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3 years ago

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.

Engineering

2 answers:

Oxana [17]3 years ago

6 0

Answer:

COP of heat pump=3.013

COP of cycle=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 =Q2/W

from equation 1

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

therefore COP =22.45/7.45 = 3.013

B) COP when cycle is reversed

COP = 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 =8.375/7.45 = 1.124

Verizon [17]3 years ago

4 0

Answer:

a) COP = 3.01

b) COP = 1.12

Explanation:

a) The heat is:

Q₂ = Q₁ + W = 15 + 7.45 = 22.45 kW

the COP of the heat pump is

COP = Q₂/W = 22.45/7.45 = 3.01

b) Operating as an air conditioner, the heat is

Q₁ = Q₂ - W = 8.375 kW

The COP is:

COP = Q₁/W = 8.375/7.45 = 1.12

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Answer:

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Explanation:

Given that Vacuum gauge pressure= 30 KPa

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We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.

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Density of $Hg=13600 \frac{kg}{m^3}$

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We know that

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But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

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