Q
T
sky
= 100 K
Chapter 7
External Forced Convection
799
Wind is blowing over the roof of a house. The rate of heat transfer through the roof and the cost of this
heat loss for 14h period are to be determined.
Assumptions
1
Steady operating conditions exist.
2
The critical Reynolds number is Re
cr
= 5
×
10
5
.
3
Air is
an ideal gas with constant properties.
4
The
pressure of air is 1 atm.
Properties
Assuming a film temperature of 10
°
C, the properties of
air are (Table A15)
7336
.
0
Pr
/s
m
10
426
.
1
C
W/m.
02439
.
0
2
5

=
×
=
υ
°
=
k
Analysis
The Reynolds number is
[
]
7
2
5
10
338
.
2
/s
m
10
426
.
1
m)
(20
m/s
)
3600
/
1000
60
(
Re
×
=
×
×
=
υ
=

∞
L
V
L
which is greater than the critical Reynolds number. Thus we have
combined laminar and turbulent flow. Then the Nusselt number and
the heat transfer coefficient are determined to be
C
.
W/m
0
.
31
)
10
542
.
2
(
m
20
C
W/m.
02439
.
0
10
542
.
2
)
7336
.
0
](
871
)
10
338
.
2
(
037
.
0
[
Pr
)
871
Re
037
.
0
(
2
4
4
3
/
1
8
.
0
7
3
/
1
8
.
0
°
=
×
°
=
=
×
=

×
=

=
=
Nu
L
k
h
k
hL
Nu
L
In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal to
the heat transfer from the roof to the surroundings (by convection and radiation), which must be equal to
the heat transfer through the roof by conduction. That is,
Q
Q
Q
Q
=
=
=
room to roof, conv+rad
roof, cond
roof to surroundings, conv+rad
Taking the inner and outer surface temperatures of the roof to be
T
s,in
and
T
s,out
, respectively, the quantities
above can be expressed as
[
]
4
,
4
4
2
8
2
,
2
2
4
,
4
,
rad
+
conv
roof,
to
room
K)
273
(
K)
273
20
(
)
.K
W/m
10
67
.
5
)(
m
300
)(
9
.
0
(
C
)
)(20
m
C)(300
.
W/m
5
(
)
(
)
(
+

+
×
+
°

°
=

+

=

in
s
in
s
in
s
room
s
in
s
room
s
i
T
T
T
T
A
T
T
A
h
Q
σ
ε
m
15
.
0
)
m
300
)(
C
W/m.
2
(
,
,
2
,
,
cond
roof,
out
s
in
s
out
s
in
s
s
T
T
L
T
T
kA
Q

°
=

=
[
]
4
4
,
4
2
8
2
,
2
2
4
4
,
,
rad
+
conv
surr,
to
roof
K)
100
(
K)
273
(
)
.K
W/m
10
67
.
5
)(
m
300
)(
9
.
0
(
C
)
10
)(
m
C)(300
.
W/m
0
.
31
(
)
(
)
(

+
×
+
°

°
=

+

=

out
s
out
s
surr
out
s
s
surr
out
s
s
o
T
T
T
T
A
T
T
A
h
Q
Solving the equations above simultaneously gives
C
5
.
3
and
C,
6
.
10
,
W
025
,
28
,
,
°
=
°
=
=
=
out
s
in
s
T
T
Q
kW
28.03
The total amount of natural gas consumption during a 14hour period is
therms
75
.
15
kJ
105,500
therm
1
85
.
0
)
s
3600
14
)(
kJ/s
03
.
28
(
85
.
0
85
.
0
=
×
=
∆
=
=
t
Q
Q
Q
total
gas
Finally, the money lost through the roof during that period is
$9.45
=
=
)
therm
/
60
.
0
$
therms)(
75
.
15
(
lost
Money
784
Air
V
∞
= 60 km/h
T
∞
= 10
°
C
T
in
= 20
°
C