Calculating Cost of 1KWhr Heat from a Reversible Heat Pump

In summary, the discussion revolved around the efficiency of a perfectly reversible heat pump, which is calculated using the coefficient of performance (COP). The COP is determined by the temperature difference between the hot and cold sides of the heat pump, and for a Carnot cycle, it is equal to 1 divided by 1 minus the ratio of the cold and hot temperatures. The cost of 1 KWhr of heat supplied by the heat pump can be calculated by dividing 1 by the COP multiplied by the cost of 1 KWhr of electricity. The confusion was caused by mistakenly using the efficiency of the electric motor instead of the efficiency of the heat pump.
  • #1
h.a.y.l.e.y
8
0
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!
 
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  • #2
h.a.y.l.e.y said:
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 [tex]COP[/tex] connected to an electric motor of efficiency [tex]\eta_e[/tex].

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

[tex]COP=\frac{Q_h}{W}=\frac{T_h}{T_h-T_c}[/tex] being [tex]W[/tex] 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:

[tex]COP_n=\frac{Q_h}{W_e}=\frac{Q_h \eta_e }{W}=\eta_e COP[/tex]

so this yield another number.

for [tex]Q_h=1KWh[/tex] the cost of the electrical power which must be supplied is:

[tex]C=\frac{1}{COP_n}c[/tex] being [tex]c[/tex] the cost of 1 KWh of electricity (6p)
 
  • #3
h.a.y.l.e.y said:
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 [itex]\degree C[/itex]

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

[tex]cop = \frac{Q_h}{W} = \frac{Q_h}{Q_h-Q_c} = \frac{1}{1 - Q_c/Q_h}[/tex]

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

[tex]Q_c/Q_h = T_c/T_h[/tex]

so:
[tex]cop = \frac{1}{1 - T_c/T_h}[/tex]

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

What is a reversible heat pump?

A reversible heat pump is a type of heating and cooling system that can both heat and cool a space by transferring heat from one location to another using electricity.

How does a reversible heat pump work?

A reversible heat pump works by using a compressor and a refrigerant to absorb heat from one location, such as the outside air or ground, and transfer it to another location, such as inside a building. This process can be reversed to cool a space by removing heat from inside and transferring it outside.

What factors affect the cost of 1KWhr heat from a reversible heat pump?

The cost of 1KWhr heat from a reversible heat pump can be affected by factors such as the efficiency of the heat pump, the cost of electricity, the outside temperature, and the size and insulation of the space being heated.

How do you calculate the cost of 1KWhr heat from a reversible heat pump?

The cost of 1KWhr heat from a reversible heat pump can be calculated by dividing the cost of electricity per KWhr by the efficiency of the heat pump. For example, if electricity costs $0.10 per KWhr and the heat pump has an efficiency of 3, the cost of 1KWhr heat would be $0.10/3 = $0.033.

Is a reversible heat pump a cost-effective heating option?

The cost-effectiveness of a reversible heat pump depends on various factors such as the cost of electricity and the efficiency of the heat pump. In general, reversible heat pumps can be a cost-effective option compared to traditional heating systems, especially in areas with mild climates where they can also be used for cooling.

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