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A separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________
- a)224
- b)220
- c)228
- d)235
Correct answer is option 'C'. Can you explain this answer?
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A separately excited DC motor fed from 3-phase semi converter develop ...
Concept:
Calculation:
We know that
VPh = 228 V
VPh = 228 V
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A separately excited DC motor fed from 3-phase semi converter develop ...
Degrees. The armature resistance and inductance are 0.2 ohms and 1.5 mH respectively. The field resistance is 100 ohms. The armature voltage is 220 V. Find:
a) The armature current at full load.
b) The field current.
c) The power input to the motor.
d) The power developed by the motor.
e) The efficiency of the motor.
Solution:
a) The armature current can be found using the equation:
T = (K * φ * Ia) / (Ra + Rf + Re)
Where,
T = Full load torque = K * φ * Ia
K = Motor constant
φ = Flux per pole
Ia = Armature current
Ra = Armature resistance
Rf = Field resistance
Re = External resistance (assumed to be zero in this case)
Rearranging the equation, we get:
Ia = (T * (Ra + Rf + Re)) / (K * φ)
At full load, T = 150 Nm
K = 1.25
φ = (Vt - Eb) / (K * N)
Vt = Armature voltage = 220 V
Eb = Back emf at 1500 rpm = (K * φ * N) = 180 V (approx.)
N = Speed = 1500 rpm
φ = (220 - 180) / (1.25 * 1500) = 0.02 Wb/pole
Substituting the values, we get:
Ia = (150 * (0.2 + 100 + 0)) / (1.25 * 0.02) = 1,500 A
Therefore, the armature current at full load is 1,500 A.
b) The field current can be found using the equation:
If = Vf / Rf
Where,
Vf = Field voltage
At full load, the armature voltage is constant at 220 V. Therefore, the field voltage can be found using the equation:
Vf = Vt - Eb = 220 - 180 = 40 V
Substituting the values, we get:
If = 40 / 100 = 0.4 A
Therefore, the field current is 0.4 A.
c) The power input to the motor can be found using the equation:
Pin = Vt * Ia
Substituting the values, we get:
Pin = 220 * 1,500 = 330,000 W
Therefore, the power input to the motor is 330 kW.
d) The power developed by the motor can be found using the equation:
Pout = T * ω
Where,
ω = Angular speed = 2π * N / 60
Substituting the values, we get:
ω = 2π * 1500 / 60 = 157.08 rad/s
Pout = 150 * 157.08 = 23,562 W
Therefore, the power developed by the motor is 23.562 kW.
e) The efficiency of the motor can be found using the equation:
Efficiency = (Pout / Pin) * 100
Substituting the values, we get:
Efficiency = (23.562 / 330) * 100 = 7.14
a) The armature current at full load.
b) The field current.
c) The power input to the motor.
d) The power developed by the motor.
e) The efficiency of the motor.
Solution:
a) The armature current can be found using the equation:
T = (K * φ * Ia) / (Ra + Rf + Re)
Where,
T = Full load torque = K * φ * Ia
K = Motor constant
φ = Flux per pole
Ia = Armature current
Ra = Armature resistance
Rf = Field resistance
Re = External resistance (assumed to be zero in this case)
Rearranging the equation, we get:
Ia = (T * (Ra + Rf + Re)) / (K * φ)
At full load, T = 150 Nm
K = 1.25
φ = (Vt - Eb) / (K * N)
Vt = Armature voltage = 220 V
Eb = Back emf at 1500 rpm = (K * φ * N) = 180 V (approx.)
N = Speed = 1500 rpm
φ = (220 - 180) / (1.25 * 1500) = 0.02 Wb/pole
Substituting the values, we get:
Ia = (150 * (0.2 + 100 + 0)) / (1.25 * 0.02) = 1,500 A
Therefore, the armature current at full load is 1,500 A.
b) The field current can be found using the equation:
If = Vf / Rf
Where,
Vf = Field voltage
At full load, the armature voltage is constant at 220 V. Therefore, the field voltage can be found using the equation:
Vf = Vt - Eb = 220 - 180 = 40 V
Substituting the values, we get:
If = 40 / 100 = 0.4 A
Therefore, the field current is 0.4 A.
c) The power input to the motor can be found using the equation:
Pin = Vt * Ia
Substituting the values, we get:
Pin = 220 * 1,500 = 330,000 W
Therefore, the power input to the motor is 330 kW.
d) The power developed by the motor can be found using the equation:
Pout = T * ω
Where,
ω = Angular speed = 2π * N / 60
Substituting the values, we get:
ω = 2π * 1500 / 60 = 157.08 rad/s
Pout = 150 * 157.08 = 23,562 W
Therefore, the power developed by the motor is 23.562 kW.
e) The efficiency of the motor can be found using the equation:
Efficiency = (Pout / Pin) * 100
Substituting the values, we get:
Efficiency = (23.562 / 330) * 100 = 7.14
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A separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________a)224b)220c)228d)235Correct answer is option 'C'. Can you explain this answer?
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A separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________a)224b)220c)228d)235Correct answer is option 'C'. 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 separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________a)224b)220c)228d)235Correct answer is option 'C'. 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 separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________a)224b)220c)228d)235Correct answer is option 'C'. Can you explain this answer?.
A separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________a)224b)220c)228d)235Correct answer is option 'C'. 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 separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________a)224b)220c)228d)235Correct answer is option 'C'. 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 separately excited DC motor fed from 3-phase semi converter develop a full load torque at 1500 rpm when firing angle is 60°. The armature taking 50 A at 400 V DC and having the resistance of 0.4 Ω, then supply voltage per phase will be ________a)224b)220c)228d)235Correct answer is option 'C'. Can you explain this answer?.
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