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A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed?
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A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armatur...
Field Current Calculation
To find the field current (If), use Ohm's law:
- Voltage across the field winding (Vf) = 220 V (supply voltage)
- Field resistance (Rf) = 110 Ω
If = Vf / Rf
If = 220 V / 110 Ω = 2 A
Copper Losses Calculation
Copper losses occur in both the armature and the field windings:
- Armature current (Ia) = 38 A
- Armature resistance (Ra) = 0.5 Ω
Armature copper losses (Pcu_armature) = Ia^2 * Ra
Pcu_armature = (38 A)^2 * 0.5 Ω = 722 W
Field copper losses (Pcu_field) = If^2 * Rf
Pcu_field = (2 A)^2 * 110 Ω = 440 W
Total copper losses (Pcu_total) = Pcu_armature + Pcu_field
Pcu_total = 722 W + 440 W = 1162 W
Speed Calculation
The output power (Po) can be calculated as:
- Input power (Pin) = 8.8 kW = 8800 W
- Iron and friction losses = 400 W
- Total losses = Copper losses + Iron and friction losses
- Total losses = 1162 W + 400 W = 1562 W
Output power (Po) = Pin - Total losses
Po = 8800 W - 1562 W = 7238 W
Now, use the motor’s equation to find speed (N):
- Total flux (Φ) = 100 mWb = 0.1 Wb
- Number of poles (P) = 6
- Armature conductors (Z) = 1300
N = (60 * Po) / (Φ * Z * Ia / P)
N = (60 * 7238 W) / (0.1 Wb * 1300 * 38 A / 6)
N ≈ 913.6 RPM
Summary
- Field Current: 2 A
- Total Copper Losses: 1162 W
- Speed: Approximately 914 RPM
To find the field current (If), use Ohm's law:
- Voltage across the field winding (Vf) = 220 V (supply voltage)
- Field resistance (Rf) = 110 Ω
If = Vf / Rf
If = 220 V / 110 Ω = 2 A
Copper Losses Calculation
Copper losses occur in both the armature and the field windings:
- Armature current (Ia) = 38 A
- Armature resistance (Ra) = 0.5 Ω
Armature copper losses (Pcu_armature) = Ia^2 * Ra
Pcu_armature = (38 A)^2 * 0.5 Ω = 722 W
Field copper losses (Pcu_field) = If^2 * Rf
Pcu_field = (2 A)^2 * 110 Ω = 440 W
Total copper losses (Pcu_total) = Pcu_armature + Pcu_field
Pcu_total = 722 W + 440 W = 1162 W
Speed Calculation
The output power (Po) can be calculated as:
- Input power (Pin) = 8.8 kW = 8800 W
- Iron and friction losses = 400 W
- Total losses = Copper losses + Iron and friction losses
- Total losses = 1162 W + 400 W = 1562 W
Output power (Po) = Pin - Total losses
Po = 8800 W - 1562 W = 7238 W
Now, use the motor’s equation to find speed (N):
- Total flux (Φ) = 100 mWb = 0.1 Wb
- Number of poles (P) = 6
- Armature conductors (Z) = 1300
N = (60 * Po) / (Φ * Z * Ia / P)
N = (60 * 7238 W) / (0.1 Wb * 1300 * 38 A / 6)
N ≈ 913.6 RPM
Summary
- Field Current: 2 A
- Total Copper Losses: 1162 W
- Speed: Approximately 914 RPM
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A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed?
Question Description
A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed? 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 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed?.
A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed? 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 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed?.
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A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed?, a detailed solution for A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed? has been provided alongside types of A 6 pole, 220 V, lap-connected DC shunt motor has totally 1300 armature conductors . The armature current is 38 A and total flux for all the poles is 100 mWb. The armature and field resistances are 0.5 Ω and 110 Ω, respectively. The input power is 8.8 KW. The iron and friction losses are 400 W. Calculate the field current , copper losses and speed? theory, EduRev gives you an
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