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ZanzabumX [31]
3 years ago
11

Problem 89:A given load is driven by a 480 V six-pole 150 hp three-phase synchronous motor with the following load and motor dat

a. Determine the voltage E௙ necessary for this operating condition. Note: assume that the rotational loss torque is negligible. Load: T௅ൌ0.05∗????௥ଶ????mMotor: E௙ൌ400 V; Xௗൌ1ΩAnswer: Eത௙ൌ400∠-17.36° V
Engineering
1 answer:
expeople1 [14]3 years ago
6 0

Answer:

E_f=400

Explanation:

From the question we are told that:

Load V=480

Poles p=6

Power P=150hp

3-Phase

Load:

Tl=0.05*\omega_s^2Nm

Motor:

Ef=400V\\\\X_d=1ohm

Generally the equation for Synchronous speed is mathematically given by

N_s=\frac{120F}{p}=\frac{120*60}{6}

N_s=1200rpm\\\\N_s=125.66 rads/sec

Therefore

Tl=0.05*\omega_s2Nm

With

\omega=N_s

We have

Tl=0.05*(125.66)^2Nm

T_l=789.52 Nm

Therefore

Load Power

P_l=T_l*\omega_s\\\\P_l=789.52*125.66

P_l=9922watts

Generally the equation for Load Power is mathematically given by

P_l=\frac{\sqrt{3}*E_f.V_t}{x_d}*sin\theta\\\\9922=\frac{\sqrt{3}*480*400}{1}*sin\theta

\theta=17.4 \textdegre3

Therefore

Voltage

E_f=400

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<u>Explanation</u>:

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