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ladessa [460]
3 years ago
5

A steam power plant with a power output of 230 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,0

00 kJ/kg, determine the overall efficiency of this plant.
Engineering
1 answer:
NARA [144]3 years ago
7 0

Answer:

\eta =46\%

Explanation:

Hello!

In this case, we compute the heat output from coal, given its heating value and the mass flow:

Q_H=60\frac{tons}{h}*\frac{1000kg}{1ton}*\frac{1h}{3600s}*\frac{30,000kJ}{kg}\\\\Q_H=500,000\frac{kJ}{s}*\frac{1MJ}{1000J} =500MW

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

\eta =\frac{230MW}{500MW}*100\% \\\\\eta =46\%

Best regards!

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In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
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Answer:

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Explanation:

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Temperature,

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Volume,

V = 1000 r

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Now,

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o,

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       =\frac{1013250\times 1}{8.3145\times 298}

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As we know,

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or,

⇒        MW=\frac{m}{n}

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3 0
3 years ago
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Answer:

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b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

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Answer:

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3 0
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