of 40 /40
1 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 h fp ) 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

# Heat Transfer - NC State University · PDF file1 Heat Transfer Heat Transfer • Introduction – Practical occurrences, applications, factors affecting heat transfer – Categories

vongoc

• View
274

4

Embed Size (px)

Citation preview

1

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)

• 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

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 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, microwave5

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

• Processing of canned soups (inactivate microorganisms & maximize nutrient content, color/flavor/texture)

6

3

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

Note: Thermal diffusivity ( = k/cp in m2/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

Manufacturer: Perkin-Elmer

Q = m1 cp(1) (T1) = m2 cp(2) (T2)

cp(2) = {m1/m2} {(T1)/ (T2)} cp(1)

Differential Scanning Calorimeter (DSC)

9

4

Thermal Conductivity (Fitch Apparatus)

t: timem, cp, A, T: For heat sink(mass, sp. ht., area, temp.)Ti: Initial temp. of heat sinkT∞: Temp. of heat sourceL: Thickness of sample

Heat Source(ice-water mix)

Q

Heat Sink(copper block)

Sample(cheese slice)

Insulation

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

This can be rewritten as:

Y X

SlopeIntercept

10Note: ‘k’ is always a positive number

Thermal Conductivity ProbeKD2 Pro Probe (Manufacturer: Decagon Devices)

Single needle probe: Can measure ‘k’

Dual needle probe: Can measure ‘k’ and ‘’ = k / ( cp)

11

SampleSample

k: Thermal conductivity (W/m K); : Thermal diffusivity (m2/s): Density (kg/m3); cp: Specific heat (J/kg K)

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

5

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

QuestionsQ: 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

6

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 cooking16

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

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

7

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 ConductionRate 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 quantityA: 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

• Slab: Q = kA (T/x)

T = T1 – T2

x

T1

T2

For the same value of Q (example: use of a heater on one side of a slab),

For insulators (low k), “T1 – T2” is largeFor 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 smallFor good conductors (high k), Q is large

Heat flow

Note: x and A are assumed to be the same in all of the above situations21

8

Conduction Across a Slab or Cylinder

• Slab: Q = kA (T/x)

• Cylinder: Q = kAlm (T/r)

x

T1

T2

k: Thermal conductivity (W/m K)A: Area across which heat transfer is taking place (m2)T = T1 – T2: Temperature difference (K)Alm: Logarithmic mean area (m2)

T1

T2

Heat flow

Heat flowr

22

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

Logarithmic Mean Area (Alm)

• 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

T1

T2

Heat flowr

riro

L

23

Q = kAlm (T/r)

T = T1 – T2

r = ro - ri

Logarithmic Mean Temp Diff (Tlm)

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

Tp(o)

Tw(o)

Tp(i)

Hot water Tw(i)

Product

T1 T2

Double Tube Heat Exchanger

24

Note: Tlm comes into play when the temperature difference across the two ends where heat transfer is taking place, is not the same

9

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∞)

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, , cp)

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)

27

10

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

– hair-solid: 5-25 W/m2 K; hwater-solid: 20-100 W/m2 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)

– hair-solid: 10-200 W/m2 K; hwater-solid: 50-10,000 W/m2 K

– hboiling water or steam to solid: 3,000-100,000 W/m2 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

QuestionQ: 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

11

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 & conduction31

Grashof (NGr) Number

f: Coefficient of volumetric thermal expansion (K-1)

g: Acceleration due to gravity (= 9.81 m/s2)

f: Density of fluid (kg/m3)

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 forces32

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 diffusivities33

12

Properties of Air

34

0.720.740.740.740.730.730.730.730.730.720.720.720.720.720.730.720.720.720.72

Properties of Water

35

Free Convection (Plate)• NNu = hdc/kf = f (NGr, NPr)

• NNu = a (NGr NPr)m; NRa = NGr NPr

• For vertical plate (dc = plate height) a = 0.59, m = 0.250 (for 104 < NRa < 109)

a = 0.10, m = 0.333 (for 109 < NRa < 1013)

• For inclined plate (for NRa < 109) Use same eqn as vertical plate & replace ‘g’ by ‘g cos’ in NGr

• For horizontal plate (dc = Area/Perimeter) Upper surface hot

• a = 0.54, m = 0.250 (for 104 < NRa < 107)

• a = 0.15, m = 0.333 (for 107 < NRa < 1011)

Lower surface hot• a = 0.27, m = 0.250 (for 105 < NRa < 1011)

36

13

Free Convection (Cylinder)

• For vertical cylinder (dc = cylinder height)

– Similar to vertical plate if D ≥ 35L/(NGr)0.25

• For horizontal cylinder (dc = cylinder diameter)

– For 10-5 < NRa < 1012

37

Note: NRa = NGr NPr

Free Convection (Sphere)

For sphere, dc = D/2

Note 1: For all free convection situations, determine properties at the film temperature {Tfilm = (Ts + T∞)/2} unless otherwise specifiedNote 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

NRa = NGr NPr

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

14

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)• NNu = 1.86 (NRe x NPr x dc/L)0.33 (b/w)0.14

• 2. Transitional flow

(2100 < NRe < 4000)– Friction factor (f)

• For smooth pipes:

• For non-smooth pipes, use Moody chart (graph of: f, NRe, /D)

dc: ID of pipe, L: Length of pipe

41

Moody Diagram

= 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

Reynolds Number (NRe)

Fri

ctio

n F

acto

r (f

)

Relative R

ough

ness (/D

)

42

15

Forced Convection in a Pipe (contd.)

3. Turbulent flow (NRe > 4000) of a Newtonian fluid in a pipe,

NNu = 0.023 (NRe)0.8 (NPr)0.33 (b/w)0.14

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

43

Note: For all forced convection situations, use bulk temperature of fluid to determine properties (unless otherwise specified)

‘hfp’ for Forced Convection over a Sphere

NNu = hdc/kf = f (NRe, NPr) – similar to flow in a pipe

NNu = 2 + 0.6 (NRe)0.5 (NPr)0.33

For 1 < NRe < 70,000 and 0.6 < NPr < 400

44

Note 1: dc is the outside diameter of the sphereNote 2: Determine all properties at the film temperature

{Tfilm = (Ts + T∞)/2}

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

16

46

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

17

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/W50

Thermal ResistancesConduction: 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

18

Electrical Analogy

V R

IA current (I) flows because there is a driving force, the potential Difference (V), across the resistance (R)

I = V/R

Q = T/(Thermal Resistance)

V

R1I

R2

I = V/(R1 + R2)

Q = T/(Sum of Thermal Resistances)

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

19

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 KConvective heat transfer coefficient: W/m2 KOverall heat transfer coefficient: W/m2 KThermal resistance to heat transfer: K/W 57

20

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

• 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

x95 °C

T

Q

Q

k

h 5 °C• A heater is used to maintain the left end of the slab at 95 °C• Ambient air on right side is at 5 °C• What factors determine the magnitude of temperature at right end of slab?

Note: The same Q flows through the slab and outsideThus, Q = kA (95 – T)/x = hA (T – 5) 59

Heat Exchangers (HX)

60

21

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

Pneumaticallyactuated

Variable Gap

Pneumatically actuated

Steam

Product

www.process-heating.com

Intense, turbulent mixing of steam and product occurs. It results in rapid heating and dilution of product. A vacuum chamber is used downstream to evaporate steam that condensed into product.

62

Steam Infusion

Product in

CIP in

Air out

Steam in

Cooling water in/out

Product out

This is a gentler process than steam injection. A vacuum chamber is used downstream to evaporate steam that condensed into product.

63

22

Batch Retorts• Static (with or without crates/baskets)

– Horizontal

– Vertical

• Rotary (axial or end-over-end rotation)

• Reciprocating– Shaka

End-over-end Rotation

Axial Rotation

Rotary, end-over-end or reciprocating motion is used to mix the product and make the temperature distribution uniform

64

Static & Rotary Retorts

www.libertyprocess.co.uk

www.jbtfoodtech.com

SuperAgi Retort

65

Crateless Retort (Semi-Continuous)

www.maloinc.comProblem: Cold spot identification66

23

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

24

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

1 X 41 X 4

PHEs (All channels in Parallel)

71

4 X 14 X 1

PHEs (All Channels in Series)

72

25

2 X 21 X 4

PHEs (Series & Parallel)

73

Regeneration in a PHE

HeaterCooler Regenerator

Raw product

Hot pasteurized product

Regeneration: Energy of hot pasteurized product is used to pre-heat cold raw product. Typical regeneration efficiency is ~90%.

Cold pasteurized product

74

Double Tube, Triple Tube, Multitube HXHeating from 1 side

Heating from 2 sides

75

26

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)

Disadvantage: Particle damage, uncertain residence time, cleaning

Product

Steam jacket

Insulation

Shaft with motor rotating it

Motor

Cross-sectional view

Steam jacket

Product

Insulation

Shaft

77

Double Tube Heat Exchanger (DTHE)

Heat TransferProduct

Hot/Cold water

Double tube heat exchangerTwo concentric tubesProduct generally flows in inner tubeHeating/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 tubeBoth streams may flow in the same or opposite directions 78

27

Heat Transfer in a Double Tube HX(Hot water heating a product)

roiTco

rii

Tho ho

hiU

ProductTci, mc, cp(c)

.

Hot waterThi, mh, cp(h)

.

L

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

Q = hoAo [Thot water - Twall (outside)]

Q = kAlm [Twall (outside) - Twall (inside)]/r

Q = hiAi [Twall (inside) - Tproduct]

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

ho

hi

U

Product

Hot water

Convective resistance (1/ho Ao)

Convective resistance (1/hi Ai)Conductive resistance (r/k Alm)

81

28

Overall Heat Transfer Coefficient (U)

Q = T / [(1/hoAo) + (r/kAlm) + (1/hiAi)]

Denominator: Total thermal resistance

1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi)

Thus, Q = UAlmTlm U: W/m2 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

29

Tubular Heat Exchanger (Co- and Counter-Current)

Ai = Inside surface area of inner pipe = 2 rii LAo = Outside surface area of inner pipe = 2 roi LAlm = 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)]

Tem

pera

ture

Tem

pera

ture

1/UAlm = (1/hoAo) + (r/kAlm) + (1/hiAi) and Q = UAlmTlm

h: Hotc: Coldi: Inleto: Outlet

h: Hotc: Coldi: Inleto: Outlet

Used only when rapid initial cooling is neededLower heat transfer efficiencyTco ≤ Tho alwaysT btwn hot & cold fluid dec. along length

Most commonly usedHigher heat transfer efficiencyTco can be greater than Tho

T btwn hot & cold fluid does not change significantly along length

85

Co-Current Arrangement

roiTco rii

Thoho

hiUProduct

Tci, mc, cp(c)

.

Thi, mh, cp(h).

L

Hot water

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,Q1 = mc cp(c) (T)c = Q2 = mh cp(h) (T)h = UAlm Tlm

with 1/UAlm = 1/hiAi + r/kAlm + 1/hoAo

Note 1: (T)c = (Tco – Tci); (T)h = (Thi – Tho)Note 2: (T)1 = (Thi – Tci); (T)2 = (Tho – Tco)

Subscripts for m, cp, T, T:‘h’ for hot, ‘c’ for cold‘i’ for inlet, ‘o’ for outletSubscripts for h, A:‘i’ for inside, ‘o’ for outside

Q1

Q3

Q2

r = roi - rii

. ..

86

Counter-Current Arrangement

roiTco rii

Tho ho

hiUProduct

Tci, mc, cp(c)

.

Thi, mh, cp(h).

L

Hot water

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,Q1 = mc cp(c) (T)c = Q2 = mh cp(h) (T)h = UAlm Tlm

with 1/UAlm = 1/hiAi + r/kAlm + 1/hoAo

Note 1: (T)c = (Tco – Tci); (T)h = (Thi – Tho)Note 2: (T)1 = (Tho – Tci); (T)2 = (Thi – Tco)

Subscripts for m, cp, T, T:‘h’ for hot, ‘c’ for cold‘i’ for inlet, ‘o’ for outletSubscripts for h, A:‘i’ for inside, ‘o’ for outside

Q1

Q3

Q2

r = roi - rii

. ..

87

30

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

r1 r2 r1r2r3

Without Insulation With Insulation

ToTo

Ti Ti

Q = T / [(1/hoAo) + (r/kAlm)pipe + (1/hiAi)] Q = T / [(1/hoA’o) + (r/kAlm)insulation + (r/kAlm)pipe + (1/hiAi)]

ho

hi

ho

hi

rpipe = r2 – r1 and Ao = 2r2L(Alm)pipe = 2L (r2 – r1) / [ln (r2/r1)]

rinsulation = r3 – r2 and A’o = 2r3L(Alm)insulation = 2L (r3 – r2) / [ln (r3/r2)]

Setting dQ/dr3 = 0, and ensuring that d2Q/dr32 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 decreasesIf r2 > kins/ho, heat loss decreases even if a small amount of insulation is added

Adding insulation inc. conductive res. & dec. convective res.

90

31

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.05 0.1 0.15 0.2 0.25 0.3

Qin

s/ Q

bar

e

Insulation Thickness (m)

k/h = 0.01 mk/h = 0.03 mk/h = 0.05 mk/h = 0.1 m

Qins/Qbare versus Thickness of Insulation

Zero benefit

Outside Radius of Pipe = 2.25 cm

91

Dimensionless Numbers in Heat Transfer

92

Dimensionless Numbers in Heat Transfer

• Reynolds number

• Nusselt number

• Prandtl number

• Grashof number

For circular cross-section pipes

93

32

Dimensionless Numbers (contd.)

• Biot number

• Fourier number

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 soliddc (for free convection): Look up notes/slides based on shape of objectdc (for forced convection pipe flow) = 4 (Across-section)/(Wetted perimeter)Dc: (for unsteady state heat transfer): Distance btwn hottest & coldest points in solid object

s = k/( cp) = Thermal diffusivity (m2/s)

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

33

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

and Fourier # (NFo = st/Dc2) Note: s = k/( cp) 97

Categories of Unsteady State Heat Transfer

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 objectDc for sphere: Radius of sphereDc for an infinitely long cylinder: Radius of cylinderDc for infinite slab with heat transfer from top & bottom: Half thickness of slabDc for an infinite slab with heat transfer from top: Thickness of slab

Categories are based on the magnitude of Biot # (NBi = hDc/ks)

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)

– The two modes of heat transfer are• Convection (external) Q = hA T = T/(1/hA)

• Conduction (internal) Q = kA T/x = T/(x/kA)

100 °C air20 °C

20 °Ch

99

34

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

• 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 air20 °C 20 °C 50 °C 49 °C

t = 0 min t = 2 min

100 °C air 100 °C air20 °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 air20 °C 20 °C 70 °C 45 °C

101

Category #1

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): Density of solid object (kg/m3)V: Volume of solid object (m3)cp: Specific heat of solid object (J/kg K)Note: V = mass of object (kg)

The above equation can be used to determine temperature, T, at time, tOR

Based on time-temperature data, the equation can be used to determine ‘h’

Shape Area Volume

Brick 2 (LW+LH+WH) LWH

Cylinder 2RL + 2R2 R2L

Sphere 4R2 (4/3)R3

L: Length of brick or cylinderW, H: Width, height of brick resp.R: Radius of cylinder or sphere 102

35

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/NBi = 0

1/NBi = 100

Note: s = k/(cp) = Thermal diffusivity of solid object in m2/s

1 2 3 4 50 10 10020 30 40 50 60 70 80 90

104

0.001

1

0.01

0.1

Heisler Chart (For Finite Sphere)

NFo (= t/Dc2)

TR

[=

(T-T

∞)/

(Ti-T

∞)]

105

36

Heisler Chart (For Infinite Cylinder)

NFo (= t/Dc2)

TR

[=

(T-T

∞)/

(Ti-T

∞)]

106

Heisler Chart (For Infinite Slab)

NFo (= t/Dc2)

TR

[=

(T-T

∞)/

(Ti-T

∞)]

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

37

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 encountered when…..

‘k’ is high ORDc is small

Neither ‘k’ nor ‘h’ are high and Dc is not too

small or too large

‘h’ is high ORDc is large

Resistance that is negligible

Conductive(Internal)

None Convective(External)

T that is small Btwn center and surface of solid

None Btwn fluid and surface of solid

Solution approach

Lumped parameter eqn.

Heisler chart(s) Heisler chart(s)(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 cylinder: Intersection ofinfinite cylinder & infinite slab

Finite brick: Intersection of 3infinite slabs

111

38

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, height112

Calculations for Finite Cylinder (Heisler Chart)Infinite Cylinder Infinite Slab

Characteristic dimension (Dc)

Biot number (NBi)NBi = h Dc/ks

1/NBi

Thermal diffusivity ()s = ks/(s cp(s))

Fourier number (NFo)NFo = st/Dc

2

Temperature ratio (TR)from Heisler chart

(based on values of NFo

& 1/NBi)

Solve for “T” from the above equation

If both NBi < 0.1, the lumped parameter method (eqn) can be used

113

Calculations for Finite Brick (Heisler Chart)Infinite Slab #1 Infinite Slab #2 Infinite Slab #3

Characteristic dimension (Dc)

Biot number (NBi)NBi = h Dc/ks

1/NBi

Thermal diffusivity (s)s = ks/(s cp(s))

Fourier number (NFo)NFo = st/Dc

2

Temperature ratio (TR)from Heisler chart

(based on values of NFo

& 1/NBi)

Solve for “T” from the above equation

If all 3 NBi < 0.1, the lumped parameter method (eqn) can be used

114

39

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

• 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 = UAlmTlm 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

40

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