Question

A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 20.0 kW of electric power (generated by doing work at the rate 20.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 20.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 3.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you get" is heat delivered. So the coefficient of performance of a heat pump is K=QH/Win. An average price for electricity is about 40 MJ per dollar. A furnace or heat pump will run typically 300 hours per month during the winter.
a. How much electric power (in kW) does the heat pump use to deliver 15 kJ/s of heat energy to the house?
b. An average price for electricity is about 40 MJ per dollar. A fumace or heat pump will run typically 200 hours per month during the winter. What does one month's heating cost in the home with a 15kW electric heater and in the home of the neighbor who uses an aquivalent heat pump?

Answer 1

Solution :

A refrigerator is a heat engine running backwards with its main aim is to extract heat from the cold reservoir. When the engine running backwards is used to exhaust or give out hot air into a hot reservoir is known as a heat pump.

Given :

Coefficient of performance of the heat pump, COP = 3

We know COP of heat pump is K  $=\frac{Q_h}{W_{in}}$

a). Therefore the COP can also be written as :

  $COP=\frac{T_h}{T_h-T_c}$

           $=\frac{Q_h}{W_{in}}$

          $=\frac{Q_h}{P_{in}}$

∴ $3=\frac{15}{P_{in}}$

  $P_{in }=\frac{15}{3}$

        = 5 kW

b). The heating cost in a home of one month with a 15 kW electric heater can be calculated as :

  1 dollar = 40 MJ (given)

∴ $1 / 40000 kWh (15 kW)(200 x 3600 s)

    = $ 270