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trasher [3.6K]
3 years ago
5

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

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

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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Describe with an example how corroded structures can lead to environment pollution? ​
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5 0
3 years ago
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

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4 0
3 years ago
1). Mention any four operations that requires airlines. 2). Explain how airflow is applicable to the above mentioned operations.
vladimir2022 [97]

Answer:

Following are the answer to this question:

Explanation:

1)

Following the four operations in the airlines:

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It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.

4 0
3 years ago
Prefix version of 6600 volts​
GenaCL600 [577]

Answer:

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Explanation:

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deci (d)   ------  10⁻¹

centi (c)   ------  10⁻²

milli (m)   ------   10⁻³

kilo (k)     ------   10³

mega (M) -----   10⁶

giga (G)   ------   10⁹

We have:

6600 volts

converting to exponential form:

=> 6.6 x 10³ volts

Thus, we know that the prefix of kilo (k) is used for 10³.

Hence,

=> <u>6.6 kilo volts = 6.6 k volts</u>

7 0
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