Question

A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.

a) Calculate the COP of the heat pump.
b) In the summer, the cycle is reversed to cool the house. Calculate the COP of the cycle when it is operated as an air-conditioner assuming the working fluid rejects 15 Btu/s to the Earth.

Answer 1

Answer:

COP of the heat pump is 3.013

OP of the cycle is  1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

     = 15 + 7.45

     = 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

Answer 2

Answer:

• 3.013
• 1.124

Explanation:

W = Q2 - Q1 ----- equation 1

W = work done

Q2 = final energy

Q1 = initial energy

A) calculate the COP of the heat pump

COP = $\frac{Q2}{W}$

from equation 1

Q2 = Q1 + W = 15 + 7.45 = 22.45 KW

therefore COP = $\frac{22.45}{7.45}$ = 3.013

B) COP when cycle is reversed

COP = $\frac{Q1}{W}$

from equation 1

Q1 + W = Q2 ------ equation 2

Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2

Q1 = 8.375 KW

COP = $\frac{8.375}{7.45}$  = 1.124