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10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.?
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10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when deliveri...
Given data:
- Power output (P) = 10 kW
- Voltage (V) = 250 V
- Speed (N) = 1000 rpm
- Number of poles (P) = 6
- Number of armature conductors (C) = 534
- Full load copper loss (Pcl) = 0.64 kW
- Total brush drop (Vb) = 1 V
Calculating flux per pole:
1. The power output of the generator can be calculated using the formula:
P = (V * I) - Pcl
where I is the armature current.
2. Rearranging the equation, we get:
I = (P + Pcl) / V
3. The armature current (Ia) can be calculated by dividing the power output by the voltage:
Ia = P / V
4. The flux per pole (Φ) can be calculated using the formula:
Φ = (V - Ia * Ra) / (N * Z * P)
where Ra is the armature resistance and Z is the number of parallel paths.
5. The armature resistance (Ra) can be calculated using the formula:
Ra = (V - Vb) / Ia
6. Substituting the values in the formula, we get:
Ra = (250 - 1) / (P / V)
7. The number of parallel paths (Z) can be calculated using the formula:
Z = C / 2
8. Substituting the values in the formula, we get:
Z = 534 / 2
9. Finally, substituting all the values in the formula for flux per pole, we get:
Φ = (250 - [(250 - 1) / (P / V)] * [(534 / 2) * 6]) / (1000 * (534 / 2) * 6)
Answer:
The flux per pole in the generator is determined to be Φ = 0.0028 Wb.
- Power output (P) = 10 kW
- Voltage (V) = 250 V
- Speed (N) = 1000 rpm
- Number of poles (P) = 6
- Number of armature conductors (C) = 534
- Full load copper loss (Pcl) = 0.64 kW
- Total brush drop (Vb) = 1 V
Calculating flux per pole:
1. The power output of the generator can be calculated using the formula:
P = (V * I) - Pcl
where I is the armature current.
2. Rearranging the equation, we get:
I = (P + Pcl) / V
3. The armature current (Ia) can be calculated by dividing the power output by the voltage:
Ia = P / V
4. The flux per pole (Φ) can be calculated using the formula:
Φ = (V - Ia * Ra) / (N * Z * P)
where Ra is the armature resistance and Z is the number of parallel paths.
5. The armature resistance (Ra) can be calculated using the formula:
Ra = (V - Vb) / Ia
6. Substituting the values in the formula, we get:
Ra = (250 - 1) / (P / V)
7. The number of parallel paths (Z) can be calculated using the formula:
Z = C / 2
8. Substituting the values in the formula, we get:
Z = 534 / 2
9. Finally, substituting all the values in the formula for flux per pole, we get:
Φ = (250 - [(250 - 1) / (P / V)] * [(534 / 2) * 6]) / (1000 * (534 / 2) * 6)
Answer:
The flux per pole in the generator is determined to be Φ = 0.0028 Wb.
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10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.?
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10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.? 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 10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.?.
10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.? 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 10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 kW, 250 V dc, 6 pole shunt generator runs at 1000 rpm when delivering full load. The armature has 534 lap connected conductors. Full load copper loss is 0.64 kW. The total brush drop is 1 volt. Determine the flux per pole. Neglect shunt current.?.
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