Electrical machines miscellaneous
- A three-phase, 11 kV, 50 Hz, 2 pole, star connected, cylindrical rotor synchronous motor is connected to an 11 kV, 50 Hz source, Its synchronous reactance is 50 Ω per phase, and i t s st at or resistance is negligible. The motor has a constant field excitation. At a particular load torque, its stator current is 100A at unity power factor. If the load torque is increased so that the stator current is 120 A, then the load angle (in degrees) at this node is _____________.
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Ef = Vt – Ia × S
= 11 kV - j100 × 50 √3
= 6350 – j5000
| Ef | = √(6350)² + (5000)²
= 8082.23
Now, (Ia × S)2 = Ef 2 + Vt 2 – 2Ef Vt cos δ
(120 × 50)2 = 8082.232 + 63502 – 2 × 8082.23 × 6350 × cos δ
∴ δ = – 47.27°
Correct Option: A
Ef = Vt – Ia × S
= 11 kV - j100 × 50 √3
= 6350 – j5000
| Ef | = √(6350)² + (5000)²
= 8082.23
Now, (Ia × S)2 = Ef 2 + Vt 2 – 2Ef Vt cos δ
(120 × 50)2 = 8082.232 + 63502 – 2 × 8082.23 × 6350 × cos δ
∴ δ = – 47.27°
- Two three-phase transformers are realized using single-phase transformers as shown in the figure.
The phase different (in degree) between voltage V1 and V2 is____________.
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Upper transformer secondary is connected in ∆ Bottom transformer secondary is connected in Y Hence, Phase angle between delta voltage & star voltage is 30°.
Correct Option: D
Upper transformer secondary is connected in ∆ Bottom transformer secondary is connected in Y Hence, Phase angle between delta voltage & star voltage is 30°.
- A 220 V, 3-phase, 4-pole, 50 Hz inductor motor of wound rotor type is supplied at rated voltage and frequency. The stator resistance, magnetizing reactance, and core loss are negligible. The maximum torque produced by the rotor is 225% of full load torque and it occurs at 15% slip. The actual rotor resistance is 0.03 Ω/phase. The value of external resistance (in Ohm) which must be inserted in a rotor phase if the maximum torque is to occur at start is ___________.
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Smt = r2 x2 ∴ 0.15 = r2 = 0.03 x2 x2
∴ x2 = 0.2 Ω
For τext = τemax∴ Tset = 2 = 1 Tem 1 + SmT smT
∵ SmT = 1Now , 1 = r2 x2
∴ r2 = x2 = 0.2 Ω
Hence external resistance = 0.2 – 0.03 = 0.17 Ω
Correct Option: B
Smt = r2 x2 ∴ 0.15 = r2 = 0.03 x2 x2
∴ x2 = 0.2 Ω
For τext = τemax∴ Tset = 2 = 1 Tem 1 + SmT smT
∵ SmT = 1Now , 1 = r2 x2
∴ r2 = x2 = 0.2 Ω
Hence external resistance = 0.2 – 0.03 = 0.17 Ω
- The figure shows the per-phase equivalent circuit of a t wo-pol e t hr ee-phase i nduct i on mot or operating at 50 Hz. The “air-gap” voltage, Vg across the magnetizing inductance, is 210V rms, and the slip, is 0.05. The torque (in Nm) produced by the motor is _______.
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VOC across j6.28 Ω = Vg = 210 V
Now from the given circuit,Rth = j6.28 × (0.4 + j0.22) (0.04 + j0.22) + j6.28
= 0.216 ∠ 80.04
= 0.0375 + j0.2127Now , I2 = Vs = 210 z (0.0375 + 1) + j(0.2127 + j0.22)
∴ |I2| = 186.81 Aτ = 3 × I22 . r2 ωs s = 3 × 186.812 × 1 = 333.27 N-m 314.15 Correct Option: C
VOC across j6.28 Ω = Vg = 210 V
Now from the given circuit,Rth = j6.28 × (0.4 + j0.22) (0.04 + j0.22) + j6.28
= 0.216 ∠ 80.04
= 0.0375 + j0.2127Now , I2 = Vs = 210 z (0.0375 + 1) + j(0.2127 + j0.22)
∴ |I2| = 186.81 Aτ = 3 × I22 . r2 ωs s = 3 × 186.812 × 1 = 333.27 N-m 314.15
- A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is 200 radians/ second. At its rated torque of 500 Nm, its speed is 180 radian/second. The motor is used to directly drive a load whose load torque TL depends on its rotational speed (in radians/second), such that TL = 2.78 × ωT. Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, when it drives this load is ______.
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Under steady state condition
load torque = motor torque
∴ 500 = 2.78 × ωT∴ ωT = 500 = 178.88 rad / sec 20 + 8 Correct Option: A
Under steady state condition
load torque = motor torque
∴ 500 = 2.78 × ωT∴ ωT = 500 = 178.88 rad / sec 20 + 8
