Electrical machines miscellaneous
- Given the maximum efficiency of a transformer of a 500 kVA, 3300/500V, 50 Hz single-phase transformer is 97% and occurs at 3/4 of full-load, unity power factor. If the impedance is 10%, the regulation at full-load, power factor 0.8 lagging will be _____ % .
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η = 0.97 = 500 × 3 × 1 4 500 × 3 + Pcore + Pcu × 9 4 16 ⇒ Pcore + Pcu × 9 = 11.60 16
⇒ Pcu = 10.31 kW (at full-load)∴ re (p.u.) = Pcu = 10.31 = 0.02062 kVArated 500
xe (p.u.) = √Ze² - re²
= √0.1² - (0.02062)² = 0.0978
∴ Regulation = re cos θ + xe sin θ
= 0.02062 × 0.8 + 0.0978 × 0.6 = 7.52%
Correct Option: C
η = 0.97 = 500 × 3 × 1 4 500 × 3 + Pcore + Pcu × 9 4 16 ⇒ Pcore + Pcu × 9 = 11.60 16
⇒ Pcu = 10.31 kW (at full-load)∴ re (p.u.) = Pcu = 10.31 = 0.02062 kVArated 500
xe (p.u.) = √Ze² - re²
= √0.1² - (0.02062)² = 0.0978
∴ Regulation = re cos θ + xe sin θ
= 0.02062 × 0.8 + 0.0978 × 0.6 = 7.52%
- A 3– phase delta/star transformer is supplied at 6000 V on the delta-connected side. The terminal voltage on the secondary side when supplying full load at 0.8 lagging power-factor is 415 V. The equivalent resistance and reactance drops for the transformer are 1% and 5% respectively. The turns ratio of the transformer is _______ .
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Turn ratio = Phase voltage on primary side Phase voltage on secondary side = 6000 × √3 = 24 415
Correct Option: B
Turn ratio = Phase voltage on primary side Phase voltage on secondary side = 6000 × √3 = 24 415
- A 400 V/ 200 V/ 200 V, 50 Hz three winding transfor mer is connected as shown in the given figure. The reading of the voltmeter,V will be _______ V.
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NA
Correct Option: A
NA
- The low voltage winding of a 400 / 230V, 1– phase, 50 Hz transformer is to be connected to a 25 Hz, the supply voltage should be _______ V .
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NA
Correct Option: C
NA
- For a p-pole machine, the relation between electrical (θe) and mechanical (θm) degree is given by
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NA
Correct Option: D
NA