Solution to Homework 27.1
22.5
Efficiency of this unit is 98%, so the electrical power required is:
98%
electrical
heat
p
Q
Q
VC
T
r
=
=
D
&
.
The volume flux is:
3
3
3
4
2 gal
3.7854
10
m
1min
m
1.2618
10
min
gal
60 s
s
V



=
=
&
and the temperature difference is:
5
(120
60)
60
33.3
9
o
o
T
F
C
=

=
=
=
D
constantpressure heat capacity is:
o
o
cal
4.184 J
1000 g
4184 J
1
g
C
cal
1 kg
kg
C
p
C
=
=
�
�
so the electrical power required is:
3
3
4
o
3
o
98%
kg
m
4184 J
1
10
1.2618
10
33.3 C
m
s
kg
C
0.98
17580 W
17939 W
17.9 kW
0.98
p
electrical
VC
T
Q
r

D
=
°

=

&
22.6
Efficiency of this unit is 70%, so the natural gas burning energy required is:
70%
NG
heat
Q
Q
=
.
From the result in (22.5),
17580 W
heat
Q
=
, and the natural gas burning energy is:
55.509 MJ
ˆ
kg
NG
methane
Q
m
H
m

=
= 

D
&
&
So the mass consumption rate is:
(
)
m
6
m
ˆ
70%
2.205 lb
17580 W
3600 s
0.7
55.509
10
J/kg
kg
hr
3.59
lb
/hr
heat
methane
Q
m
H
=

� D
=


&
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22.9
traveling miles
total combustion energy
required energy
fuel economy =
tank volume
tank volume
mile
specific heat of combustion
total mass
required energy
tank volume
mile
specific heat of combustion
densit
=

=
=
required energy
y
mile
o
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 Spring '08
 V
 Thermodynamics, Heat, Flux, Combustion, Heat of combustion, Qheat

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