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A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.?
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A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor i...
Given:
Power input at no load (Pnl) = 660 kW
Power input at full load (Pfl) = 11.20 kW
Line current at no load (Inl) = 8 A
Line current at full load (Ifl) = 18 A
Stator effective resistance (Rs) = 1.2 ohm
Friction and winding loss (W) = 420 W
Number of poles (p) = 4
Frequency (f) = 50 Hz
Phase voltage (Vp) = 400/√3 = 230.94 V
(i) Stator core loss:
At no load, the stator copper loss is negligible. Therefore, the input power at no load is equal to the stator core loss.
Pnl = Pcore = 660 kW
(ii) Slip at full load:
At full load, the input power can be calculated as:
Pfl = 3VpIflcos(θ)
where cos(θ) is the power factor. Assuming a power factor of 0.8, we get:
Pfl = 3 x 230.94 x 18 x 0.8 = 11.20 kW
The output power can be calculated as:
Pout = Pfl - W = 11.20 - 0.42 = 10.78 kW
The slip can be calculated as:
s = (Pin - Pout)/Pin
where Pin is the input power at full load. Substituting the values, we get:
s = (11.20 - 10.78)/11.20 = 0.037
(iii) Total rotor ohmic losses at full load:
Since the rotor ohmic losses are negligible at no load, we can assume that they are negligible at full load as well. Therefore, the total rotor losses at full load are zero.
(iv) Full load speed:
The synchronous speed can be calculated as:
Ns = 120f/p = 1500 rpm
The actual speed at full load can be calculated as:
N = (1 - s)Ns
Substituting the values, we get:
N = (1 - 0.037) x 1500 = 1448 rpm
Power input at no load (Pnl) = 660 kW
Power input at full load (Pfl) = 11.20 kW
Line current at no load (Inl) = 8 A
Line current at full load (Ifl) = 18 A
Stator effective resistance (Rs) = 1.2 ohm
Friction and winding loss (W) = 420 W
Number of poles (p) = 4
Frequency (f) = 50 Hz
Phase voltage (Vp) = 400/√3 = 230.94 V
(i) Stator core loss:
At no load, the stator copper loss is negligible. Therefore, the input power at no load is equal to the stator core loss.
Pnl = Pcore = 660 kW
(ii) Slip at full load:
At full load, the input power can be calculated as:
Pfl = 3VpIflcos(θ)
where cos(θ) is the power factor. Assuming a power factor of 0.8, we get:
Pfl = 3 x 230.94 x 18 x 0.8 = 11.20 kW
The output power can be calculated as:
Pout = Pfl - W = 11.20 - 0.42 = 10.78 kW
The slip can be calculated as:
s = (Pin - Pout)/Pin
where Pin is the input power at full load. Substituting the values, we get:
s = (11.20 - 10.78)/11.20 = 0.037
(iii) Total rotor ohmic losses at full load:
Since the rotor ohmic losses are negligible at no load, we can assume that they are negligible at full load as well. Therefore, the total rotor losses at full load are zero.
(iv) Full load speed:
The synchronous speed can be calculated as:
Ns = 120f/p = 1500 rpm
The actual speed at full load can be calculated as:
N = (1 - s)Ns
Substituting the values, we get:
N = (1 - 0.037) x 1500 = 1448 rpm
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A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.?
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A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load 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 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.?.
A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load 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 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.?.
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A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.?, a detailed solution for A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.? has been provided alongside types of A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.? theory, EduRev gives you an
ample number of questions to practice A 10kw, 400V, 3-phase, 4-pole, 50 HZ delta connected induction motor is running at no lode with a line current of 8A and an input power is 660 kW. At full load, the line current is 18 A and the input power is 11.20 kW. Stator effective resistance per phase is 1.2 ohm and friction, winding loss is 420 watts. For negligible rotor ohmic losses at no load, calculate (i) stator core loss (ii) slip at full load (iii) total rotor ohmic losses at full load (iv) full load speed.? tests, examples and also practice Electrical Engineering (EE) tests.
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