This document provides an introduction to heat transfer in food processing. It defines key terms like convective heat transfer coefficient and overall heat transfer coefficient. It describes methods for estimating these values, such as empirical correlations for forced and free convection in pipes. Equations are provided for calculating heat transfer through tubular heat exchangers based on temperature differences, heat transfer rates, and surface areas. An example problem demonstrates using these equations to determine the exit temperature of a fluid in countercurrent and parallel flow heat exchange.

1. INTRODUCTION TO FOOD ENGINEERING
Lecture 5
HEAT TRANSFER IN
FOOD PROCESSING

2. Objectives
• Calculate convective heat transfer coefficient
• Calculate overall heat transfer coefficient
• Calculate heat transfer area in tubular heat exchanger

3. Estimation of Convective Heat-Transfer
Coefficient
• h is predicted from empirical correlation for
Newtonian fluids only
• Forced convection

4. Forced Convection
 
Pr
Re
Nu N
,
N
f
N 
k
hD

NNu = Nusselt number
NRe = Reynold number
NPr = Prandtl number
μ
D
u
ρ
_

k
μcp


5. (4.
38)
100,
L
D
N
N Pr
Re 





 

14
.
0
w
b
66
.
0
Pr
Re
Pr
Re
Nu
L
D
N
N
045
.
0
1
L
D
N
N
085
.
0
3.66
N 














 







 



Larminar flow in pipes
NRe < 2100
For
b = bulk, w = wall

6. (4.
39)
100,
L
D
N
N Pr
Re 








14
.
0
w
b
0.33
Pr
Re
Nu
L
D
N
N
1.86
N 














 



For

7. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

8. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.
41)
imeter
wetted per
area
free
4
De


(4.
42)
 70,000
N
1
400
N
6
.
0
3
1
Pr
0.5
Re
Nu
Re
Pr
for
N
0.60N
2
N 







9. Free Convection
(4.
43)
 m
Pr
Gr
Nu N
N
a
k
hD
N 


10. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

11. Example
• Water flowing at 0.02 kg/s is heated from 20 to 60 C
in a horizontal pipe (D = 2.5 cm). Inside T = 90 C.
Estimate h if the pipe is 1 m long.
– Average T = (20+60)/2 = 40 C
–  = 992.2 kg/m3, cp = 4.175 kJ/kg C
– k = 0.633 W/m C,  = 658.026 x 10-6 Pa.s
– NPr = cp/k = 4.3, w is  at 90 C

12. D
m
D
u
N



.
_
Re
4


= 1547.9 laminar flow
)
025
.
0
)(
3
.
4
)(
9
.
1547
(
)
( Pr
Re 


L
D
N
N
= 166.4 > 100
NNu = 11.2
14
.
0
6
6
Nu
10
909
.
308
10
026
.
658
33
.
0
)
4
.
166
(
86
.
1 







 

N

13. (0.025m)
C)
33W/m
(11.2)(0.6
Nu 


D
k
N
h
= 284 W/m2 C

14. Turbulent flow in pipes
14
.
0
w
b
33
.
0
Pr
8
.
0
Re
Nu
μ
μ
023
.
0 









 N
N
N

15. Estimation of Overall Heat-Transfer
Coefficient
• Conduction + Convection

16. • If temperature of fluid in pipe is higher
– Heat flows to outside
– Ti > T
Ui = overall heat transfer coefficient
based on inside area
 

 T
-
T
A
U
q i
i
i

17. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
 
1
i
i
i T
-
T
A
h

q
 
 
1
2
2
1
lm
r
-
r
T
-
T
A
k
q 
 

 T
-
T
A
h
q 2
0
0
Convection from inside
Conduction
Convection to outside

18. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.
48)
 
l
i
i
i
T
-
T
A
h
q

(4.
49)
 
2
l
lm
1
2
T
-
T
A
k
r
-
r
q

(4.
50)
 

 T
-
T
A
h
q
2
0
0

19. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.
51)

 T
-
T
A
U
q
i
i
i
(4.
52)
 
0
0
lm
1
2
i
i
i
i A
h
q
A
k
r
-
r
q
A
h
q
A
U
q



(4.
53)
 
0
0
lm
1
2
i
i
i
i A
h
1
A
k
r
-
r
A
h
1
A
U
1



(4.

20. Example
• A steel pipe (k = 43 W/mC) inside D = 2.5 cm, 0.5
cm thick, conveys liquid food at 80 C. Inside h = 10
W/m2C. Outside temp = 20 C, outside h = 100
W/m2C. Calculate overall heat transfer coefficient
and heat loss from 1 m length of pipe.
 
0
0
lm
1
2
i
i
i
i A
h
1
A
k
r
-
r
A
h
1
A
U
1



o
o
i
lm
i
i
o
i
i r
h
r
kr
)r
r
(r
h
1
U
1





21. – ro = 0.0175 m
– Ri = 0.0125 m
– rlm = 0.01486 m
– 1/Ui = 0.10724 m2 C/W
– Ui = 9.32 W/m2 C
• Heat loss
– q = UiAi(80 – 20)
– = 43.9 W
• Uo = 6.66 W/m2 C
– q = 43.9 W

22. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
6.Role of Insulation in Reducing Heat Loss from
Process Equipment
(4.
55)
Lh
r
2
1
r
r
ln
Lk
2
l
T
-
T
q
0
0
i
0
i



 
(4.
56)
 
 
  0
r
h
k
-
r
l
r
h
k
r
r
ln
T
-
T
kL
2
-
dr
dq
2
0
0
0
2
0
0
i
0
b
i
0













23. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.
57)
0
r
h
k
-
r
l
2
0
0
0









(4.
58)
0
c
h
k
r 

24. Design of a Tubular Heat Exchanger
• Determine desired heat-transfer area for a given
application. Assuming
– Steady-state conditions
– Overall heat-transfer coefficient is constant
throughout the pipe length
– No axial conduction of heat in metal pipe
– Well insulated, negligible heat loss

25. (4.
59)
  i
overall
i dA
T
U
dq 

(4.
60)
  i
c
h
i
h
ph
h
c
pc
c
q dA
T
-
T
U
dT
c
m
dT
c
m
d 


(4.
61)
q
T
-
T
dq
T
d l
2 



Design of Tubular Heat Exchanger
• Heat transfer from one fluid to another
• Energy balance for double-pipe heat exchanger

26. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

27. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.
62)
q
T
-
T
1 l
2 









 
 i
i dA
T
d
T
U
(4.
63)
 









i
2
l
A
0
i
l
2
T
T
i
dA
q
T
-
T
T
T
U
1
Slope of T line

28. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.
64)
 
 
l
2
l
2
i
i
T
T
ln
T
-
T
A
U
q





(4.
65)
 
  difference
rature
mean tempe
log
T
T
ln
T
-
T
l
2
l
2






29. Example
• A liquid food (Cp = 4.0 kJ/kgC) flows in inner pipe
of heat exchanger. The food enters at 20 C and exits
at 60 C. Flow rate = 0.5 kg/s. Hot water at 90 C enters
and flows countercurrently at 1 kg/s. Average Cp of
water is 4.18 kJ/kgC.
– Calculate exit temp of water
– Calculate log-mean temperature difference
– If U = 2000 W/m2C and Di = 5 cm calculate L.
– Repeat calculations for parallel flow.

30. • Liquid food
– Inlet temp = 20 C
– Exit temp = 60 C
– Cp = 4.0 kJ/kg C
– Flow rate = 0.5 kg/s
• Water
– Inlet temp = 90 C exit temp = ?
– Cp = 4.18 kJ/kgC
– Flow rate = 1.0 kg/s

31. • q = mcCpc Tc = mhCph  Th
• Tc = 70.9 C
• Tlm = 39.5 C
• q = UA(T)lm = UDiL(T)lm
• = mCp T = 80 kJ/s
• L = 6.45 m
• For parallel flow L = 8 m