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dmitriy555 [2]
3 years ago
15

A food department is kept at 2128C by a refrigerator in an environment at 308C. The total heat gain to the food department is es

timated to be 3300 kJ/h and the heat rejection in the condenser is 4800 kJ/h. Determine the power input to the compressor, in kW and the COP of the refrigerator.
Physics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

2.2

Explanation:

Q_u = Heat rejection in the condenser = 3300 kJ/h

Q_L = Heat gain to the food department = 4800 kJ/h

Power output is given by

W=Q_u-Q_L\\\Rightarrow W=4800-3300\\\Rightarrow W=1500\ kJ/h

COP of a refrigerator is given by

COP=\frac{Desired\ effect}{Work}\\\Rightarrow COP=\frac{Q_L}{W}\\\Rightarrow COP=\frac{3300}{1500}\\\Rightarrow COP=2.2

The COP of the refrigerator is 2.2

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An electric motor has an effective resistance of 29. 4 ω and an inductive reactance of 42. 6 ω. When working under load. the rms
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An electric motor has an effective resistance of 29. 4 ω and an inductive reactance of 42. 6 ω. When working under load. the rms voltage across the alternating source is 442 v.  The rms current will be  8.54 A

AC stands for “Alternating Current,” meaning voltage or current that changes polarity or direction, respectively, over time. AC electromechanical generators, known as alternators, are of simpler construction than DC electromechanical generators.

RMS or root mean square current/voltage of the alternating current/voltage represents the D.C current/voltage that dissipates the same amount of power as the average power dissipated by the alternating current/voltage. For sinusoidal oscillations, the RMS value equals peak value divided by the square root of 2.

I (RMS) = RMS voltage / \sqrt{R^{2}+ X_{L} { ^{2}  }

           =  442 / \sqrt{29.4^{2} + 42.6^{2}  }

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           = 442 / \sqrt{2679.12}

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To learn more about Alternating Current here

brainly.com/question/11673552

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