Chapter 8
Internal Forced Convection
853
Hot air enters a sheet metal duct located in a basement. The exit temperature of hot air and the rate
of heat loss are to be determined.
Assumptions
1
Steady flow conditions exist.
2
The inner surfaces of the duct are smooth.
3
The thermal
resistance of the duct is negligible.
4
Air is an ideal gas with constant properties.
5
The pressure of air is 1
atm.
Properties
We expect the air temperature to drop somewhat, and evaluate the air properties at 1 atm and
the estimated bulk mean temperature of 50
C (Table A15),
7228
.
0
Pr
C
J/kg.
1007
/s;
m
10
797
.
1
C
W/m.
02735
.
0
;
kg/m
092
.
1
2
5

3
p
C
k
Analysis
The surface area and the Reynolds number are
m
9.6
m)
m)(12
2
.
0
(
4
4
aL
A
s
D
A
p
a
a
a
h
c
4
4
4
0 2
2
.
m
509
,
44
/s
m
10
797
.
1
m)
m/s)(0.20
(4
Re
2
5
h
m
D
V
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are
roughly
m
0
.
2
m)
2
.
0
(
10
10
h
t
h
D
L
L
which is much shorter than the total length of the duct. Therefore, we can assume fully developed
turbulent flow for the entire duct, and determine the Nusselt number from
2
.
109
)
7228
.
0
(
)
509
,
44
(
023
.
0
Pr
Re
023
.
0
3
.
0
8
.
0
3
.
0
8
.
0
k
hD
Nu
h
and
C
.
W/m
93
.
14
)
2
.
109
(
m
2
.
0
C
W/m.
02735
.
0
2
Nu
D
k
h
h
The mass flow rate of air is
kg/s
0.1748
m/s)
(4
0.2)m
)(0.2
kg/m
092
.
1
(
2
3
V
A
m
c
In steady operation, heat transfer from hot air to the duct must be equal to the heat transfer from the duct
to the surrounding (by convection and radiation), which must be equal to the energy loss of the hot air in
the duct. That is,
Q
Q
Q
E
conv,in
conv+rad,out
hot air
Assuming the duct to be at an average temperature of
T
s
, the quantities above can be expressed as
:
Q
conv,in
60
ln
60
)
m
6
.
9
)(
C
.
W/m
93
.
14
(
ln
2
2
ln
s
e
s
e
i
s
e
s
i
e
s
i
s
i
T
T
T
T
Q
T
T
T
T
T
T
A
h
T
A
h
Q
:
Q
conv+rad,out
4
4
4
4
2
8
2
2
2
4
4
)
273
10
(
)
273
(
)
.
W/m
10
67
.
5
)(
m
0.3(9.6
+
C
)
10
)(
m
C)(9.6
.
W/m
10
(
)
(
K
T
K
T
Q
T
T
A
T
T
A
h
Q
s
s
o
s
s
o
s
s
o
:
E
hot air
C
)
C)(60
J/kg.
7
kg/s)(100
1748
.
0
(
)
(
e
i
e
p
T
Q
T
T
C
m
Q
This is a system of three equations with three unknowns whose solution is
C
3
.
33
and
,
,
s
e
T
T
Q
C
45.1
W
2622
832
Air
60
C
4 m/s
L
= 12 m
= 0.3
Air duct
20 cm
20 cm
T
=
10
C
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Chapter 8
Internal Forced Convection
Therefore, the hot air will lose heat at a rate of 2622 W and exit the duct at 45.1
C.
854
"!PROBLEM 854"
"GIVEN"
T_i=60
"[C]"
L=12
"[m]"
side=0.20
"[m]"
Vel=4
"[m/s], parameter to be varied"
"epsilon=0.3 parameter to be varied"
T_o=10
"[C]"
h_o=10
"[W/m^2C]"
T_surr=10
"[C]"
"PROPERTIES"
Fluid$='air'
C_p=CP(Fluid$, T=T_ave)*Convert(kJ/kgC, J/kgC)
k=Conductivity(Fluid$, T=T_ave)
Pr=Prandtl(Fluid$, T=T_ave)
rho=Density(Fluid$, T=T_ave, P=101.3)
mu=Viscosity(Fluid$, T=T_ave)
nu=mu/rho
T_ave=T_i10
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 Spring '08
 BENARD
 Heat Transfer

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