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MrRa [10]
2 years ago
13

Why the power factor is Low in no load test in induction motor ?​

Engineering
1 answer:
marshall27 [118]2 years ago
5 0

Answer:

At no load, the Cosine of the phase angle between the stator current(I1) and the stator applied voltage(V1) is the power factor of an induction motor. The reason of the low power factor is high rotor reactance at lower loading. The rotor reactance value gets decreased with increased loading on induction motor..','.

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A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co
podryga [215]

Answer:

a) T_{H} = 1.967\,^{\circ}F, b) COP_{R} = 9.105, c) T_{H} = 115.934\,^{\circ}F, d) COP_{R} = 6.995, e) T_{H} = 25.129\,^{\circ}C

Explanation:

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COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}

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10\cdot T_{H} - 4196.7 = 419.67

T_{H} = 461.637\,R

T_{H} = 1.967\,^{\circ}F

b) The coefficient of performance is:

COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}

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c) The temperature of the hot reservoir can be determined with the help of the following relation:

\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}

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5 = \frac{479.67\,R}{T_{H}-479.67\,R}

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COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}

COP_{R} = 6.995

e) The temperature of the cold reservoir is:

8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}

8.9\cdot T_{H} - 2386.535 = 268.15

T_{H} = 298.279\,K

T_{H} = 25.129\,^{\circ}C

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