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3 years ago

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A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

between the network voltage and the generatorinduced voltage is 60 degrees. Calculate the generator induced voltage if the network receives 300 MW of power.

1 answer:

Aleks04 [339]3 years ago

7 0

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$

the network voltage phase is

$V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV$

the power transmitted is equal to:

$|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV$

the line induced voltage is

$|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV$

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Answer:

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b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( $\frac{M1}{100}$ )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

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3 years ago

For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep

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Answer:

MRR = 1.984

Explanation:

Given that

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Now the metal removal rate given as

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V= Speed

f=Feed

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By putting the values

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3 years ago

A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

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Answer:

The distance between the station A and B will be:

$x_{A-B}=55.620\: km$

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$x_{1}=x_{0}+v_{0}t+0.5at^{2}$

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

$x_{1}=\frac{1}{2}(0.6)(60)^{2}$

$x_{1}=1080\: m$

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

$v_{1}=v_{0}+at$

$v_{1}=(0.6)(60)=36\: m/s$

Then the second distance will be:

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$x_{2}=(36)(1500)=54000\: m$

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$v_{f}^{2}=v_{1}^{2}-2ax_{3}$

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

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$x_{3}=\frac{36^{2}}{2(1.2)}$

$x_{3}=540\: m$

Therefore, the distance between the station A and B will be:

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I hope it helps you!

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3 years ago

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Answer:

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Explanation:

We are given:

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• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

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• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

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•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

$F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}$

Substituting figures in the equation we have:

$= \frac{1}{1+0.1(2.5-1)+0(2-1)}$

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$D = \frac{Vp}{FFS}$

$D = \frac{2777.7}{70}$

D = 39.685 Pc/mi/en

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Answer:

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