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Ira Lisetskai [31]
2 years ago
12

Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool

ed water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T
Engineering
1 answer:
MariettaO [177]2 years ago
5 0

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T  = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

<u>Determine temperature of the cooled water exiting the cooling tower</u>

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 +  (0.01273) (33.94) = 29.577 KJ/kmol°C

<u>First step : calculate the value of Q </u>

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) +  43.13 (18.26089)

Q =  95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

<u>Hence the temperature of the cooled water can be calculated using the equation below</u>

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000.    ∴ T  = 43.477° C

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Answer:

  83.6%

Explanation:

<h3>(a)</h3>

On its current schedule, the plant can theoretically produce ...

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  = (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

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<h3>(b)</h3>

On the proposed schedule, this production becomes ...

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The increase in capacity is ...

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_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

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A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i
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Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

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This is the value of heat transferred watt per foot length.

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