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 9 - Gas Power Cycles - Problems - Page 540: 9-36E

Answer

$W_{\text {net }}=38.0\text{hp}$

Work Step by Step

From the data specified in the problem statement, $$ r=\frac{\boldsymbol{v}_1}{\boldsymbol{v}_2}=\frac{\boldsymbol{v}_1}{0.098 \boldsymbol{v}_1}=10.20 $$ Since the compression and expansion processes are isentropic,$$ \begin{aligned} & T_2=T_1\left(\frac{\boldsymbol{v}_1}{\boldsymbol{v}_2}\right)^{k-1}=T_1 r^{k-1}=(565 \mathrm{R})(10.20)^{1.4-1}=1430.7 \mathrm{R} \\ & T_4=T_3\left(\frac{\boldsymbol{v}_3}{\boldsymbol{v}_4}\right)^{k-1}=T_3\left(\frac{1}{r}\right)^{k-1}=(2860 \mathrm{R})\left(\frac{1}{10.20}\right)^{1.4-1}=1129.4 \mathrm{R} \end{aligned} $$ Application of the first law to the compression and expansion processes gives $$ \begin{aligned} w_{\text {net }} & =c_v\left(T_3-T_4\right)-c_v\left(T_2-T_1\right) \\ & =(0.171 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(2860-1129.4) \mathrm{R}-(0.171 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(1430.7-565) \mathrm{R} \\ & =147.9\ \mathrm{Btu} / \mathrm{lbm} \end{aligned} $$ When each cylinder is charged with the air-fuel mixture, $$ \nu_1=\frac{R T_1}{P_1}=\frac{\left(0.3704 \mathrm{psia} \cdot \mathrm{f}^3 / \mathrm{bm} \cdot \mathrm{R}\right)(565 \mathrm{R})}{14 \mathrm{psia}}=14.95\ \mathrm{f}^3 / \mathrm{lbm} $$ The total air mass taken by all 6 cylinders when they are charged is $$ m=N_{\mathrm{cyl}} \frac{\Delta V}{v_1}=N_{\mathrm{cy} 1} \frac{\pi B^2 S / 4}{v_1}=(6) \frac{\pi(3.5 / 12 \mathrm{ff})^2(3.9 / 12 \mathrm{f}) / 4}{14.95 \mathrm{ft}^3 / \mathrm{lbm}}=0.008716\ \mathrm{lbm} $$ The net work produced per cycle is $$ W_{\text {net }}=m w_{\text {met }}=(0.008716 \mathrm{lbm})(147.9 \mathrm{Btu} / \mathrm{lbm})=1.289\ \mathrm{Btu} / \mathrm{cycle} $$ The power produced is determined from $$ \dot{W}_{\text {net }}=\frac{W_{\text {tet }} \dot{n}}{N_{\text {rev }}}=\frac{(1.289 \mathrm{Btu} / \text { cycle })(2500 / 60 \mathrm{rev} / \mathrm{s})}{2 \mathrm{rev} / \mathrm{cycle}}\left(\frac{1 \mathrm{hp}}{0.7068\ \mathrm{Btu} / \mathrm{s}}\right)=\mathbf{3 8 . 0 \mathrm { hp }} $$ since there are two revolutions per cycle in a four-stroke engine.
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