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TIGHTENING THE ENVELOPE eto 2 Section- R The building envelope includes windows, doors, walls, the roof, and the foundation. Heat always flows from the warmer side of the building shell to the colder side. The most commonly BUILDING discussed parameters of heat flow through the building envelope, in or out, are conduction, infiltration, and solar ENVELOPE radiation. Conduction is heat flow through a material from hot to cold. This phenomenon explains why the handle on a stove pot SECTION R becomes hot, and why people insulate walls. Infiltration is a form of convection in which heat flows via air movement. This phenomenon explains why occupants feel cold when the door is open on a winter day, and why caulking small cracks around windows improves comfort. Radiation is heat flow over a distance from hot to cold, the way the Sun's heat reaches Earth. Building occupants use window shades to block radiation.
Conduction (roof, walls, windows). WINDOWS Conductivity depends on INFILTRATION the materials used in the eto 3 Section- R 4 Section- R building shell. Insulation In older buildings, heat often leaks through breaks in slows, but does not stop, insulation or around windows. This infiltration can greatly heat flow through walls reduce the insulation's effectiveness, so R-values alone do and roofs. R-value not fully describe the energy efficiency of a wall or roof. indicates how well an All buildings allow some level of uncontrolled airflow insulation barrier through the building envelope. Infiltration paths include impedes heat flow—the seals around operable windows, cracks or seams in exterior larger the R-value, the panels, doorjambs, and shell penetrations such as holes for less heat flows through a wiring or roof curbs for HVAC equipment. wall or roof by conduction in a given amount of time. Air flowing into or out of these leakage paths is driven by Windows typically have a pressure differences caused by HVAC equipment between very low R-value the inside and outside of the building, between windward and leeward sides of the building, and between upper and lower floors (natural convection, commonly called the chimney effect)
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EFFECTIVE BUILDING SHADING WINDOW OPTIONS eto 5 Section- R 6 Section- R Interior shading. Spectrally selective glass. This type of glass can maximize Venetian blinds and other or minimize solar gain and shading depending on the operable shades are low- chosen selectivity. cost and effective solutions Double-glazed, low-e systems. Layers of low-e film are for keeping out sunlight. stretched across the interior air space between glass panes, Shades can be installed and windows with this feature offer R-values as high as 8. between two panes of Gas filled windows. Using argon or krypton gas between window glazing to glass panes, this technology minimizes the convection automatically open and currents and conduction through the gas-filled space, close shades in response to reducing overall heat transfer through the window. light. Electrochromic windows. When integrated with a Low-emissivity (low-e) daylighting control system, these windows can preserve the coatings Low-e coatings view outside while varying their tint to modulate insulate better than bare transmitted light, glare, and solar heat gain. Sensors that adjust tint can automatically balance comfortable lighting windows, while allowing as with energy efficiency. much solar heat gain as possible.
WINDOW FILM HEATING REDUCTION SHADING COEFFICIENT EXAMPLE eto 7 Section- R 8 Section- R Performance of window films are measured by a Savings from window films can be found using (1 - SC). shading coefficient, SC. SC = 0 means no heat passes. The annual solar heat gain for windows facing west in a SC = 1 means it does nothing to stop heat. building has been calculated to be165,000 kJ/m2 . If a shading film with shading coefficient 0. 6 can be applied to these windows at a cost of $20.00 per square metre, Therefore, the fractional heat flow reduction due to the window film is given by (1-SC). calculate the simple payback. The COP of the air conditioner is 2.7 and electricity costs $0.06 per kWh.
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ROOFING OPTIONS eto 9 Section- R 10 Section- R Reduced heat gain = (0.4)x 165,000 kJ/m2/yr Measures that can be employed to reduce heat flow into and out Find EER = COP x 3600 = kJ/kWh of a building through the roof Use Railroad Track Method to solve for the cost include roof insulation, cool roofs, to remove this reduced heat gain and green roofs.
Much of a building's heat losses and gains occur through the roof, Finally, solve for SPP so there are often significant energy-savings opportunities related to roof efficiency.
HEAT LOSS AND GAIN COOL CASE STUDY Heat is lost and gained through walls and
eto 11 Section- R ceilings. 12 Section- R Installing a reflective roof membrane on a 100,000-square- foot Target retail store in Austin, Texas, reduced the Consider a two cm (20 mm) plywood wall. average summer daily maximum roof-surface temperature from 168° to 126° Fahrenheit. T in This temperature reduction cut the building's total air- wall conditioning energy use by 11 percent and peak air- inside conditioning demand by 14 percent. Researchers estimate surface that this cool roof installation will save about $65,000 over of wall the course of its useful life. Tout According to the building manager, the difference in labor inside outside and materials costs for installing a white thermoplastic air film surface roof instead of a black rubber roof was negligible, so that the payback for this system was immediate. outside air film
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BASIC HEAT FLOW EQUATIONS BASIC HEAT FLOW EQUATIONS
Conduction heat losses through walls and ceilings The equation often is written A×T q = []W q = U × A × T []W ΣR Heat loss is proportional to the area A and the temperature difference ΔT between inside and where U is the overall thermal conductance. outside. ΣR is the sum of the resistances of everything that 1 1 W resists heat flow. U = = 2 2 m C R R Total m C R = R + R + R inside air film plywood outside air film W
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SURFACE AIR FILM RESISTANCE (BUILDINGS)* WALL AND INSULATION RESISTANCES
Wall or roof Direction of Rs R can be obtained from the conductance C, given for position heat flow (m2·°C/W) a specified thickness of material, Still air 1 R Horizontal up 0.11 C Horizontal down 0.16 If the conductivity k is given, R can be calculated Vertical horizontal 0.12 knowing k and the material thickness t in metres. Moving air 24 km/h (winter) All 0.030 t R 12 km/h (summer) All 0.044 k
*Data from 2001 ASHRAE Fundamentals Handbook, American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc., Atlanta, Georgia. p. 25.2.
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CONDUCTANCES AND CONDUCTIVITIES Material Description Conductivity Conductance BASIC HEAT FLOW PROBLEM k C or U
Building Gypsum 12.7 (0.16) 12.6 What is the rate of heat loss through a two centimetre 18 Section- R boards mm plywood wall if the inside temperature is 16°C and the Plywood 0.11 outside temperature is 0°C and windy? The area of the wall is Plywood 19 (0.116) 6.1 mm 100 square metres. Include surface films. Insulating Fibreglass, 0.036 Solution start – materials batt, blanket, Rpw = t/K = 0.02/0.11 = 0.182 board Mineral wool, 0.039 batt, blanket, loose Vermiculite 0.066 (expanded)
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BASIC HEAT FLOW EQUATIONS COMPOSITE WALL EXAMPLE How could we reduce the heat flow? 20 Section- R A 160 m2 wall is exposed to an inside temperature of 24 Often the answer is insulation—a simple and °C and an outside winter temperature of 4 °C in windy inexpensive way of adding more resistance to the weather. The wall consists of 12.7 mm plywood, 90 mm denominator of the equation of fiberglass insulation, and 12.7 mm gypsum board. How much heat is lost through the wall? Include surface films. A×T q = []W ΣR
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BASIC HEAT FLOW EQUATIONS DEGREE DAYS EXAMPLE
Temperature varies with time of day and season, so we For three days the outside temperature averages 10 C 22 Section- R often resort to this heat conduction equation each day. The number of HDD for this three-day period is: Q = U× A ×DD × 24 [Wh/yr] HDD = (18.3 C – 10 C) 3 days C days where DD can be HDD or CDD. Annual units are = 25 degree days yr
UxA is also known as the conduction part of the overall Building Load Coefficient BLC. Other parts include infiltration, ventilation, and slab-on-grade factor.
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EXAMPLE AIR HEAT FLOW PROBLEMS A wall has an area of 100 m2 and has a thermal conductance of 1.4 W/m2·C. If there are 3000 degree- days in the annual heating season, what is the total q = Mh []W General amount of heat that must be supplied by the heating q = M C T []W Sensible heat only system? p Solution: Q = U x A x DD/yr x 24 h/day Wh/yr M = Mass Flow Rate (kg/h) h = Enthalpy Difference (kJ/kg)
Cp = Heat Capacity (kJ/kg·C) T = Temperature Difference (C)
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AIR HEAT FLOW PROBLEMS AIR WATER HEAT FLOW PROBLEMS
Air: Sensible Heat Only Air: General
1.204 kg .001m3 1.006 kJ 1.204 kg .001 m3 q = LPS × 3 × × × T q = LPS × × × h m L kg °C m3 L
q = LPS ×1.2 × ∆T W q = LPS ×1.2 × ∆h []W
Sensible heat only
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DUCT LOSS EXAMPLE WATER: SENSIBLE HEAT ONLY 5,000 LPS of air leaves an air handler at 10 C. It is
delivered to a room at 18 C. There was no air volume 27 Section- R loss due to duct air leaks. No moisture was added, or taken away from the air in the duct. How many kW heat 1000 kg 1 m3 4.2 kJ gain occurred because of heat transfer by conduction? q = LPS × × × × T m3 1000 L kg C (A) 48 kW (B) 20 kW (C) 36 kW q = LPS × 4.2 × T []kW (D) 60 kW (E) 3 kW Sensible heat only
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CENTRIFUGAL FAN AND PUMP LAWS CHILLED WATER FLOW EXAMPLE Flow and Speed
A chiller supplies cold water with a T of 6 C. 29 Section- R 30 Section- R How many LPS of this water is needed to provide LPS RPM new = new one kW of air conditioning? LPSold RPMold Pressure (Head) and Speed
2 2 Pnew [RPMnew ] [LPSnew ] = 2 = 2 Pold []RPM []LPS Power and Speed old old
3 kWnew []LPSnew = 3 kWold []LPSold
THE PSYCHROMETRIC CHART RACTICE XAMPLE P E The Psychrometric Chart graphically represents the ACE Industries presently has a 10 kW ventilating fan 31 Section- R steam tables for moisture in air at conditions we 32 Section- R that draws warm air from a production area. The encounter in HVAC work. motor recently failed, and they think they can replace The Psych Chart allows complex problems to be it with a smaller motor. They have determined that worked out easily, and provides a feel for common they can reduce the amount of ventilation by one- HVAC processes that we are interested in. third. ((,Now, the new amount of ventilation will be two thirds of the old amount.) The standard ASHRAE Psych Chart has a horizontal What size motor is needed now? axis for dry bulb temperature, and a vertical axis for humidity ratio in pounds of moisture per kg of dry air. Other parameters on the chart are: relative humidity, wet bulb temperature, enthalpy, specific volume, and Answer: 2.97 kW (round up to 3 kW) saturation temperature.
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COMMON HVAC PROCESSES eto 33 Section- R 34 Section- R
HEATING EXAMPLE eto 36 Section- R For air, q = LPS ×1.2 × ∆h []W
Air at 20C dry bulb and 50% relative humidity flows at 3200 LPS and is heated to 32C dry bulb and humidified to 40% RH. Approximately how many kW is required in this process?
(A) 61 kW (B) 96 kW (C) 183 kW (D) 351 kW
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COOLING EXAMPLE eto 37 Section- R 38 Section- R How many kW of air conditioning is required to cool 500 LPS of air at 32 °C and 60 % relative humidity (RH) to 15 °C and 100 % RH?
CEM REVIEW PROBLEMS An absorption chiller with a COP of 0.8 is powered by hot CEM REVIEW PROBLEMS
water that enters at 90C and exits at 80C at a rate of 2 39 Section- R The conduction part of the Building Load Coefficient 40 Section- R LPS. The chilled water operates on a 5 C temperature (U*A) for a building is 3500 W per degree C. difference. Calculate the chilled water flow rate. Solution of this problem does not require a knowledge of how Estimate the seasonal energy consumption for absorption chillers work internally. heating if the heating season has 1800 degree days. (A) 0.8 LPS The heating unit efficiency is 80%. (B) 1. 6 LPS (C) 3.2 LPS (A) 700 GJ/yr (D) 3.6 LPS (B) 350 GJ/yr (E) 2.4 LPS (C) 462 GJ/yr (D) 720 GJ/yr
(E) 680 GJ/yr
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APPENDIX eto 41 Section- R Conversions from IP to SI
Rip x 0.1761 = Rsi END OF SECTION R Kip x 0.1442 = Ksi ( IP K uses inches thick.) Uip x 5.68 = Usi
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