Heat Transfer Heat Transfer • Introduction – Practical occurrences, applications, factors affecting heat transfer – Categories and modes of heat transfer • Conduction – In a slab and across a pipe • Convection – Free (natural) and forced (in a pipe and over a solid object)
– Determination of convective heat transfer coefficient (h and hfp) • Radiation • Thermal resistances to heat transfer • Overall heat transfer coefficient (U) • Steady state heat transfer – In a tubular heat exchanger (without and with insulation) • Dimensionless numbers in heat transfer – Steady: Reynolds #, Prandtl #, Nusselt #, Grashof #; Unsteady: Fourier #, Biot # • Unsteady state heat transfer – For conduction/convection driven heat transfer; Heisler chart 2 Introduction
3 Practical Occurrences • Is a metallic park bench colder than a wooden park bench? • What is wind-chill factor? What is heat index? • Why dress in layers during winter? • How does a fan provide cooling effect? Does it blow cold air? • What is the insulation used in houses? Is it for winter or summer? • Why does our skin dry-up in a heated room? • What time of the day and why do we get sea-breeze? • Why are higher altitude places colder? • Does hot water freeze faster than cold water? • In winter, do hot or cold water pipes burst first? • What is greenhouse effect? What is the principle behind it? • Can you lose weight by drinking cold water? • Why are “fins” present on the outside of the radiator of a car? • “Bridge freezes before road” -- Why? • Why is salt used to melt ice on the road? When is sand used? • How does an igloo keep an Eskimo warm? • Why do you see cars breakdown or pull over to the shoulder of a highway during traffic jams? Do traffic jams cause breakdowns or do breakdowns cause traffic jams? 4 Heat Transfer in Various Industries
• Automobile: Radiator and engine coolant • Electronics: Cooling of motherboard/CPU by fan • Pharmaceutical: Freeze drying of vaccines • Metallurgical: Heating/cooling during steel manufacture • Chemical: Condensation, boiling, distillation of chemicals
Home: Refrigerator, AC, heater, dryer, stove, microwave 5 Heat Transfer in the Food Industry • Melting: Thawing of a frozen food (turkey) • Freezing: Freezing of ice-cream mix • Drying: Drying of fruits • Evaporation: Spray drying of coffee or concentration of juices • Sublimation: Freeze drying of coffee • Heating/cooling of milk • Baking of bread • Processing of canned soups (inactivate microorganisms & maximize nutrient content,
color/flavor/texture) 6 What Factors Affect Rate of Heat Transfer? •Thermal
– Specific heat (cp in J/kg K) • Measured using Differential scanning calorimeter (DSC) – Thermal conductivity (k in W/m K) • Measured using Fitch apparatus or thermal conductivity probe (Lab #5) • Physical – Density ( in kg/m3) • Measured using pycnometer • Rheological (measured using rheometer/viscometer) – Viscosity ( in Pa s) for Newtonian fluids OR – Consistency coefficient (K in Pa sn) and flow behavior index (n) for power-law fluids
2 Note: Thermal diffusivity ( = k/cp in m /s) combines the effect of several factors7 Specific Heat, Thermal Conductivity, and Thermal Diffusivity
• Specific heat (cp) – A measure of how much energy is required to raise the temperature of an object • Thermal conductivity (k) – A measure of how quickly heat gets conducted from one part of an object to another • Thermal diffusivity () – It combines the effects of specific heat, thermal conductivity, and density of a material. Thus, this one quantity can be used to determine how temperature changes at various points within an object.
8 Specific Heat (DSC Method) Heat flux held constant & temperature diff. measured
Q = m1 cp(1) (T1) = m2 cp(2) (T2)
cp(2) = {m1/m2} {(T1)/ (T2)} cp(1)
Differential Scanning Calorimeter (DSC)
Manufacturer: Perkin-Elmer 9 Thermal Conductivity (Fitch Apparatus)
Slope Heat Source (ice-water mix) Intercept Q This can be rewritten as: Sample t: time (cheese slice) m, cp, A, T: For heat sink (mass, sp. ht., area, temp.) T : Initial temp. of heat sink Heat Sink i (copper block) T∞: Temp. of heat source Insulation L: Thickness of sample Y X
Plot on y-axis versus t on x-axis & set intercept = 0
Slope = -kA/(m cp L); Solve for k: k = - (Slope) (m cp L)/A
Note: ‘k’ is always a positive number 10 Thermal Conductivity Probe KD2 Pro Probe (Manufacturer: Decagon Devices) Single needle probe: Can measure ‘k’
Dual needle probe: Can measure ‘k’ and ‘’ = k / ( cp)
Sample Sample
k: Thermal conductivity (W/m K); : Thermal diffusivity (m2/s) 3 : Density (kg/m ); cp: Specific heat (J/kg K) 11 Values of Thermal Conductivity (k) • Good conductors of heat have high k values – Cu: 401 W/m K – Al: 250 W/m K – Fe: 80 W/m K – Stainless steel: 16 W/m K • Insulators have very low (but positive) k values – Paper: 0.05 W/m K – Cork, fiberglass: 0.04 W/m K – Cotton, styrofoam, expanded polystyrene: 0.03 W/m K – Air: 0.024 W/m K (lower k than insulators!) • Foods and other materials have intermediate to low k values – Foods: 0.3 to 0.6 W/m K (water: ~0.6 W/m K at room temperature) – Glass: 1.05 W/m K; Brick: 0.7 - 1.3 W/m K; Concrete: 0.4 - 1.7 W/m K – Plastics (commonly used): 0.15 - 0.6 W/m K
• Thermally conductive plastics may have k > 20 W/m K 12 Empirical Correlations
cp = 4.187 (Xw) + 1.549 (Xp) + 1.424 (Xc) + 1.675 (Xf) + 0.837 (Xa) Heldman & Singh, 1981
k = 0.61 (Xw) + 0.20 (Xp) + 0.205 (Xc) + 0.175 (Xf) + 0.135 (Xa) Choi & Okos, 1984
w: water, p: protein, c: carbohydrates, f: fat, a: ash
13 Effect of Temperature on k, , , cp)
14 Questions Q: When the same heating source is used to heat identical quantities of water and butter, which will be hotter after a certain time? Ans: Butter; because it has a lower specific heat
Q: In winter, is a metallic park bench colder than a nearby wooden park bench? Ans: NO. A metallic bench has a higher thermal conductivity and hence conducts heat very well, thereby taking away the heat generated by our body very fast and making us feel colder. 15 Categories of Heat Transfer • Steady state – Temperatures at all points within the system remain constant over time – The temperatures at different locations within the system may be different, but they do not change over time – Strictly speaking, steady state conditions are uncommon • Conditions are often approximated to be steady state – Eg.: Temperature inside a room or refrigerator • Unsteady state – Temperature(s) at one or more points in the system change(s) over time
– Eg.: Temperature inside a canned food during cooking 16 Modes of Heat Transfer • Conduction – Translation of vibration of molecules as they acquire thermal energy • Occurs in solids, liquids, and gases – Heat transfer from hot plate to vessel/pot – Heat transfer from surface of turkey to its center • Convection – Fluid currents developed due to temp. differences {within a fluid (liquid/gas) or between a fluid and a solid} or the use of a pump/fan • Occurs in liquids and gases – Heat transfer from hot vessel/pot to soup in it • Radiation – Emission & absorption of electromagnetic radiation between two surfaces (can occur in vacuum too) • Occurs in solids, liquids, and gases
– Radiation from sun; reflective thermos flask; IR heating of buffet food 17 Conduction
18 Basics of Conduction
• Conduction involves the translation of vibration of molecules along a temperature gradient as they acquire thermal energy (mainly analyzed within solids; however, it takes place in liquids and gases also) – Actual movement of particles does not occur • Good conductors of electricity are generally good conductors of heat • Thermal conductivity (k) is used to quantify the ability of a material to conduct heat
19 Fourier’s Law of Heat Conduction Rate of heat transfer by conduction is given by Fourier’s law of heat conduction as follows: Q = - kA (T/x) The negative sign is used to denote/determine the direction of heat transfer (Left to right or right to left)
Q: Energy transferred per unit time (W) k: Thermal conductivity (W/m K); it is a +ve quantity A: Area of heat transfer (m2) T: Temperature difference across the ends of solid (K) x: Distance across which heat transfer is taking place (m) Q/A: Heat flux (W/m2) 20 Temperature Difference Across a Slab Heat flow T1 •Slab: Q = kA (T/x) x T T = T1 –T2 2 For the same value of Q (example: use of a heater on one side of a slab),
For insulators (low k), “T1 –T2” is large For good conductors (high k), “T1 –T2” is small
For the same value of “T1 –T2” (example: fixed inside temperature of room and outside air temperature), For insulators (low k), Q is small For good conductors (high k), Q is large Note: x and A are assumed to be the same in all of the above situations 21 Conduction Across a Slab or Cylinder Heat flow T1 •Slab: Q = kA (T/x) x
T2
T1 Heat flow r T2 • Cylinder: Q = kAlm (T/r) k: Thermal conductivity (W/m K) A: Area across which heat transfer is taking place (m2)
T = T1 –T2: Temperature difference (K) 2 Alm: Logarithmic mean area (m )
Note: Alm comes into play when the area for heat transfer at the two ends across which heat transfer is taking place, is not the same 22 Logarithmic Mean Area (Alm) r T1 Heat flow r Q = kA (T/r) T2 o lm ri T = T1 –T2 r= ro -ri L • Slab: Area for heat transfer is same at both ends • Cylinder
– Area at one end (outside) is Ao (= 2roL)
– Area at other end (inside) is Ai (= 2riL) – Which area should be used in determining Q?
–Alm = (Ao –Ai) / ln (Ao/Ai) = 2L (ro –ri) / [ln (ro/ri)]
–Note: Ao > Alm > Ai 23 Logarithmic Mean Temp Diff (Tlm) Double Tube Heat Exchanger
Tw(o) Hot water Tw(i) T1 T2
Product Tp(o) Tp(i)
T is NOT constant across the length of tube
T1 = Tw(o) –Tp(i) , T2 = Tw(i) –Tp(o)
Tlm = (1 – 2) / [ln (1 / 2)]
Note: Tlm lies between T1 and T2 Subscripts: ‘w’ for water; ‘p’ for product, ‘i’ for inlet, ‘o’ for outlet
Note: Tlm comes into play when the temperature difference across the two ends where heat transfer is taking place, is not the same 24 Convection
25 Basics of Convection • It involves transfer of heat by movement of molecules of fluid (liquid or gas) due to – Temperatures differences within a fluid or between a fluid and a solid object OR – An external agency such as a pump or a fan • Convection is a combination of – Diffusion (microscopic/molecular level) • Random Brownian motion due to temperature gradient – Advection (macroscopic level) • Heat is transferred from one place to another by fluid movement
26 Newton’s Law of Cooling for Convection
Rate of heat transfer by convection (for heating or cooling) is given by Newton’s law of cooling as follows:
Q = h A (Ts -T∞) Q: Energy transferred per unit time (W) h: Convective heat transfer coefficient -- CHTC (W/m2 K) A: Surface area available for heat transfer (m2)
T = Ts –T∞ : Temperature difference (K) Ts: Surface temperature of solid object (K) T∞: Free stream (or bulk fluid) temperature of fluid (K) CHTC (h): Measure of rate of heat transfer by convection; NOT a property; depends on fluid velocity, surface characteristics (shape, size, smoothness), fluid properties (, k, , c ) p 27 Categories of Convection • Free (or natural) convection – Does not involve any external agency in causing flow – Heat transfer between bottom of vessel and fluid in it – Cooling of human body – Cooling of radiator fluid in car engine during idling 2 2 –hair-solid: 5-25 W/m K; hwater-solid: 20-100 W/m K • Forced convection – External agency such as fan/pump causes flow – Cooling of radiator fluid in car engine during motion – Ice-cream freezer (Blast air) – Stirring a pot of soup – Heat transferred from computers (fan) 2 2 –hair-solid: 10-200 W/m K; hwater-solid: 50-10,000 W/m K 2 –hboiling water or steam to solid: 3,000-100,000 W/m K 28 Free Convection • Fluid comes into contact with hot solid • Fluid temperature near solid increases • Fluid density near solid decreases • This results in a buoyancy force that causes flow • Rate of heat transfer (Q & h) depends on – Temperature difference between fluid and surface of solid
– Properties (, , k, cp) of fluid – Dimensions and surface characteristics (smoothness) of solid
NNu = hdc/kf = f (NGr , NPr) 29 Question Q: What is wind-chill factor? In winter, a thermometer reads -20 °C when air is stationary. All of a sudden, a gust of wind blows. What will the thermometer read? Ans: -20 °C. As wind speed increases, more heat is removed from our body due to an increase in ‘h’ and hence ‘Q’. Thus, we feel colder than when the air is stationary. The air is NOT colder, we just feel colder since more heat is removed from our body and our body is unable to generate enough heat to replace the energy lost to the surroundings. 30 Nusselt Number (NNu)
h: Convective heat transfer coefficient (W/m2 K)
dc: Characteristic dimension (m)
kf: Thermal conductivity of fluid (W/m K)
Nusselt number represents the ratio of heat transfer by convection & conduction 31 Grashof (NGr) Number
-1 f: Coefficient of volumetric thermal expansion (K ) g: Acceleration due to gravity (= 9.81 m/s2) 3 f: Density of fluid (kg/m )
Ts: Surface temperature of solid object (K)
T∞: Free stream temperature of fluid (K) dc: Characteristic dimension of solid object (m) (Obtained from tables based on shape & orientation of solid object)
f: Viscosity of surrounding fluid (Pa s)
Grashof number represents the ratio of buoyancy and viscous forces 32 Prandtl Number (NPr)
cp(f): Specific heat of fluid (J/kg K)
f: Viscosity of fluid (Pa s)
kf: Thermal conductivity of fluid (W/m K)
Prandtl number represents the ratio of momentum and thermal diffusivities 33 Properties of Air
0.72 0.74 0.74 0.74 0.73 0.73 0.73 0.73 0.73 0.72 0.72 0.72 0.72 0.72 0.73 0.72 0.72 0.72 0.72
34 Properties of Water
35 Free Convection (Plate)
•NNu = hdc/kf = f (NGr, NPr)
m •NNu = a (NGr NPr) ; NRa = NGr NPr
• For vertical plate (dc = plate height) 4 9 . a = 0.59, m = 0.250 (for 10 < NRa < 10 ) 9 13 . a = 0.10, m = 0.333 (for 10 < NRa < 10 ) 9 • For inclined plate (for NRa < 10 )
. Use same eqn as vertical plate & replace ‘g’ by ‘g cos’ in NGr
• For horizontal plate (dc = Area/Perimeter) . Upper surface hot 4 7 • a = 0.54, m = 0.250 (for 10 < NRa < 10 ) 7 11 • a = 0.15, m = 0.333 (for 10 < NRa < 10 ) . Lower surface hot • a = 0.27, m = 0.250 (for 105 < N < 1011) Ra 36 Free Convection (Cylinder)
• For vertical cylinder (dc = cylinder height) 0.25 – Similar to vertical plate if D ≥ 35L/(NGr)
• For horizontal cylinder (dc = cylinder diameter)
-5 12 – For 10 < NRa < 10
Note: NRa = NGr NPr 37 Free Convection (Sphere)
NRa = NGr NPr
For sphere, dc = D/2
Note 1: For all free convection situations, determine properties at the film temperature {Tfilm = (Ts + T∞)/2} unless otherwise specified Note 2: For all free convection scenarios, as the T between the fluid and surface of solid increases, NGr increases. Thus, NNu and ‘h’ increase. 38 Forced Convection
• Fluid is forced to move by an external force (pump/fan) • Rate of heat transfer (Q & h) depends on
– Properties (, , k, cp) of fluid – Dimensions and surface characteristics (smoothness) of solid • ‘h’ does NOT depend on – Temperature difference between fluid and surface of solid • ‘h’ strongly depends on Reynolds number – When all system and product parameters are kept constant, it is flow rate (a process parameter) that strongly affects ‘h’
NNu = hdc/kf = f (NRe , NPr) 39 Categories of Convective Heat Transfer Coefficient for Forced Convection
• Between a moving fluid and a stationary solid object – Transfer of heat from hot pipe to a fluid flowing in a pipe – Generally depicted by ‘h’
• Between a moving fluid and a moving particle – Transfer of heat from a hot fluid to a freely flowing particle in a suspension (particulate/multiphase food)
– Generally depicted by ‘hfp’
40 Forced Convection in a Pipe •NNu = hdc/kf = f (NRe, NPr) • Three sub-categories of forced convection exist…..
• 1. Laminar flow (NRe < 2100) – A. Constant surface temperature of pipe
•NNu = 3.66 (for fully developed conditions) – B. Constant surface heat flux
•NNu = 4.36 (for fully developed conditions) – C. Other situations (for entry region & fully developed) 0.33 0.14 •NNu = 1.86 (NRe x NPr x dc/L) (b/w) dc: ID of pipe, L: Length of pipe • 2. Transitional flow
(2100 < NRe < 4000) – Friction factor (f) • For smooth pipes:
• For non-smooth pipes, use Moody chart (graph of: f, NRe, /D) 41 Moody Diagram Relative Roughness ( Friction Factor (f) Friction Factor /D)
Reynolds Number (NRe) = 259 x 10-6 m for cast iron; 1.5235 x 10-6 m for drawn tubing; 152 x 10-6 m for galvanized iron; 45.7 x 10-6 m for steel or wrought iron 42 Forced Convection in a Pipe (contd.)
3. Turbulent flow (NRe > 4000) of a Newtonian fluid in a pipe, 0.8 0.33 0.14 NNu = 0.023 (NRe) (NPr) (b/w)
b: Viscosity of fluid based on bulk fluid temperature
w: Viscosity of fluid based on wall temperature
The term “(b/w)” is called the viscosity correction factor and can be approximated to “1.0” in the absence of information on wall temperature
Note: For flow in an annulus, use same eqn with dc = 4 (Acs/Wp) = dio –doi dio: Inside diameter of outside pipe doi: Outside diameter of inner pipe Note: For all forced convection situations, use bulk temperature of fluid to determine properties (unless otherwise specified) 43 ‘hfp’ for Forced Convection over a Sphere
NNu = hdc/kf = f (NRe, NPr) – similar to flow in a pipe
0.5 0.33 NNu = 2 + 0.6 (NRe) (NPr)
For 1 < NRe < 70,000 and 0.6 < NPr < 400
Note 1: dc is the outside diameter of the sphere Note 2: Determine all properties at the film temperature
{Tfilm = (Ts + T∞)/2} 44 Comparison of Free and Forced Convection
• Free convection [Q = hA T; NNu = hdc/kf = f (NGr, NPr)] – Does not involve any external agency in causing flow • Temperature difference (T) causes density difference; this causes flow – Q & h depend on • Temperature difference between surface of solid and surrounding fluid (T)
• Properties (, , k, cp) of fluid • Dimensions and surface characteristics (smoothness) of solid
• Forced convection [Q = hA T; NNu = hdc/kf = f (NRe, NPr)] – External agency such as fan/pump causes flow – Q & h depend on
• Properties (, , k, cp) of fluid • Dimensions and surface characteristics (smoothness) of solid – Only ‘Q’ and NOT ‘h’ depends on temperature difference between surface of solid and surrounding fluid (T) 45 Radiation
46 Basics of Radiation Heat Transfer
Rate of heat transfer by radiation is given by Stefan- Boltzmann law as follows: Q = σ A ε T4
Q: Energy transferred per unit time (W) σ: Stefan-Boltzmann constant (= 5.669 x 10-8 W/m2 K4) A: Surface area of object (m2) ε: Emissivity of surface (ranges from 0 to 1.0) T: Temperature (K)
47 Infrared Thermometer
• Infrared thermometer can be used to non-invasively and remotely determine the surface temperature of an object • Care should be exercised in ensuring that ONLY emitted energy is measured and NOT reflected energy (may have to use non-reflecting tape on metallic surfaces) • The emissivity of some infrared thermometers can be adjusted; for others, a pre-set value of 0.95 is commonly programmed
48 Thermal Resistances to Heat Transfer
49 Thermal Resistances to Heat Transfer
• Conduction – Slab: Q = kA (T/x) = T/[(x/kA)]
– Cylinder: Q = kAlm (T/r) = T/[(r/kAlm)] – Driving force for heat transfer: T
– Thermal resistance to heat transfer: (x/kA) or (r/kAlm) • Convection •Q = hA(T) = T/[(/hA)] – Driving force for heat transfer: T – Thermal resistance to heat transfer: (/hA)
Units of thermal resistance to heat transfer: K/W 50 Thermal Resistances Conduction: Q = kA T/x Single slab: Q = T/[(x/kA)]
Multiple slabs: Q = T/[(x1/k1A) + (x2/k2A) + …]
Cylindrical shell: Q = T/[(r/kAlm)]
Multiple cylindrical shells: Q = T/[(r1/k1Alm(1)) + (r2/k2Alm(2))] Convection: Q = hA T Single convection: Q = T/[(/hA)]
Multiple convections: Q = T/[(/h1A1) + (/h2A2) + …] Combination of conduction and convection Multiple slabs
Q = T/[(x1/k1A) + (x2/k2A) + (/h1A1) + (/h2A2) + …] Multiple cylindrical shells
Q = T/[(/h1A1) + (r1/k1Alm(1)) + (r2/k2Alm(2)) + (/h2A2)]
Units of thermal resistance to heat transfer: K/W 51 Electrical Analogy A current (I) flows because there is a driving I force, the potential Difference (V), across the resistance (R) V R I = V/R
Q = T/(Thermal Resistance)
I R1 I = V/(R1 + R2) V Q = T/(Sum of Thermal Resistances) R2
Thermal resistances are additive (similar to electrical resistances) 52 k, h, U, Resistances, and Temperatures
• As thermal conductivity (k) increases, thermal resistance due to conduction (x/kA) decreases – Thus, temperature difference between center and surface of object decreases • As convective heat transfer coefficient (h) increases, thermal resistance due to convection (1/hA) decreases – Thus, temperature difference between the fluid and surface of the solid object decreases
53 Overall Heat Transfer Coefficient (OHTC)
54 What is Overall Heat Transfer Coefficient?
• OHTC (denoted by the symbol ‘U’) refers to a single quantity that can be used to quantify the effect of all forms (conduction and convection) of heat transfer taking place in a system • It facilitates the use of one equation (instead of individual equations for each conductive and convective heat transfer in the system) to determine the total heat transfer taking place in the system – All thermal resistances in the system are added in order to facilitate this process
55 OHTC (or U) in Different Scenarios
• Three conductive heat transfers
. 1/(UA) = x1/(k1A) + x2/(k2A) + x3/(k3A) • Two convective heat transfers
. 1/(UA) = 1/(h1A1) + 1/(h2A2) • One conductive and one convective heat transfer
. 1/(UA) = x1/(k1A) + 1/(hA) Note 1: 1/UA > 1/hA; Thus, U < h Note 2: If there is no conductive resistance, U = h
U: Overall heat transfer coefficient (W/m2 K) “1/UA”: Overall thermal resistance (K/W) 56 k, h, U, Resistances, and Temperatures • As thermal conductivity (k) increases, thermal resistance due to conduction (x/kA) decreases – Thus, temperature difference between center and surface of object decreases • As convective heat transfer coefficient (h) increases, thermal resistance due to convection (1/hA) decreases – Thus, temperature difference between the fluid and surface of the solid object decreases • As overall heat transfer coefficient (U) increases, overall thermal resistance (1/UA) decreases – Thus, temperature difference between the two points across which heat transfer is taking place, decreases Thermal conductivity: W/m K Convective heat transfer coefficient: W/m2 K Overall heat transfer coefficient: W/m2 K Thermal resistance to heat transfer: K/W 57 Why Dress in “Layers” in Winter? • Single jacket of thickness x: –Q = T/[(x/kA)] • Two jackets, each of thickness x/2: –Q = T/[{(x/2)}/kA) + {(x/2)}/kA)] = T/[(x/kA)]
• If the above two expressions are identical, why is it better to wear two jackets, each of thickness x/2?
Air trapped between the two jackets adds a convective thermal resistance. Thus, Q = T/[{(x/2)}/kA) + {(x/2)}/kA) + (1/hA)] Thus, total thermal resistance increases and Q decreases 58 Effect of Resistance on Temperature Q • A heater is used to maintain the Q h 5 °C left end of the slab at 95 °C • Ambient air on right side is at 5 °C k • What factors determine the 95 °C magnitude of temperature at right T x end of slab? • T is affected by 95 °C at left AND 5 °C at right • Resistance to heat transfer from left (by conduction) is x/kA • Resistance to heat transfer from right (by convection) is 1/hA • If both resistances are equal, T = (95 + 5)/2 = 50 °C • If conductive resistance is less (occurs when ‘k’ is high), T > 50 °C • If convective resistance is less (occurs when ‘h’ is high), T < 50 °C Note: The same Q flows through the slab and outside
Thus, Q = kA (95 – T)/x= hA(T –5) 59 Heat Exchangers (HX)
60 Types of Heating Equipment • Direct contact – Steam injection, steam infusion • Indirect contact (Other than plate, tubular, Shell & tube, SSHE) – Retorts (Using hot water, steam, or steam-air for heating) • Batch (Agitation: None, axial or end-over-end): With our w/o basket/crate • Continuous (With agitation): Conventional, Hydrostatic – Plate: Series, parallel, series-parallel – Tubular: Double tube, triple tube, multi-tube – Shell & tube: Single pass, multiple pass, cross-flow – Scraped surface heat exchanger (SSHE) • Alternative/Novel/Emerging Technologies – Microwave and radio frequency heating • Uses electromagnetic radiation; polar molecules heat up – Ohmic heating
• Electric current in food causes heating; ions in food, cause heating 61 Steam Injection Intense, turbulent mixing of steam and product occurs. It Steam results in rapid heating and dilution of product. A vacuum Pneumatically chamber is used downstream actuated to evaporate steam that condensed into product.
Pneumatically actuated
Variable Gap Product
www.process-heating.com 62 Steam Infusion Air out CIP in Steam in Product in
Cooling water in/out This is a gentler process than steam injection. A vacuum chamber is used downstream to evaporate steam that condensed into product.
Product out 63 Batch Retorts
• Static (with or without crates/baskets) Rotary, end-over-end or reciprocating – Horizontal motion is used to mix – Vertical the product and make the temperature • Rotary (axial or end-over-end rotation) distribution uniform End-over-end Rotation
Axial Rotation
• Reciprocating – Shaka
64 Static & Rotary Retorts
Raw canned Horizontal Basket Retort foods are placed in basket
SuperAgi Retort
www.libertyprocess.co.uk
www.jbtfoodtech.com 65 Crateless Retort (Semi-Continuous)
Problem: Cold spot identification www.maloinc.com 66 Conventional Continuous Rotary Retort
67 Hydrostatic Retort
Height of water column provides enough pressure to prevent water from boiling at temperatures well above 100 °C
68 Hydrostatic Retort/Sterilizer
69 Plate Heat Exchanger (PHE)
Hot fluid flows on one side of plate and cold fluid flows on other side of plate. Heat transfer occurs across each plate.
Simple, efficient, inexpensive; used for not too viscous fluids
70 PHEs (All channels in Parallel)
Advantage Low pressure drop
Disadvantage 1 X 4 Low heat transfer 1 X 4
71 PHEs (All Channels in Series)
Advantage High heat transfer
4 X 1 Disadvantage 4 X 1 High pressure drop
72 PHEs (Series & Parallel)
Advantage Optimized pressure drop and heat transfer
2 X 2 1 X 4
73 Regeneration in a PHE
Hot pasteurized product
Cold pasteurized product
Cooler Regenerator Heater
Raw product
Regeneration: Energy of hot pasteurized product is used to pre-heat cold raw product. Typical regeneration efficiency is ~90%. 74 Double Tube, Triple Tube, Multitube HX
Heating from 1 side
Heating from 2 sides
75 Shell & Tube: (One & Two Pass)
Note: Presence of baffles creates cross-flow pattern (product and heating/cooling medium flow at right angles to one another) with uniform heat transfer throughout the heat exchanger. Baffles prevent short- circuiting of heating medium directly from the inlet port to the outlet port. 76 Scraped Surface Heat Exchanger (SSHE) Motor Shaft with motor rotating it Blade Blade Insulation
Steam jacket Shaft
Product Cross-sectional view Insulation
Steam Advantage: Mixing of viscous foods jacket Disadvantage: Particle damage, uncertain residence time, cleaning Product 77 Double Tube Heat Exchanger (DTHE)
Hot/Cold water
Heat Transfer Product
Double tube heat exchanger Two concentric tubes Product generally flows in inner tube Heating/cooling medium generally flows in outer tube (annulus) One stream gains heat while the other stream loses heat (hence, heat exchanger) Heat transfer takes place across wall of inner tube
Both streams may flow in the same or opposite directions 78 Heat Transfer in a Double Tube HX (Hot water heating a product)
Q = hoAo [Thot water -Twall (outside)] Hot water Tho ho . Thi, mh, cp(h)
Q = kAlm [Twall (outside) -Twall (inside)]/r Q = h A [T -T ] h U i i wall (inside) product Product i rii roi . T Tci, mc, cp(c) co
L
Subscripts for T: ‘c’ for cold, ‘h’ for hot, ‘i’ for inlet, ‘o’ for outlet 79 Heat Transfer from Hot Water to Product
• Convection . From hot water to outside surface of inner tube
. Q = hoAo [Thot water -Twall (outside)] • Conduction . From outside of inner tube to inside of inner tube
. Q = kAlm [Twall (outside) -Twall (inside)]/r • Convection . From inside surface of inner tube to bulk of product
. Q = hiAi [Twall (inside) -Tproduct]
80 Resistances to Heat Transfer from Hot Water to Product
Hot water ho Convective resistance (1/ho Ao) Conductive resistance (r/k Alm) Convective resistance (1/hi Ai) Product hi U
81 Overall Heat Transfer Coefficient (U)
Q = T / [(1/hoAo) + (r/kAlm) + (1/hiAi)] Thermal resistances have been added
Denominator: Total thermal resistance
1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi) 2 Thus, Q = UAlm Tlm U: W/m K
U: Accounts for all modes of heat transfer from hot water to product U is NOT a property; it is NOT fixed for a HX; it depends on material properties, system dimensions, and process parameters
82 Determination of U: Theoretical Method
•1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi)
•hi and ho are usually determined using empirical correlations • ‘k’ is a material property of the tube of HX
•Ai, Ao, and Alm are determined based on dimensions (length & radii) of heat exchanger tubes
• Once Ai, Ao, Alm, hi, ho, and k are known, U is calculated using the above equation
83 Determination of U: Experimental Method (Hot Water as Heating Medium) . . Q = mc cp(c) Tc = mh cp(h) Th
= UAlm Tlm Assumption: Heat loss = zero
Once the mass flow rates and temperatures of the product and hot water are experimentally determined, U can be calculated
If there is heat loss,
Qlost by hot water = Qgained by product + Qlost to outside 84 Tubular Heat Exchanger (Co- and Counter-Current) 1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi)andQ = UAlm Tlm
h: Hot c: Cold h: Hot i: Inlet c: Cold o: Outlet i: Inlet o: Outlet Temperature Temperature
Used only when rapid initial cooling is needed Most commonly used Lower heat transfer efficiency Higher heat transfer efficiency
Tco ≤ Tho always Tco can be greater than Tho T btwn hot & cold fluid dec. along length T btwn hot & cold fluid does not Ai = Inside surface area of inner pipe = 2 rii L change significantly along length Ao = Outside surface area of inner pipe = 2 roi L Alm = Logarithmic mean area of inner pipe = (Ao –Ai) / [ln (Ao/Ai)] = 2L (roi –rii)/[ln (roi/rii)] Tlm = Logarithmic mean temperature difference = (T1 – T2)/[ln(T1/T2)] 85 Co-Current Arrangement Q3 Hot water . Thi, mh, cp(h) ho Q2 Tho Product U . h Q T , m , c i 1 Tco ci c p(c) rii roi
L r = roi -rii
Q1: Energy transferred from heating medium to product (= energy gained by product) Q2: Energy lost by heating medium (= energy gained by product and surroundings) Q3: Energy transferred from heating medium to surroundings (= energy gained by surroundings) Note: Q2 = Q1 + Q3
Common approximation: Q3 = 0 (valid if HX is insulated) In this case, . . . Subscripts for m, cp, T, T: Q1 = mc cp(c) (T)c = Q2 = mh cp(h) (T)h = UAlm Tlm ‘h’ for hot, ‘c’ for cold with 1/UAlm = 1/hiAi + r/kAlm + 1/hoAo ‘i’ for inlet, ‘o’ for outlet Note 1: (T)c = (Tco –Tci); (T)h = (Thi –Tho) Subscripts for h, A: Note 2: (T)1 = (Thi –Tci); (T)2 = (Tho –Tco) ‘i’ for inside, ‘o’ for outside 86 Counter-Current Arrangement Q3 . Hot water Tho Q2 ho Thi, mh, cp(h) Product U . Q h T , m , c 1 i Tco ci c p(c) rii roi
L r = roi -rii
Q1: Energy transferred from heating medium to product (= energy gained by product) Q2: Energy lost by heating medium (= energy gained by product and surroundings) Q3: Energy transferred from heating medium to surroundings (= energy gained by surroundings) Note: Q2 = Q1 + Q3
Common approximation: Q3 = 0 (valid if HX is insulated) In this case, . . . Subscripts for m, cp, T, T: Q1 = mc cp(c) (T)c = Q2 = mh cp(h) (T)h = UAlm Tlm ‘h’ for hot, ‘c’ for cold with 1/UAlm = 1/hiAi + r/kAlm + 1/hoAo ‘i’ for inlet, ‘o’ for outlet Note 1: (T)c = (Tco –Tci); (T)h = (Thi –Tho) Subscripts for h, A: Note 2: (T)1 = (Tho –Tci); (T)2 = (Thi –Tco) ‘i’ for inside, ‘o’ for outside 87 Insulation
88 Effect of Insulation • Air has a lower ‘k’ value than most insulating materials. Why do we use insulation then, to minimize heat loss from a heated pipe? • Does the addition of insulation around a hot pipe surrounded by a cold fluid always decrease the heat loss from the pipe? –Not always! • Addition of insulation – Increases the thermal resistance to heat transfer by conduction (x/kA) – Decreases the thermal resistance to heat transfer by convection (1/hA)
– The net effect may be an increase or decrease in heat loss 89 Heat Loss (Q) without and with Insulation Without Insulation With Insulation ho h To To o
h h Ti i r1 r2 r3 r2 r1 i Ti
Q = T / [(1/h A ) + (r/kA ) + (1/h A )] o o lm pipe i i Q = T / [(1/hoA’ o) + (r/kAlm)insulation + (r/kAlm)pipe + (1/hiAi)] r = r –r and A = 2r L pipe 2 1 o 2 rinsulation = r3 –r2 and A’o = 2r3L (A ) = 2L (r –r) / [ln (r /r )] lm pipe 2 1 2 1 (Alm)insulation = 2L (r3 –r2) / [ln (r3/r2)] Adding insulation inc. conductive res. & dec. convective res. 2 2 Setting dQ/dr3 = 0, and ensuring that d Q/dr3 is –ve, yields Qmax This happens when r3 = kins/ho = rcritical = rc
Thus, as insulation is added, heat loss increases till r3 = kins/ho; then it decreases If r2 > kins/ho, heat loss decreases even if a small amount of insulation is added 90 Qins/Qbare versus Thickness of Insulation Outside Radius of Pipe = 2.25 cm 2
1.8
1.6 k/h = 0.01 m 1.4 k/h = 0.03 m k/h = 0.05 m 1.2 k/h = 0.1 m
bare 1 / Q 0.8 ins Q 0.6
0.4
0.2 Zero benefit 0 0 0.05 0.1 0.15 0.2 0.25 0.3
Insulation Thickness (m) 91 Dimensionless Numbers in Heat Transfer
92 Dimensionless Numbers in Heat Transfer
• Reynolds number
• Nusselt number For circular cross-section pipes
• Prandtl number
• Grashof number
93 Dimensionless Numbers (contd.)
• Biot number
s = k/( cp) • Fourier number = Thermal diffusivity (m2/s)
Reynolds number: Ratio of inertial & viscous forces Nusselt number: Ratio of heat transfer by convection & conduction Prandtl number: Ratio of momentum & thermal diffusivities Grashof number: Ratio of buoyancy & viscous forces Biot number: Ratio of internal & external resistance to heat transfer Fourier number: Ratio of heat conduction & heat storage
Subscripts: ‘f’ for fluid & ‘s’ for solid
dc (for free convection): Look up notes/slides based on shape of object dc (for forced convection pipe flow) = 4 (Across-section)/(Wetted perimeter) Dc: (for unsteady state heat transfer): Distance btwn hottest & coldest points in solid object 94 Nusselt # (NNu) and Biot # (NBi)
• Both are denoted by hd/k • Nusselt # – Used in STEADY state heat transfer (to determine ‘h’)
–dc: Characteristic dimension (= pipe diameter for flow in a pipe)
–kf: Thermal conductivity of FLUID •Biot# – Used in UNSTEADY state heat transfer (to determine the relative importance of conduction versus convection heat transfer)
–Dc: Distance between hottest and coldest point in solid object
–ks: Thermal conductivity of SOLID 95 Unsteady State Heat Transfer (Heat Conduction to the Center of a Solid Object)
96 Basics of Unsteady State heat Transfer • Temperature at one or more points in the system changes as a function of time • Goal of unsteady state heat transfer – Determine time taken for an object to attain a certain temperature OR – Determine temperature of an object after a certain time
– Sometimes, it is used to determine ‘h’ • The dimensionless numbers that come into play for
unsteady state heat transfer are Biot # (NBi = hDc/ks 2 and Fourier # (NFo = st/Dc ) Note: s = k/( cp) 97 Categories of Unsteady State Heat Transfer
Categories are based on the magnitude of Biot # (NBi = hDc/ks) 1. Negligible internal (conductive) resistance
. NBi < 0.1 (also called lumped capacitance/parameter method) 2. Finite internal and external resistances
. 0.1 < NBi < 40 3. Negligible external (convective) resistance
. NBi > 40
Dc for unsteady state heat transfer: Distance between points of maximum temperature difference within solid object
Dc for sphere: Radius of sphere Dc for an infinitely long cylinder: Radius of cylinder Dc for infinite slab with heat transfer from top & bottom: Half thickness of slab Dc for an infinite slab with heat transfer from top: Thickness of slab 98 Modes of Heat Transfer from Air to the Center of a Sphere • Consider hot air (at 100 °C) being blown over a cold sphere (at 20 °C) h 20 °C
20 °C 100 °C air – The two modes of heat transfer are • Convection (external) Q = hA T= T/(1/hA) • Conduction (internal) Q = kA T/x= T/(x/kA)
99 Significance of Magnitude of NBi
•NBi < 0.1 (Cat. #1) => Conductive resistance is low
– Occurs when ks is very high (metals) OR Dc is very small t = 0 min t = 2 min
100 °C air 100 °C air 20 °C 20 °C 50 °C 49 °C
•NBi > 40 (Cat. #3) => Convective resistance is low
– Occurs when ‘h’ is very high (NRe is high) OR Dc is large t = 0 min t = 2 min
100 °C air 100 °C air 20 °C 20 °C 95 °C 55 °C 100 Significance of Magnitude of NBi (contd.)
• 0.1 < NBi < 40 (Cat. #2) => Neither conductive nor convective resistance is negligible (both are of the same order of magnitude) – Occurs when neither ‘h’ nor ‘k’ is very high t = 0 min t = 2 min
100 °C air 100 °C air 20 °C 20 °C 70 °C 45 °C
101 Category #1
The above equation can be used to determine temperature, T, at time, t OR Based on time-temperature data, the equation can be used to determine ‘h’
Ti: Initial temperature of solid object (K) T∞: Temperature of surrounding fluid (K) h: Convective heat transfer coefficient (W/m2 K) A: Surface area for heat transfer (m2) Shape Area Volume : Density of solid object (kg/m3) Brick 2 (LW+LH+WH) LWH V: Volume of solid object (m3) Cylinder 2RL + 2R2 R2L 2 3 cp: Specific heat of solid object (J/kg K) Sphere 4R (4/3)R Note: V = mass of object (kg) L: Length of brick or cylinder W, H: Width, height of brick resp. R: Radius of cylinder or sphere 102 Category #2 • Need to use Heisler charts
– TR on y-axis and Fourier number (NFo) on x-axis
• Several straight lines based on different values of 1/NBi
• Knowing temperature, T (and thus TR), and value of
1/NBi, we can determine x-axis value (or NFo) and hence time, t OR
• Knowing time, t (and hence NFo or x-axis value), and value of 1/NBi, we can determine y-axis value (or TR) and hence temperature, T
103 Sample Heisler Chart
1
1/NBi = 100 0.1
0.01
1/NBi = 0 0.001
0 1 2 3 4 5 1020 30 40 50 60 70 80 90 100
2 Note: s = k/(cp) = Thermal diffusivity of solid object in m /s 104 Heisler Chart (For Finite Sphere) )] ∞ -T i )/(T ∞ TR [=(T-T
2 NFo (= t/Dc ) 105 Heisler Chart (For Infinite Cylinder) )] ∞ -T i )/(T ∞ TR [=(T-T
2 NFo (= t/Dc ) 106 Heisler Chart (For Infinite Slab) )] ∞ -T i )/(T ∞ TR [=(T-T
2 NFo (= t/Dc ) 107 Category #3
• Need to use Heisler charts to determine time- temperature relation – These charts are a way to approximate the exact solution (equation) that represents how temperature (T) changes as a function of time (t) • Similar approach as category #2
• Since NBi > 40, 1/NBi is very small (~ 0) • Thus, we use Heisler charts with the line
corresponding to 1/NBi = 0
–Note: 1/NBi = k/(hDc) 108 Summary of Categories of Unsteady State Heat Transfer
Category 1 Category 2 Category 3
NBi NBi < 0.1 0.1 < NBi < 40 NBi > 40 This category is ‘k’ is high OR Neither ‘k’ nor ‘h’ are ‘h’ is high OR encountered Dc is small high and Dc is not too Dc is large when….. small or too large Resistance that Conductive None Convective is negligible (Internal) (External) T that is small Btwn center and None Btwn fluid and surface of solid surface of solid Solution Lumped Heisler chart(s) Heisler chart(s) approach parameter eqn. (with 1/NBi = 0) 109 Heisler Charts • For a finite sphere • For an “infinitely” long cylinder • For a slab “infinitely” long in two dimensions & finite in one dimension
• Heat transfer in one dimension/direction only • Temperature at ONLY center of object can be determined – Use Gurney-Lurie charts for temperatures at other locations within object
Rule of thumb: If one dimension of an object is at least 10 times another of its dimension, the first dimension is considered to be “infinite” in comparison to the other 110 Finite Objects
Finite brick: Intersection of 3 infinite slabs
Finite cylinder: Intersection of infinite cylinder & infinite slab
111 Finite Objects • Finite objects (such as a cylinder or brick can be obtained as an intersection of infinite objects)
Heisler charts have to be used twice or thrice respectively to determine temperatures for finite cylinder (food in a can) and finite brick (food in a tray)
Note: (t)finite cylinder ≠ (t)infinite cylinder + (t)infinite slab (t)finite brick ≠ (t)infinite slab, width + (t)infinite slab, depth + (t)infinite slab, height 112 Calculations for Finite Cylinder (Heisler Chart) Infinite Cylinder Infinite Slab Characteristic
dimension (Dc)
Biot number (NBi) If both NBi < 0.1, the lumped parameter method (eqn) can be used NBi = h Dc/ks
1/NBi Thermal diffusivity ()
s = ks/(s cp(s))
Fourier number (NFo) 2 NFo = st/Dc Temperature ratio (TR) from Heisler chart
(based on values of NFo & 1/NBi)
Solve for “T” from the above equation 113 Calculations for Finite Brick (Heisler Chart) Infinite Slab #1 Infinite Slab #2 Infinite Slab #3 Characteristic
dimension (Dc)
Biot number (NBi) If all 3 NBi < 0.1, the lumped parameter method (eqn) can be used NBi = h Dc/ks
1/NBi
Thermal diffusivity (s) s = ks/(s cp(s))
Fourier number (NFo) 2 NFo = st/Dc Temperature ratio (TR) from Heisler chart
(based on values of NFo & 1/NBi)
Solve for “T” from the above equation 114 Temperature Ratio (TR)
• At time t = 0, T = Ti and hence TR = 1
• At time t = ∞, T = T∞ and hence TR = 0 • Thus, TR starts off at 1.0 and can at best reach 0.0
• Low values of TR (closer to 0.0) indicate a significant
change in temperature (from Ti) of the object • High values of TR (closer to 1.0) indicate a minimal
change in temperature (from Ti) of the object 115 Summary • Categories of steady state heat transfer . Conduction, convection, radiation • Conduction: Fourier’s law of heat conduction
. Q = - kA (T/x); replace A & T by Alm & Tlm resp. for cyl. • Logarithmic mean area
. Alm = (Ao –Ai) / ln (Ao/Ai) = 2L (ro –ri) / [ln (ro/ri)] • Logarithmic mean temperature difference
. Tlm = (1 – 2) / [ln (1 / 2)] • Convection: Newton’s law of cooling . Q = h A (T) • Free convection: Due to density differences within a fluid; Forced convection: Due to external agency (fan/pump)
. NNu = f(NGr & NPr) for free; NNu = f(NRe & NPr) for forced 116 Summary (contd.) • Thermal resistances to heat transfer by conduction (x/kA) and convection (1/hA) are additive • Overall heat transfer coefficient (U) combines the effect of all forms of heat transfer taking place between any two points in a system
•Q = UAlm Tlm is the most generic form of equation for steady state heat transfer involving conduction and/or convection • Tubular heat exchanger calculations are based on: . . . Q = mp cp(p) Tproduct = mhw cp(hw) Thot water (for no heat loss)
= UAlm Tlm 117 Summary (contd.) • Thermal properties . Specific heat: Important in determining T of an object . Latent heat: Important in determining energy required for phase change . Thermal conductivity: Important in determining rate of heat conduction in an object • For unsteady state heat transfer, the lumped capacitance
method (for NBi < 0.1) or Heisler charts (for NBi > 0.1) are used to establish time-temperature relations . Heisler charts are applicable only for 1-D heat transfer to determine center temperature . Use multiple 1-D objects to create 2-D or 3-D objects . Characteristic dimension in unsteady state heat transfer
• Distance between the hottest and coldest point in object 118