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Chapter 15: Q29P (page 412)

Question:(II) A Carnot engine performs work at the rate of 520 kW with an input of 950 kcal of heat per second. If the temperature of the heat source is 520掳C, at what temperature is the waste heat exhausted?

Short Answer

Expert verified

The temperature of waste heat exhausted is \(T = 689.9\;{\rm{K}}\).

Step by step solution

01

Given Data

The temperature of heat source is \({T_0} = 520^\circ {\rm{C}}\).

The input given to engine is \(\frac{Q}{t} = 950\;{\rm{kcal/s}}\).

The power output is \(P = 520\;{\rm{kW}}\).

02

Perfect Carnot engine efficiency

In this problem, to determine the exhaust temperature, use the two relations of Carnot engine efficiency and equate them to obtain the result.

03

Calculation of the temperature of the exhausted waste heat

The relation of efficiency is given by,

\(\begin{aligned}{c}{\eta _{\rm{C}}} &= 1 - \frac{T}{{{T_{\rm{0}}}}}\\\frac{W}{Q} &= 1 - \frac{T}{{{T_{\rm{0}}}}}\\\left( {\frac{{\frac{W}{t}}}{{\frac{Q}{t}}}} \right) &= 1 - \frac{T}{{{T_{\rm{0}}}}}\\\left( {\frac{P}{{\frac{Q}{t}}}} \right) &= 1 - \frac{T}{{{T_{\rm{0}}}}}\end{aligned}\)

On plugging the values in the above relation.

\(\begin{aligned}{c}\left( {\frac{{520\;{\rm{kW}} \times \frac{{1000\;{\rm{W}}}}{{1\;{\rm{kW}}}}}}{{\left( {950\;{\rm{kcal/s}} \times \frac{{4186\;{\rm{J}}}}{{1\;{\rm{kcal}}}}} \right)}}} \right) &= 1 - \left( {\frac{T}{{\left( {520^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}} \right)\\T &= 689.9\;{\rm{K}}\end{aligned}\)

Thus, \(T = 689.9\;{\rm{K}}\) is the required exhaust temperature.

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