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3 years ago

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An industrial firm has two electrical loads connected in parallel across the power source. Power is supplied to the firm at 4000

V rms. One load is 30 kW of heating use, and the other load is a set of motors that together operate as a load at 0.6 lagging power factor and at 150 kVA. Determine the total current and the plant power factor.

1 answer:

zvonat [6]3 years ago

5 0

Answer:

I = 27.65A < 40.59°

PowerFactor = 0.76

Explanation:

Current on the heating load is:

I1 = 30KW / 4KV = 7.5A < 0°

Current on the inductive load:

I2 = (150KVA*0.6) /4KV = 22.5A with an angle of acos(0.6)=53.1°

The sum of both currents is:

It = I1 + I2 = 7.5<0° + 22.5<53.1° = 27.65<40.59°

Now, the power factor will be:

pf = cos (40.59°) = 0.76

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Answer with Explanation:

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$1mm=10^{-3} m$

Radius of solenoid,r'=1 cm= $1\times 10^{-2} m$

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Where $A=\pi r^2$

Using the formula

$4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}$

$l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m$

Number of turns of wire= $\frac{l}{2\pi r'}$

Number of turns of wire= $\frac{50.26}{2\pi(1\times 10^{-2}}=800$

Hence, the number of turns of the solenoid,N=799

Magnetic field in solenoid,B= $\mu_0 nI$

$3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6$

$n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}$

$n=6404 turns/m$

$n=\frac{N}{L}$

$L=\frac{N}{n}$

$L=\frac{799}{6404}$

$L=0.125 m=0.125\times 100=12.5 cm$

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D ) sketch of the cycle on p-V coordinates

attached below

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