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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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b

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Answer:

Explanation:

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\dfrac{\partial^2T}{\partial x^2}=  \ 0  \  ;  \ if \  T = f(x)  \\ \\ \dfrac{\partial^2T}{\partial y^2}=  \ 0  \  ;  \ if \  T = f(y)  \\ \\ \dfrac{\partial^2T}{\partial z^2}=  \ 0  \  ;  \ if \  T = f(z)

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}

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c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0

where;

The radial directional term = \dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) and the axial directional term is \dfrac{\partial^2 T}{\partial z^2 }

d) The heat equation for a wire going through a furnace is:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]

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\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}

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