Electrical machines miscellaneous
- A 300 kVA, 3300/400 V, three-phase transformer has its primary and secondary windings connected in delta for a dummy load test, for circulating full-load current, the magnitude of the injected voltage in open delta of HV windings is 360 V. The leakage impedance in per -unit system is _______
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Voltage across each phase of h.v. winding
= 360 = 120 V = VSC 3
VPh = 3300 V ;IPh = 300 × 103 = 100 = 30.3 amp. 3 × 3300 3.3 ZeH = VSC = 120 = 3.96 Ω ISC 30.3 Zbase = Vbase = 3300 = 108.9 Ω Ibase 30.3 ∴ ZeH (p.u.) = 3.96 = 0.0364 p.u. 108.9
Correct Option: D
Voltage across each phase of h.v. winding
= 360 = 120 V = VSC 3
VPh = 3300 V ;IPh = 300 × 103 = 100 = 30.3 amp. 3 × 3300 3.3 ZeH = VSC = 120 = 3.96 Ω ISC 30.3 Zbase = Vbase = 3300 = 108.9 Ω Ibase 30.3 ∴ ZeH (p.u.) = 3.96 = 0.0364 p.u. 108.9
- The self inductance of the primary winding of a single phase, 50 Hz, transformer is 800 mH, and that of the secondary winding is 600 mH. The mutual inductance between these two windings i s 480 mH. The secondar y winding of this transformer is short circuited and the primary winding is connected to a 50 Hz, single phase, sinusoidal voltage source. The current flowing in both the winding is less than their respective rated currents. The resistance of both windings can be neglected. In this connection, what is the effective inductance (in mH) seen by the source?
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Zin = V1 = (R1 + jX1) + ω2 M2 I1 R2 + jX2 + ZL
Given, L1 = 800 mH
L2 = 600 mH
M = 480 mH
W = 314 rad/sec
ZL = 0
R1 R2 neglected∴ Zin = jX1 + ω2 M2 = j X1 - ω2 M2 jX2 jX2 = j 314 × 0.8 - 3142 × 0.482 0.6 × 314
= j[251.32 – 120.576]
= j | 30.744 = jw Leff = j314.Leff
∴ Leff = 0.416 = 416 mHCorrect Option: A
Zin = V1 = (R1 + jX1) + ω2 M2 I1 R2 + jX2 + ZL
Given, L1 = 800 mH
L2 = 600 mH
M = 480 mH
W = 314 rad/sec
ZL = 0
R1 R2 neglected∴ Zin = jX1 + ω2 M2 = j X1 - ω2 M2 jX2 jX2 = j 314 × 0.8 - 3142 × 0.482 0.6 × 314
= j[251.32 – 120.576]
= j | 30.744 = jw Leff = j314.Leff
∴ Leff = 0.416 = 416 mH
- An induction motor having full load torque of 60 Nm when delta-connected develops a starting torque of 120 Nm. For the same supply voltage, if motor is changed to star -connection, then starting torque developed will be _______ Nm .
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With Delta Connection :
Starting current per phase = E √R² + X² ∴ Starting torque = 3RI2 = 3 RI2 s since , s = 1 = 3 RE2 R2 + X2
With Star Connection :Starting current / phase = E / √3 √R² + X²
∴ Starting torque = 3 RI2= 3 × R × E2 = RE2 3 (R2 + X2) R2 + X2 ∴ New starting torque = 120 = 40 Nm 3
Correct Option: A
With Delta Connection :
Starting current per phase = E √R² + X² ∴ Starting torque = 3RI2 = 3 RI2 s since , s = 1 = 3 RE2 R2 + X2
With Star Connection :Starting current / phase = E / √3 √R² + X²
∴ Starting torque = 3 RI2= 3 × R × E2 = RE2 3 (R2 + X2) R2 + X2 ∴ New starting torque = 120 = 40 Nm 3
- A 6-pole lap-connected dc generator with 480 conductors has armature resistance of 0.06 ohm. If the conductors are reconnected to form a wave winding, other things remaining unchanged, the value of the armature resistance will be _______ ohm .
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Rlap = Z . ρ L p2 A and Rwave = z . ρ L 4 A Rlap = 4 = 4 = 4 = 1 Rwave p2 62 36 9
∴ Rwave = Rlap × 9 = 0.06 × 9 = 0.54 Ω
Correct Option: C
Rlap = Z . ρ L p2 A and Rwave = z . ρ L 4 A Rlap = 4 = 4 = 4 = 1 Rwave p2 62 36 9
∴ Rwave = Rlap × 9 = 0.06 × 9 = 0.54 Ω
- Find the transformer ratios a and b that the impedance (Zin) is resistive and equal 2.5 Ω when the network is excited with a sine wave voltage of angular frequency of 5000 rad/s.
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From given data :
Xc = – j20 Ω
XL = j5 ΩNow , -20 + 5 = 0 b2
b = 0.5Also , Zin = 1 2.5 + j5 - j20 b2 a2 ∴ 2.5 = 2.5 a2 b2 ∴ 2.5 = 2.5 a2 (0.5)2 a2 = 1 (0.5)2
a2 = 4
∴ a = 2
Option (b) is correct.Correct Option: B
From given data :
Xc = – j20 Ω
XL = j5 ΩNow , -20 + 5 = 0 b2
b = 0.5Also , Zin = 1 2.5 + j5 - j20 b2 a2 ∴ 2.5 = 2.5 a2 b2 ∴ 2.5 = 2.5 a2 (0.5)2 a2 = 1 (0.5)2
a2 = 4
∴ a = 2
Option (b) is correct.