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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:

Anna007 [38]3 years ago

4 0

Answer:

COP of the heat pump is 3.013

OP of the cycle is 1.124

Explanation:

W = Q₂ - Q₁

Given

Q₂ = Q₁ + W

= 15 + 7.45

= 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

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

GuDViN [60]3 years ago

3 0

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

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Keith_Richards [23]

Answer:

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. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C

MrRissso [65]

Answer:

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Explanation:

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Assumption,

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