27 temperature at design load, and the Figure 4-2 total heat output of the distribution 50,000 Btu/hr heat system at that same condition. This dissipation supply temperature information should be determined as at thermal part of any accurate design process. equilibrium = 99 ºF Once a pairing of supply water temperature and corresponding 8 circuit of 1/2" PEX, each 350 feet heat output from the distribution long, embedded in system is established, a graph concrete slab. can be created based on the heat output of a distribution system being approximately proportional limit! to the difference between supply controller! water temperature and room air set for ! temperature. 140 ºF Figure 4-3 shows this relationship for a hypothetical distribution 50,000 Btu/hr system that’s designed to release boiler output 50,000 Btu/hr into a space at 70ºF air temperature, when supplied with 180ºF water. stabilizes at 99ºF. The burner remains in continuous The vertical axis of Figure 4-3 operation, but the water temperature leaving the boiler indicates the system’s heat output. The lower horizontal doesn’t rise above 99ºF. The installer thinks there’s axis shows the difference between supply water something wrong because the boiler is not reaching the temperature and room air temperature. This temperature temperature (140ºF) that was set on the boiler’s difference can be thought of as the “driving ∆T” that high limit controller. After several more hours of continuous operation, the boiler’s outlet temperature Figure 4-3 remains at 99ºF. The installer now thinks that the design load condition high limit controller is defective, removes it, and heads to the wholesaler for a replacement. supply water temperature (ºF) (assuming room air temp. = 70 ºF) There’s nothing wrong with the boiler’s high limit 70 90 110 130 150 170 190 controller! The system simply reached thermal 50000 equilibrium at a water temperature well below 45000 the boiler’s high limit setting. The floor heating 40000 subsystem was capable of releasing 50,000 35000 Btu/hr of heat when supplied with 99ºF water. Thermodynamically, there was no need for the 30000 water temperature to rise higher. Because the limit 25000 controller was set well above this temperature, it (Btu/hr) 20000 could not intervene, and thus could not affect the operating conditions. Its presence is irrelevant, as 15000 far as “controlling” this system. heat output of source = 10000 5000 PREDICTING THERMAL EQUILIBRIUM heat dissipation from distribution system It’s possible to estimate the supply water 0 temperature at which a specified distribution system 0 20 40 60 80 100 120 supply water temp. - room air temp. (ºF) will reach thermal equilibrium. What’s needed is an accurate relationship between the supply water no load condition

28 “pushes” heat from the water out though the heat emitters and into the room air. Figure 4-4

The yellow dot at the top right of the graph represents 10000 design load conditions (e.g., a supply water temperature 9000 of 180ºF, and a corresponding heat dissipation of 50,000 Btu/hr from the distribution system). The yellow dot in the 8000 lower left represents a “no load” condition. Supplying 70ºF 7000 water to any distribution system contained within a space at 70ºF air temperature would yield zero heat transfer. The 6000 red line represents the proportionality between these two 5000 extreme conditions. 4000 The supply water temperature necessary to dissipate 3000 any rate of heat output between 0 and 50,000 Btu/ Heat output (Btu/hr) hr can be estimated by finding that rate on the vertical 2000 axis, drawing a horizontal line to intersect the red line on 1000 Room air temperature the graph, and then a vertical line from this intersection assumed to be 70 ºF to the lower horizontal axis. The number read from the 0 horizontal axis is the ∆T between the necessary supply 70 80 90 100 110 120 water temperature and the room air temperature. Supply water temperature (ºF)

For example, for the distribution system represented The greater the total surface area of heat emitters in a in Figure 4-3 to release 20,000 Btu/hr, the ∆T between system, the lower the required supply water temperature. supply water temperature and room air temperature needs to be approximately 44ºF. If the room temperature LOWERING WATER TEMPERATURE was 70ºF, the supply water temperature would need to be OF EXISTING SYSTEMS approximately 44 + 70 = 114ºF. If the room temperature Many hydronic heating systems in pre-1980 North was to be 65ºF, the necessary supply water temperature American buildings were designed when energy was for the same rate of heat output would be 44 + 65 = 109ºF. relatively inexpensive. Boilers burning fossil fuels were the dominant heat source, and their efficiency was SUPPLY WATER TEMPERATURE of lesser concern than it is today. Designers typically vs. HEAT EMITTER SURFACE AREA specified heat emitters, such as finned-tube baseboard, The slope of the line of heat output versus the difference that required high supply water temperatures (180+ºF) at between supply water temperature and room air design load conditions. High water temperatures allowed temperature depends on the total surface area of the the heat emitters to be relatively small, which reduced heat emitters in the distribution system. The greater the installation cost. Very few hydronic heating systems were total surface area of heat emitter, the steeper the slope of designed around low-temperature heat emitters. the line, as shown in Figure 4-4. Today, these “legacy” distribution systems present Steeper sloped lines imply greater heat output rates at a challenge when an older heat source needs to lower supply water temperature. be replaced or supplemented by a modern lower temperature heat source, such as a mod/con boiler or a Figure 4-4 assumes that panel radiators are used as the hydronic heat pump. heat emitters. The larger the panel radiator’s surface area, the greater it’s heat output at a given supply water Although a mod/con boiler can produce the high water temperature. temperatures required by older distribution systems, they do so with a significant loss in thermal efficiency. When a modern heat source such as a modulating/ High-temperature distribution systems do not create condensing boiler or hydronic heat pump is used, a critical conditions that allow flue gases to condense inside the design objective is to keep the required supply water boiler. Without flue gas condensation, a supposedly high- temperature to the distribution system as low as possible. performance mod/con boiler will yield minimal efficiency Doing so improves the thermal efficiency of the heat gains over a lower cost conventional boiler. source, and thus minimizes the primary energy required.

29 Most hydronic heat pumps cannot reliably produce water For example, assume an existing building has a design temperatures over 125ºF. Those that can reach higher heating load of 100,000 Btu/hr, based on maintaining temperatures often do so with a major decrease in their an interior temperature of 70ºF. The existingDRAFT hydronic (4-16-18) coefficient of performance. distribution system uses standard finned-tube baseboard DRAFTand (4-16-18) requires a supply water temperature of 180ºF at In summary, nearly all contemporary hydronic heat designconditions. load conditions. Also assume Also, assumethat improvements that improvements to the building envelope have reduced the design sources perform better when matched with a distribution to the building envelope have reduced the design load system that1. Reduceoperates theat a design lower water load temperature.of the building What envelope from loadthrough100,000 from improvementsBtu/hr100,000 to Btu/hr70,000 suchto Btu/hr.70,000 as added TheBtu/hr. new The supply new supply water temperature to the existing follows are concepts and methods for lowering the water waterdistribution temperature system to the under existing design distribution load conditions system can be estimated using Formula 4-1: temperatureinsulation, of existing better higher windows temperature and reduced hydronic air leakage. under design load conditions can be estimated using distribution systems. Formula 4-1:

The two fundamentals2. Add heat emitters ways to toreduce the existing the supply system. water ⎛ 70,000 ⎞ Tnew = 70 + ⎜ ⎟ × (180 − 70) = 147º F temperature of any hydronic heating system are: ⎝ 100,000⎠

1. ReduceCombinations the design ofload these of thetwo buildingapproaches envelope are also A graphpossible. of the supply water temperature versus outdoor through improvements such as added insulation, better temperatureA graph offor the both supply the waterexisting temperature building versus(e.g., with outdoor temperature for both the existing windows and reduced air leakage. designbuilding load of(e.g., 100,000 with design Btu/hr), load and of after100,000 the Btu/hr),envelope and after the envelope improvements which Building envelope improvements reduce the designimprovements heating loadwhich of reduced the building. the design After loadsuch to 70,000 2. Add heat emitters to the existing system. Btu/hr,reduced is shown the in design Figure load 4-5. to 70,000 Btu/hr, is shown in Figure 4-5. improvements are made, the original hydronic distribution system can meet the reduced design Combinations of these two approaches are also possible. In this example, reducing the design heating load from load while operating at lower supply water temperatures.100,000[insert Btu/hr ﬁgure to 4-5] 70,000 Btu/hr reduced the required Building envelope improvements reduce the design supply water temperature from 180ºF to 147ºF. heating load of the building. After such improvements are made, theThe original change hydronic in supply distribution water systemtemperature can meet depends on the change180 ºF supply in design @ design heating load load. The (existing system) the reducednew design supply load water while temperature operating at can lower be supplydetermined Figure based 4-5 on the same concepts used for water temperatures. 147 ºF supply @ design load (after envelope improvements) outdoor reset control. It can be calculated using formula 1804-1 :ºF supply @ design load The change in supply water temperature depends on 190(existing system) the change in design heating load. The new supply 147 ºF supply @ design load water temperatureFormula 4-1 can: be determined based on the 170 same concepts used for outdoor reset control. It can be (after envelope improvements) calculated using Formula 4-1: 150 190 Formula 4-1: 130 170 ⎛ ⎞ 110 Qnew Tnew = Tin + ⎜ ⎟ × (TDe − Tin ) 150 90 ⎝ Qexisting ⎠ Supply water temperature (ºF)

70 130 Where: 70 60 50 40 30 20 10 0 Outdoor temperature ( ºF) Tnew = supply water temperature at design load after building envelopeWhere: improvements (ºF) 110 Tin = desired indoor air temperature (ºF) Qnew = designTnew =heating supply load water after temperature building envelope at design load after building envelope improvements (ºF) 90 improvements (Btu/hr) Supply water temperature (ºF) Tin = desired indoor air temperature (ºF) In this example, reducing the design heating load from 100,000 Btu/hr to 70,000 Btu/hr reduced Qexisting = existing design heating load (before improvements) (Btu/hr) the required supply water temperature from 180 ºF to 147 ºF. Qnew = design heating load after building envelope improvements70 (Btu/hr) T = existing supply water temperature at design load De 70 60 50 40 30 20 10 0 (before improvements)Qexisting = existing (ºF) design heating load (before improvements) (Btu/hr) Outdoor temperature ( ºF) TDe = existing supply water temperature at design load (before improvements) (Btu/hr)

For example, assume an existing building has a design heating load of 100,000 Btu/hr, based Page "70 of "128 30 on maintaining an interior temperature of 70 ºF. The existing hydronic distribution system uses standard ﬁn-tube baseboard and requires a supply water temperature of 180 ºF at design load

Page "69 of "128 ADDING HEAT EMITTERS TO tubing that doesn’t have fins on it. The existing finned LOWER SUPPLY WATER TEMPERATURE tube length will be designated as Le. If reducing the building’s design heating load is not an option, or does not lower the required supply water Step 3: Determine the desired (lower) supply water temperature to the desired value, it will be necessary to temperature for which the system is to supply design add heat emitters to the system. load output.

A wide variety of heat emitters could be added. For Step 4: Estimate the lower average circuit water example, an existing high-temperature baseboard system temperature by subtracting 5 to 10ºF from the supply could have more baseboard added to it, assuming water temperature determined in Step 3. In circuits with sufficient wall space is available. Another option might more than 2 gallons per minute flow rate through the be to change out some existing baseboard emitters with baseboard, assume the average water temperature will new baseboard emitters that have higher heat output be 5ºF below the supply water temperature. In circuits ratings. It may also be possible to add a different type of with less than 2 gallons per minute flow rate through the heat emitter to the system. An example would be adding baseboard, assume the average water temperature will panel radiators to an existing baseboard system. Another be 10ºF below the lower supply water temperature. would be adding some areas of radiant panel heating (floor, wall or ceiling) to the existing system. Step 5: Find the new average circuit water temperature on the horizontal axis of the graph in Figure 4-6. Draw a The choice of which type of heat emitter to add will vertical line up from this point until it intersects the red depend on several factors, including: curve. Draw a horizontal line from this intersection to the vertical axis of the graph, and read the heat output of the 1. Availability of different makes/models of heat emitters. finned tube at the lower average circuit water temperature. 2. Cost of the new heat emitters. This number is designated as qL. The green lines and 3. How difficult it is to integrate the new heat emitters into numbers in Figure 4-6 show how qL is determined for an the building. average circuit water temperature of 115ºF. 4. Aesthetic preferences. 5. Floor coverings (in the case of radiant floor panels). 6. Surface temperature limitations (in the case of radiant Figure 4-6 panels). 7. The specific supply water temperature that is to supply Note: This graph is based on design load output in the renovated distribution system. standard residential finned-tube baseboard with 2.25" square fins and 3/4" tubing, operating at 1 gpm flow Each project has to be evaluated individually based on rate. It does NOT include the 15% these factors. heating effect factor allowance included in some I=B=R rating tables. ADDING MORE FINNED-TUBE BASEBOARD 900 One way to lower the required supply water temperature of an existing baseboard system is to add more baseboard. 800 65 ºF air temperature entering baseboard 700 The following procedure can be used to calculate the amount of finned-tube baseboard to be added to reduce 600 the supply water temperature at design load to a pre- q =146 Btu/hr/ft 500 L determined value. It assumes that the baseboard being added is the same make and model as the existing 400 baseboard. It also assumes that the existing baseboard is a standard residential-grade product with nominal 300 2.25” square aluminum fins with an I=B=R rated output of 200 approximately 600 Btu/hr/ft at 200ºF water temperature. Heat output per foot (Btu/hr/ft) 115 ºF 100 Step 1: Accurately determine the building’s design heat load. 0 65 85 105 125 145 165 185 205 220 Step 2: Determine the total length of finned tube in the Water temperature (ºF) existing distribution system. Do not include the length of

31 DRAFT (4-16-18)

The green lines and numbers in Figure 4-6 show how qL is determined for an average circuit water temperature of 115ºF.

[insert ﬁgure 4-6]

Note: This graph is based on standard residential ﬁn-tube baseboard with 2.25" square ﬁns and 3/4" tubing. 900

800 65 ºF air temperature entering baseboard 700 600 q =146 Btu/hr/ft 500 L 400 300 200

Heat output per foot (Btu/hr/ft) 115 ºF 100 0 65 85 105 125 145 165 185 205 220 Water temperature (ºF)

Step 6: Determine the length of baseboard to be added using Formula 4-2.

Step 6: Determine the length of baseboard to be added In most buildings, adding this much baseboard is not a using Formula 4-2.Formula 4-2: practical solution, especially in kitchens or bathrooms. Alternatives include using baseboard with higher heat Formula 4-2: output or using other types of heat emitters to achieve design load the necessary design load output. Ladded = − Le qL One option is to consider adding “high output” finned- Where: tube baseboard rather than standard baseboard. Figure Ladded = length Where:of finned tube of same make/model 4-7 shows the heat available from one model of high baseboard to be added (feet) output baseboard (shown as the blue curve), and for design load = designLadded heating = length load of finned-tubeof building (Btu/hr)of same make/modelcomparison, baseboard standard to be residential added (feet) baseboard (shown as qL = output of baseboarddesign load at = thedesign lower heatingDRAFT average load (4-16-18) circuitof building the (Btu/hr) red curve). water temperature (Btu/hr/ft) qL = output of baseboard at the lower average circuit water temperature (Btu/hr/ft) Le = total existing length of baseboard in system (feet) The steps of the previous procedure can be modified For example: AssumeLe = total a building existing has length a calculated of baseboard design in load systemto of determine 40,000 (feet) Btu/hr, the and amount its of high output finned-tube For example: Assume a building has a calculated baseboard that is required to reduce the supply water distribution system contains 120 feet of standard residential finned-tube baseboard. The goal is design load of 40,000 Btu/hr, and its distribution system temperature to the system under design load. containsto reduce 120 the feetsupply of water standard temperature residential to 120 finned-tube ºF at design conditions, using more of the same baseboard. The goal is to reduce the supply water Steps 1-4: Same baseboard. Assume the temperature drop of the distributionPage system "73 ofis 10"128 ºF. Determine the temperature to 120ºF at design conditions, using more of amountthe same of baseboardbaseboard. that Assume must be the added: temperature drop of Step 5: Determine the output of high-output baseboard the distribution system is 10ºF. Determine the amount of at the average circuit water temperature using Figure baseboard that must be added: 4-7 (or manufacturer’s literature for a specific make and Solution: model). Solution: Step 1: The design load has been calculated as 40,000 Btu/hr.Step 6: The required length of high output baseboard to Step 1: The design load has been calculated as 40,000 add to the system is found using Formula 4-3. Btu/hr. Step 2: The total amount of finned-tube in the system is 120 feet. Step 2: The total amount of finned tube in the system is Figure 4-7 120 feet. Step 3: The lower supply water temperature at design load will be 120ºF. high output standard Step 3: The lower supply water temperature at design fin-tube fin-tube load will be 120ºF. baseboard baseboard Step 4: The lower average circuit water temperature will be 120 - (10/2)500 = 115ºF. Step 4: The lower average circuit water temperature will 450 65 ºF air temperature be 120 - (10/2) = 115ºF. entering baseboard Step 5: The output of the finned-tube at an average circuit water temperature400 of 115 ºF is Stepdetermined 5: The output from Figure of the 4-6finned as 146 tube Btu/hr/ft. at an average circuit 350 water temperature of 115ºF is determined from Figure 4-6 as 146 Btu/hr/ft. 300 Step 6: The required additional length of baseboard is now calculated using Formula 4-2: 250 Step 6: The required additional length of baseboard is now calculated using Formula 4-2: 200 150 ⎡ Btu ⎤ Heat output (Btu/hr/ft) 100 design load ⎢ 40,000 ⎥ L = − L = ⎢ hr −120⎥ = 154 ft added e Btu 50 qL ⎢ 146 ⎥ ⎢ hr ⋅ ft ⎥ ⎣ ⎦ 0 65 75 85 95 105 115 125 135 145 155 Although it might be possible to add 154 feet of baseboard Water temperature (ºF) to Althoughthe system, it might it would be possible require tolots add of 154available feet of wall baseboard space. to the system, it would require lots of available wall space. In most buildings, adding this much baseboard is not a practical solution, especially in kitchens or bathrooms. Alternatives include using baseboard with higher heat 32output or using other types of heat emitters to achieve the necessary design load output.

Page "74 of "128 DRAFT (4-16-18)

Step 6: The required length of high output baseboard to add to the system is found using Formula 4-3.

Formula 4-3:

Formula 4-3: still a substantial length. The building must be carefully design load-(q )(L ) evaluated to see if this additional length of baseboard can L = L e ho q be accommodated. Where: ho Lho = length of high-output finned-tube baseboard to be If the added length of high-output baseboard cannot be added (feet) accommodated, another option is to raise the supply water design load = designWhere: heating load of building (Btu/hr) temperature constraint from 120 to 130ºF under design qL = output of existingLho = length baseboard of high output at the finned-tube lower average baseboard load to be conditions. added (feet) This would reduce the amount of added water temperature (Btu/hr/ft) high-output baseboard in the previous example to 40 feet. design load = design heating load of building (Btu/hr) Le = total existing length of baseboard in system (feet) qho = output ofqL =high-output output of existing baseboard baseboard at atthe the lower lower averageOTHER water HEAT temp EMITTERerature (Btu/hr/ft) OPTIONS average water temperatureLe = total existing (Btu/hr/ft) length of baseboard in system If(feet) the amount of finned tube that must be added is beyond what can be accommodated, consider other For example: Assumeqho = output a building of high has output a calculated baseboard designat the lower added average heat water emitter temperature options. (Btu/hr/ft) They include panel radiators, load of 40,000 Btu/hr, and its distribution system contains fan-coils or areas of radiant floor, radiant wall or radiant 120 feet of standardFor example: residential Assume finned-tube a building baseboard.has a calculated ceiling design panels. load of 40,000In each Btu/hr, case, and the its selection of these new The goal is to reduce the supply water temperature at heat emitters should be based on a selected supply water distribution system contains 120 feet of standard residential finned-tube baseboard. The goal is design load to 120ºF. Additional high-output baseboard will temperature at design load, along with a “credit” for the DRAFT (4-16-18) be added to allowto thisreduce lower the watersupply temperature water temperature operation. at design existing load to 120heat ºF. emitters Additional in the high system output operating at the lower Assume that thebaseboard temperature will be dropadded of to allowthe distributionthis lower water supply temperature water operation. temperature. Assume The that fundamental the concept is system at design load is 10ºF, and this existing baseboard given by FormulaFormula 4-4. 4-4: has the same outputtemperature as in drop the of previous the distribution example system (146 at design load is 10 ºF, and this existing baseboard Btu/hr/ft at averagehas thecircuit same water output temperature as in the previous of 115ºF). example Formula(146 Btu/hr/ft 4-4: at average circuit water Determine the amount of high-output baseboard required temperature of 115 ºF). Determine the amount of high output baseboard required based on the based on the thermal performance of the high-output Qn = design load - Q baseboard shownthermal in Figure performance 4-7. shown in figure 4-7. Where: Qn = required heat output of the new heat emitters at Solution: lower supply waterWhere: temperature (Btu/hr) Solution: design load = the design heating load of the building Step 1: The design load has been calculated as 40,000 (Btu/hr) Qn = required heat output of the new heat emitters at lower supply water temperature (Btu/hr) Btu/hr. Step 1: The design load has been calculated as Q40,000e = heat Btu/hr outputdesign of existing load = heatthe designemitters heating at the lowerload of the building (Btu/hr) supply water temperature (Btu/hr) Step 2: The total amount of existing DRAFTfinned (4-16-18)tube in the Qe = heat output of existing heat emitters at the lower supply water temperature (Btu/hr) Step 2: The total amount of existing finned-tube in the system is 120 feet system is 120 feet. Once the value of Qn is determined, designers can Step 3: The new lower supply water temperature at design loaduse will betables 120 ºF or graphs from manufacturers to determine Step 3: The new lower supply water temperature at the heat outputOnce of specific the value heat of emitters Qn is determined, based on the designers can use tables or graphs from manufacturers to design load will be 120ºF. average water temperature within them. Remember that Step 4: The new lower average circuit water temperature will be 120 - 5 = 115 ºF Page "76the of "128average waterdetermine temperature the heatwill typically output ofbe specific5 to 10ºF heat emitters based on the average water temperature Step 4: The new lower average circuit water temperature lower than the supplywithin waterthem. temperature. Remember that the average water temperature will be 5 to 10 ºF lower than the will beStep 120 5: -The 5 = output 115ºF. of high output finned-tube at an average water temperature of 115 ºF is The goal is to selectsupply a grouping water temperature. of new heat emitters with determined from figure 4-7 as 335 Btu/hr/ft. Step 5: The output of high-output finned tube at an a total heat output that’s approximately equal to the value average water temperature of 115ºF is determined from of Qn in Formula 4-4. FigureStep 4-7 6: as The 335 required Btu/hr/ft. length of high output baseboard to add to the system is foundThe using goal is to select a grouping of new heat emitters with a total heat output that’s Formula 4-3. Assume a building has a calculated design load of 40,000 Step 6: The required length of high-output baseboard to Btu/hr, and its approximatelydistribution system equal contains to the value120 feet of Qofn in Formula 4-4. add to the system is found using Formula 4-3. standard residential finned-tube baseboard. The goal is to reduce the supply water temperature to 120ºF under design load-(q )(L ) 40,000-(146)(120) design load conditions.Assume aPanel building radiators has a calculatedare available design load of 40,000 Btu/hr, and its distribution system L = L e = = 67 ft ho q 335 in 24” x 72” size that can release 4,233 Btu/hr when ho operated at an containsaverage 120water feet temperature of standard of residential115ºF in finned-tube baseboard. The goal is to reduce the Although this is a reduction compared to the 154 feet of rooms with 70ºFsupply interior water temperature. temperature How many to 120 of ºFthese under design load conditions. Panel radiators are available added baseboard required in the previous example, it is radiators are necessary to meet the design load? in 24” x 72” size that can release 4,233 Btu/hr when operated at an average water temperature Although this is a reduction compared to the 154 feet of added baseboard required in the previous example, it is still a substantial length. The building must be carefully evaluatedof 115 to ºFsee in rooms with 70 ºF interior temperature. How many of these radiator are necessary to if this additional length of baseboard can be accommodated. meet the design load? 33

If the added length of high output baseboard cannot be accommodated, another option is to raise the supply water temperature constraint from 120 to 130 ºF under design loadSolution: conditions. First, use Formula 4-4 to determine the output required of the new radiators. This would reduce the amount of added high output baseboard in the previous example to 40 feet.

Qn = design load - Qe = design load - (qL )(Le ) = 40,000- (146)(120) = 22,480Btu / hr OTHER HEAT EMITTER OPTIONS:

If the amount of ﬁnned-tube that must be added is beyond what can be accommodated,The number of radiators needed is then found as follows: consider other added heat emitter options. They include panel radiators, fan-coils, or areas of radiant ﬂoor, radiant wall, or radiant ceiling panels. In each case, the selection of these new 22,480Btu / hr heat emitters should be based on a selected supply water temperature at design load, along = 5.3 radiators Btu / hr with a “credit” for the existing heat emitters in the system operating at the lower supply42 3water3 temperature. The fundamental concept is given by Formula 4-4 radiator Page "78 of "128

Page "77 of "128 DRAFT (4-16-18)

Formula 4-4:

Qn = design load - Q

Where:

Qn = required heat output of the new heat emitters at lower supply water temperature (Btu/hr) design load = the design heating load of the building (Btu/hr)

Qe = heat output of existing heat emitters at the lower supply water temperature (Btu/hr)

Once the value of Qn is determined, designers can use tables or graphs from manufacturers to determine the heat output of speciﬁc heat emitters based on the average water temperature DRAFT (4-16-18) within them. Remember that the average water temperature will be 5 to 10 ºF lower than the Formula 4-4: supply water temperature.

Qn = design load - Q

Where: The goal is to select a grouping of new heat emitters with a total heat output that’s

Qn = required heat output of the new heat emitters at lower supply water temperature (Btu/hr) approximately equal to the value of Qn in Formula 4-4. design load = the design heating load of the building (Btu/hr)

Qe = heat output of existing heat emitters at the lower supply water temperature (Btu/hr) Assume a building has a calculated design load of 40,000 Btu/hr, and its distribution system Once the value of Qn is determined, designers can use tables or graphs from manufacturers to determine the heat outputcontains of speciﬁc 120 heat feetemitters of based standard on the average residential water temperature ﬁnned-tube baseboard. The goal is to reduce the within them. Remember that the average water temperature will be 5 to 10 ºF lower than the supply water temperature.supply water temperature to 120 ºF under design load conditions. Panel radiators are available in 24” x 72” size that can release 4,233 Btu/hr when operated at an average water temperature The goal is to select a grouping of new heat emitters with a total heat output that’s

approximately equalof to the115 value ºF of inQn inrooms Formula 4-4with. 70 ºF interior temperature. How many of these radiator are necessary to meet the design load? Assume a building has a calculated design load of 40,000 Btu/hr, and its distribution system contains 120 feet of standard residential ﬁnned-tube baseboard. The goal is to reduce the supply water temperature to 120 ºF under design load conditions. Panel radiators are available in 24” x 72” size thatSolution: can release 4,233 First, Btu/hr usewhen operatedFormula at an 4-4average to waterdetermine temperature the output required of the new radiators. of 115 ºF in rooms with 70 ºF interior temperature. How many of these radiator are necessary to meet the design load? Solution: First, use Formula 4-4 to determine the output 6. What type of control will be used to regulate heat Solution: First, use FormulaQ = d4-4es toig determinen load the - Qoutput= requireddesig ofn thelo newad radiators.- (q )(L ) = 40,000- (146)(120) = 22,480Btu / hr required of the newn radiators. e L outpute to each zone of the distribution system?

Qn = design load - Qe = design load - (qL )(Le ) = 40,000- (146)(120) = 22,480Btu / hr There is no one best approach. Every situation must be The number of radiators needed is then foundevaluated as follows: individually while weighing these factors to determine the best fit for that project. The number of radiators needed is then found as follows: The number of radiators needed is then found as follows: CONVERTING SERIES LOOPS 22,480Btu / hr = 5.3 2ra2di,at4or8s 0Btu / hr Btu / hr = 5.3 radiators TO PARALLEL BRANCHES 4233 Btu / hr radiator 4233 Many residential hydronic systems have finned-tube radiatPageor "78 of "128 baseboards connected in series or “split-series” circuits, as shown in Figures 4-8 and 4-9. The designer could either add 6 of these panel radiators,Page "78 of "128 or choose a slightly higher supply water temperature and When several heat emitters will to be added to a system use 5 radiators. using series or split-series connected baseboards, they should NOT be simply cut into the series circuit. Doing Another option is to use panel radiators of different sizes, so could substantially increase the flow resistance of that provided that their total output at the lower supply water circuit, which reduces flow, assuming the same circulator temperature could meet the value of Qn. In this example, is used. Adding heat emitters in series also increases the the 6 new radiators would be combined with the 120 feet temperature drop of the circuit. This reduces the heat output of existing baseboard to provide the 40,000 Btu/hr design of heat emitters near the end of the circuit, especially when load at a supply temperature of 120ºF. the supply temperature to that circuit is lowered.

A similar calculation could be made for fan-coils, air One approach is to make strategic cuts into the series handlers or other heat emitters. circuit where it is easiest to access, and reconnect the segments back into a parallel distribution system. In the case of radiant panels, the designer needs to determine the output of each square foot of panel based on Figure 4-8 the lower average circuit temperature, and the specific construction of the panel. The total required panel area is VENT Auxiliary found by dividing this number into the boiler value of Qn.

PIPING FOR SUPPLEMENTAL series circuit HEAT EMITTERS existing boiler There are several factors that could influence how the supplemental heat emitters are piped into the system. They include: Figure 4-9

1. Are the existing heat emitters piped in a series circuit? 2. Are the existing heat emitters piped as individual parallel circuits?

3. Where is the piping between existing VENT Auxiliary heat emitters easiest to access? boiler 4. Are there multiple heat emitters within split series circuit a given space? 5. What are the flow resistance characteristic of the supplemental heat emitters relative to those of the existing existing boiler heat emitters?

34 These cuts could make each room a separate parallel In this example the existing series circuit was divided into circuit. They might also make a group of two rooms into four branch circuits. Supplemental heat emitters were a new zone. added to each of these branch circuits. Two of the branch circuits received additional finned-tube baseboard, and One of the easiest ways to divide an existing series loop the other two received panel radiators. These heat emitter or split-series distribution system into multiple parallel selections illustrate that multiple types of supplemental circuits is by creating a homerun distribution system. Each heat emitters can be used depending on available wall heat emitter, or grouping of an existing heat emitter and space, budget and aesthetic preferences. a supplemental heat emitter, is supplied by a separate circuit of PEX or PEX-AL-PEX tubing. This tubing is easy In some branch circuits, the supplemental heat emitters to route through cavities or along framing. The homerun were added upstream of the existing baseboard. In others, circuits begin and end at a manifold station. The concept they were added downstream of the existing baseboard. is shown in Figure 4-10. The choice depends on the available wall space and

Figure 4-10

EXISTING DISTRIBUTION SYSTEM

existing series circuit

MODIFIED DISTRIBUTION SYSTEM

existing baseboard added baseboard

added panel rad.

1/2" PEX pr PEX-AL-PEX added panel rad. tubing (typical)

manifold station

35 placement of the existing baseboard within each room. Designers should estimate the water temperature in Figure 4-11 the circuit at the location where the supplemental heat baseboard enclosure emitter will be placed, and size it accordingly. 3/4" copper x 1/2" PEX-AL-PEX vent vent ell fitting The ¾” copper tubing in the existing circuit was cut at radiator valve locations that preserved a reasonable amount of existing 3/4" finned-tube element tubing, but also allow convenient transition to ½” PEX or PEX-AL-PEX tubing. Adapter fittings for transitioning from ¾” copper to ½” PEX or PEX-AL-PEX tubing are readily available. The ½” PEX or PEX-AL-PEX supplies and returns are routed back to a manifold station. That thermostatic manifold station should be equipped with individual operator circuit-balancing valves that allow the flow rate through each of the new branch circuits to be adjusted.

All four branch circuits in Figure 4-10 operate simultaneously 1/2" MPT (e.g., they are not configured as individual zones). As 1/2" x 1/2" PEX-AL-PEX 1/2" such, this distribution system presents a constant flow PEX-AL-PEX adapter fitting PEX-AL-PEX tubing tubing resistance. Due to the parallel versus series configuration, that flow resistance of the modified distribution system is likely to be lower than that of the original series circuit. This 90º turn as it passes through the valve, and flows into should be verified by calculating the head loss or pressure the baseboard element. The other end of the finned- drop of the branch with the greatest flow resistance using tube element can be equipped with a transition adapter standard hydronic pipe analysis methods. If the head (straight or 90º) to convert from ¾” copper to ½” PEX or loss and total flow rate through the modified distribution PEX-AL-PEX. system is significantly lower than that of the original series circuit, consider a replacement circulator that operates Another type of thermostatic radiator valve allows the with reduced power input. valve body and actuator to be mounted within the baseboard enclosure, while the adjustment knob is A significant benefit of a parallel distribution system is mounted at normal thermostat height on the wall. The that all branches receive water at approximately the same adjustment knob connects to the valve actuator using a temperature. This is likely to boost the heat output of capillary tube, as seen in Figure 4-12. some existing baseboards that were previously located near the end of the series circuit.

CREATING NEW ZONES Figure 4-12 Another advantage of parallel distribution systems is the ease of creating multiple zones. If the existing series circuit is converted to multiple branches, each of those branches could be equipped with a thermostatic radiator valve. These non-electric valves automatically modulate to vary the flow rate in each branch in response to the room temperature. As room temperature begins to drop, the thermostatic valve opens to increase flow through that branch circuit to boost heat output, and vice versa.

Thermostatic radiator valves are available in several configurations. One is known as an “angle pattern” supply valve. It can be mounted on the inlet of a finned- tube baseboard element, as shown in Figure 4-11.

The thermostatic actuator on the valve projects through a hole in the end cap of the baseboard. Heated water enters the port of the valve facing the floor, makes a

36 Figure 4-13 MODIFIED DISTRIBUTION SYSTEM

TRV existing baseboard added baseboard

TRV

existing baseboard added baseboard TRV

added panel rad.

TRV

pressure added regulated panel circulator radiator 1/2" PEX pr PEX-AL-PEX tubing (typical)

manifold station

thermostats Figure 4-14 MODIFIED DISTRIBUTION SYSTEM

existing baseboard added baseboard added panel radiator

existing baseboard added baseboard

pressure regulated circulator 1/2" PEX pr PEX-AL-PEX added tubing (typical) panel radiator

manifold station

manifold valve actuators

37 It’s also possible to use panel radiators with built-in necessary supplemental heat emitters. However, if the thermostatic radiator valves. piping modifications to do this are comparable in cost/ time to creating two independent zones, then the latter is Figure 4-13 shows how the distribution system can be arguably a better choice. modified using thermostatic radiator valves to create a distribution system with four independently controlled 4. Once all the supplemental heat emitters have been zones. This adds flexibility for adjusting interior comfort selected, and the proposed modifications to the conditions far beyond that of the original series loop system. distribution system have been sketched, always run a flow and head loss analysis for the modified system. This Another possibility is to install a low-voltage (24 VAC) is used to confirm sufficient flow to each branch, and to manifold valve actuator on each circuit valve at the determine a suitable circulator for the modified system. manifold station. These actuators are wired to four new thermostats, one for each zone. This option is shown in 5. If an existing conventional boiler will be used as Figure 4-14. the heat source for the modified system, which now operates at significantly lower water temperatures, be The systems shown in Figures 4-13 and 4-14 use valves sure the boiler is protected against sustained flue gas for zoning. In Figure 4-13, non-electric thermostatic condensation by installing a thermostatic mixing valve radiator valves regulate the flow through each parallel near the boiler, as shown in Figure 4-15. branch circuit. In Figure 4-14, low-voltage manifold actuators are attached to valves within the return manifold. In both cases, the original circulator has been replaced with a variable-speed pressure-regulated Figure 4-15 circulator. These circulators automatically adjust their distribution speed and power input as the valves open, close or closely system modulate flow. This helps stabilize the differential pressure spaced across the manifold station, and maintain consistent flow tees rates within each zone circuit, regardless of what other zones are operating. These modern circulators also have significantly lower electrical energy consumption relative to standard wet rotor circulators with permanent split thermostatic capacitor (PSC) motors. mixing valve with high Cv. Minimum DESIGN GUIDELINES setting = 130 ºF The modifications shown to convert a series baseboard circuit into parallel branch circuits are just a few of many boiler possibilities. Each conversion situation must consider circulator the exact layout of the existing heat emitters, and the practicality of modifying the system into parallel branches. Designers should follow these guidelines.

1. Always determine what type of supplemental heat existing boiler emitter will be used in each room, and where it will be located before modifying the piping.

2. From the standpoint of cost and installation time, it’s best to use as much of the existing (and accessible) piping as possible.

3. Always consider the benefit versus cost of creating new zones when modifying the existing system. For example, if two bedrooms are typically maintained at the same temperature, and the existing system has accomplished this, it’s likely best to keep these two bedrooms together on the same zone after adding the

38 5. HEAT EXCHANGER FUNDAMENTALS heat exchanger is almost always limited by the natural convection at its outer surface. Many hydronic heating and cooling systems need to move heat from one fluid to another without mixing those Figure 5-1 illustrates a typical immersion heat exchanger fluids. One example is heat moving from hot “system coil within a tank. water” that passes through a cast iron boiler into cooler domestic water that is being heated for showers, washing, etc. Another example is heat moving from an antifreeze Figure 5-1 tank solution that flows through a solar collector to water within a thermal storage tank. The fluids exchanging heat may be at significantly different temperatures and immersion heat bouyancy- pressures, and must be fully isolated from each other. exchanger coil driven natural Such processes require a heat exchanger. hot ﬂuid within tank convection inlet There are many types of heat exchangers. Nearly all of them separate the fluids exchanging heat using metal surfaces. Common materials used to separate the fluids include copper, cupronickel, stainless steel, carbon steel and titanium/stainless steel alloys. These metals all have relatively high thermal conductivity, and thus present minimal conduction resistance. hot ﬂuid EXTERNAL vs. IMMERSION HEAT EXCHANGERS outlet The types of heat exchangers commonly used in hydronic systems can be categorized as external or immersion.

External heat exchangers are standalone devices with a minimum of four piping connections. They have an inlet This heat exchanger geometry is commonly used for and outlet connection for each of the two fluids that pass indirect domestic water heaters that heat cold domestic through the heat exchanger. water in the tank by passing hot water from a boiler or other heat source through the immersion coil. An immersion heat exchanger (also known as an “internal” heat exchanger) usually takes the form of a helical coil Figure 5-2 illustrates the types of the external heat made of copper, stainless steel or carbon steel. It’s exchangers most commonly used in hydronic heating permanently mounted inside a tank, with only the inlet and cooling systems. and outlet connections accessible outside that tank. The outer surface of an immersion heat exchanger is surrounded by the fluid within the tank. Figure 5-2

The flow through each side of an external heat exchanger is typically driven by a circulator. This implies that forced convection heat transfer occurs between each fluid and the surfaces separating the fluids. Recall that forced convection results in higher rates of heat transfer in comparison to natural convection.

The flow through the internal side of an immersion heat exchanger is typically driven by a circulator, and thus heat transfer between this inner surface and the fluid is by forced convection. However, flow over the external side of an immersion heat exchanger is usually created by buoyancy differences in the fluid surrounding the heat exchanger. The heat exchanged between the outer surface and the surrounding fluid is by natural convection. The rate of heat transfer from an immersion

39 All external heat exchanger designs create two pressure- when operated at the same inlet fluid temperatures and tight cavities that separate the fluids exchanging heat. temperature drops. The larger heat exchanger will also They also create an internal surface across which heat have a lower head loss than the smaller exchanger when moves by conduction through the metal separating the operating at the same flow rate. two fluids. Small brazed plate heat exchangers, such as 3” x 8” FLAT PLATE HEAT EXCHANGERS x 20 plates, are often used in residential domestic Flat plate heat exchangers water-heating applications. Larger brazed plate heat Figure 5-3 are the most commonly exchangers, such as 10” x 20” x 80 plates, are capable of used heat exchangers in transferring heat at several hundred thousand Btu/hr, and modern hydronic systems. are used in applications such as snowmelting or district In smaller sizes, these heat heating for small commercial buildings. exchangers are made by stacking specially formed The specifications for brazed plate heat exchangers stainless steel plates, as often list a certain rate of heat transfer. It’s very important shown in Figure 5-3. to realize that this rating is based on specific entering fluid temperatures and flow rates. For example, a certain The narrow spaces heat exchanger may be listed as capable of transferring between the pre-formed 150,000 Btu/hr based on one fluid entering at 180ºF and a plates form channels flow rate of 15 gpm, and the other fluid entering at 100ºF through which each fluid and at 15 gpm. However, any difference between these passes. The surface rating conditions and the actual operating conditions of patterns on the plates the heat exchanger in a given application could have a create turbulence that major impact on heat transfer rate. In general, the greater Courtesy of GEA PHE Systems enhances convection. the difference between the entering fluid temperatures, These patterns also and the higher the flow rates, the higher the rate of heat create a flow distribution manifold at each of the four exchange. This will be discussed later in this section. piping connections. If the channels between the plates were numbered in the same order in which the plates are Large flat plate heat stacked, one fluid flows through channels 1, 3, 5, 7, etc., Figure 5-4 exchangers are also while the other fluid flows through channels 2, 4, 6, 8, etc. fabricated as a stack of stainless steel plates. The plates provide a very large surface area between However, the plates are the two fluids, while also creating a compact device not brazed together. with low fluid volume. The combination of high surface Instead they are stacked area and low volume allows flat plate heat exchangers together with elastomeric to respond very quickly to changes in fluid temperatures gaskets in between. or flow rates. It also allows flat plate heat exchangers to The plate stack is then be significantly smaller than other types of external heat mechanically compressed exchangers of comparable thermal performance. using large threaded steel shafts and nuts. In small to medium sizes, the stainless steel plates are This compression causes brazed together at their perimeters to form a “brazed plate” the gaskets to form the heat exchanger. Once brazed, the plates are permanently perimeter seal as well as sealed in place, and cannot be disassembled for cleaning. internal seals between Brazed plate heat exchangers are usually made in one adjacent plates. The stack

of three common plate sizes (3” x 8”, 5” x 12”, or 10” x Courtesy of B&G of plates and gaskets are 20”). Other plate sizes such as 5” x 20” are sometimes capped at each end by available. The thermal and hydraulic performance of thick steel plates capable a brazed plate heat exchanger is determined by the of withstanding high number of plates used. For example, a 5” x 12” x 40 pressures if necessary. The internal plates and end plates plate heat exchanger has twice as many plates as a 5” x are held in alignment by a structural steel frame. This 12” x 20 plate unit. The 40-plate heat exchanger will be type of heat exchanger is called a “plate & frame” heat capable of approximately twice the rate of heat transfer exchanger. An example is shown in Figure 5-4.

40 One benefit of plate & frame heat exchangers is that FLOW DIRECTION THROUGH HEAT EXCHANGERS the frame is often built so that additional plates can In heat exchangers that operate with liquids, there are be added to the stack if ever needed. An example two possible flow configurations: would be when the rate of heat exchange has to be maintained using a new heat source that supplies lower • Both liquids pass in the same direction through the heat temperature hot water to the heat exchanger. Another exchanger. would be when the rate of heat exchange needs to be • The liquids flow in opposite directions through the heat increased to accommodate expansion of an existing exchanger. heating system. It’s also possible, to disassemble a plate & frame heat exchanger if required for cleaning When the two liquids move in the same direction, the or maintenance. configuration is called “parallel flow.” When the two fluids move in opposite directions, the configuration is called Figure 5-5 shows the rear side of a large plate & frame “counterflow.” heat exchanger used in a district heating application. Notice that the threaded steel rods that hold the plate stack together extend well beyond the rear end Figure 5-6 plate. If more plates are to be added, the nuts on the Thot treaded shafts are removed, allowing the end plate in to slide backwards on the steel frame. The additional plates and their associated gaskets are placed into the stack. The rear end plate is then moved back into hot fluid position, and the nuts retightened to a specified torque, Thot resealing the larger plate stack. out

Tcold out Figure 5-5 cold fluid

Tcold in

Thot in Thot out

Tcold Tcold in parallel flow out

Thot in

hot fluid

Tcold out

cold fluid Thot out

Tcold in

Thot in

Thot out

Tcold out Tcold counterflow in

41 Figure 5-6 show how the temperature of the two liquid Recall from previous discussions that the rate of streams would change as they flow through a heat conduction heat transfer is directly proportional to the exchanger piped for parallel flow versus one piped for temperature difference between the outer surfaces of counterflow. a material through which heat is passing. The higher this temperature difference, the greater the rate of heat In a parallel flow configuration, the temperature of the transfer by conduction. cooler liquid absorbing heat can never meet or exceed the temperature of the exiting hot liquid. All other factors Similarly, the rate of convective heat transfer is directly being equal, the rate of heat exchange is limited by the proportional to the temperature difference between the temperature of the exiting hot fluid stream. temperature in the bulk of the fluid stream (e.g., away from the boundary layer), and the temperature of the solid When a counterflow configuration is used, the leaving material the fluid is passing over. It’s also dependent on temperature of the stream absorbing heat can be higher the convection coefficient between the fluid and surface. than the leaving temperature of the stream providing heat. These relationships suggest that the rate of heat transfer The rate of heat exchange will always be higher when a between two fluids passing through any heat exchanger heat exchanger is configured for counterflow operation. would depend on the temperature difference between the two fluids. However, the temperature difference between Counterflow should also be used with immersion heat the two streams is not the same at all locations on the exchangers. Figure 5-7 shows how the flow through heat exchanger. Each fluid is either losing or gaining heat an immersion coil heat exchanger that’s adding heat to as it passes through the heat exchanger, and thus the the tank (on left) is piped so that flow is in the opposite temperature difference between the fluids varies depending direction of natural convection currents inside the tank on where it is measured within the heat exchanger. (e.g., flow is from top to bottom of coil). Heat transfer theory defines a specific temperature Another immersion heat exchanger that removes heat difference based on the fluid temperatures at all four from the right side of the tank is also piped for counterflow connections of the heat exchanger. That difference (e.g., flow is from bottom to top of coil). is called the log mean temperature difference, and is defined by Formula 5-1. RATE OF HEAT EXCHANGE The rate of heat transfer between two fluids passing through a heat Figure 5-7 exchanger is dependent on several check valves prevent thermosiphoning factors, including: unpressurized vent • The internal surface area that thermal storage tank separates the two fluids. air at heat load • The thermal conductivity and atmospheric source thickness of the plate material. pressure • The convection coefficients developed between each fluid and the internal surfaces. • The flow rate on each side of the heat exchanger. • The specific heat, density and viscosity of each fluid. • The insulation (if any) surrounding heat the heat exchanger. input heat • The direction of flow of one fluid coil extraction relative to another. coil • The “log mean temperature difference” between the two fluids. general direction of flow driven by natural convection in vicinity of vertical coil

42 Figure 5-8 Figure 5-9 (∆T) 1 140 ºF 100 ºF 140 ºF 100 ºF (∆ T ) − (∆ T ) LMTD = 1 2 ⎡ (∆ T ) ⎤ DRAFT (4-16-18) ln ⎢ 1 ⎥ ⎣(∆ T )2 ⎦

122 ºF 77 ºF

DRAFT (4-16-18) (∆ T ) − (∆ T ) 40 − 60 −20 counterﬂow LMTD = 1 2 = = = 49.3º F ⎡ ⎤ ⎡ 40 ⎤ − (∆ T )1 0.405 side absorbing heat ln ln − − − − − − − side supplying heat supplying side ⎢ ⎥ ⎢ ⎥ (Tho Tci ) (Thi Tco ) (122 77) (104 68) 45 36 9 (∆ T ) ⎣ 60 ⎦ LMTD = = = = = 40.3º F ⎣ 2 ⎦ ⎡ − ⎤ ⎡ − ⎤ ⎡ 45 ⎤ side absorbing heat (Tho Tci ) (122 77) 0.223

side supplying heat supplying side ln ln ln Heat transfer theory defines a specific temperature difference based on the fluid temperatures ⎢ ⎥ ⎢ − ⎥ ⎣⎢ 36 ⎦⎥ ⎣(Thi − Tco )⎦ ⎣(104 68)⎦ at all four connections of the heat exchanger. That difference is called the log mean

120 ºF 60 ºF side absorbing heat temperature difference, and is deﬁned by formula 5-1. heat supplying side 120 ºF 60 ºF (∆T) 2 104 ºF 68 ºF Formula 5-1:

Formula 5-1: exchanger, minus the temperature122 ºF 68 difference ºF between (T − T ) − (T − T ) the fluid at the other end, divided by the natural logarithm LMTD = ho ci hi co of the ratio of these two temperature differences. This is parallel flow ⎡(Tho − Tci )⎤ ln ⎢ ⎥ illustrated in Figure 5-9. DRAFT (4-16-18) ⎣(Thi − Tco )⎦ (Tho − Tci ) − (Thi − Tco ) (122 − 68) − (104 − 77) 54 − 27 27 LMTD = = = = = 39.0º F If (∆T) and (∆T) are the same, the LMTD, as it ⎡would(T − T )⎤ ⎡(122 − 68) ⎤ ⎡54 ⎤ 0.693 1 2 ln ho ci ln ln ⎢ ⎥ ⎢ − ⎥ ⎣⎢ 27 ⎦⎥ Where: be calculated using Formula 5-1 or in Figure 5-9,⎣(Thi −isTco )⎦ ⎣(104 77)⎦ LMTD = log mean temperature difference (ºF) undefined mathematically. In these cases, the LMTD = side absorbing heat T140ho =ºF temperature100 ºF Where: of the fluid supplying heat at the outlet (∆T)1 = (∆T)2. heat supplying side of the heat exchanger (ºF) LMTD = log mean temperature difference (ºF) T = temperature of the fluid supplying heat at the inlet of The LMTD of a heat exchanger depends on the relative hi 104 ºF 77 ºF the heat exchangerTho =(ºF) temperature of the fluid supplying heat at theflow outlet directions of the heat exchangerof the two (ºF) fluids passing through the T = temperature of the fluid absorbing heat at the outlet heat exchanger. Figure 5-10 shows two identical heat co Thi = temperature of the fluid supplying heat at the inlet of the heat exchanger (ºF) of the heat exchanger (ºF) exchangers, one piped for parallel flow and the other Tco = temperature of the fluid absorbing heat at the outlet of the heat exchanger (ºF) Tci = temperature of the fluid absorbing heat at the inlet piped for counterflow.It can The be LMTD shown forthat eachthe LMTD situation will always has be greater when the two fluids exchanging heat pass

of the heat exchangerside absorbing heat Tci = temperature (ºF) of the fluid absorbing heat the inletbeen of the calculated heat exchanger using (ºF)Formula 5-1. side supplying heat supplying side in opposite directions through the heat exchanger. Thus, to attain the highest rate of heat ln [ ] = the natural logarithm of the quantity within brackets. ln [ ] = the natural logarithm of quantity within brackets.It can be shown thatexchange, the LMTD heat willexchangers always shouldbe greater always be piped for counterflow. Example:120 ºF Calculate60 ºF the log mean temperature difference when the two fluids exchanging heat pass in opposite for the heat exchanger shown in Figure 5-8. directions through the heat exchanger. Thus, to attain the Example: Calculate the the log mean temperature difference for the heat exchanger shown in highest rate of heat Onceexchange, the LMTD heat is calculated,exchangers the rateshould of heat exchange across the heat exchanger can figure 5-8. always be piped for counterflow.(theoretically) be calculated using formula 5-2: (Tho − Tci ) − (Thi − Tco ) (140 −100) − (120 − 60) 40 − 60 −20 LMTD = = = = = 49.3º F ⎡(T − T )⎤ ⎡(140 −100)⎤ ⎡ 40 ⎤ −0.405 ln ho ci ln ln ⎢ ⎥ ⎢ − ⎥ ⎣⎢ 60 ⎦⎥ ⎣(Thi − Tco )⎦ ⎣ (120 60) ⎦ Once the LMTD is calculated, the rate of heat exchange [inset figure 5-8] across the heat exchangerFormula can 5-2: (theoretically) be calculated Formula 5-1 can be used for either parallel flow or using Formula 5-2: Formula 5-1 can be used for either parallel flow or counterflow heat exchanger configurations. counterflow heat exchanger configurations. Formula 5-2: The LMTD LMTD can also can be thought also of beas the thought temperature of difference as the between temperature the fluids at one end of the heat exchanger, minus the temperature difference between the fluid at the other end, difference between the fluids at one end of the heat Q = UA(LMTD) divided by the natural logarithm of the ratio of these two temperature differences. This is Page 104" of 128" illustrated in figure 5-9.

[insert ﬁgure 5-9] 43

Page 101" of 128"

Page 102" of 128" Figure 5-10

122 ºF 77 ºF counterﬂow

(Tho − Tci ) − (Thi − Tco ) (122 − 77) − (104 − 68) 45 − 36 9 LMTD = = = = = 40.3º F ⎡(T − T )⎤ ⎡(122 − 77)⎤ ⎡ 45 ⎤ 0.223 ln ho ci ln ln ⎢ ⎥ ⎢ − ⎥ ⎣⎢ 36 ⎦⎥ ⎣(Thi − Tco )⎦ ⎣(104 68)⎦ side absorbing heat side supplying heat supplying side

104 ºF 68 ºF

122 ºF 68 ºF parallel ﬂow

DRAFT (4-16-18) (Tho − Tci ) − (Thi − Tco ) (122 − 68) − (104 − 77) 54 − 27 27 LMTD = = = = = 39.0º F ⎡(T − T )⎤ ⎡(122 − 68) ⎤ ⎡54 ⎤ 0.693 ln ho ci ln ln ⎢ ⎥ ⎢ − ⎥ ⎣⎢ 27 ⎦⎥ ⎣(Thi − Tco )⎦ ⎣(104 77)⎦ side absorbing heat side supplying heat supplying side Where:

104 ºFQ = rate77 of ºFheat transfer through heat exchanger (Btu/hr) U = overall heat transfer coefﬁcient of the heat exchanger (Btu/hr/ft2/ºF)

Where: A = area of surface(s) separating fluids insideWhere: the heat exchanger (ft2) Q = rate of heat transfer through heat exchanger (Btu/hr) U = overall heat transfer coefficient of the heat U = overall heat transferLMTD =coefficient log mean of temperature the heat differenceexchanger across the (Btu/hr/ft heat exchanger2/ºF) (ºF) 2 exchanger (Btu/hr/ft /ºF) hc = convection coefficient for the fluid absorbing heat A = area of surface(s) separating fluids inside the heat (Btu/hr•ft2•ºF) 2 exchanger (ft ) The value of “U” depends on the convectionhh = coefficientsconvection coefficient for each fluid, for the as fluid well releasing as the thickness, heat LMTD = log mean temperature difference across the (Btu/hr•ft2•ºF) heat exchanger (ºF)shape, and thermal conductivity of the surfacesd = thickness separating of wall the separating fluids. Ifthe is fluidsdefined (ft) by formula 5-3. k = thermal conductivity of wall separating the fluids The value of “U” depends on the convection coefficients (Btu/hr•ft•ºF) for each fluid, as well as the thickness, shape and thermal conductivity of the surfaces separating the fluids. It is The values of the convection coefficients hh and hc can defined by FormulaFormula 5-3. 5-3: vary widely depending on the fluid, its density, specific heat and viscosity, as well as the fluid’s flow rate. These Formula 5-3: 1 conditions need to be evaluated on both sides of the U = heat exchanger. The theoretical determination of the ⎡ d ⎤ 1 + + 1 convection coefficients is mathematically complex. It ⎢ ⎥ is usually based on empirical correlations of operating ⎣hc k hh ⎦

44 Where: U = overall heat transfer coefﬁcient of the heat exchanger (Btu/hr/ft2/ºF)

2 hc = convection coefﬁcient for the ﬂuid absorbing heat (Btu/hr•ft •ºF)

2 hh = convection coefficient for the fluid releasing heat (Btu/hr•ft •ºF) d = thickness of wall separating the fluids (ft) k = thermal conductivity of wall separating the fluids (Btu/hr•ft•ºF)

The values of the convection coefficients hh and hc can vary widely depending on the fluid, its density, specific heat, density, viscosity, as well as the fluid’s flow rate. These conditions need to be evaluated on both side of the heat exchanger. The theoretical determination of the convection coefficients is mathematically complex. It is usually based on empirical correlations of operating conditions that are expressed in the form of dimensionless numbers such as the Reynolds number, Nusselt number, and Prandtl number . Many heat transfer textbooks

Page 105" of 128" conditions that are expressed in the form of dimensionless numbers, Figure 5-11 such as the Reynolds number, Nusselt number and Prandtl number. Many heat transfer textbooks summarize methods for estimating convection coefficients for heat exchangers of specific types and geometries. ATD = 120-110 = 10 ºF 1 Typical values of U in water-to-water heat exchangers are in the range of 100 to 300 (Btu/hr/ft2/ºF). 120 ºF 110 ºF APPROACH TEMPERATURE DIFFERENCE Another term that’s commonly used in describing the operating conditions of a heat exchanger is called the “approach temperature difference” (ATD). Since all heat exchangers have two inlets and two outlets, there are technically two approach temperature differences, as shown in Figure 5-11.

Customarily, the approach temperature difference refers to the incoming temperature of the fluid supplying the heat minus the leaving temperature of the fluid absorbing the heat. For the heat exchanger shown in Figure 5-11, this would be 10ºF. side absorbing heat side supplying heat supplying side Approach temperature difference can be thought of as a “thermal penalty” that’s present simply because heat has to move through a heat exchanger. A temperature difference must be present between the 100 ºF 60 ºF two sides of any heat exchanger to create heat transfer. A hypothetical heat exchanger with infinite internal surface area and no external heat loss would have an approach temperature difference of 0ºF, and thus ATD = 100-60 = 40 ºF induce no thermal penalty. While this is not possible for any “real” heat 2 exchanger, approach temperature differences on the order of 2 to 3ºF are achievable in some situations.

Figure 5-12a

45 Figure 5-12b Sizing and selection software allows designers to enter values for flow rates, temperatures, type of fluids being used and required rates of heat transfer. The software uses these inputs to select one or more specific heat exchanger configurations that can meet or slightly exceed the required thermal performance. The software also DRAFT (4-16-18) provides pressure drop or head loss information, which is used for the hydraulic design of FOULINGthe circuit FACTORS containing FORthe heatHEAT EXCHANGERS exchanger.

FOULING FACTORS FOR HEATThe EXCHANGERS listed thermal and hydraulic performance of heat exchangers is usually based on the The listed thermalassumption and hydraulic that the performance heat exchanger is free of any internal scaling or debris. A relatively thin of heat exchangers is usually based on the assumptionaccumulation that the of heatscaling exchanger on the surfaces that separate the two fluids can significantly lower the is free of any internal scaling or debris. A relativelyheat thin exchanger’s accumulation thermal of dirt performance. or For optimal performance and long life it’s important to scaling onkeep the the surfaces surfaces that of heat separate exchangers clean. the two fluids can significantly lower the heat exchanger’s thermal performance. For optimal performance and long life, it’s important Theto keepaffect theof scale surfaces accumulation of heat is sometimes expressed as a fouling factor, which is defined by exchangersFormula clean. 5-4: The effect of scale accumulation is sometimes expressed as a fouling factor, which is definedFormula by 5-4:Formula 5-4:

Formula 5-4:

1 1 Rf = − Udirty Uclean

As the heat exchanger size increases, and the incoming Where: fluid temperatures and flow rates remain constant, the 2 approach temperature difference decreases. However, Rf = fouling factor (ºF•hr•ftWhere:/Btu) U = overall heat transfer coefficient for the surfaces as the approach temperature difference decreases dirty 2 Rf = fouling factor (ºF•hr•ft /Btu) (desirable), the size and cost of the heat exchanger that are fouled (Btu/hr/ft2/ºF) U = overall heat transfer coefficient for the surface increases (undesirable). Heat exchangers are sometimes clean Udirty = overall heat transfer coefficient for the surfaces that are fouled (Btu/hr/ft2/ºF) sized based on achieving a maximum approach that is clean (Btu/hr/ft2/ºF) temperature difference for a specific application. Uclean = overall heat transfer coefficient for the surface that is clean (Btu/hr/ft2/ºF) The fouling factor can be thought of as an additional SIZING HEAT EXCHANGERS “R-value” added to the surfaces of the material separating Although it is possible to theoretically estimate the the two fluids. The greater the thermal resistance of any thermal performance of heat exchangers, the calculations accumulated dirt or scale,The fouling the lower factor the can overall be thought heat of as an additional “R-value” added the surfaces of the transfer rate through the heat exchanger, with all other involved are complex and time consuming. Today, many material separating the two fluids. The greater the thermal resistance of any accumulated dirt or heat exchanger manufacturers offer sizing and selection factors being equal. The value of Rf is determined software, either for downloading or to be used online. An experimentally by measuringscale, the the lower rate ofthe heat overall transfer heat transfer rate through the heat exchanger, all other factors being example of the user interface from one such software offering through a heat exchanger that has been fouled based on equal. The value of Rf is determined experimentally by measuring the rate of heat transfer is shown in Figure 5-12. specific operating conditions and service time, versus through a heat exchanger that has been fouled based on specific operating conditions and service time, versus the rate of heat exchange through an identical new heat exchanger 46 operating with the same fluids, inlet flow rates, and temperatures.

Fouling factors are generally small numbers in the range of 0.0005 to 0.005 (hr•ft2•ºF/Btu). Values for fouling factors can be obtained in the following reference: [Kakac,S. and H. Liu, Heat

Page "110 of "128 the rate of heat exchange through an identical new heat Figure 5-13 exchanger operating with the same fluids, inlet flow rates and temperatures.

Fouling factors are generally small numbers in the range of 0.0005 to 0.005 (hr•ft2•ºF/Btu). Values for fouling factors can be obtained in the following reference: [Kakac,S. and H. Liu, Heat Exchangers - Selection, Rating and Thermal Design, CRC Press, (1998)]. Some sizing and selection software allows different fouling factors to be used on each side of the heat exchanger’s internal surface.

KEEPING HEAT EXCHANGERS CLEAN Fouling factors can be reduced by keeping the internal surfaces of a heat exchanger free of scale and debris.

Scale formation is generally caused by calcium or magnesium salts dissolved in water. So-called “hard” water can have a high concentration of these mineral Dirt and iron oxides such as magnetite can be minimized salts. The ability of these mineral salts to come out of in hydronic systems through use of low velocity zone dirt solution and attach to internal heat exchanger surfaces separators or magnetic separators. The latter are capable increases with increasing water temperature. of separating both non-magnetic and magnetic particles from a flow stream. Fouling can also be caused by residual soldering flux or cutting oil left in the system. The chemical breakdown of Best practice is to provide a dirt separator on each flow glycol-based antifreeze solutions can also cause films stream entering a heat exchanger, as shown in Figure to form on heat exchanger surfaces. Corrosion due to 5-14. oxidation of ferrous metals also creates sludge that can lodge within heat exchangers. Figure 5-14 One of the best ways to minimize the potential scale DIRTMAG formation due to mineral salts is to demineralize the separator water used in hydronic heating or cooling systems. Demineralization can be done by passing the untreated water through a bed of resin beads that absorb the DIRTMAG ions formed by the mineral salts. For hydronic system separator applications, demineralization can reduce the total dissolved solids content of the water into an ideal range heat of 10-30 parts per million (PPM). exchanger

idronics #18 (Water Quality in Hydronic Systems) provides a full discussion of demineralization techniques and hardware.

Dirt or other debris that lodges within heat exchangers can also reduce heat transfer. In extreme cases, it can render common heat exchangers, which cannot be disassembled for cleaning, useless, leaving no alternative but to replace the heat exchanger. Figure 5-13 shows an example of a heavily fouled plate that was removed from a plate & frame heat exchanger.

47 Caleffi DirtMag separators are recommended if either In addition to external or immersion heat exchangers that circuit passing through the heat exchanger contains may be in the system, the heat exchanger within the heat components made of carbon steel or cast iron, both of source, be it a boiler, heat pump or other source, should which can be a source of magnetite. DirtMag separators be kept as clean as possible to maximize heat transfer to are also recommended if a circulator with a permanent system water and maintain high efficiency. magnet motor (e.g., ECM) is used in either circuit. Figure 5-15 shows a Caleffi Dirtmag separator with the magnet removed, and being “blown down” to eliminate iron oxide Figure 5-16 (magnetite) from the system. domestic water Figure 5-15 heating subassembly

DHW to/from space heating

combination buffer tank isolation/flush valves

Figure 5-17

idronics #15 (Separation in Hydronic Systems) provides a full discussion of dirt separation techniques and hardware.

Another useful detail for heat exchangers that handle domestic water (which may contain dissolved mineral salts) is to install a combination isolation/flushing valve on the piping that carries domestic water into and out of the heat exchanger. Figure 5-16 shows a typical detail.

The combination isolation/flush valves allow the domestic water side of the heat exchanger to be isolated from the remainder of the system. The hose bibs on the side ports of these valves can be connected to a flushing assembly that circulates a mild acid solution through the heat Courtesy of Apollo Unlimited exchanger to dissolve scaling caused by precipitation of mineral salts. Once the scale has been dissolved and Figure 5-17 shows a severely corroded boiler heat removed, fresh water can be passed through the heat exchanger. The scaling is the result of poor water exchanger to remove any remaining cleaning solution. quality in combination with repeated thermal expansion The isolation valves are then reopened to put the heat and contraction of the heat exchanger surfaces as the exchanger back in service. boiler operates. This scale can significant reduce boiler efficiency.

48 Figure 5-18 SUMMARY:

The transfer of heat at predictable rates and in many different locations is critical to the proper operation of hydronic heating or cooling systems. Understanding how heat is transferred by conduction, convection and radiation, and having reasonable methods of predicting the rate of heat transfer, can identify potential “bottlenecks” in the path from a heat source to one or more heat emitters, or between a source of chilled water and one or more terminal units.

Many of the performance problems associated with improperly designed, installed or commissioned hydronic systems can be traced to inadequate heat transfer in at least one area of the system. It could be a boiler with a fouled heat exchanger, an undersized heat emitter or insufficient flow rate caused by an improper circulator selection. It could be a thick carpet placed over a heated floor slab, a finned-tube element within a baseboard that’s covered with pet hair or a heat exchanger that’s piped for parallel flow rather than counterflow. It could be

Courtesy of Apollo Unlimited an assumption that natural convection is as effective as Systems equipped with high-performance air and dirt forced convection, or that the characteristics of the fluid separators, and operating with properly demineralized circulating through the system have no significant effect water, provide a strong defense against loss of boiler on the rate of heat transfer. heat exchanger performance due to scaling over the life of the system. The list of “it could be” scenarios extends to hundreds of possibilities in which inadequate heat transfer is the Heat emitters are also heat exchangers in a generic sense. underlying problem. Designers with a solid understanding Dust, or physical damaged to the fins on baseboard of heat transfer fundamentals, and the ability to estimate elements, radiators, or the coil of an air handler will heat transfer rates using the formulas and methods decrease heat transfer. Figure 5-18 shows the finned- discussed in this issue of idronics, are far less likely to tube element in a wall convector with an accumulation encounter these “it could be” scenarios. of dust, food waste, wrappers, gum, and other debris. In addition to being a hygiene issue, the rate of heat emission from this baseboard will only be a fraction of its intended output.

Annual inspections of heating systems should include an examination of all accessible heat exchanger surfaces. Fins that are bent can usually be straighten using a fin comb. Dust and pet hair can be removed using a brush and shop vac.

49 APPENDIX A: PIPING SYMBOL LEGEND

GENERIC COMPONENTS

3-way motorized circulator mixing valve blower compressor ﬁlter circulator w/ 4-way motorized mixing valve coil reversing! valve union

circulator w/ internal check valve swing check valve comp. RV evaporator TXV condenser evaporator

spring-loaded water-to-air gate valve check valve heat pump TXV (in heating mode) water-to-water globe valves purging valve heat pump blower (in heating mode) ball valve

pressure gauge ﬁlter compressor primary/secondary coil fitting pressure relief valve reversing! hose bib valve pressure & comp. RV drain valve condenser TXV temperature relief valve

diverter tee water-to-air heat pump condenser cap metered (in cooling mode) evaporator balancing valve TXV

water-to-water heat pump diaphragm-type brazed- (in cooling mode) expansion tank

plate heating mode heat reversing exchanger valve condenser evaporator

panel radiator TXV with dual isolation valve reversible water-to-water heat pump

Comp.

Modulating conventional boiler tankless water heater

solar water tank (with upper coil)

solar collector Modulating / condensing boiler

wood-fired boiler solar collector array

indirect water heater (with trim) solar water tank (with electric element)

50 APPENDIX A: PIPING SYMBOL LEGEND

CALEFFI COMPONENTS

Float - type air vent inline check valve

3-way thermostatic mixing valve

manifold station with DISCAL balancing valves air separators Quicksetter+ balancing valve

DIRTCAL dirt separators pressure- reducing valve distribution (3/4") station

Autoﬁll backﬂow geothermal preventer manifold comb station DIRTMAG HYDROFILL dirt separators

backﬂow preventer

pressure- reducing valve HYDROFILL NA10045 boiler tim kit DISCALDIRTMAG DISCALDIRT multi-zone controls air & dirt air & dirt separator separator NA10045

differential pressure bypass valve

single zone control zone valve ThermoCon buffer tank (2 way)

zone valve (3 way)

LEGIOMIX thermoelectric zone valve (2 way)

Hydro Separator Hydro Separator TANKMIXER motorized ball valve (2 way)

Hydro Separator motorized THERMOSETTER ball valve adjustable balancing (3 way) valve

variable oriﬁce balancing valve solar thermostatic circulation FLOWCAL radiator valve station balancing valve thermostatic FLOWCAL radiator valve balancing valve

dual isolation SEP4 valve for SEP4 fixed orifice panel radiators SEP4 balancing valve high-temperature solar 3-way thermostatic QuickSetter high-temperature mixing valve balancing solar DISCAL valve w/ air separators HydroBlock flowmeter isolar mixing units differential high-temperature temperature solar pressure controller boiler relief valve protection valve high-temperature solar air vent high- temperature ThermoBlock solar high-temperature expansion shut-off valve for tank solar air vent Hydrolink (4 configurations) symbols are in Visio library @ www.caleffi.com

51 DRAFT (4-16-18) DRAFT (4-16-18) B = heat output of the baseboard at 200 °F average water temperature, 1 gpm from manufacturer’s literature* (Btu/hr/ft) APPENDIX B: THERMAL MODEL FOR A FINNED TUBE The values 0.04, -0.4172, -2.3969319 and 1.4172 are all exponents. APPENDIX B: THERMAL MODEL FOR A The values 0.04, -0.4172, -2.3969319 and 1.4172 are all FINNED-TUBEBASEBOARD BASEBOARD exponents. Once the outlet temperature for a baseboard is determined, Formula B-2 can use it along with This appendix provides a detailed analytical model for Once the outletthe temperatureinlet temperature for aand baseboard flow rate tois determined,determine the total heat output. theThis thermalappendix providesperformance a detailed of analytical a finned-tube model for the baseboard thermal performance Formula of a finned-tubeB-2 can use it along with the inlet temperature heatbaseboard emitter. heat Theemitter. model The model allows allows the the temperature temperature of ofthe thewater passingand flow through rate the to determine the total heat output. waterbaseboard passing to be determinedthrough theat any baseboard location along to the be element, determined based on the inlet water Formula B-2: at any location along the element, based on the inlet Formula B-2: temperature, room air temperature, fluid, and flow rate. When the length of the baseboard is water temperature, room air temperature, fluid and flow rate.known When the outlet the fluid length temperature of the can baseboard be calculated. is Once known, the fluid the outlet temperature is " fluiddetermined outlet, ittemperature can be combined can with be the calculated. inlet temperature Once and the flow fluid rate to determine the total outletheat output temperature of the heat emitter.is determined, it can be combined Where: with the inlet temperature and flow rate to determine the Toutlet = outletWhere: temperature of fluid leaving baseboard (ºF) total heat output of the heat emitter. T = fluid temperature at inlet of baseboard (ºF) The fluid temperature at a given location along a finned-tube baseboard caninlet be found using D = density of fluid within baseboard (lb/ft3) Formula B-1: The fluid temperature at a given location along a finned- c = specific Theatoutlet =of outlet fluid temperaturewithin baseboard of fluid leaving(Btu/lb/ºF) baseboard (ºF) tube baseboard can be found using Formula B-1: = fluid flow rate through the baseboard (gpm) ƒ Tinlet = fluid temperature at inlet of baseboard (ºF) Formula B-1: Formula B-1: For example:D Assume= density aof finned-tubefluid within baseboard baseboard (lb/ft is3 )20 feet long, locatedc =in specific a room heat with of fluida floor-level within baseboard air temperature (Btu/lb/ºF) of 65ºF, and supplied with water at 4 gpm and 180ºF. The baseboard’sƒ manufacturer= fluid flow rate ratesthrough its the heat baseboard output (gpm) at 500 " Btu/hr/ft when operating with 200ºF water and a flow rate Where: of 1 gpm. Determine the heat output of this baseboard at For example: Assume a finned-tube baseboard is 20 feet long, located in a room with a floor- TWhere:f = fluid temperature at some location L along length of the stated conditions, and compare it to the output if the finned-tube element (ºF) flow rate werelevel reduced air temperature to 0.5 gpm. of 65 ºF, and supplied with water at 4 gpm and 180 ºF. The baseboard’s Troom = room air temperature entering the baseboard (°F) Tf = fluid temperature at some location L along length of finned-tube element (ºF) manufacturer rates its heat output at 500 Btu/hr/ft when operating with 200 ºF water and a flow Tinlet = fluid temperature at inlet of baseboard (ºF) The outlet temperature of the baseboard can be rate of 1 gpm. DetermineDRAFT the (4-16-18) heat output of this baseboard at the stated conditions, and DT room= density = room airof temperaturefluid within entering baseboard the baseboard (lb/ft3 )(°F) determined using Formula B-1: cT =inlet specific = fluid temperature heat of fluid at inlet within of baseboard baseboard (ºF) (Btu/lb/ºF) compare it to the output if the flow rate were reduced to 0.5 gpm. = fluid flow rate through the baseboard (gpm) ƒ 3 LD = = position density of alongfluid within baseboard baseboard beginning (lb/ft ) from inlet (ft) Bc == specificheat output heat of offluid the within baseboard baseboard at (Btu/lb/ºF) 200°F average water The outlet temperature of the baseboard can be determined using Formula B-1: temperature,ƒ = fluid flow rate 1 throughgpm, from the baseboard manufacturer’s (gpm) literature* (Btu/ hr/ft) " L = position along baseboard beginning from inlet (ft)

The total heat released from the baseboard can now be calculated using Formula B-2: Figure B-1 heat output:! Page 121" of 128" a. 8,282 Btu/hr!

b. 6,522 Btu/hr" Page 122" of 128" Using the same formulas at the reduced flow rate of 0.5 gpm yields an outlet temperature of 180 ºF 153.48ºF, and a total heat output of 6,522 Btu/hr. A comparison of the two operating conditions flow rate:! is shown in figure B-1. outlet temperature:! a. 4 gpm! a. 175.79 ºF! 20 feet of fin-tube [insertbaseboard figure B-1] ! b. 0.5 gpm b. 153.48 ºF rated at 500 Btu/hr/ft when! heat output:! supplied with 200 ºF water at 1 gpma. 8,282 Btu/hr! b. 6,522 Btu/hr

180 ºF ﬂow rate:! outlet temperature:! a. 4 gpm! a. 175.79 ºF! 20 feet of ﬁn-tube baseboard! b. 0.5 gpm b. 153.48 ºF 52 rated at 500 Btu/hr/ft when! " supplied with 200 ºF water at 1 gpm

By repeating these calculations, it is possible to plot heat output of this baseboard as a function of ﬂow rate. Figure B-2 shows the results over a wider range of ﬂow rate.

[insert ﬁgure B-2]

Page 123" of 128" DRAFT (4-16-18)

The total heat released from the baseboard can now be calculated The total heat released from the baseboard can now be calculated using Formula B-2: using Formula B-2:

Using the same formulas at the reduced flow rate of 0.5 gpm yields an outletUsing temperature the same formulas of 153.48ºF, at the reduced and a totalflow rate heat of output0.5 gpm ofyiel 6,522ds an outletBtu/hr. temperature of A comparison153.48ºF, and of a thetotal two heat operating output of 6,522 conditions Btu/hr. A comparisonis shown inof Figurethe two operatingB-1. conditions is shown in figure B-1. By repeating these calculations, it is possible to plot heat output of this baseboard as a function of flow rate. Figure B-2 shows the results over[insert a wider figure range B-1] of flow rate.

This graphs shows a rapid increaseheat output: in heat! transfer at low flow rates, and a relatively slight gaina. in8,282 heat Btu/hrtransfer! as flow rates rise above approximately 2 gpm. b. 6,522 Btu/hr

A similar180 ºFanalytical model for the thermal performance of a radiant panel circuit is given in Appendix C. flow rate:! outlet temperature:! a. 4 gpm! a. 175.79 ºF! 20 feet of fin-tube baseboard! b. 0.5 gpm b. 153.48 ºF rated at 500 Btu/hr/ft when! supplied with 200 ºF water at 1 gpm Figure" B-2 20 feet of fin-tube baseboard! rated at 500 Btu/hr/ft when! By repeating thesesupplied calculations, with it is200 possible ºF water to plot heatat 1ou gpmtput of this baseboard as a function of flow rate. Figure B-2 shows the results over a wider range of flow rate. 180 ºF fixed[insert supplyfigure B-2 ] temperature 9000 8000 Page 123" of 128" 7000 6000

5000

4000

3000 heat output (Btu/hr) 2000

1000

0 0 1 2 3 4 5 6 7 8 ﬂow rate (gpm)

53 DRAFT (4-16-18)

Where:

Toutlet = outlet temperature of ﬂuid leaving radiant panel (ºF)

Tinlet = fluid temperature at inlet of radiant panel (ºF) D = density of fluid within baseboard (lb/ft3) c = specific heat of fluid within baseboard (Btu/lb/ºF) ƒ = fluid flow rate through the baseboard (gpm)

DRAFT (4-16-18) The value of “a” in Formula C-1 is based on the exact construction of the radiant panel, including tube spacing, tubing embedment method, finish floor coverings, and underside APPENDIX C: THERMAL MODEL FOR A RADIANTinsulation. PANEL This value can be determined based on a known relationship between heat output, DRAFT (4-16-18) CIRCUIT and a corresponding circuit water temperature and room air temperature. The relationship is APPENDIX C: THERMAL MODEL FOR A RADIANTgiven PANELas Formula C-3. Formula C-1 can be used to determine the outlet temperature of a specific radiant panel based CIRCUIT on its construction and the conditions under which it operates: Formula C-3 APPENDIX C: THERMAL MODEL FOR A RADIANTFormulaFormula C-1 PANEL can C-1: be CIRCUIT used to determine the outlet temperatureFormula of aC-3 specific radiant panel based ⎛ ⎞ on its construction and the conditions under which it operates: ⎛ s ⎞ qup + qdown Formula C-1 can be used to determine the outlet " a = ⎜ ⎟⎜ ⎟ ⎝ ⎠ − temperature of a specific radiant panel based⎛ onal its⎞ 12 ⎝ Tw Troom ⎠ construction and the conditions under which it− ⎜operates:⎟ Formula C-1: ⎝ f (8.01Dc)⎠ " Tout = Troom + (Tin − Troom ) × e Where: Formula C-1: s = tubing spacing (inches) 2 ⎛ ⎞ qup = upward Where:heat flux from panel (Btu/hr/ft ) al 2 −⎜ ⎟ q = downward heat flux from panel (Btu/hr/ft ) Where: ⎝ f (8.01Dc)⎠ down s =tubing spacing (inches) " Tout = Troom + (Tin − Troom ) × e Tw = water temperature in circuit corresponding to the 2 above heat outputsqup = upward(ºF) heat flux from panel (Btu/hr/ft ) Where: Tout = fluid outlet temperature for baseboard (ºF) Troom = room air temperature corresponding to the 2 above heat outputsqdown =(ºF) downward heat flux from panel (Btu/hr/ft ) Where:Tin = fluid inlet temperature to baseboard (ºF) T = fluid outlet temperature for baseboard (ºF) out Tw = water temperature in circuit corresponding to the above heat outputs (ºF) Tin = fluid inletT roomtemperature = temperature to bas of eboardroom air (ºF) entering baseboardFor (ºF) example, a radiant floor panel with 12” tube spacing 2 Troom = temperature of room air entering baseboard (ºF) creates an averageTroom =upward room airheat temperature output of 25 corresponding Btu/hr/ft , to the above heat outputs (ºF) Tout = afluid = thermal outlet temperatureoutput parameter for baseboard for a specific (ºF) panel construction 2 a = thermal output parameter for a specific panel with a corresponding downward heat loss ofDRAFT 2 Btu/hr/ft (4-16-18). constructionTin = fluidl = length inlet temperature of radiant panel to baseboard circuit (ft) (ºF) The average water temperature in the radiant panel circuit l = lengthT roomof radiant =c =temperature specific panel heat circuit of ofroom fluid (ft) air flowing entering through baseboard circuit (ºF)basedat these on its conditionsaverageFor temperature example, is 110ºF, and A(Btu/lb/ºF) radiant the room floor air panel temperature with 12” tube spacing creates an average upward heat output c = specific heat of fluid flowing through circuit based on above the panel is 70ºF. Determine the value of “a” for use a = thermalD = density output of parameter fluid in circuit for a based specific on panelit average construction temperature (lb/ft3) DRAFT (4-16-18) its average temperature (Btu/lb/ºF) in Formula C-3.of 25 Btu/hr/ft2, with a corresponding downward heat loss of 2 Btu/hr/ft2. The average water D = densityl = length f of= flow offluid radiant rate in through circuitpanel circuitcircuitbased (ft)(gpm) on its average Solution: 3 temperature in the radiant panel circuit at these conditions is 110 ºF, and the room air temperaturec = specific (lb/fte = 2.71828) heat of (base fluid flowingof natural through logarithm circuit system) based onSolution: its average temperature (Btu/lb/ºF) f = flow rate through circuit (gpm) Solution: D = density of fluid in circuit based on it average temperature (lb/ft3) ⎛temperature⎞ above the panel is 70ºF. Determine the value of “a” for use in Formula C-3. e = 2.71828 (base of natural logarithm system) ⎛ s ⎞ qup + qdown ⎛12 ⎞⎛ 25 + 2 ⎞ " a = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ = 0.675 f = flowOnce rate the through outlet circuit temperature (gpm) for the panel is determined, it can⎛ be combined⎞ with the inlet ⎛ s⎝12⎞ q⎠up⎝+Tqdown− T ⎛12⎠⎞⎛ ⎝2512+⎠2⎝110⎞ − 70 ⎠ Once the outlet temperature for the panel is determined, " a = ⎜ ⎟⎜ w ⎟ =room⎜ ⎟⎜ ⎟ = 0.675 e = 2.71828temperature (base andof natural flow rate logarithm to determine system) the total heat output.⎝12 ⎠ ⎝UTse− FormulaT ⎠ ⎝ 12C-2⎠⎝ for110 this− 70 ⎠ it can be combined with the inlet temperature and flow w room rate to determinecalculation. the total heat output. Use Formula C-2 Assume a radiant floor panel consists of a single 300-foot- for this calculation.Once the outlet temperature for the panel is determined,long it canAssume circuit be combined ofa radiant½” PEXwith floor thetubing, inletpanel spaced consists 12 of inches a single apart. 300-foot-long circuit of ½” PEX tubing, spaced Assume a radiant floor panel consists of a single 300-foot-long circuitPage of ½ 126"” PEX oftubing, 128" spaced When operating at an average water temperature of temperatureFormula and C-2: flow rate to determine the total heat output. Use Formula C-2 for this Formula C-2: 110ºF,12-inches12-inches this panel’sapart. apart. When average When operating upward operating at an averageheat at outputan water average temperaturis 25 Btu/watere of temperatur 110ºF this panel’se of 110ºF this panel’s calculation. hr averageintoaverage a room upward upward at heat 70ºF, outputheat and outputis 25its Btu/hr downward is 25 into Btu/hr a room heat into at 70ºF,output a room and is its at downward 70ºF, and heat its output downward is heat output is 2.52.5 Btu/hr. Btu/hr. Further Further assumeassume that that the radiantthe radiant panel circuit panel is suppliedcircuit with water at a temperature 2.5 Btu/hr. Further assume that the radiant panel circuit is supplied with water at a temperature " is supplied with water at a temperature of 120ºF and flow Formula C-2: of 120ºF and flow rate of 0.8 gpm. Determine: rate ofof 120ºF 0.8 gpm. and Determine: flow rate of 0.8 gpm. Determine: Where: a. Thea. The “a” “a” value value of thethe panel panel Page 125" of 128" Toutlet = " outlet temperature of fluid leaving radiant panel b. Theb.a. The Thetotal total “a”heat heat value transfer of thefrom from panel the theradiant radiant panel panel (ºF) c. The total upward heat transfer from the radiant panel c.b. The The total total upward heat heat transfer transfer fromfrom the the radiant radiant panel panel Tinlet = fluid temperature at inlet of radiant panel (ºF) D = density of fluid within baseboard (lb/ft3) Solution:c. The The total value upward of “a” heat can transfer be determined from the radiant from the panel c = specific heat of fluid within baseboard (Btu/lb/ºF)Page 125" of 128" givenSolution: data Theusing value Formula of “a” can C-3: be determined from the given data using Formula C-3: ƒ = fluid flow rate through the baseboard (gpm) Solution: The value of “a” can be determined from the given data using Formula C-3: The value of “a” in Formula C-1 is based on the exact construction of the radiant panel, including tube spacing, " tubing embedment method, finish floor coverings and underside insulation. This value can be determined It is necessary to determine the density and specific based on a known relationship between heat output and heatIt is of necessary the water to determineat its average the density temperature and specific before heat of usingthe water at its average a corresponding circuit water temperature and room air Formulatemperature" C-1. beforeHowever, using theFormula average C-1. However,water temperature the average water of temperature of the circuit temperature. The relationship is given as Formula C-3. theis circuit not known is not at this known point. at As this an approximation,point. As an approximation,assume a temperature drop along the circuit of 20ºF. Under this assumption, the average water temperature would be 120-(20/2) = 110ºF. The It is necessary to determine the density and specific heat of the water at its average density of water at 110ºF is 61.8 lb/ft3. The specific heat of water at this temperature is 1.00 Btu/ lb/ºF.temperature This data, alongbefore with using the other Formula stated andC-1. calculated However, numbers the averageare now used water in Formula temperature of the circuit 54 C-1is notto determine known theat thiscircuit’s point. outlet As temperature. an approximation, assume a temperature drop along the circuit of 20ºF. Under this assumption, the average water temperature would be 120-(20/2) = 110ºF. The density of water at 110ºF is 61.8 lb/ft3. The specific heat of water at this temperature is 1.00 Btu/ lb/ºF. This data, along with the other stated and calculated numbers are now used in Formula C-1 to determine the circuit’s outlet temperature. Page 127" of 128"

Page 127" of 128" assume a temperature drop along the circuit of 20ºF. Under this assumption, the average water temperature would be 120 - (20/2) = 110ºF. The density of water at 110ºF is 61.8 lb/ft3. The specific heat of water at this temperature is 1.00 Btu/lb/ºF. This data, along with the DRAFT (4-16-18) other stated and calculated numbers, is nowDRAFT used (4-16-18) in Formula C-1 to determine the circuit’s outletDRAFT temperature. (4-16-18)

" " " The total heat output of the circuit can now be determined using Formula C-2: The total heat output of the circuit can now be determined usingTheThe total Formula total heat heat C-2:output output of the of circuit the circuit can now can benow determined be determined using Formulausing Formula C-2: C-2:

" " Finally, the total upward heat transfer can be estimated by multiplying the total heat transfer by theFinally,Finally, ratio theofthe upward totaltotal upward heat flux heatheat to transfer totaltransfer heat can canflux. be be estimated estimated by multiplying the total heat transfer by by multiplyingFinally, the thetotal totalupward heat heat transfer transfer by can the be ratio estimated of by multiplying the total heat transfer by the ratio of upward heat flux to total heat flux. upwardthe heatratio fluxof upward to total heat heat flux flux. to total heat flux.

⎛ ⎞ qup ⎛ 25 ⎞ "Qup = Qtotal ⎜ ⎟ = 8043⎜ ⎟ = 7,312Btu / hr ⎛ + ⎞ ⎝ + ⎠ ⎝ qup qqupdown ⎠ ⎛25 252.5 ⎞ Q = Q ⎛ q = 8043⎞ = 7,312Btu / hr " up total ⎜ up⎟ ⎝⎜ ⎛ 25⎠⎟ ⎞ Q = Q⎝ qup + qdown ⎠ = 804325 + 2.5 = 7,312Btu / hr " up total ⎜ ⎟ ⎝⎜ ⎠⎟ ⎝ qup + qdown ⎠ 25 + 2.5 END OF MANUSCRIPT INSERTEND OF PRODUCT MANUSCRIPT PAGES END OF MANUSCRIPT INSERT PRODUCT PAGES INSERT PRODUCT PAGES

Page 128" of 128" 55 Page 128" of 128" Page 128" of 128" QuickSetter™ Balancing valve with flow meter

132 series CALEFFI

Function

The 132 series balancing valve accurately sets the flow rate of heating and cooling transfer fluid supplied to fan coils and terminal units. Proper hydronic system balancing ensures that the system operates according to design specifications, providing satisfactory thermal comfort with low energy consumption. The flow meter is housed in a bypass circuit on the valve body and can be shut off during normal operation. The flow meter permits fast and easy circuit balancing without added differential pressure gauges and reference charts. The threaded version is furnished with a hot pre-formed insulation shell to optimize thermal performance for both hot and cold water systems. Patent pending.

Product range 132 series Balancing valve with flow meter, threaded connections ½”, ¾”, 1”, 1 ¼”, 1 ½”, and 2" NPT female 132 series Balancing valve with flow meter, flanged connections 2 ½”, 3”, 4” ANSI 125

Series 132 Threaded 132 Flanged Materials Valve Body: brass cast iron Ball: brass brass Ball control stem: brass, chrome-plated brass, chrome-plated Ball seal seat: PTFE R-PTFE Control stem guide: PSU PTFE Seals: peroxide-cured EPDM peroxide-cured EPDM

Flow meter Body: brass brass Bypass valve stem: brass, chrome-plated brass, chrome-plated Springs: stainless steel stainless steel Seals: peroxide-cured EPDM peroxide-cured EPDM Flow meter float and indicator cover: PSU PSU Performance Suitable Fluids: water, glycol solutions water, glycol solutions Max. percentage of glycol: 50% 50% Max. working pressure: 150 psi (10 bar) 150 psi (10 bar) Working temperature range: 14 - 230º F (-10 - 110º C) 14 - 230º F (-10 - 110º C) Flow rate range unit of measurement: gpm gpm Accuracy: ±10% ±10%

Control stem angle of rotation: 90º 90º Control stem adjustment wrench: ½" - 1¼": 9 mm with 5½" diameter handwheel 1½" - 2": 12 mm Connections: ½" - 2": NPT female 2½", 3" and 4": ANSl B16.1 125 CLASS RF flanged

Flow rate correction factor: 20% - 30% glycol solutions: 0.9 20% - 30% glycol solutions: 0.9 40% - 50% glycol solutions: 0.8 40% - 50% glycol solutions: 0.8 Insulation Material: closed cell expanded PE-X Thickness: 25/64 inch (10 mm) Density: - inner part: 1.9 lb/ft3 (30 kg/m3) - outer part: 3.1 lb/ft3 (50 kg/m3) Thermal conductivity (DIN 52612): - at 32°F (0°C): 0.263 BTU·in/hr·ft2·°F (0.038 W/(m·K)) - at 104°F (40°C): 0.312 BTU·in/hr·ft2·°F (0.045 W/(m·K)) Coefficient of resistance to water vapor (DIN 52615): < 1,300 Working temperature range: 32 - 212º F (0 - 100º C) Reaction to fire (DIN 4102): class B2

--56 Dimensions

A C

3/4” all sizes C

D CALEFFI 15 bar

7 D E B 6 5 4 3 2

4 bolt holes for 2” and 3”. 8 bolt holes for 4”.

A A B

Code A B C D Weight (lb) Code A B C D Wt (lb/kg) Bolt Wt 132 5/16 13/16 3/4 Code A B C D E circle 432A132432A 1/2” ½” 3 3 5/”16" 11 13/16"” 5 5¾ " ” 2.0/0.92.0 dia (lb/kg) 132552A 3/4” 3 5/16” 1 13/16” 5 3/4” 1.8 132552A ¾” 3 5/16" 1 13/16" 5 ¾" 1.8/0.8 132060A 2 ½” 11 7⁄16" 6 31⁄32" 3 7⁄8" 7" 5 ½" 32/15 132622A 1” 3 3/8” 1 7/8” 6 1/4” 2.4 132662A 1” 3 3/8" 1 7/8" 6 ¼" 2.4/1.1 132080A 3" 12 7⁄32" 7 9⁄32" 3 7⁄8" 7 ½" 6" 40/18 132772A 1 1/4” 3 1/2” 2” 6 1/2” 2.8 132772A 1¼” 3 ½" 2" 6 ½" 2.8/1.3 25 29 7 132882A 1 1/2” 3 5/8” 2 1/4” 6 3/4” 3.4 132100A 4" 13 ⁄32" 7 ⁄32" 3 ⁄8" 9" 7½" 57/26 5 132992A132882A 2” 1½” 3 3/43 ”/8" 22 1/2 ¼"” 6 ¾"7” 3.4/1.54.4 132992A 2” 3 ¾" 2 ½" 7" 4.4/2.0

Flow rate ranges Flow rate ranges

Flow rate Flow rate Code Connection Full open Cv Code Connection Full open Cv (GPM) (GPM) 132432A ½” NPT ½ – 1¾ 1.0 132060A 2 ½” flange 30 - 105 87 132552A ¾” NPT 2.0 – 7.0 6.3 132080A 3” flange 38 - 148 164 132662A 1” NPT 3.0 – 10.0 8.3 132100A 4” flange 55 - 210 242 132772A 1¼” NPT 5.0 – 19.0 15.2 132882A 1½” NPT 8.0 – 32.0 32.3 132992A 2” NPT 12.0 – 50.0 53.7

-- Advantages of balanced circuits Operating principle

Balanced circuits have the following principal benefits: The balancing valve is a hydraulic device that controls the flow rate of the heating/cooling transfer fluid. 1. The system emitters operate properly in heating, cooling and The control mechanism is a ball valve (1), operated by a control dehumidification, saving energy and providing greater comfort. stem (2). The flow rate is manually and properly set by use of the 2. The zone circuit pumps operate at maximum efficiency, reducing convenient onboard flow meter (3) housed in a bypass circuit the risk of overheating and excessive wear. on the valve body. This circuit is automatically shut off during 3. High fluid velocities which can result in noise and abrasion are normal operation. The flow rate is indicated by a metal ball (4) avoided. sliding inside a transparent channel (5) with an integral graduated 4. The differential pressures acting on the circuit control valves are scale (6). reduced preventing faulty operation.

1 3 Control valve 15 1 2 7 5 6 5 3 7 6 5 4 2 3 4 3 6 2

5 3 2 7 6 5 4 4

5 6 7 Manual valve 2

Construction details Flow meter bypass valve The bypass valve (1) opens and closes the circuit between the flow Flow meter meter and the valve. The bypass valve is easily opened by pulling the When activated, the flow rate operating ring (2), and is automatically closed by the internal return is indicated on the flow meter spring (3) when finished reading the flow rate. The spring and the housed in a bypass circuit on EPDM seal (4) provide a reliable seal to isolate the flow meter during the valve body. When finished normal operation, protecting potential debris from interfering with reading the flow rate, the flow spring/magnetic disc mechanism. meter is automatically shut The operating ring (2) material has low thermal conductivity to avoid off, isolating it during normal burns if the flow meter is opened while hot fluid is passing through the operation. valve. Use of a flow meter greatly simplifies the process of system 1 balancing since the flow rate can be measured and controlled at any time without differential pressure gauges or reference charts. The onboard flow 4 3 2 meter eliminates the need to calculate valve settings during Ball/magnet indicator system setup. Additionally, the 2 The metal ball (4) that indicates the flow rate is not unique onboard flow meter in direct contact with the heating/cooling transfer offers unprecedented time and fluid passing through the flow meter. cost savings by eliminating the This is an effective and innovative measuring 5 long and difficult procedure system in which the ball slides up and down inside of calculating pre-settings a transparent channel (5) that is isolated from the 7 associated with using traditional fluid flowing through the body of the flow meter. balancing devices. The ball is moved by a magnet (6) connected to a 6 The by-pass circuit easily float (7). In this way the flow rate indication system detaches for cleaning. remains perfectly clean and provides reliable 4 readings over time.

-- 58 Complete closing and opening of the valve Hydraulic characteristics at 100% open The valve can be completely closed and opened. A slot on the control ∆p (psi) (psi) (bar) (feet of head) stem indicates the valve position. When the control stem is turned 1/2” 3/4” 1” 1 1/4” 1 1/2” 2” 20 20 46.20 fully clockwise (the slot is 1.0 23.10 perpendicular to the axis Completely closed Completely open 10 10 of the valve), the valve is 0.5 0.4 fully closed (A). When the 5 5 11.55 4 4 0.3 9.24 control stem is turned 3 3 0.2 6.93 4.62 fully counter-clockwise 2 2 (the slot is parallel to the 0.1 axis of the valve), the 1 1 2.31 valve is fully open (B). 0.05 0.04 1.16 A B 0.5 0.5 0.924 0.4 0.4 0.03 0.693 0.3 0.3 0.02 0.462 For the flanged version, rotate the adjusting handwheel 90º for the 0.2 0.2 0.01 complete opening and closing of the valve as shown in (A) and (B). 0.231 When at the desired position, lock the adjustment screw. 0.1 0.1 ) 0 0 00 0.2 0.5 1 2 5 10 2 5 100 2

Completely closed Completely open gpm ( G

0 5 (l/h) 200 100 500 1000 2000 5000 10000 20000 Flow rate Code Connection Full open Cv (GPM) 132432A ½” NPT ½ – 1¾ 1.0 A B 132552A ¾” NPT 2.0 – 7.0 6.3 132662A 1” NPT 3.0 – 10.0 8.3 Insulation 132772A 1¼” NPT 5.0 – 19.0 15.2 The 132 series threaded 132882A 1½” NPT 8.0 – 32.0 32.3 version,is supplied with a hot pre-formed insulating shell. This 132992A 2” NPT 12.0 – 50.0 53.7 system ensures perfect heat insulation and keeps out water (feet of 2½” vapor from the environment. ∆p (psi) (psi) (bar) head) Additionally, this type of insulation 20 20 46.00 is ideal in cold water circuits as 1.0 it prevents condensation from 10 10 2½” 20.00 forming on the surface of the 0.5 valve body. 0.4 5 5 10.00 0.3 4 3” 4 9.00 3 3 0.2 7.00 5.00 2 2 Installation 0.1 3.00 Install the balancing valve in a location that ensures free access to 1 1 2.00 the flow meter shutoff valve, control stem and flow rate indicator. To 0.05 4” 1.50 ensure accurate flow measurement, straight sections of pipe installed 0.04 0.5 0.5 as shown is recommended. 0.4 0.4 0.03 1.00 3 2 4 5 6 7 3 2 4 5 6 7 0.3 0.3 0.02 0.70

1327 series 7 132 series 6 6 0.50 5 5 4 0.2 0.2 4 3 3 2 2 Pump 0.01 0.30 0.1 0.1 0.005 0.20 5D 10D 0.05 0.05 0.003 0.11 ) 00 00 00 0 0

The valve can be installed in any position with respect to the flow 10 2 5 100 2 3 4 gpm ( G direction shown on the valve body. Additionally, the valve can be

installed either horizontally or vertically. (l/h) 2000 5000 25000 45000 90000 10000 20000 3 2 4 5 6 7

3 2 4 5 6 7

2 2

3 3 4

4 CALEFFI CALEFFI

5 CALEFFI CALEFFI 5

6 6

7 7 7 7 7 7 7 7 6 6 6 6 6 6 5 5 5 5 5 5 4 4 4 4 4 4 3 3 3 3 3 4 5 6 7 3 2 3 4 5 6 7 2 2 3 2 2 2 2 2

Flow rate Code Connection Full open Cv (GPM) 132060A 2 ½” flange 30 - 105 87 132080A 3” flange 38 - 148 164

132100A 4” flange 55 - 210 242

4 CALEFFI CALEFFI

5 CALEFFI CALEFFI 5

6 --

7 7 7 7 7 7 6 6 6 6 5 5 5 5 4 4 4 4 3 3 3 3 4 5 6 7 2 3 2 2 4 5 6 7 2 3 59 2 2 QuickSetterTM Pull - Adjust - Release • Industry unique design; simply pull, adjust, and release for quick and accurate circuit balancing. • No need for differential pressure tools or cross-reference charts. • Built-in flow meter with memory pointer provides direct readout in GPM. • Flow window scale never clouds; indicator ball travels within transparent isolated glass channel. • Full range of pipe sizes from 1/2" to 2" NPT and 2-1/2" to 4" flanged to fit most commercial projects. • Low lead models for plumbing applications are perfect for DHW recirculation.

Components for today's modern hydronic systems

Heating & Cooling

www.caleffi.com - Milwaukee, WI USA 60

Quicksetter Family 8.375 x 11.indd 1 4/16/18 2:32 PM