Calculating Cost of 1KWhr Heat from a Reversible Heat Pump

[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!

 

[Clausius2]

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!

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)

 

[Andrew Mason]

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!

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, \Delta S = Qh/Th - Qc/Tc = 0[/tex] so:<br /> <br /> Q_c/Q_h = T_c/T_h<br /> <br /> so:<br /> cop = \frac{1}{1 - T_c/T_h}<br /> <br /> 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.<br /> <br /> AM

 

[h.a.y.l.e.y]

V Helpful, I can now get the correct answer!
Thanks for your time guys.