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andrew-mc [135]
3 years ago
14

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:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

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Explanation:

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7. P = 250 , θ = 20°

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8. P = 8 , θ = (-42)°

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9. P = 35 , θ = (-65)°

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y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

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