Electrical machines miscellaneous
- A separately excited 300 V DC shunt motor under no load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 Ω. and leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then the speed in rpm is _____.
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Given 300 V1
N1 = 900 rpm
Ra = 0.5 Ω
Ia1 = 2 A;
Ia2 = 15 A
∴ Eb1 = 300 – 2 × 0.5 = 299 V
Eb2 = 300 – 15 × 0.5 = 292.5 V∴ N2 = Eb2 N1 Eb1 ⇒ N2 = 900 × 2925 = 880 rpm 29 Correct Option: B
Given 300 V1
N1 = 900 rpm
Ra = 0.5 Ω
Ia1 = 2 A;
Ia2 = 15 A
∴ Eb1 = 300 – 2 × 0.5 = 299 V
Eb2 = 300 – 15 × 0.5 = 292.5 V∴ N2 = Eb2 N1 Eb1 ⇒ N2 = 900 × 2925 = 880 rpm 29
- A non-salient pole synchronous generator having synchronous reactance of 0.8 pu is supplying 1 pu power to a unity power factor load at a terminal volt age of 1.1 pu. Neglecting the armature resistance, the angle of the voltage behind the synchronous reactance with respect to the angle of the terminal voltage in degrees is ________.
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Given, P = 1 P.u ;
Vt = 1.1 P.u ;
Xs = 0.8 P.u
Pf = 1∴ Ia = P = 1 = 0.91 V cos φ 1.1 × 1
∴ E = √(V cos φ + Ia Ra)² + (V sin φ + Ia Xs)²
E = √(1.1 × 1 + 0)² + (0 + 0.91 × 0.8)² = 1.314 V
∴ From power equation,P = E V sin δ Xs ⇒ 1 = 1.314 × 1.1 sin δ 0.8
⇒ δ = 33.61 °
Correct Option: A
Given, P = 1 P.u ;
Vt = 1.1 P.u ;
Xs = 0.8 P.u
Pf = 1∴ Ia = P = 1 = 0.91 V cos φ 1.1 × 1
∴ E = √(V cos φ + Ia Ra)² + (V sin φ + Ia Xs)²
E = √(1.1 × 1 + 0)² + (0 + 0.91 × 0.8)² = 1.314 V
∴ From power equation,P = E V sin δ Xs ⇒ 1 = 1.314 × 1.1 sin δ 0.8
⇒ δ = 33.61 °
- The no-load speed of a 230 V separately excited dc motor is 1400 rpm. The armature resistance drop and the brush drop are neglected. The field current is kept constant at rated value. The torque of the motor in Nm for an armature current of 8 A is ____________.
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Given, V = 230V, N =1400rpm
Ra = 0 , Ia = 8A∴ Torque , T = V . Ia = 230 × 8 = 12.5 Nm W 2π × 1400 60
Correct Option: C
Given, V = 230V, N =1400rpm
Ra = 0 , Ia = 8A∴ Torque , T = V . Ia = 230 × 8 = 12.5 Nm W 2π × 1400 60
- A synchronous generator is connected to an infinite bus with excitation voltage Ef = 1.3 pu. The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6 pu to the bus. Assume the infinite bus voltage to be 1.0 pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is _________.
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P = Ef Vt sin δ Xs ⇒ 0.6 = 1.3 × 1 sin δ 1.1
⇒ sin δ = 0.5077
⇒ δ = 30.5°
Reactive power Q (in p.u.) supplied by the generator= Ef Vt cos δ - Vt2 Xs Xs = 1.3 × 1 × 0.8616 - 1 = 0.109 p.u. 1.1 1.1 Correct Option: D
P = Ef Vt sin δ Xs ⇒ 0.6 = 1.3 × 1 sin δ 1.1
⇒ sin δ = 0.5077
⇒ δ = 30.5°
Reactive power Q (in p.u.) supplied by the generator= Ef Vt cos δ - Vt2 Xs Xs = 1.3 × 1 × 0.8616 - 1 = 0.109 p.u. 1.1 1.1
- A single phase induction motor draws 12 MW power at 0.6 lagging power. A capacitor is connected in parallel to the motor to improve the power factor of the combination of motor and capacitor to 0.8 lagging. Assuming that the real and reactive power drawn by the motor remains same as before, the reactive power delivered by the capacitor in MAR is __________.
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P1 = 12 MW
cos φ1 = 0.6
Q1 = P1 tan φ1= P1 
8 
= 16 MVAR 6 
When p.f. becomes 0.8
⇒ cos θ2 = 0.8
Then, Q2 = P1 tan φ2 [real power remain same]= 12 6 = 9 MVAR 8
Since, reactive power drawn by motor remain the same, reactive power supplied by capacitor
QC = Q1 – Q2 = 16 – 9 = 7 MVAR
Correct Option: B
P1 = 12 MW
cos φ1 = 0.6
Q1 = P1 tan φ1= P1 8 = 16 MVAR 6
When p.f. becomes 0.8
⇒ cos θ2 = 0.8
Then, Q2 = P1 tan φ2 [real power remain same]= 12 6 = 9 MVAR 8
Since, reactive power drawn by motor remain the same, reactive power supplied by capacitor
QC = Q1 – Q2 = 16 – 9 = 7 MVAR
