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UBC MECH327 A4 Q4.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \usepackage{cancel} \begin{document} {\large UBC MECH327 Assignment 4 Q4} \begin{align*} &\text{a) \underline {Determine the work produced by the turbine in kJ/kg$steam$.}}\\ \\ &P_{in}=3\text{ MPa},\quad T_{in}=673.15\text{ K},\quad V_{in}=160\text{ m/s}.\\ &\text{From the superheated steam table:}\quad %v_g=0.0993\text{ m$^3$/kg},\quad %u_g=2933\text{ kJ/kg},\quad h_g=3231\text{ kJ/kg},\quad s_g=6.921\text{ kJ/kg K}.\\ \\ &P_{out}=101.325\text{ KPa}\text{ (1 atm)},\quad T_{out}=373.15\text{ K},\quad V_{out}=100\text{ m/s}.\\ &\text{From the saturated steam table:}\quad %v_g=1.673\text{ m$^3$/kg},\quad h_g=2675.8\text{ kJ/kg},\quad s_g=7.355\text{ kJ/kg K}.\\ \\ &\text{The turbine lost heat at 30 kJ/kg from its surface at 400 K.}\\ \\ &\text{Steady State:}\quad \dot{Q}_{in}-\dot{W}_{out} +\sum_{in}\dot{m}_{in}\left(h_{in}+\frac{V_{in}^2}{2}+gz_{in}\right) -\sum_{out}\dot{m}_{out}\left(h_{out}+\frac{V_{out}^2}{2}+gz_{out}\right)=0.\\ &\text{Assuming $\dot{m}_{in}=\dot{m}_{out}=\dot{m}$ and $z_{in}=z_{out}=0$,}\quad \dot{Q}_{in}-\dot{W}_{out} +\dot{m}\left[\Big(h_{in}-h_{out}\Big)+\left(\frac{V_{in}^2}{2}-\frac{V_{out}^2}{2}\right)\right]=0.\\ &\frac{\dot{W}_{out}}{\dot{m}}=\frac{\dot{Q}_{in}}{\dot{m}} +\left[\Big(h_{in}-h_{out}\Big)+\left(\frac{V_{in}^2}{2}-\frac{V_{out}^2}{2}\right)\right].\\ \\ &\text{Given }\frac{\dot{Q}_{in}}{\dot{m}}=\frac{Q_{in}}{m}=-30\text{ kJ/kg},\quad \frac{W_{out}}{m}=\frac{\dot{W}_{out}}{\dot{m}} =-30+\left[\Big(3231-2675.8\Big)+\left(\frac{160^2}{2}-\frac{100^2}{2}\right)\div 1000\right]=543\text{ kJ/kg}.\\ \\ &\text{Note 1: This is work output per Kg, not per second.}\\ \\ &\text{Note 2: From the formula sheet, internal energy change has been accounted for in the enthalpy ($h$) term.}\\ \\ \\ &\text{b) \underline {What is the rate of entropy production if the control volume includes only the turbine and its contents?}}\\ &\text{Formula:}\quad \frac{dS}{dt}=\sum\frac{\dot{Q}_k}{T_k}+\sum_{in}\dot{m}_{in}s_{in}-\sum_{out}\dot{m}_{out}s_{out}+\dot{S}_{gen}=0\quad\text{(Steady State)}.\\ &\text{Swap time rate with mass rate:}\quad \frac{Q_{in}}{m}\cdot\frac{1}{T}+\sum_{in}s_{in}-\sum_{out}s_{out}+\frac{S_{gen}}{m}=0.\\ &\frac{S_{gen}}{m}=-\frac{Q_{in}}{m}\cdot\frac{1}{T}-\sum_{in}s_{in}+\sum_{out}s_{out} =-(-30)\frac{1}{400}-6.291+7.355=1.139\text{ kJ/kg K}.\\ \\ \\ &\text{c) \underline {What is the rate of entropy production if the control volume is expanded to include some of the surroundings,}}\\ &\qquad\text{\underline {such that the heat transfer is assumed to take place at the temperature of the surroundings (27◦C)?}}\\ &\text{Use surrounding temperature $T_{sur}$=300 K (27$^\circ$C) instead of the turbine surface temperature 400 K:}\\ &\frac{S_{gen}}{m}=-\frac{Q_{in}}{m}\cdot\frac{1}{T_{sur}}-\sum_{in}s_{in}+\sum_{out}s_{out} =-(-30)\frac{1}{300}-6.291+7.355=1.164\text{ kJ/kg K}.\\ \\ \\ &\text{d) \underline {Comment on the difference in entropy production for the two control volumes.}}\\ &\text{Extra entropy has been produced around the surface of the turbine as the heat transfers through the}\\ &\text{cooler environment to reach the lower temperature of 27$^\circ$C.}\\ \end{align*} \end{document}