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The ratio of starting to full load current for a 15 kW, 300 V and 3ϕ induction motor with DOL starter is _________. For 3 phase inductor motor has full load efficiency 0.90 the full load PF is 0.8 and short-circuit current is 30A at 100V.
Correct answer is between '2.2,2.3'. Can you explain this answer?
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Verified Answer
The ratio of starting to full load current for a 15 kW, 300 V and 3 ...
Ist = 30 A
Starting current when it work at 300V
Ist = 30 × 3 = 90 A
∴ Ratio of starting current to full load current
= 90/40.09 = 2.244
Starting current when it work at 300V
Ist = 30 × 3 = 90 A
∴ Ratio of starting current to full load current
= 90/40.09 = 2.244
Most Upvoted Answer
The ratio of starting to full load current for a 15 kW, 300 V and 3 ...
Given data:
- Power rating of the motor (P) = 15 kW
- Voltage rating of the motor (V) = 300 V
- Number of phases (N) = 3
- Full load efficiency (η) = 0.90
- Power factor (PF) = 0.8
- Short-circuit current (Isc) = 30 A
- Short-circuit voltage (Vsc) = 100 V
1. Finding the full load current (Ifl):
The full load current (Ifl) can be calculated using the formula:
Ifl = P / (V * √3)
Substituting the given values:
Ifl = 15,000 W / (300 V * √3)
Ifl ≈ 27.39 A
2. Finding the starting current (Istart):
The starting current (Istart) for a direct-on-line (DOL) starter is typically 6 to 7 times the full load current (Ifl). Therefore, we can assume Istart = 6 * Ifl.
Istart ≈ 6 * 27.39 A
Istart ≈ 164.34 A
3. Finding the ratio of starting to full load current:
The ratio of starting to full load current is given by:
Ratio = Istart / Ifl
Substituting the values:
Ratio ≈ 164.34 A / 27.39 A
Ratio ≈ 5.997
4. Analyzing the correct answer:
The correct answer is stated to be between 2.2 and 2.3. However, the calculated ratio is approximately 5.997, which is not within the given range. Therefore, the given answer is incorrect.
Conclusion:
The ratio of starting to full load current for the 15 kW, 300 V, 3-phase induction motor with a DOL starter is approximately 5.997. The correct answer is not between 2.2 and 2.3, as stated.
- Power rating of the motor (P) = 15 kW
- Voltage rating of the motor (V) = 300 V
- Number of phases (N) = 3
- Full load efficiency (η) = 0.90
- Power factor (PF) = 0.8
- Short-circuit current (Isc) = 30 A
- Short-circuit voltage (Vsc) = 100 V
1. Finding the full load current (Ifl):
The full load current (Ifl) can be calculated using the formula:
Ifl = P / (V * √3)
Substituting the given values:
Ifl = 15,000 W / (300 V * √3)
Ifl ≈ 27.39 A
2. Finding the starting current (Istart):
The starting current (Istart) for a direct-on-line (DOL) starter is typically 6 to 7 times the full load current (Ifl). Therefore, we can assume Istart = 6 * Ifl.
Istart ≈ 6 * 27.39 A
Istart ≈ 164.34 A
3. Finding the ratio of starting to full load current:
The ratio of starting to full load current is given by:
Ratio = Istart / Ifl
Substituting the values:
Ratio ≈ 164.34 A / 27.39 A
Ratio ≈ 5.997
4. Analyzing the correct answer:
The correct answer is stated to be between 2.2 and 2.3. However, the calculated ratio is approximately 5.997, which is not within the given range. Therefore, the given answer is incorrect.
Conclusion:
The ratio of starting to full load current for the 15 kW, 300 V, 3-phase induction motor with a DOL starter is approximately 5.997. The correct answer is not between 2.2 and 2.3, as stated.
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The ratio of starting to full load current for a 15 kW, 300 V and 3ϕ induction motor with DOL starter is _________. For 3 phase inductor motor has full load efficiency 0.90 the full load PF is 0.8 and short-circuit current is 30A at 100V.Correct answer is between '2.2,2.3'. Can you explain this answer?
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The ratio of starting to full load current for a 15 kW, 300 V and 3ϕ induction motor with DOL starter is _________. For 3 phase inductor motor has full load efficiency 0.90 the full load PF is 0.8 and short-circuit current is 30A at 100V.Correct answer is between '2.2,2.3'. 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 The ratio of starting to full load current for a 15 kW, 300 V and 3ϕ induction motor with DOL starter is _________. For 3 phase inductor motor has full load efficiency 0.90 the full load PF is 0.8 and short-circuit current is 30A at 100V.Correct answer is between '2.2,2.3'. 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 The ratio of starting to full load current for a 15 kW, 300 V and 3ϕ induction motor with DOL starter is _________. For 3 phase inductor motor has full load efficiency 0.90 the full load PF is 0.8 and short-circuit current is 30A at 100V.Correct answer is between '2.2,2.3'. Can you explain this answer?.
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