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viktelen [127]
4 years ago
8

You are an engineer in an electric-generation station. You know that the flames in the boiler reach a temperature of 1200 K and

that cooling water at 300 K is available from a nearby river. What is the maximum efficiency your plant will ever achieve?
Engineering
1 answer:
olga nikolaevna [1]4 years ago
5 0

Answer:

<em>The maximum efficiency the plant will ever achieve is 75%</em>

<em>Explanation:</em>

From the question given, we recall the following:

<em>Th flames in the boiler reaches a temperature of = 1200K</em>

<em>the cooling water is = 300K</em>

<em>The maximum efficiency the plant will achieve is defined as:</em>

Let nmax = 1 - Tmin /Tmax

Where,

Tmin = Minimum Temperature in plants

Tmax = Maximum Temperature in plants

The temperature of the cooling water  = Tmin = 300K

The temperature of the flames in boiler = Tmax = 1200k=K

The maximum efficiency becomes:

nmax =  1 - Tmin /Tmax

nmax =  1 - 300 /1200

nmax =  1-1/4 =0.75

nmax =  75%

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Answer:

length of starter crack = 1.056

Explanation:

given data

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solution  

we use here A 20 table so we get here value that is

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Kic = 117.5 MPa \sqrt{m}

so length of starter crack  will be

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put here value and we will get

a =  \frac{1}{\pi } (\frac{117.5}{2.5\times 0.6 \times 1360})^2  

solve it we get

length of starter crack = 1.056

5 0
3 years ago
8. Air at 25C, 100 kPa and air at 50C, 200 kPa at 1 to 1 volume ratio are mixed inside an adiabatic compressor to 55C, 500 kPa a
quester [9]

Answer:

Workdone, w = 68.935 kJ

Explanation:

m1h1 + m2h2 + Q = m3h3 - Wc

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p1V1 = m1R1T1

p2V2 = m2R2T2

Since they are at 1:1,

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m1/m2 = 0.542

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m2 = 3.243kg

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Wc = m3h3 - m1h1 - m2h2

h = Cp x T, since air is an ideal gas

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8 0
4 years ago
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4 0
3 years ago
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bagirrra123 [75]

Answer: um wuh anyways thxs for the points!

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Answer:

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6 0
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