Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 6 - The Second Law of Thermodynamics - Problems - Page 326: 6-154

Answer

See answer below.

Work Step by Step

$\dot{Q}_H=(hA)_H(T_H-T_H^*)$ $\eta = \dot{W}_e/\dot{Q}_H=1-\frac{T_L^*}{T_H^*}$ $\dot{W}_e=\left (1-\frac{T_L^*}{T_H^*}\right)(hA)_HT_H\left (1-\frac{T_H^*}{T_H}\right)$ Defining $x=1-\frac{T_H^*}{T_H},\ y=\frac{T_L^*}{T_H^*}$ $\dot{W}_e=(hA)_H(1-y)x$ $\left (\frac{\dot{Q}_H}{\dot{Q}_L}\right)_{rev}=\dfrac{T_H^*}{T_L^*}$ $\dfrac1y=\dfrac{(hA)_H(T_H-T_H^*)}{(hA)_L(T_L^*-T_L)}=\dfrac{(hA)_H}{(hA)_L}\dfrac{\left (1-\frac{T_H^*}{T_H}\right)}{\left (\frac{T_L^*}{T_H^*}\frac{T_H^*}{T_H}-\frac{T_L}{T_H}\right)}$ Defining $k=\dfrac{(hA)_H}{(hA)_L}$ $\dfrac1y=k\dfrac{x}{y(1-x)-\frac{T_L}{T_H}}$ $x=\dfrac{y-\frac{T_L}{T_H}}{y(k+1)}$ $\dot{W}_e=(hA)_H(1-y)\dfrac{y-\frac{T_L}{T_H}}{y(k+1)}$ $\dfrac{\partial\dot{W}_e}{\partial y}=0$ $y=\frac{T_L^*}{T_H^*}=\left(\frac{T_L}{T_H}\right)^{1/2}$ Substituting back: $\dot{W}_{max}=\dfrac{(hA)_HT_H}{1+(hA)_H/(hA)_L}\left[1-\frac{T_L}{T_H}\right]^2$
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