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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
 PrReNu N,NfN 
k
hD
NNu = Nusselt number
NRe = Reynold number
NPr = Prandtl number
μ
Duρ
_

k
μcp


5. (4.38)
100,
L
D
NN PrRe 




 
14.0
w
b
66.0
PrRe
PrRe
Nu
L
D
NN045.01
L
D
NN085.0
3.66N 












 





 

Larminar flow in pipes
NRe < 2100
For
b = bulk, w = wall

6. (4.39)
100,
L
D
NN PrRe 






14.0
w
b
0.33
PrReNu
L
D
NN1.86N 












 
For

7. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

8. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.41)
imeterwetted per
areafree4
De


(4.42) 70,000N1
400N6.0
31
Pr
0.5
ReNu
Re
Pr
forN0.60N2N 


9. Free Convection
(4.43)
 m
PrGrNu NNa
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
mDu
N


._
Re
4

= 1547.9 laminar flow
)025.0)(3.4)(9.1547()( PrRe 
L
D
NN
= 166.4 > 100
NNu = 11.2
14.0
6
6
Nu
10909.308
10026.658
33.0)4.166(86.1 







 

N

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

14. Turbulent flow in pipes
14.0
w
b33.0
Pr
8.0
ReNu
μ
μ
023.0 





 NNN

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-TAUq iii

17. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
 1iii T-TAhq
 
 12
21
lm
r-r
T-T
Akq 
  T-TAhq 200
Convection from inside
Conduction
Convection to outside

18. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.48) li
ii
T-T
Ah
q

(4.49) 
2l
lm
12
T-T
Ak
r-rq

(4.50)  T-T
Ah
q
2
00

19. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.51) T-T
AU
q
i
ii
(4.52)
 
00lm
12
iiii Ah
q
Ak
r-rq
Ah
q
AU
q

(4.53)
 
00lm
12
iiii Ah
1
Ak
r-r
Ah
1
AU
1

(4.54)
CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.51) T-T
AU
q
i
ii
(4.52)
 
00lm
12
iiii Ah
q
Ak
r-rq
Ah
q
AU
q

(4.53)
(4.54)

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.
 
00lm
12
iiii Ah
1
Ak
r-r
Ah
1
AU
1

oo
i
lm
iio
ii rh
r
kr
)rr(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)
Lhr2
1
r
r
ln
Lk2
l
T-T
q
00i
0
i



 
(4.56)
 
   0
rh
k
-
r
l
rhkrrln
T-TkL2
-
dr
dq
2
000
2
00i0
bi
0










23. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING
(4.57)0
rh
k
-
r
l
2
000






(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)  ioveralli dATUdq 
(4.60)  ichihphhcpccq dAT-TUdTcmdTcmd 
(4.61)
q
T-T
dq
Td l2 


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-T1 l2 





 
 ii dA
Td
TU
(4.63)
 






i2
l
A
0
i
l2
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)
 
 l2
l2
ii
TTln
T-T
AUq



(4.65)
 
  differenceraturemeantempelog
TTln
T-T
l2
l2




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