Topic 7 Heat Transfer

FOOD1360/1577 Principles of Food Engineering

Robert Driscoll

FST 2010 1/76

Heat Transfer

S&H Chapter 4 Notes on electric circuits Sect 3.2

 Movement of heat energy from one point to another by virtue of temperature gradient (difference) hot A Heat flux cold FST 2010 2/76

Use in unit operations:

  

Used in most unit operations in food industry Why do we heat?

  Cooking Sterilising    Pasteurising Blanching Browning Why do we cool?

   Preservation Texture (icecream!) Stop cooking quickly (see S&H Sect 4.1: Heat Exchangers).

FST 2010 3/76

Modes of heat transfer

1.

2.

3.

Conduction Convection Radiation These notes follow S&H Sect 4.3. So look at his examples!

FST 2010 4/76

1. Conduction

 Heat transfer between neighbouring molecules Esp. for solids, but also for fluids!

 vibration of molecules, or  drift of electrons (in metals)  no net movement of molecules FST 2010 5/76

Fourier’s equation

 Rate of flow = driving force / resistance

dQ



dt



kA dT dx

  The negative indicates heat flows from hot to cold.

Logical: rate of heat flow depends on     Temperature difference Material properties Distance to flow Cross-sectional area for flow.

FST 2010 6/76

Terms used

      dQ/dt = rate of heat flow (J/s or Watts) dx = thickness (m). Use D x for finite differences.

A = area normal to the flow direction (m 2 ) k = thermal conductivity of the material (in J/ o C.s.m or W/m.K) dT = temperature difference ( o C) (use D T = T 1 -T 2 for finite difference).

For steady state heat flow through a material,

dT dx

 D

T

D

x

FST 2010 7/76

T

1

Example: slab

T

2

A

 

kA



T

1  D

x T

2  FST 2010 8/76

Worked Example

  Use Fourier’s equation to find D T=T 1 -T 2 (see pic).

Solution:  D

T

   

kA kA

D  D

x T

D

x

 30  0 .

025 0 .

2  0 .

1  37 .

5 o

C

 Why is D x negative?

T

1  30

W

A=0.1m

2

T

2 FST 2010 9/76

Thermal Conductivity, k

 thermal conductivity: a property of a material   good conductors of heat: metals bad conductors of heat: insulators  What are typical values?

FST 2010 10/76

Some examples

Metals: Glass: Alloys: Air: Water: Ice: Insulations: 50-400 J/ o C.s.m

0.51 J/ o C.s.m

10-120 J/ o C.s.m

0.024 J/ o C.s.m

0.7 J/ o C.s.m

2.3 J/ o C.s.m

0.026-0.052 J/ o C.s.m

(cork, foamed plastics, expanded rubber etc.)  Air is a good insulator!

FST 2010 11/76

Thermal conductivity of Foods

    Close to k for water for higher moisture foods For fruits and vegs with W >60%: k = 0.148 + 0.00493 W W = water content in % Meats with W=60-80%: k = 0.08 + 0.0052 W For solid foods, low W, see Toledo, Singh and Heldman, Fellows etc.

FST 2010 12/76

Example

 What is the thermal conductivity of a block of meat with a moisture content of 75%?

 How would this information be useful?

c

 0 .

08  0 .

0052  75  0 .

47

W/m.K

FST 2010 13/76

Notes:

 Heat flux = heat transfer rate per unit area = dQ/Adt (in J/s.m

2 )  Use dx for

differential

, D x for

difference

.    Power is measured in Watt (W) Energy is measured in Joule (J) Power is energy per second (W = J/s) FST 2010 14/76

Example

 A cork slab 10 cm thick has one face at 12 o C and the other at 21 o C. If the thermal conductivity of cork is 0.042 W/mK, estimate the heat transfer per square meter.

Solution: ‘Per square meter’ means find heat flux. So

A

 0 .

042  ( 21   12 ) / 0 .

1 which gives 13.9 W/m 2 .

FST 2010 15/76

Similarity to electrical circuits

dQ dt

 

kA dT dx

  D

T

 D

x kA

  From physics - electrical circuits:    heat flow corresponds to electric current D x/kA corresponds to resistance D T corresponds to voltage (driving force)  This is a useful analogy.

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Electrical Circuit

 For a simple circuit, V R A I

I



V R

FST 2010 17/76

Apply to heat

dT   D

T

 D

x kA

 R=dx/kA FST 2010 So we have defined a heat resistance, R 18/76

Flow through layers

 If heat flows through a series of layers of material, we call it

series

heat transfer.

dT T 1 k 1 k 2 T 2

R

1 

dx

1

k

1

A R

2 

dx

2

k

2

A

dx 1 REAL dx 2 FST 2010 ANALOGY 19/76

  With the electric circuit, total resistance is R 1 + R 2 .

So total heat resistance is:

R T



dx

1

k

1

A



dx

2

k

2

A

 1

UA

 Setting R T =1/UA is convenient:   D

T R T

 

UA

D

T

 U is the overall heat transfer coefficient.

Defines heat flow resistance.

Defines heat flow in terms of U FST 2010 20/76

2. Convection

 Heat transfer by the net movement of molecules in a fluid is called a convection current: Fluid Flow Cold Body Hot Body FST 2010 21/76

Just for one surface

 Governing equation: 

hA



T s



T

 

T

s Hot Body Fluid Flow

T

 FST 2010 22/76

Two Types of Convection

 Natural convection (slow heat transfer):  induced by temperature gradient  current driven by density difference  Forced convection (rapid heat transfer):   movement of molecules by external energy e.g. a stirrer, fan, agitator, pump etc.

FST 2010 23/76

Which is faster?

 In liquids, convection is much faster than conduction! Soup heats faster than dog food.

Heat Diffuses FST 2010 Convection Mixing 24/76

Newton’s law of cooling



hA



T s



T

   where    h = convective heat transfer coefficient A = area of heat transfer D T= temperature difference.

 h is also called the surface heat transfer coefficient (J/ o C.s.m

2 or W/m 2 K) FST 2010 25/76

Convective resistance

 Since dQ/dt=hA D T, then  D

T

 1

hA

Again comparing with our electrical circuit analogy:

I



V R

 we can identify a convective resistance as R=1/hA.

FST 2010 26/76

Examples of htc

    ‘htc’ = heat transfer coefficient Values determined experimentally Air: Free convection Forced convection Water: Free convection Forced convection Boiling water Condensing water vapour 5-25 J/ o C.s.m

2 10-200 20-100 50-10,000 3,000 –100,000 5000 – 100,000 FST 2010 27/76

Example

 Estimate the rate of heat loss from a hot plate with a surface area of 1m 2 . The plate surface temperature is 120 o C and the ambient temperature is 20 o C. Convective heat transfer is estimated to be 10J/ o C.s.m

2   

hA

10 

T s

 1  

T

  120   20   1000 W FST 2010 28/76

3. Radiation

     heat transfer by electromagnetic waves similar to transfer of light Important in baking ovens requires no physical medium for propagation can occur in vacuum (unlike conduction and convection) FST 2010 29/76

Radiation equation

 q = s e A T 4 where  s = Stefan-Boltzmann constant   = 5.669x10

-8 J/m 2 .s.K

4 e = emissivity, relative to a black body  T = temperature of emitting body in K FST 2010 30/76

Example calculation

 Calculate the rate of heat energy emitted by 10 m 2 of a polished stainless steel tube (emissivity=0.15). The temperature of the tube is 100 o C.

  se

AT K

4 5 .

669  10  8  0 .

15  10   273 .

15  100  4  1 .

65 kW FST 2010 31/76

Combined modes

 For resistances in series, add resistances.

h 1 ,A h 2 ,A k,x,A T 1 T 2 R 1 R 2 FST 2010 R 3 32/76

Finding U:

 Adding the resistances:  Solving: 1

UA

 1

h

1

A



x kA

 1

h

2

A

 where U is the overall htc for the system.



UA

(

T

1 

T

2 ) Note: UA is the reciprocal of R T FST 2010 33/76

Worked Problem

    Water boiling in a Cu saucepan (ID=200 mm).

For copper, k=400 W/m.K

Convective resistance (base to water) = 0.0105 K/W Find rate of heat transfer.

FST 2010 100 o C 250 o C base 3 mm 34/76

Solution

   Two resistances:  conduction through copper saucepan base  convection from base to water.

So find each resistance, then add:

R

1

R

2 

x kA

 400  0 .

 003   0 .

1 2    0 .

0105 K/W (given) 2 .

4  10  4 K/W So heat flow is:  D

T

/ 

R

1 

R

2   250  100 0.0105

 0.00024

 14.0

kW FST 2010 Later problem: other types 35/76

Heat Flow Thru Pipes

r 2 r 1 Exterior, T 2 Length, L Product, T 1 FST 2010 36/76

T 1 r 1 dr

End View

T 2  Apply Fourier’s equation to heat flows through a thin annulus, dr.

r 2 FST 2010 37/76

 Since

Derivation

 

kA dT dr

 

k

2 

rL dT dr

 

dr

 2 

L k dT r

 

r

 2

r

1 1

dr r

 2 

L k

(

T

1 

T

2 ) FST 2010 38/76

Solving:

 ln 

r

2

r

1   (

T

1 

T

 2

Q

) 2 

L k

 2 

L k

(

T

1 ln

r

2

r

1 

T

2 ) FST 2010 39/76

Cylinder thermal resistance

 Comparing with the electrical circuit analogy again, we can identify R cyl :

R cyl

 ln 2    2

L k r

1  This resistance acts like the slab and convective resistances seen before.

FST 2010 40/76

Example: Two Conductive Layers

 Calculate the rate of heat loss through a fibreglass insulated steam pipe. Ignore convection.

T=35 o C 150 o C fibreglass iron pipe ID = 2.1 cm MD = 2.67 cm OD = 3.335 cm k fibreglass = 0.0351 W/m.K

k pipe = 45 W/m.K

FST 2010 41/76

Solving for heat flow

Two layers, two resistances to add.

R pipe = ln(1.335/1.05)/(2  45) (per unit length) = 0.00085 R insul = ln(1.667/1.335)/(2  0.0351) = 1.007

Total resistance Thus rate of heat flow = (150-35)/(1.008) = 114 W/m FST 2010 42/76

Example: Conduction / Convection

h 2 , T  k r i h 1 , T 1 r o  Data: h 1 =6.5 W/m 2 K h 2 =4.5 W/m 2 K T 1 =135 o C T  =20 o C k 1 =202 W/mK r i =4.88 cm r o =5.08 cm FST 2010 43/76

Solution

   Inner convective resistance:

R i

 1 

h

1

A

1 6 .

5  ( 2  1 0 .

0488 ) Outer convective resistance:

R o

 1

h

2

A

2  4 .

5  1 ( 2  0 .

0508 )  0 .

502  0 .

696 Pipe wall conductive resistance:

R o

 ln 

r o

2 

k r i

 ln  5 .

08 2  4 .

( 202 ) 88   0 .

00003 FST 2010 44/76

  So total resistance = 0.502+0.696+0=1.20

So heat loss is:

q



L



T

1 

T



R T

 135  20 1 .

20  95 .

8

W

/

m

 Note that heat losses are written

per unit length.

Why?

FST 2010 45/76

Insulation

  Insulation has low thermal conductivity Often used for

lagging

– covering pipes and vessels to reduce heat flow  Important over  long lengths or  large surface areas.

FST 2010 46/76

Choice of insulation

 Depends on:  Cost of materials vs. rate of heat loss  Conductivity of the insulation (may vary with age)  The finish (coating) of the insulation  Critical thickness!

FST 2010 47/76

Critical Thickness

  Adding insulation may increase heat losses!

Why? – increase in surface area.

R T

 1

hA

 ln

r o

2

r i



kL

 where r 0 is outer radius – main variable.

FST 2010 48/76

Minimum R

T

 Differentiate wrt r o :

dR T dr o

  1

h

2 

r o

2

L

 1

r i r i r o

1 2 

kL

 Solving, R T has a minimum at outer radius r c =k/h FST 2010 49/76

Plotting R

T Effect of Increasing Outer Radius (k=0.17, h=3)

2.5

2 1.5

1 0.5

0 0 0.02

0.04

0.06

Outer Radius, m 0.08

0.1

FST 2010 50/76

Combined modes

 Conduction and convection combined T 1 k q SOLID x h T 2 q FLUID FST 2010 T  51/76

Energy conserved:

 For steady state, the rate of energy flow through the solid must balance the rate of energy removal by the fluid.

q

  

kA



T

2

x



T

1  

hA

(

T

2 

T

 )  We can use this fact to check that the electrical analogy works: FST 2010 52/76

 So

Solving

q

 

T

1

x kA



T

  1

hA

 Total resistance is:

R T



x kA

 1

hA

FST 2010 53/76

Estimating h

  h is determined empirically h is affected by   type and velocity of fluid (water, air, oil etc.) physical properties of fluid (density, viscosity, thermal conductivity, specific heat, thermal expansion coefficient)   temperature difference geometrical shape of the system (D, L, r)  gravity (density differences generate natural convection) FST 2010 54/76

Using DA for h

 Nu (Nusselt number)

hL k

 Pr (Prandtl number)  Gr (Grashof number)

g

 

T w v

2 

T

 

x

3  Length to Diameter ratio  Re (Reynolds number) 

D u

 FST 2010

c p



k L D

55/76

For forced convection:

  Typically: Nu = f (Re, Pr) E.g. for turbulent flow in pipes:

Nu D

 0 .

027 Re 0 .

8

D

Pr 1 3    

w

  0 .

14  where all properties are evaluated at the bulk fluid temperature, except  w at the wall temperature.

FST 2010 56/76

Summary

      Models for conduction/convection Analogy to electrical circuits - R T Combined modes Insulation (eg. critical thickness) Cylindrical geometry Values for h and k FST 2010 57/76

Problems

   On the following pages are some sample problems for interactive solution in class.

Each problem will be shown and a short time given to attempt a solution. Then the solution will be revealed in stages.

This tutorial is only of benefit if you actively participate. It is the exercise of thinking through

how to solve

which is of greatest benefit.

FST 2010 58/76

Problem 1

 Compare the rates of heat transfer for boiling water in a saucepan of ID=200mm, base 3mm, convective resistance 0.0105 K/W for bases of copper, aluminium, cast iron, stainless steel and pyrex.

Values for k, W/m.K

Cu 386 Al 204 base 3 mm (compare earlier problem).

100 o C 250 o C Fe SS Glass 73 14 0.76

FST 2010 59/76

Solution 1

 Same method as before: add the two resistances:  conduction through saucepan base  convection from base to water.

 D

T

/ 

R

1 

R

2   250  100 0

.

0105  0

.

003

/k

 0 .

1 2 Base thickness Base area Material k W/m.K

heat flow, kW copper alumin ium 386 204 13.9

13.7

cast iron 73 stainless steel 14 pyrex 0.76

12.7

8.7

1.1

FST 2010 60/76

Problem 2

 Calculate the rate of heat loss per meter from an insulated steam pipe if      Convective resistance (steam to pipe) is 1.63x10

-3 K/W Convective resistance (insulation to air) is 2.09x10

-3 K/W Steam temperature is 150 o C.

Air temperature is 30 o C.

The pipe is a 4” pipe (schedule 40 steel pipe).

 The 5cm thick insulation has a thermal conductivity of 0.1 W/mK.

FST 2010 61/76

Diagram

 Schedule 40 means ID=4.026”, OD=4.5”, and is a standard pipe size.

 Convert to SI (meter): ID=0.10226 OD=0.11430

air 150 o C  As insulation is 5 cm thick, insulation is from 0.05715 to 0.10715 m (use radii). insulation steel FST 2010 62/76

Solution 2

 Find resistances:   Steam to pipe Pipe

R

2  ln  

r i

2 

kL



R

1 ln  0 .

1143 0 .

10226  2  ( 73 ) 1   1 .

63  10  3 K/W 2 .

43  10  4 K/W  Insulation

R

3  ln 2

r

 

r i



kL

 ln  0 .

10715 0 .

05715 2  ( 0 .

1 ) 1   1 .

000 K/W  Insulation to air

R

4  2 .

09  10  3 K/W FST 2010 63/76

Ctd

 So heat losses per unit length are:  D

R T

 150 1 .

 30 004  120 W/m FST 2010 64/76

Problem 3

 A temperature difference of 85 o C is imposed on a fibre glass layer with k=0.035 W/m.K, 130 mm thick. Find the heat transferred per hour per square meter.

 SOLUTION: 

kA

D

T

D

x

Hence  0 .

035  1  85 0 .

130  22 .

9 W/m 2

Q

 D

t

 22 .

9  3600 / 1000  82 .

4 kJ/m 2 FST 2010 65/76

Problem 4

 A composite wall is constructed from 20 mm of material (k=1.3 W/m.K) and insulated with material of k=0.035 W/m.K so that the heat

flux

is kept under 1.83 kW. If the inside temperature is 1300 o C and outside is 30 o C, find the required thickness of insulation.

A=1 m 2 1300 o C 30 o C 1.83 kW FST 2010 66/76

Solution 4

 This is an example of resistances in series again.

R tot



R wall



R ins

D T  

x w k w A w



x i k i A i R i



x i k i A R w



x w k w A

 0 .

02 1 .

3   D

T

1 /  0 .

035

R tot



x i

 1  1300  30   0 .

0154  28 .

6

x i

  1 .

83 kW /m 2  Solving for x i :

x i

 2 .

37 cm FST 2010 67/76

Problem 5

     A brewery fermentation tank of 13 m diameter and 18 m high is placed in a room with average temperature 15 o C. The tank is constructed with 9 mm steel plate lined with 12.5 mm glass internal coating. Internal convection can be ignored, and the convective heat transfer coefficient on the outside of the tank is 10 W/m 2 .K. If the beer is kept at 37 o C, estimate the heater size required to maintain temperature.

How quickly is the beer cooling without a heater (assume tank is full)?

FST 2010 68/76

Solution 5

 Let’s start with a diagram or two.

steel glass 15 o C 37 o C 13 m OD TANK H=18 m h=10 k steel =15 W/m.K

k glass =0.73 W/m.K

FST 2010

Q

69/76

Ctd

  Next, let’s calculate the heat losses thru walls (ignore top and bottom of tank in this example).

Since thickness << diameter, don’t need cylinder form!

R tot



R steel



R glass



R conv



x s k s A s



x g k g A g

 1

hA o

   0 .

009 15    13  18  0 .

0125 0 .

78    13  18  10  1   13  18 8 .

162  10  7  D

T

/

R tot

  2 .

180   37  15  10   5  1 .

586 1 .

36  10   4  10   4 138 .

7 kW FST 2010 70/76

Ctd (optional)

  (Do this bit after heat exchanger work!) From our heat exchanger work, we can equate this to the heat loss from the beer:

dQ

 

c w dT dt dt

density  1000   13 2  139 .

2 kW 2 vol  18  c w 4 .

18 

dT dt



dT dt

 0 .

05 deg/hour FST 2010 71/76

Problem 6

    Dryness is a measure of how useful steam is.

As steam condenses to water, it gives up its latent heat.

So the steam vapour is far more useful than the condensed water.

The heat content of the steam is a function of dryness:

h d



h l



d



v

 So  if d=0, then the ‘steam’ is just water.

 If d=1, the steam is totally vapour.

FST 2010 72/76

Ctd

   A 30 m pipe (OD=75 mm) conveys steam at 1000 kg/hr. The steam pressure is 198.53 kPa. The steam enters with a dryness factor of 0.98 and must leave with a dryness factor of 0.95 or higher.

For an ambient temperature of 20 o C, how thick an insulation layer (k=0.2 W/m.K) will I need?

Q d i =0.98

d o =0.95

steam FST 2010 73/76

Solution 6

   Ignore convective heat transfer, and the pipe conduction.

At P=198.53, the steam temperature is 120 o C.

Heat losses (from the dryness factor) are: 

S

 

h l



d



v



inlet

 

h l



d



v



outlet

   

d i

  0 .

98

d o

  0 .



v

95   1000 3600  2203  1000 3600  18 .

4 kW FST 2010 74/76

Ctd

 Use Fourier’s equation for a cylinder:  2 ln 

kL

D

T



r o r i

  66 kW  2  ln 

r o

 

r o

 0 .

2  30 0 .

0375   120 0 .

046 m   20 46 mm   18 , 400  So required thickness is 46-37.5=8.5 mm FST 2010 75/76

End Problems

FST 2010 76/76