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Q31PE

Expert-verifiedFound in: Page 551

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Steam locomotives have an efficiency of \(17\% \)** **and operate with a hot steam temperature of \(425{\;^o}C\). (a) What would the cold reservoir temperature be if this were a Carnot engine? (b) What would the maximum efficiency of this steam engine be if its cold reservoir temperature were \(150{\;^o}C\)?**

For a Carnot’s engine, the temperature of the cold reservoir is \(306.34\;^\circ {\rm{C}}\). Maximum efficiency can be achieved, when the temperature of the cold reservoir is \(150\;^\circ {\rm{C}}\), is \(39.4\% \).

**We calculate the temperature of the hot reservoir by using the formula for efficiency of the Carnot engine when the efficiency and temperature of the cold reservoir are given. We further find the efficiency when the temperature of the cold reservoir is 150 ^{o}C**.

Efficiency of engine\(\eta = 17\% \)

The temperature of the hot reservoir \({T_h} = 425\;^\circ {\rm{C}} = 698\;{\rm{K}}\)

The efficiency of the Carnot engine \(h = 1 - \frac{{{T_c}}}{{{T_h}}}\)

Here,

\(\eta \) - efficiency of the engine.

\({T_c}\)- the temperature of the cold reservoir.

\({T_h}\)- temperature of hot reservoir.

The temperature of cold reservoir is calculated as

\(\begin{aligned}{}h &= 1 - \frac{{{T_c}}}{{{T_h}}}\\0.17 &= 1 - \frac{{{T_c}}}{{698\;{\rm{K}}}}\\\frac{{{T_c}}}{{698\;{\rm{K}}}} = 1 - 0.17\end{aligned}\)

\(\begin{aligned}{}{T_c} &= 698\;{\rm{K}} \times 0.83\\ &= 579.34\;{\rm{K}}\\ &= 306.34{\;^o}{\rm{C}}\end{aligned}\)

Temperature of cold reservoir \({T_c} = 150\;^\circ {\rm{C}}\)

Efficiency of engine is

\(\begin{aligned}{}h &= 1 - \frac{{{T_c}}}{{{T_h}}}\\ &= 1 - \frac{{423\;{\rm{K}}}}{{698\;{\rm{K}}}}\\ &= 0.394\\ &= 39.4\% \end{aligned}\)

Therefore, the temperature of cold reservoir is \(306.34{\;^o}{\rm{C}}\) for Carnot engine. Maximum efficiency is \(39.4\% \)when temperature of cold reservoir is \(150\;^\circ {\rm{C}}\).

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