Question 3.22: Air [R = 1716, cp = 6003 ft = lbf/(slug . °R)] flows steadil...

Part (a) The inlet and exit densities can be computed from the perfect-gas law:
\rho_{1}=\frac{p_{1}}{RT_{1}}=\frac{150(144)}{1716(460+300)}=0.0166 slug/ft^3

\rho_{2}=\frac{p_{2}}{RT_{2}}=\frac{40(144)}{1716(460+35)} =0.00679 slug/ft^3
The mass flow is determined by the inlet conditions
\dot{m}=\rho_{1}A_{1}V_{1} =(0.0166)\frac{\pi}{4}(\frac{6}{12})^2(100)=0.325 slug/s Knowing mass flow, we compute the exit velocity
\dot{m}=0.325=\rho_{2}A_{2}V_{2}=(0.00679)\frac{\pi}{4}(\frac{6}{12})^2 V_{2}
or V_{2}=244ft/s
Part (b) :The steady flow energy equation (3.69)

applies with \dot{W}_{v}=0, z_{1}=z_{2} and \hat{h}=c_{p}T

\underline{\dot{Q}}=\dot{W}_{s}=\dot{m}(c_{p}T_{2}+\frac{1}{2}V^{2}_{2}-c_{p}T_{1}+\frac{1}{2}V^{2}_{1})
Convert the turbine work to foot-pounds-force per second with the conversion factor 1hp =550 ft = lbf/s. The turbine work \dot{W}_{s} is positive \dot{Q}-700(550)= 0.3253[6003(495)+\frac{1}{2}(244)^2- 6003(760)-\frac{1}{2} (100)^2 ]

=-510,000 ft\cdot lbf/s
or \dot{Q}=(-125,000 ft\cdot lbf/s)
onvert this to British thermal units as follows:
\dot{Q}=(-125,000 ft\cdot lbf/s)\frac{3600 s/h}{778.2 ft \cdot lbf/Btu}

=- 578,000 Btu/h
The negative sign indicates that this heat transfer is a loss from the control volume.