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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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(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e
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Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

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Explanation:

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5 0
3 years ago
1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con
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3 years ago
When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b
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Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\tau_{ave} = \frac{T}{2tA_{m}}

A_m = 100 mm × 200 mm = 20000 mm² = 0.02 m²

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T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

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Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, A_m = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\tau_{ave} = \frac{T}{2tA_{m}}

Plugging in then values, gives;

\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa

The shear stress will be 80,000,000 Pa or 80 MPa.

7 0
3 years ago
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