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1 Heat Transfer Introduction Heat Transfer in Various Industries Heat Transfer in the Food Industry

1 Heat Transfer Introduction Heat Transfer in Various Industries Heat Transfer in the Food Industry

• Introduction – Practical occurrences, applications, factors affecting heat transfer – Categories and modes of heat transfer • Conduction – In a slab and across a pipe • – Free (natural) and forced (in a pipe and over a solid object) Heat Transfer – 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 (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

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 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? Introduction • 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 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? 3 4

Heat Transfer in the Food Industry Heat Transfer in Various Industries • Melting: Thawing of a frozen food (turkey) • Freezing: Freezing of ice-cream mix • Automobile: Radiator and engine • Drying: Drying of fruits • Electronics: Cooling of motherboard/CPU by fan • Evaporation: Spray drying of coffee or • Pharmaceutical: Freeze drying of vaccines of juices • Metallurgical: Heating/cooling during steel • Sublimation: Freeze drying of coffee manufacture • Heating/cooling of milk • Chemical: Condensation, boiling, distillation of chemicals • Baking of bread • Processing of canned soups (inactivate microorganisms & maximize nutrient content, Home: Refrigerator, AC, heater, dryer, stove, microwave 5 color/flavor/texture) 6

1 What Factors Affect Rate of Heat Transfer? Specific Heat, Thermal Conductivity, •Thermal and Thermal Diffusivity – Specific heat (cp in J/kg K) • Measured using Differential scanning calorimeter (DSC) • Specific heat (cp) – Thermal conductivity (k in W/m K) – A measure of how much energy is required to raise the • Measured using Fitch apparatus or thermal conductivity probe (Lab #5) temperature of an object • Physical • Thermal conductivity (k) – Density ( in kg/m3) – A measure of how quickly heat gets conducted from one • Measured using pycnometer part of an object to another • Rheological (measured using rheometer/viscometer) • Thermal diffusivity () – Viscosity ( in Pa s) for Newtonian fluids – It combines the effects of specific heat, thermal OR conductivity, and density of a material. Thus, this one – Consistency coefficient (K in Pa sn) and flow behavior index (n) quantity can be used to determine how temperature for power-law fluids changes at various points within an object. 2 Note: Thermal diffusivity ( = k/cp in m /s) combines the effect of several factors7 8

Thermal Conductivity (Fitch Apparatus) Specific Heat (DSC Method) Slope Heat flux held constant & temperature diff. measured Heat Source (ice-water mix) Intercept Q = m c (T ) = m c (T ) Q 1 p(1) 1 2 p(2) 2 This can be rewritten as: Sample cp(2) = {m1/m2} {(T1)/ (T2)} cp(1) t: time (cheese slice) m, cp, A, T: For heat sink Differential Scanning Calorimeter (DSC) (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 Manufacturer: Perkin-Elmer 9 10

Thermal Conductivity Probe Values of Thermal Conductivity (k) • Good conductors of heat have high k values KD2 Pro Probe (Manufacturer: Decagon Devices) – Cu: 401 W/m K Single needle probe: Can measure ‘k’ – Al: 250 W/m K – Fe: 80 W/m K Dual needle probe: Can measure ‘k’ and ‘’  = k / ( cp) – 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 Sample – Air: 0.024 W/m K (lower k than insulators!) Sample • 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 ) – Glass: 1.05 W/m K; Brick: 0.7 - 1.3 W/m K; Concrete: 0.4 - 1.7 W/m K k: Thermal conductivity (W/m K); : Thermal diffusivity (m2/s) 3 – Plastics (commonly used): 0.15 - 0.6 W/m K : Density (kg/m ); cp: Specific heat (J/kg K) 11 • Thermally conductive plastics may have k > 20 W/m K 12

2 Effect of Temperature on k, , , cp) 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

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Questions Categories of Heat Transfer Q: When the same heating source is used to heat • Steady state identical quantities of water and butter, which will be – Temperatures at all points within the system remain hotter after a certain time? constant over time Ans: Butter; because it has a lower specific heat – 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 Q: In winter, is a metallic park bench colder than a • Conditions are often approximated to be steady state nearby wooden park bench? – Eg.: Temperature inside a room or refrigerator Ans: NO. A metallic bench has a higher thermal • Unsteady state conductivity and hence conducts heat very well, – Temperature(s) at one or more points in the system thereby taking away the heat generated by our body change(s) over time very fast and making us feel colder. 15 – 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 – 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 Conduction (liquid/) 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 18

3 Basics of Conduction Fourier’s Law of Heat Conduction Rate of heat transfer by conduction is given by • Conduction involves the translation of vibration of Fourier’s law of heat conduction as follows: molecules along a temperature gradient as they acquire Q = - kA (T/x) thermal energy (mainly analyzed within solids; however, it takes place in liquids and gases also) The negative sign is used to denote/determine the – Actual movement of particles does not occur direction of heat transfer (Left to right or right to left) • Good conductors of electricity are generally good Q: Energy transferred per unit time (W) conductors of heat k: Thermal conductivity (W/m K); it is a +ve quantity • Thermal conductivity (k) is used to quantify the ability A: Area of heat transfer (m2) of a material to conduct heat 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) 19 20

Temperature Difference Across a Slab Conduction Across a Slab or Cylinder Heat flow Heat flow T1 T1 • Slab: Q = kA (T/x) • Slab: Q = kA (T/x) x x T T2 T = T1 –T2 2 For the same value of Q (example: use of a heater on one T1 Heat flow r side of a slab), T2 For insulators (low k), “T –T” is large 1 2 • Cylinder: Q = kAlm (T/r) For good conductors (high k), “T1 –T2” is small k: Thermal conductivity (W/m K) 2 For the same value of “T1 –T2” (example: fixed inside A: Area across which heat transfer is taking place (m ) temperature of room and outside air temperature), T = T1 –T2: Temperature difference (K) 2 For insulators (low k), Q is small Alm: Logarithmic mean area (m ) For good conductors (high k), Q is large Note: Alm comes into play when the area for heat transfer at the two ends across Note: x and A are assumed to be the same in all of the above situations which heat transfer is taking place, is not the same 21 22

Logarithmic Mean Area (Alm) Logarithmic Mean Temp Diff (Tlm) Double Tube Heat Exchanger r T1 Heat flow

r Q = kAlm (T/r) T Hot water T T2 r o w(o) w(i) i T T T = T1 –T2 1 2 r= r -r o i T Product Tp(o) L p(i) • Slab: Area for heat transfer is same at both ends • Cylinder T is NOT constant across the length of tube

T1 = Tw(o) –Tp(i) , T2 = Tw(i) –Tp(o) – Area at one end (outside) is Ao (= 2roL) Tlm = (1 – 2) / [ln (1 / 2)] – Area at other end (inside) is Ai (= 2riL)

– Which area should be used in determining Q? Note: Tlm lies between T1 and T2 Subscripts: ‘w’ for water; ‘p’ for product, ‘i’ for inlet, ‘o’ for outlet –Alm = (Ao –Ai) / ln (Ao/Ai) = 2L (ro –ri) / [ln (ro/ri)] Note: Tlm comes into play when the temperature difference across the two ends –Note: Ao > Alm > Ai 23 where heat transfer is taking place, is not the same 24

4 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 Convection 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

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Categories of Convection Newton’s Law of Cooling for Convection • Free (or natural) convection – Does not involve any external agency in causing flow Rate of heat transfer by convection (for heating or – Heat transfer between bottom of vessel and fluid in it cooling) is given by Newton’s law of cooling as follows: – Cooling of human body

Q = h A (Ts -T∞) – Cooling of radiator fluid in car engine during idling –h : 5-25 W/m2 K; h : 20-100 W/m2 K Q: Energy transferred per unit time (W) air-solid water-solid h: Convective heat transfer coefficient -- CHTC (W/m2 K) • A: Surface area available for heat transfer (m2) – External agency such as fan/pump causes flow

T = Ts –T∞ : Temperature difference (K) – Cooling of radiator fluid in car engine during motion Ts: Surface temperature of solid object (K) – Ice-cream freezer (Blast air) T∞: Free stream (or bulk fluid) temperature of fluid (K) – Stirring a pot of soup CHTC (h): Measure of rate of heat transfer by convection; NOT a property; – Heat transferred from computers (fan) 2 2 depends on fluid velocity, surface characteristics (shape, size, smoothness), –hair-solid: 10-200 W/m K; hwater-solid: 50-10,000 W/m K fluid properties (, k, , c ) 2 p 27 –hboiling water or steam to solid: 3,000-100,000 W/m K 28

Question Free Convection Q: What is wind-chill factor? In winter, a • Fluid comes into contact with hot solid thermometer reads -20 °C when air is stationary. All of a sudden, a gust of wind blows. What will the • Fluid temperature near solid increases thermometer read? • Fluid density near solid decreases • This results in a buoyancy force that causes flow Ans: -20 °C. As wind speed increases, more heat is removed from our body due to an increase in ‘h’ • Rate of heat transfer (Q & h) depends on and hence ‘Q’. Thus, we feel colder than when the – Temperature difference between fluid and surface of solid air is stationary. The air is NOT colder, we just feel – Properties (, , k, c ) of fluid p colder since more heat is removed from our body – Dimensions and surface characteristics (smoothness) of solid and our body is unable to generate enough heat to replace the energy lost to the surroundings. NNu = hdc/kf = f (NGr , NPr) 29 30

5 Grashof (NGr) Number Nusselt Number (NNu)

-1 f: Coefficient of volumetric thermal expansion (K ) g: Acceleration due to gravity (= 9.81 m/s2) 2 3 h: Convective heat transfer coefficient (W/m K) f: Density of fluid (kg/m )

Ts: Surface temperature of solid object (K) dc: Characteristic dimension (m) T∞: Free stream temperature of fluid (K) k : Thermal conductivity of fluid (W/m K) f dc: Characteristic dimension of solid object (m) (Obtained from tables based on shape & orientation of solid object)

f: Viscosity of surrounding fluid (Pa s)

Nusselt number represents the ratio of heat transfer by convection & conduction Grashof number represents the ratio of buoyancy and viscous forces 31 32

Properties of Air Prandtl Number (NPr)

0.72 0.74 0.74 0.74 0.73 0.73 cp(f): Specific heat of fluid (J/kg K) 0.73 0.73 0.73 f: Viscosity of fluid (Pa s) 0.72 0.72 k : Thermal conductivity of fluid (W/m K) 0.72 f 0.72 0.72 0.73 0.72 0.72 0.72 0.72 Prandtl number represents the ratio of momentum and thermal diffusivities 33 34

Properties of Water 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) 35 Ra 36

6 Free Convection (Cylinder) Free Convection (Sphere)

• For vertical cylinder (dc = cylinder height) 0.25 – Similar to vertical plate if D ≥ 35L/(NGr) NRa = NGr NPr

• For horizontal cylinder (dc = cylinder diameter) For sphere, dc = D/2

-5 12 –For 10 < NRa < 10 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 Note: NRa = NGr NPr 37 surface of solid increases, NGr increases. Thus, NNu and ‘h’ increase. 38

Forced Convection Categories of Convective Heat Transfer • Fluid is forced to move by an external force (pump/fan) Coefficient for Forced Convection • Rate of heat transfer (Q & h) depends on • Between a moving fluid and a stationary solid object – Properties (, , k, cp) of fluid – Dimensions and surface characteristics (smoothness) of solid – Transfer of heat from hot pipe to a fluid flowing in a pipe • ‘h’ does NOT depend on – Generally depicted by ‘h’ – Temperature difference between fluid and surface of solid • ‘h’ strongly depends on Reynolds number • Between a moving fluid and a moving particle – When all system and product parameters are kept constant, it – Transfer of heat from a hot fluid to a freely flowing is flow rate (a process parameter) that strongly affects ‘h’ particle in a (particulate/multiphase food)

– Generally depicted by ‘hfp’

NNu = hdc/kf = f (NRe , NPr) 39 40

Forced Convection in a Pipe Moody Diagram •NNu = hdc/kf = f (NRe, NPr) • Three sub-categories of forced convection exist…..

• 1. Laminar flow (NRe < 2100)

– A. Constant surface temperature of pipe ( Roughness Relative

•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 (f) Factor Friction •N = 1.86 (N x N x d /L) ( / )  Nu Re Pr c b w /D) dc: ID of pipe, L: Length of pipe • 2. Transitional flow

(2100 < NRe < 4000) – Friction factor (f) Reynolds Number (NRe) • For smooth pipes:  = 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 • For non-smooth pipes, use Moody chart (graph of: f, NRe, /D) 41 42

7 Forced Convection in a Pipe (contd.) ‘hfp’ for Forced Convection over a Sphere

3. Turbulent flow (NRe > 4000) of a Newtonian fluid in a pipe, NNu = hdc/kf = f (NRe, NPr) – similar to flow in a pipe 0.8 0.33 0.14 NNu = 0.023 (NRe) (NPr) (b/w) 0.5 0.33 NNu = 2 + 0.6 (NRe) (NPr) b: Viscosity of fluid based on bulk fluid temperature For 1 < N < 70,000 and 0.6 < N < 400 w: Viscosity of fluid based on wall temperature Re Pr

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 1: dc is the outside diameter of the sphere Note 2: Determine all properties at the film temperature Note: For all forced convection situations, use bulk temperature of fluid {Tfilm = (Ts + T∞)/2} to determine properties (unless otherwise specified) 43 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 Radiation

• 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 46

Basics of Radiation Heat Transfer Infrared Thermometer

• Infrared thermometer can be used to non-invasively Rate of heat transfer by radiation is given by Stefan- and remotely determine the surface temperature of Boltzmann law as follows: an object Q = σ A ε T4 • Care should be exercised in ensuring that ONLY emitted energy is measured and NOT reflected Q: Energy transferred per unit time (W) energy (may have to use non-reflecting tape on σ: Stefan-Boltzmann constant (= 5.669 x 10-8 W/m2 K4) metallic surfaces) A: Surface area of object (m2) • The emissivity of some infrared thermometers can ε: Emissivity of surface (ranges from 0 to 1.0) be adjusted; for others, a pre-set value of 0.95 is T: Temperature (K) commonly programmed

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8 Thermal Resistances to Heat Transfer

• Conduction – Slab: Q = kA (T/x) = T/[(x/kA)]

– Cylinder: Q = kAlm (T/r) = T/[(r/kAlm)] Thermal Resistances to Heat Transfer – 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 49 50

Thermal Resistances Electrical Analogy Conduction: Q = kA T/x A current (I) flows because there is a driving Single slab: Q = T/[(x/kA)] I force, the potential Difference (V), across the

Multiple slabs: Q = T/[(x1/k1A) + (x2/k2A) + …] resistance (R) R Cylindrical shell: Q = T/[(r/kAlm)] V I = V/R Multiple cylindrical shells: Q = T/[(r1/k1Alm(1)) + (r2/k2Alm(2))] Convection: Q = hA T Q = T/(Thermal Resistance) Single convection: Q = T/[(/hA)] Multiple convections: Q = T/[(/h A ) + (/h A ) + …] 1 1 2 2 I R1 Combination of conduction and convection I = V/(R1 + R2) Multiple slabs V Q = T/(Sum of Thermal Resistances) Q = T/[(x1/k1A) + (x2/k2A) + (/h1A1) + (/h2A2) + …] R2 Multiple cylindrical shells

Q = T/[(/h1A1) + (r1/k1Alm(1)) + (r2/k2Alm(2)) + (/h2A2)] Thermal resistances are additive (similar to electrical resistances) Units of thermal resistance to heat transfer: K/W 51 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 Overall Heat Transfer Coefficient (OHTC) • 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

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9 What is Overall Heat Transfer Coefficient? OHTC (or U) in Different Scenarios

• OHTC (denoted by the symbol ‘U’) refers to a • Three conductive heat transfers

single quantity that can be used to quantify the effect . 1/(UA) = x1/(k1A) + x2/(k2A) + x3/(k3A) of all forms (conduction and convection) of heat • Two convective heat transfers transfer taking place in a system . 1/(UA) = 1/(h1A1) + 1/(h2A2) • It facilitates the use of one equation (instead of • One conductive and one convective heat transfer individual equations for each conductive and . 1/(UA) = x1/(k1A) + 1/(hA) convective heat transfer in the system) to determine Note 1: 1/UA > 1/hA; Thus, U < h the total heat transfer taking place in the system Note 2: If there is no conductive resistance, U = h – All thermal resistances in the system are added in order to facilitate this process U: Overall heat transfer coefficient (W/m2 K) “1/UA”: Overall thermal resistance (K/W) 55 56

k, h, U, Resistances, and Temperatures Why Dress in “Layers” in Winter? • As thermal conductivity (k) increases, thermal resistance due to conduction (x/kA) decreases • Single jacket of thickness x: – Thus, temperature difference between center and surface of object – Q = T/[(x/kA)] decreases • Two jackets, each of thickness x/2: • As convective heat transfer coefficient (h) increases, thermal resistance due to convection (1/hA) decreases – Q = T/[{(x/2)}/kA) + {(x/2)}/kA)] = T/[(x/kA)] – Thus, temperature difference between the fluid and surface of the solid object decreases • If the above two expressions are identical, why is it • As overall heat transfer coefficient (U) increases, overall thermal resistance (1/UA) decreases better to wear two jackets, each of thickness x/2? – Thus, temperature difference between the two points across which heat transfer is taking place, decreases Air trapped between the two jackets adds a convective Thermal conductivity: W/m K thermal resistance. Thus, Convective heat transfer coefficient: W/m2 K Q = T/[{(x/2)}/kA) + {(x/2)}/kA) + (1/hA)] Overall heat transfer coefficient: W/m2 K Thus, total thermal resistance increases and Q decreases Thermal resistance to heat transfer: K/W 57 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 Heat Exchangers (HX) • 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 60

10 Types of Heating Equipment Steam Injection • Direct contact Intense, turbulent mixing of steam and product occurs. It – Steam injection, steam infusion Steam results in rapid heating and • Indirect contact (Other than plate, tubular, Shell & tube, SSHE) dilution of product. A vacuum chamber is used downstream – Retorts (Using hot water, steam, or steam-air for heating) Pneumatically actuated to evaporate steam that • Batch (Agitation: None, axial or end-over-end): With our w/o basket/crate condensed into product. • 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) Pneumatically actuated

• Alternative/Novel/Emerging Technologies Variable Gap – Microwave and radio frequency heating Product • Uses electromagnetic radiation; polar molecules heat up – Ohmic heating www.process-heating.com • Electric current in food causes heating; ions in food, cause heating 61 62

Steam Infusion Batch Retorts Air out CIP in • Static (with or without crates/baskets) Rotary, end-over-end Steam in or reciprocating Product in – 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

Cooling water in/out This is a gentler process than steam injection. A vacuum • Reciprocating chamber is used downstream to – Shaka evaporate steam that condensed into product.

Product out 63 64

Static & Rotary Retorts Crateless Retort (Semi-Continuous)

Raw canned Horizontal Basket Retort foods are placed in basket

SuperAgi Retort

www.libertyprocess.co.uk

www.jbtfoodtech.com Problem: Cold spot identification www.maloinc.com 65 66

11 Conventional Continuous Rotary Retort Hydrostatic Retort

Height of water column provides enough pressure to prevent water from boiling at temperatures well above 100 °C

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Hydrostatic Retort/Sterilizer 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

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PHEs (All channels in Parallel) PHEs (All Channels in Series)

Advantage Advantage Low pressure drop High heat transfer

Disadvantage Disadvantage 1 X 4 4 X 1 Low heat transfer High pressure drop 1 X 4 4 X 1

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12 Regeneration in a PHE PHEs (Series & Parallel)

Hot pasteurized product

Advantage Cold pasteurized product Optimized pressure drop and heat transfer

Cooler Regenerator Heater

Raw product 2 X 2 1 X 4

Regeneration: Energy of hot pasteurized product is used to pre-heat

73 cold raw product. Typical regeneration efficiency is ~90%. 74

Double Tube, Triple Tube, Multitube HX Shell & Tube: (One & Two Pass)

Heating from 1 side

Heating from 2 sides

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. 75 76

Scraped Surface Heat Exchanger Double Tube Heat Exchanger (DTHE) (SSHE) Motor Hot/Cold water Shaft with motor rotating it Heat Transfer Blade Blade Insulation Product

Steam jacket Shaft

Product Cross-sectional view Insulation Double tube heat exchanger Two concentric tubes Steam Product generally flows in inner tube Advantage: Mixing of viscous foods jacket Heating/cooling medium generally flows in outer tube (annulus) Disadvantage: Particle damage, One stream gains heat while the other stream loses heat (hence, heat exchanger) Heat transfer takes place across wall of inner tube uncertain residence time, cleaning Product 77 Both streams may flow in the same or opposite directions 78

13 Heat Transfer in a Double Tube HX Heat Transfer from Hot Water to Product (Hot water heating a product)

Q = hoAo [Thot water -Twall (outside)] Hot water • Convection Tho ho . Thi, mh, cp(h) . From hot water to outside surface of inner tube Q = kA [T -T ]/r lm wall (outside) wall (inside) . Q = hoAo [Thot water -Twall (outside)] U Q = hiAi [Twall (inside) -Tproduct] h Product i rii roi • Conduction . T Tci, mc, cp(c) co . 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

L . Q = hiAi [Twall (inside) -Tproduct]

Subscripts for T: ‘c’ for cold, ‘h’ for hot, ‘i’ for inlet, ‘o’ for outlet 79 80

Resistances to Heat Transfer from Hot Overall Heat Transfer Coefficient (U) Water to Product

Q = T / [(1/hoAo) + (r/kAlm) + (1/hiAi)] Hot water Thermal resistances have been added ho Convective resistance (1/ho Ao)

Conductive resistance (r/k Alm) Denominator: Total thermal resistance Convective resistance (1/hi Ai) Product hi 1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi) 2 U 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

81 82

Determination of U: Theoretical Method Determination of U: Experimental Method (Hot Water as Heating Medium) . . •1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi) Q = mc cp(c) Tc = mh cp(h) Th

•hi and ho are usually determined using empirical = UAlm Tlm correlations Assumption: Heat loss = zero • ‘k’ is a material property of the tube of HX Once the mass flow rates and temperatures of the •Ai, Ao, and Alm are determined based on dimensions (length & radii) of heat exchanger tubes product and hot water are experimentally determined, U can be calculated • Once Ai, Ao, Alm, hi, ho, and k are known, U is calculated using the above equation If there is heat loss,

Qlost by hot water = Qgained by product + Qlost to outside 83 84

14 Tubular Heat Exchanger (Co- and Counter-Current) Co-Current Arrangement 1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi)andQ = UAlm Tlm Q . 3 Hot water T , m , c h Q T h: Hot hi h p(h) o 2 ho c: Cold h: Hot c: Cold Product U i: Inlet . h Q o: Outlet i: Inlet T , m , c i 1 Tco o: Outlet ci c p(c) rii roi Temperature Temperature 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) Used only when rapid initial cooling is needed Most commonly used Q3: Energy transferred from heating medium to surroundings (= energy gained by surroundings) Lower heat transfer efficiency Higher heat transfer efficiency Note: Q2 = Q1 + Q3 Tco ≤ Tho always Tco can be greater than Tho T btwn hot & cold fluid dec. along length T btwn hot & cold fluid does not Common approximation: Q3 = 0 (valid if HX is insulated) In this case, . Ai = Inside surface area of inner pipe = 2  rii L change significantly along length . . Subscripts for m, cp, T, T: Q = m c (T) = Q = m c (T) = UA T Ao = Outside surface area of inner pipe = 2  roi L 1 c p(c) c 2 h p(h) h lm lm ‘h’ for hot, ‘c’ for cold with 1/UA = 1/h A + r/kA + 1/h A Alm = Logarithmic mean area of inner pipe = (Ao –Ai) / [ln (Ao/Ai)] lm i i lm o o ‘i’ for inlet, ‘o’ for outlet Note 1: (T) = (T –T ); (T) = (T –T ) = 2L (roi –rii)/[ln (roi/rii)] c co ci h hi ho Subscripts for h, A: Note 2: (T) = (T –T ); (T) = (T –T ) Tlm = Logarithmic mean temperature difference = (T1 – T2)/[ln(T1/T2)] 85 1 hi ci 2 ho co ‘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 = r -r oi ii Insulation 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 88

Effect of Insulation Heat Loss (Q) without and with Insulation Without Insulation With Insulation • Air has a lower ‘k’ value than most insulating materials. ho h To Why do we use insulation then, to minimize heat loss To o from a heated pipe? • Does the addition of insulation around a hot pipe T hi r r r r r hi T surrounded by a cold fluid always decrease the heat loss i 1 2 3 2 1 i from the pipe? –Not always! 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)] • Addition of insulation r = r –r and A = 2r L pipe 2 1 o 2 rinsulation = r3 –r2 and A’o = 2r3L (A ) = 2L (r –r) / [ln (r /r )] – Increases the thermal resistance to heat transfer by conduction lm pipe 2 1 2 1 (Alm)insulation = 2L (r3 –r2) / [ln (r3/r2)] (x/kA) Adding insulation inc. conductive res. & dec. convective res. 2 2 – Decreases the thermal resistance to heat transfer by Setting dQ/dr3 = 0, and ensuring that d Q/dr3 is –ve, yields Qmax This happens when r = k /h = r = r convection (1/hA) 3 ins o critical c Thus, as insulation is added, heat loss increases till r3 = kins/ho; then it decreases – The net effect may be an increase or decrease in heat loss 89 If r2 > kins/ho, heat loss decreases even if a small amount of insulation is added 90

15 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 Dimensionless Numbers in Heat Transfer

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 92

Dimensionless Numbers in Heat Transfer Dimensionless Numbers (contd.) • Biot number • Reynolds number

s = k/( cp) • Fourier number = Thermal diffusivity (m2/s)

• Nusselt number For circular Reynolds number: Ratio of inertial & viscous forces cross-section pipes Nusselt number: Ratio of heat transfer by convection & conduction Prandtl number: Ratio of momentum & thermal diffusivities Grashof number: Ratio of buoyancy & viscous forces • Prandtl number 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 • Grashof number 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 93 94

Nusselt # (NNu) and Biot # (NBi)

• Both are denoted by hd/k • Nusselt # Unsteady State Heat Transfer – Used in STEADY state heat transfer (to determine ‘h’) (Heat Conduction to the Center of a Solid Object) –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 96

16 Basics of Unsteady State heat Transfer Categories of Unsteady State Heat Transfer

• Temperature at one or more points in the system Categories are based on the magnitude of Biot # (NBi = hDc/ks) changes as a function of time 1. Negligible internal (conductive) resistance

• Goal of unsteady state heat transfer . NBi < 0.1 (also called lumped capacitance/parameter method) – Determine time taken for an object to attain a certain 2. Finite internal and external resistances temperature . 0.1 < NBi < 40 OR 3. Negligible external (convective) resistance – Determine temperature of an object after a certain time . NBi > 40 D for unsteady state heat transfer: Distance between points of maximum – Sometimes, it is used to determine ‘h’ c temperature difference within solid object • The dimensionless numbers that come into play for Dc for sphere: Radius of sphere Dc for an infinitely long cylinder: Radius of cylinder unsteady state heat transfer are Biot # (NBi = hDc/ks D for infinite slab with heat transfer from top & bottom: Half thickness of slab 2 c and Fourier # (NFo = st/Dc ) Note: s = k/( cp) 97 Dc for an infinite slab with heat transfer from top: Thickness of slab 98

Modes of Heat Transfer from Air to the Significance of Magnitude of NBi Center of a Sphere • Consider hot air (at 100 °C) being blown over a cold •NBi < 0.1 (Cat. #1) => Conductive resistance is low sphere (at 20 °C) – Occurs when ks is very high (metals) OR Dc is very small h 20 °C t = 0 min t = 2 min

20 °C 100 °C air 100 °C air 100 °C air 20 °C 20 °C 50 °C 49 °C – The two modes of heat transfer are •NBi > 40 (Cat. #3) => Convective resistance is low • Convection (external) Q = hA T= T/(1/hA) – Occurs when ‘h’ is very high (NRe is high) OR Dc is large • Conduction (internal) Q = kA T/x= T/(x/kA) t = 0 min t = 2 min

100 °C air 100 °C air 99 20 °C 20 °C 95 °C 55 °C 100

Category #1 Significance of Magnitude of NBi (contd.)

The above equation can be used to determine temperature, T, at time, t OR • 0.1 < NBi < 40 (Cat. #2) => Neither conductive nor Based on time-temperature data, the equation can be used to determine ‘h’ convective resistance is negligible (both are of the same Ti: Initial temperature of solid object (K) order of magnitude) T∞: Temperature of surrounding fluid (K) – Occurs when neither ‘h’ nor ‘k’ is very high h: Convective heat transfer coefficient (W/m2 K) 2 t = 0 min t = 2 min A: Surface area for heat transfer (m ) Shape Area Volume : Density of solid object (kg/m3) Brick 2 (LW+LH+WH) LWH V: Volume of solid object (m3) Cylinder 2RL + 2R2 R2L 2 3 100 °C air 100 °C air cp: Specific heat of solid object (J/kg K) Sphere 4R (4/3)R 20 °C 20 °C 70 °C 45 °C Note: V = mass of object (kg) L: Length of brick or cylinder W, H: Width, height of brick resp. 101 R: Radius of cylinder or sphere 102

17 Category #2 Sample Heisler Chart • Need to use Heisler charts 1

– TR on y-axis and Fourier number (NFo) on x-axis 1/NBi = 100 • Several straight lines based on different values of 1/NBi 0.1

• Knowing temperature, T (and thus TR), and value of 0.01 1/NBi, we can determine x-axis value (or NFo) and hence time, t 1/NBi = 0 OR 0.001 0 1 2 3 4 5 1020 30 40 50 60 70 80 90 100 • 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 2 103 Note: s = k/(cp) = Thermal diffusivity of solid object in m /s 104

Heisler Chart (For Finite Sphere) Heisler Chart (For Infinite Cylinder) )] )] ∞ ∞ -T -T i i )/(T )/(T ∞ ∞ TR [=(T-T TR [=(T-T

2 2 NFo (= t/Dc ) 105 NFo (= t/Dc ) 106

Heisler Chart (For Infinite Slab) Category #3

• Need to use Heisler charts to determine time-

)] temperature relation ∞ -T

i – These charts are a way to approximate the exact (equation) that represents how temperature (T) changes )/(T ∞ as a function of time (t) • Similar approach as category #2

TR [=(T-T • 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) 2 NFo (= t/Dc ) 107 108

18 Summary of Categories of Unsteady Heisler Charts State Heat Transfer • For a finite sphere • For an “infinitely” long cylinder • For a slab “infinitely” long in two dimensions & Category 1 Category 2 Category 3 finite in one dimension 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 • Heat transfer in one dimension/direction only small or too large when….. • Temperature at ONLY center of object can be Resistance that Conductive None Convective is negligible (Internal) (External) determined T that is small Btwn center and None Btwn fluid and – Use Gurney-Lurie charts for temperatures at other surface of solid surface of solid locations within object Solution Lumped Heisler chart(s) Heisler chart(s) approach parameter eqn. (with 1/NBi = 0) Rule of thumb: If one dimension of an object is at least 10 times another of its dimension, 109 the first dimension is considered to be “infinite” in comparison to the other 110

Finite Objects Finite Objects • Finite objects (such as a cylinder or brick can be Finite brick: Intersection of 3 obtained as an intersection of infinite objects) infinite slabs

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 Finite cylinder: Intersection of a tray) infinite cylinder & infinite slab

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 111 112

Calculations for Finite Cylinder (Heisler Chart) Calculations for Finite Brick (Heisler Chart) Infinite Cylinder Infinite Slab Infinite Slab #1 Infinite Slab #2 Infinite Slab #3 Characteristic Characteristic dimension (D ) c dimension (Dc) Biot number (NBi) Biot number (N ) If both NBi < 0.1, the lumped parameter method (eqn) can be used Bi N = h D /k If all 3 NBi < 0.1, the lumped parameter method (eqn) can be used Bi c s NBi = h Dc/ks 1/N Bi 1/NBi Thermal diffusivity () Thermal diffusivity (s)  = k /( c ) s s s p(s) s = ks/(s cp(s)) Fourier number (N ) Fo Fourier number (NFo) N =  t/D 2 2 Fo s c NFo = st/Dc Temperature ratio (TR) Temperature ratio (TR) from Heisler chart from Heisler chart

(based on values of NFo (based on values of NFo & 1/NBi) & 1/NBi)

Solve for “T” from the above equation 113 Solve for “T” from the above equation 114

19 Temperature Ratio (TR) Summary • Categories of steady state heat transfer . Conduction, convection, radiation • Conduction: Fourier’s law of heat conduction . • At time t = 0, T = T and hence TR = 1 Q = - kA (T/x); replace A & T by Alm & Tlm resp. for cyl. i • Logarithmic mean area • At time t = ∞, T = T∞ and hence TR = 0 . Alm = (Ao –Ai) / ln (Ao/Ai) = 2L (ro –ri) / [ln (ro/ri)] • Thus, TR starts off at 1.0 and can at best reach 0.0 • Logarithmic mean temperature difference

. Tlm = (1 – 2) / [ln (1 / 2)] • Low values of TR (closer to 0.0) indicate a significant • Convection: Newton’s law of cooling . Q = h A (T) change in temperature (from Ti) of the object • High values of TR (closer to 1.0) indicate a minimal • Free convection: Due to density differences within a fluid; Forced convection: Due to external agency (fan/pump) change in temperature (from Ti) of the object 115 . NNu = f(NGr & NPr) for free; NNu = f(NRe & NPr) for forced 116

Summary (contd.) Summary (contd.) • Thermal properties • Thermal resistances to heat transfer by conduction . Specific heat: Important in determining T of an object (x/kA) and convection (1/hA) are additive . : Important in determining energy required for • Overall heat transfer coefficient (U) combines the phase change effect of all forms of heat transfer taking place . Thermal conductivity: Important in determining rate of heat between any two points in a system conduction in an object • For unsteady state heat transfer, the lumped capacitance • Q = UAlm Tlm is the most generic form of equation method (for NBi < 0.1) or Heisler charts (for NBi > 0.1) for steady state heat transfer involving conduction are used to establish time-temperature relations and/or convection . Heisler charts are applicable only for 1-D heat transfer to • Tubular heat exchanger calculations are based on: determine center temperature . . . Use multiple 1-D objects to create 2-D or 3-D objects . Q = mp cp(p) Tproduct = mhw cp(hw) Thot water (for no heat loss) . Characteristic dimension in unsteady state heat transfer = UAlm Tlm 117 • Distance between the hottest and coldest point in object 118

20