A well-insulated house requires 138 MJ/h to keep the indoor
temperature comfortable on a cold day. Under this load, the heat pump compressor uses 7.7 kW of electrical power.
a.) Determine the COP of the heat pump
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b.) If the heat pump operates 125 hours in a winter month, what will the
homeowner spend on electric heat that month? Residential electricity costs
$0.11/kW-h.
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c.) How much would the
homeowner spend that month if she had electrical resistance heating
instead of a heat pump? |
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Read : |
This is a
straightforward application of the definition of the COP of a heat pump. |
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Given: |
QH |
138 |
MJ/h |
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Find: |
COPHP |
??? |
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WHP
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7.7 |
kW |
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Price |
0.11 |
$/kW-h |
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OpTime |
125 |
h/month |
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Diagram: |
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Assumptions: |
1 - |
The heat pump operates at steady-state. |
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2 - |
There is no loss of efficiency when the heat pump is started up or shut
down by the thermostatic control system. |
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Equations
/ Data / Solve: |
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Begin by writing the
definition for the COP of a heat pump : |
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Eqn 1 |
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We can use Eqn 1 to evaluate the COPHP. Watch the units ! |
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QH |
38.33 |
kW |
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COPHP |
4.98 |
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First let's see how
much it would cost to deliver
QH to the home using an electrical
resistance heater. |
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An electrical
resistance heater, at best, converts all of the electrical work supplied, W, into heat released into the home, QH. Therefore, in order to get 38.33 kW into your home, you must
buy 38.33 kW of electrical power. |
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Wresist |
38.33 |
kW |
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Now, we can determine
how much it would cost to
operate the resistance heater for 125 hr/month. |
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Eqn 2 |
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Costresist |
527.08 |
$/month |
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Now, we can apply Eqn 2 to determine the cost of
operating the heat pump for a month. |
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CostHP |
105.88 |
$/month |
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The difference between
these two costs is the monthly savings: |
Savings |
421.21 |
$/month |
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Verify: |
We cannot verify the steady-state assumption or the
assumption about the thermostatic control system based on the information
given in the problem statement. |
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Answers : |
COPHP |
5.0 |
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Savings |
421 |
$/month |
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Electrical
resistance heaters are not very popular, especially in cold climates. |
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The thermal efficiency of a heat pump drops significantly as the outside temperature falls. |
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When the outside temperature drops far enough that the COPHP ~ 1, it becomes more practical to use the resistance heater ! |
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