10 kW, 6-pole, 50 Hz, 3- phase induction motor has linear torque-slip characteristicsbetween zero torque and maximum torque. The slip at which maximum torque of 520 Nmoccurs is 0.2. For mechanical losses of 600 W, find the speed at which motor would run whiledelivering rated shaft power
Question
10 kW, 6-pole, 50 Hz, 3- phase induction motor has linear torque-slip characteristicsbetween zero torque and maximum torque. The slip at which maximum torque of 520 Nmoccurs is 0.2. For mechanical losses of 600 W, find the speed at which motor would run whiledelivering rated shaft power
Solution 1
To solve this problem, we need to follow these steps:
- First, we need to calculate the synchronous speed of the motor. The formula for
Solution 2
To solve this problem, we need to follow these steps:
-
First, we need to calculate the synchronous speed of the motor. The formula for synchronous speed is Ns = 120f/P, where f is the frequency and P is the number of poles. Substituting the given values, we get Ns = 120*50/6 = 1000 RPM.
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The slip at which maximum torque occurs is given as 0.2. So, the speed at maximum torque, Nm, can be calculated as Nm = Ns*(1-s), where s is the slip. Substituting the given values, we get Nm = 1000*(1-0.2) = 800 RPM.
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The mechanical power output at maximum torque can be calculated using the formula Pm = TN/60, where T is the torque and N is the speed. Substituting the given values, we get Pm = 520800/60 = 6933.33 W.
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The mechanical losses are given as 600 W. So, the net mechanical power output, Pn, can be calculated as Pn = Pm - mechanical losses. Substituting the given values, we get Pn = 6933.33 - 600 = 6333.33 W.
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The rated shaft power is given as 10 kW or 10000 W. The speed at which the motor would run while delivering rated shaft power, Nr, can be calculated using the formula Nr = Pn60/T. Substituting the given values, we get Nr = 6333.3360/520 = 730.77 RPM.
So, the motor would run at approximately 731 RPM while delivering rated shaft power.
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