Question

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 data. 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

Answer 1

Answer:

$E_f=400<-17.4volts$

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

$E_f=400<-17.4volts$