Example Problem with Complete Solution

6B-1 : Home Heat Pump COP 6 pts
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
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.
c.) How much would the homeowner spend that month if she had electrical resistance heating instead of a heat pump?
Read : This is a straightforward application of the definition of the COP of a heat pump.
Given: QH 138 MJ/h Find: COPHP ???
WHP = 7.7 kW
Price 0.11 $/kW-h
OpTime 125 h/month
Assumptions: 1 - The heat pump operates at steady-state.
2 - There is no loss of efficiency when the heat pump is started up or shut down by the thermostatic control system.
Equations / Data / Solve:
Begin by writing the definition for the COP of a heat pump :
Eqn 1
We can use Eqn 1 to evaluate the COPHP.  Watch the units !
QH 38.33 kW COPHP 4.98
First let's see how much it would cost to deliver QH to the home using an electrical resistance heater.
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.
Wresist 38.33 kW
Now, we can determine how much it would cost to operate the resistance heater for 125 hr/month.
Eqn 2
Costresist 527.08 $/month
Now, we can apply Eqn 2 to determine the cost of operating the heat pump for a month.
CostHP 105.88 $/month
The difference between these two costs is the monthly savings: Savings 421.21 $/month
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.
Answers : COPHP 5.0 Savings 421 $/month
Electrical resistance heaters are not very popular, especially in cold climates.
The thermal efficiency of a heat pump drops significantly as the outside temperature falls.
When the outside temperature drops far enough that the COPHP ~ 1, it becomes more practical to use the resistance heater !