A 415 V- 3- -connected synchronous motor runs at rated voltage and with an excitation emf of 500 V- Its synchronous impedance per phase is 0-5 -j2 - If friction- winding and iron losses are 1000 W- then the value of maximum shaft power output will be

Z_a=(0.5+j2) Ω =2.06∠75.96^∘ Ω; V_t=415 V, E_f=500 V Maximum developed power =E_fV_tZ_s E^2_fZ^2_s× R_a P_dev=500×4152.06 (5002.06)^2×0.5 =71.272 kW This ...

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Standard X

Physics

Turns Ratio

A 415 V, 3-ϕ,...

Question

A 415 V, $3 - ϕ$ , $Δ$ -connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its synchronous impedance per phase is $(0.5 + j 2) Ω$ . If friction, winding and iron losses are 1000 W, then the value of maximum shaft power output will be

247.14 kW

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165 kW

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112.12 kW

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212.81 kW

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Solution

The correct option is D 212.81 kW
$Z_{a} = (0.5 + j 2) Ω$

$= 2.06 ∠ {75.96}^{\circ} Ω; V_{t} = 415 V, E_{f} = 500 V$

Maximum developed power $= \frac{E_{f} V_{t}}{Z_{s}} - \frac{E_{f}^{2}}{Z_{s}^{2}} \times R_{a}$

$P_{d e v} = \frac{500 \times 415}{2.06} - {(\frac{500}{2.06})}^{2} \times 0.5$

$= 71.272 k W$

This is per phase power.

$∴$ Shaft power output $= [3 \times 71.272 - 1] k W$

$= 212.81 k W$

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A 415 V, 3−ϕ, Δ-connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its synchronous impedance per phase is (0.5+j2)Ω. If friction, winding and iron losses are 1000 W, then the value of maximum shaft power output will be

The correct option is D 212.81 kWZa=(0.5+j2)Ω =2.06∠75.96∘Ω;Vt=415V,Ef=500V Maximum developed power =EfVtZs−E2fZ2s×Ra Pdev=500×4152.06−(5002.06)2×0.5 =71.272kW This is per phase power. ∴ Shaft power output =[3×71.272−1]kW =212.81kW

A 415 V, $3 - ϕ$ , $Δ$ -connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its synchronous impedance per phase is $(0.5 + j 2) Ω$ . If friction, winding and iron losses are 1000 W, then the value of maximum shaft power output will be