# Homework Help: Heat Pumps

1. May 22, 2005

### h.a.y.l.e.y

A perfectly reversible heat pump heats a building at 20deg from the atmosphere at 5deg. If the heat pump is run by an electrical motor whose efficiency is 80%, what is the cost of 1KWhr of heat supplied? (1KWhr of electricity is 6p)

I am confused by this qn as I thought the efficiency, e = 1-(T1/T2), which in this case would be 0.75 and not 0.8?!
Could someone give me a clue on where I'm going wrong please!

2. May 22, 2005

### Clausius2

Let's see. You have a perfect reversible heat pump of coefficent of operation $$COP$$ connected to an electric motor of efficiency $$\eta_e$$.

The coefficient of operation of the heat pump is defined as:

$$COP=\frac{Q_h}{W}=\frac{T_h}{T_h-T_c}$$ being $$W$$ the work exerted on the cooling fluid.

So that you can work out the COP because you have all temperatures. The net COP of the whole would be:

$$COP_n=\frac{Q_h}{W_e}=\frac{Q_h \eta_e }{W}=\eta_e COP$$

so this yield another number.

for $$Q_h=1KWh$$ the cost of the electrical power which must be supplied is:

$$C=\frac{1}{COP_n}c$$ being $$c$$ the cost of 1 KWh of electricity (6p)

3. May 22, 2005

### Andrew Mason

You are confusing motor efficiency with the efficiency or coefficient of performance of the heat pump. They are quite different and independent concepts. Also you have to use K not $\degree C$

The efficiency of the heat pump is given by the coefficient of performance:

$$cop = \frac{Q_h}{W} = \frac{Q_h}{Q_h-Q_c} = \frac{1}{1 - Q_c/Q_h}$$

For a Carnot (reversible) cycle, [itex]\Delta S = Qh/Th - Qc/Tc = 0[/tex] so:

$$Q_c/Q_h = T_c/T_h$$

so:
$$cop = \frac{1}{1 - T_c/T_h}$$

Using K, work out the cop. From that you can determine the amount of heat delivered to the building for each joule of work supplied. Since the motor is 80% efficient, for each joule of electrical energy consumed it delivers .8 J of work to the heat pump.

AM

4. May 22, 2005