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ZanzabumX [31]
2 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]2 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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Explanation:

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clearing the equation

V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}

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for the entrance is 400°C and 4 MPa is superheated steam and v is : 0,7343 m^3/kg

In the outlet we have saturated vapor with quality (x) of 80%. In this case we get the specific saturated volume for the liquid (vf) and the specific volume for the saturated  (vg) gas from the thermodynamic tables. we use the next equation to get  (v) for the condition of interest, in this case 80% quality.

v= vf +x*(vg - vf)

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The area is the one for a circle

\pi *r^{2}

r1 = 0,1 m^2 for area 1

r2=0,5 m^2 for area 2

A1 = 0,0314 m^2

A2 = 0,7853 m^2

we know that  V1 is 20 m/s

replacing these values in the equation

V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}

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