05_convec (1).ppt

INTRODUCTION TO FOOD ENGINEERING
Lecture 5
HEAT TRANSFER IN
FOOD PROCESSING

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

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

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


(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

(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

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 







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

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

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

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

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

Estimation of Overall Heat-Transfer
Coefficient
• Conduction + Convection

• 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

 
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

(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

(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.

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





– 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

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













(4.
57)
0
r
h
k
-
r
l
2
0
0
0









(4.
58)
0
c
h
k
r 

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

(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

(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

(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






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.

• 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

• 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

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