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: 69

Answer

a) $\rm 13.2\; km^3/day$ b) $73\;\rm km^2$

Work Step by Step

a) To find the volume of air that is heated per day, we first need to find the amount of heat delivered to this air. $$Q=Q_L=m_{air}c_{air}\Delta T_{air}$$ whereas $Q$ is the amount of heat absorbed by air which is the amount of heat exhausted from the cooling towers. We also know, from the density law, that $m=\rho V$. So that, $$Q_L=\rho_{air} V_{air}c_{air}\Delta T_{air}$$ Solving for $V_{air}$; $$V_{air}=\dfrac{Q_L}{\rho_{air} c_{air}\Delta T_{air}} \tag 1$$ Now we need to find $Q_L$ which we can find from the COP of the power plant. $$e=\dfrac{W}{Q_H}=\dfrac{W}{W+Q_L}$$ Solving for $Q_L$; $$W=eW+eQ_L$$ $$Q_L=\dfrac{W(1-e)}{e}$$ Plugging into (1) and then plugging the known; $$V_{air}=\dfrac{W(1-e)}{e\rho_{air} c_{air}\Delta T_{air}} $$ Divide both sides by $t$; $$\dfrac{V_{air}}{t}=\dfrac{\dfrac{W}{t}(1-e)}{e\rho_{air} c_{air}\Delta T_{air}} $$ $$\dfrac{V_{air}}{t}=\dfrac{850\times10^6 \cdot (1-0.38)}{0.38\cdot 1.3 \cdot (1\times10^3) \cdot 7 }=1.524\times 10^5\;\rm m^3/s $$ $$\dfrac{V_{air}}{t}=\rm \dfrac{1.524\times 10^5\;\rm m^3}{1\;s}\cdot \dfrac{3600\;s}{1\;h}\cdot \dfrac{24\;h}{1\;day}\cdot \dfrac{1^3\;km^3}{1000^3\;m^3} $$ $$\dfrac{V_{air}}{t}=\color{red}{\bf 13.2}\;\rm km^3/day$$ b) We know that the volume is given by the area times the height. $$V_{air}=h_{air}A_{air}$$ whereas $h$ is the thickness of the air layer. Thus, $$A_{air}=\dfrac{V_{air}}{h_{air}}=\dfrac{13.2\;\rm km^3}{180\times10^{-3}\;\rm km}=\color{red}{\bf 73.3}\;\rm km^2$$
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