Department of Mechanical Science and Engineering
ME 300 Thermodynamics Section D Fall 2010
Homework #3  Solution
1
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Problem 3.1
Known:
The specifications of a heat pump.
Find:
The energy the heat pump provides to the apartment per hour. The coefficient of
performance.
The required amount of energy transfer from the surroundings to the heat
pump. Annual cost of running the heat pump. Comparison of the annual cost between heat
pump and electrical heater.
Engr. Model/Assumptions:
The heat transfer rates and power of the heat pump are
constant. The work input into the electric heater is equal to the energy released into the
room.
Analysis:
(a) To determine the energy the heat pump provides to the room we need to integrate the
heat transfer rate over time:
Q
out
=
Z
˙
Q
out
dt
= 4
kJ
s
·
3600
s
= 14400
kJ
(b) Intergrating the power over time gives as the energy per hour the heat pump consumes:
W
cycle
=
Z
˙
W
cycle
dt
= 1
.
6
kJ
s
·
3600
s
= 5760
kJ
The coefficient of performance is then:
γ
=
Q
out
W
cycle
=
14400
kJ
5760
kJ
= 2
.
5
(c) To determine the necessary heat transfer from the surroundings to the heat pump when
Q
out
= 14400
kJ
we rearrange the coefficient of performance equation to solve for
Q
in
:
γ
=
Q
out
Q
out

Q
in
→
Q
in
=
Q
out
(
γ

1)
γ
=
14400
kJ
·
(2
.
5

1)
2
.
5
= 8640
kJ
(d) The cost for running the heat pump 1500 hours a year is:
cost
= 1
.
6
kW
·
1500
h
year
·
$0
.
06
kW
·
h
=
$144
.
00
year
(e) The cost for running the electric heater 1500 hours a year is:
cost
= 4
kW
·
1500
h
year
·
$0
.
06
kW
·
h
=
$360
.
00
year
The annual savings by using the heat pump instead of an electrical heater is $216.00 per
year. Therefore, after four years the savings will exceed the initial costs. Besides the po
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 Spring '08
 Staff
 Thermodynamics, Energy, Heat, Heat Pump

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