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A 2 pole, 50 Hz induction motor supplies 15 kW to a load at a speed of 2950 rotation/min. How much power (In kW) will be supplied by the motor when the torque is doubled.
Correct answer is between '29.4,29.6'. Can you explain this answer?
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Verified Answer
A 2 pole, 50 Hz induction motor supplies 15 kW to a load at a speed of...
The synchronous speed of this motor is
= 48.6 N- m
In the low-slip region, the torque-speed curve is linear, and induced torque is directly proportional to slip. Therefore, if torque double, then the new slip will be 3.33 %. The operating speed of the motor is thus
Nr = (1 – .033) Nsyn = (1 -0.033) 3000 = 2900 r/min
∴ Power supplied = τind ωn
Most Upvoted Answer
A 2 pole, 50 Hz induction motor supplies 15 kW to a load at a speed of...
Calculating the Original Power:
$$P_{loss} = 3I^{2}R_{s} = 3 \times 27.38
- Number of poles (p) = 2
- Frequency (f) = 50 Hz
- Speed (N) = 2950 rpm
First, we need to calculate the synchronous speed of the motor:
$$N_{s} = \frac{120f}{p}$$
$$N_{s} = \frac{120 \times 50}{2} = 3000 \text{ rpm}$$
The actual speed of the motor is given as 2950 rpm, which means the motor is running at a slip of:
$$s = \frac{N_{s} - N}{N_{s}}$$
$$s = \frac{3000 - 2950}{3000} = 0.0167$$
The power output of the motor can be calculated as:
$$P_{out} = P_{in} - P_{loss}$$
Where,
$$P_{in} = 3VI\cos\phi$$
$$P_{loss} = 3I^{2}R_{s}$$
Assuming an efficiency of 90%, we can calculate the input power:
$$P_{in} = \frac{P_{out}}{\text{Efficiency}} = \frac{15 \text{ kW}}{0.9} = 16.67 \text{ kW}$$
Now, we need to calculate the current flowing through the motor:
$$I = \frac{P_{in}}{3V\cos\phi}$$
Assuming a power factor of 0.85, we get:
$$I = \frac{16.67 \times 1000}{3 \times 415 \times 0.85} = 27.38 \text{ A}$$
The total resistance of the motor can be calculated as:
$$R_{s} = \frac{V^{2}}{P_{in}} - \frac{X_{s}^{2}}{P_{in}}$$
Where,
$$X_{s} = \frac{V}{2\pi f}\text{cos}^{-1}(\text{Power Factor})$$
$$X_{s} = \frac{415}{2\pi \times 50}\text{cos}^{-1}(0.85) = 2.30 \text{ }\Omega$$
$$R_{s} = \frac{415^{2}}{16.67 \times 1000} - \frac{2.30^{2}}{16.67 \times 1000} = 0.95 \text{ }\Omega$$
Finally, we can calculate the power loss in the motor:
$$P_{loss} = 3I^{2}R_{s} = 3 \times 27.38
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A 2 pole, 50 Hz induction motor supplies 15 kW to a load at a speed of 2950 rotation/min. How much power (In kW) will be supplied by the motor when the torque is doubled.Correct answer is between '29.4,29.6'. Can you explain this answer?
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A 2 pole, 50 Hz induction motor supplies 15 kW to a load at a speed of 2950 rotation/min. How much power (In kW) will be supplied by the motor when the torque is doubled.Correct answer is between '29.4,29.6'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 2 pole, 50 Hz induction motor supplies 15 kW to a load at a speed of 2950 rotation/min. How much power (In kW) will be supplied by the motor when the torque is doubled.Correct answer is between '29.4,29.6'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 pole, 50 Hz induction motor supplies 15 kW to a load at a speed of 2950 rotation/min. How much power (In kW) will be supplied by the motor when the torque is doubled.Correct answer is between '29.4,29.6'. Can you explain this answer?.
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