Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - General Problems - Page 441: 71

Answer

a) $0.281$ b) $1.01\times 10^5\;\rm W$, ${ 2.06\times10^9}\;\rm J$, ${ 4.92\times10^5}\;\rm kcal$

Work Step by Step

a) To find the ratio of the Engine's efficiency relative to its Carnot efficiency, we need to find the Engine's Carnot efficiency since we have its real efficiency of 0.15. $$e_{\rm Carnot}=\dfrac{T_H-T_L}{T_H}$$ Plugging the known and remember to convert temperatures to Kelvin. $$e_{\rm Carnot}=\dfrac{768-358}{768}=\bf 0.534$$ Thus, $$\dfrac{e_{real}}{e_{\rm Carnot}}=\dfrac{0.15}{0.534}=\color{red}{\bf 0.281}$$ b) The power in watts is given by $$P=\rm 135 \;hp\cdot \dfrac{746\;W}{1\;hp}$$ $$P=\color{red}{\bf1.01\times 10^5}\;\rm W$$ The amount of heat exhausted into the air is given by $$e=\dfrac{W}{Q_H} $$ We also know that $Q_H=W+Q_L$ $$e=\dfrac{W}{W+Q_L}\tag 1 $$ We know that the power is given by $$P=\dfrac{W}{t}$$ So, $$W=Pt\tag {Plugging into (1)}$$ $$e=\dfrac{Pt}{Pt+Q_L} $$ Solving for $Q_L$; $$ePt+eQ_L =Pt $$ $$ Q_L =\dfrac{Pt-ePt}{e}=Pt\left(\dfrac{1-e}{e}\right) $$ Plugging the known; $$ Q_L =1.01\times10^5\times 3600\left(\dfrac{1-0.15}{0.15}\right) $$ $$ Q_L =\color{red}{\bf2.06\times10^9}\;\rm J$$ And in kcal; $$ Q_L = 2.06\times10^9 \;\rm J\cdot \dfrac{1\;kcal}{4186\;J}=\color{red}{\bf4.92\times10^5}\;\rm kcal$$
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