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

`E_f=400<-17.4volts`

Step-by-step 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=(120F)/(p)=(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=(√(3)*E_f.V_t)/(x_d)*sin\theta\\\\9922=(√(3)*480*400)/(1)*sin\theta`

`\theta=17.4 \textdegre3`

Therefore

Voltage

`E_f=400<-17.4volts`