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xeze [42]
3 years ago
10

The town of Mustang, TX is concerned that waste heat discharged from a new up- stream power plant will decimate the minnow popul

ation in its local stream. The power plant is 35% efficient with a rated output of 350 MW, and 75% of the waste heat is discharged into the stream, which flows at 150 m3/s. Before the power plant starts operations, the stream temperature was 20◦C. If the minnows can handle max- imum stream temperatures of 22◦C, will the water be safe at the point of cooling water discharge? (Assume that the waste heat is discharged to the flow of the whole stream.)
Engineering
1 answer:
vitfil [10]3 years ago
7 0

Answer:

Yes the water will be safe at the point of cooling water discharge

Explanation:

Power losses in plant= 350- 350×0.35=227.5MW

Rate of heat rejection to stream= 0.75× 227.5= 170.625MW

Rate of heat rejection= rate of flow of water× c × ΔT

170625000= 150000000× 4.186 × (Final temperature- 20)

Final temperature= 20.3 ◦C

The final temperature of stream will be 20.3 ◦C. Thechange is very small so the minnows will be able to handle this temperature.

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Answer:

Overall ideal mechanical advantage of the machine = 40

Explanation:

Given:

Ideal mechanical advantage of three machine = 2, 4, 5

Find:

Overall ideal mechanical advantage of the machine

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Overall ideal mechanical advantage of the machine = 2 × 4× 5

Overall ideal mechanical advantage of the machine = 40

3 0
3 years ago
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
6-What is the difference between the critical point and the triple point?
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Answer:

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Explanation:

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3 years ago
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3 ohms hope this helps :D ❤

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Answer:

See explaination

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