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VladimirAG [237]
3 years ago
15

You have removed a very large, thick plate of steel (AISI 1010) from your heat treat oven and have placed it on a large insulate

d surface; the steel is now at a uniform initial temperature of 600.0°C. You now cool the exposed top surface of the steel using high-velocity cooling fluid; which keeps the cooled surface at a constant 30.0°C (assume the other surface is not losing any heat). How long will you need to run the coolant to cool the steel at a depth of 4.20cm to a temperature of 60.0°C?
Engineering
1 answer:
Andreas93 [3]3 years ago
6 0

Answer:

for got sorry

Explanation:

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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
Aleks04 [339]

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

7 0
3 years ago
The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.
ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

So

we can say that acceleration= 0.6 x 9.81

acceleration,a=5.88\ \frac{m}{s^2}

We know that

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a=\omega ^2A

5.88=\omega ^2 \0.15\times 10^{-3}

\omega =198.09\ \frac{rad}{s}

We know that

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198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

                                               

       

4 0
3 years ago
I gave 15 min to finish this java program
lisov135 [29]

Answer:

class TriangleNumbers

{

public static void main (String[] args)

{

 for (int number = 1; number <= 10; ++number) {

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  }

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}

}

Explanation:

We need to run the code for each of the 10 lines. Each time we sum  numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.

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