Answer:
a. 969.1 R
b. 2237 ft/s
Explanation:
First the apparent specific heats are determined from the mass fractions of the gases:
[tex]c_{p} &=\left(\mathrm{mf} c_{p}\right)_{\mathrm{N}_{2}}+\left(\mathrm{mf} c_{p}\right) \mathrm{CO}_{2} \\ &=(0.65 \cdot 0.248+0.35 \cdot 0.203) \frac{\mathrm{Btu}}{\mathrm{lbm} \mathrm{R}} \\ &=0.232 \frac{\mathrm{Btu}}{\mathrm{lbmR}} \\ c_{v} &=\left(\mathrm{mf} c_{v}\right)_{\mathrm{N}_{2}}+\left(\mathrm{mf} c_{v}\right)_{\mathrm{CO}_{2}} \\ &=(0.65 \cdot 0.177+0.35 \cdot 0.158) \frac{\mathrm{Btu}}{\mathrm{lbmR}} \\ &=0.170 \frac{\mathrm{Btu}}{\mathrm{lbmR}}[/tex]
The isentropic coefficient then is:
[tex]k &=\frac{c_{p}}{c_{v}} \\ &=\frac{0.232}{0.17} \\ &=1.365[/tex]
The final temperature is determined from the isentropic nozzle efficiency relation:
[tex]T_{2} &=T_{1}-\eta_{N}\left(T_{1}-T_{2 s}\right) \\ &=T_{1}\left(1-\eta_{N}\left(1-\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}\right)\right) \\ &=1400\left(1-0.88\left(1-\left(\frac{800}{100}\right)^{(1.365-1) / 1.365}\right)\right) \mathrm{R} \\ &=969.1 \mathrm{R}[/tex]
b. The outlet velocity is determined from the energy balance:
[tex]h_{1} &+\frac{v_{1}^{2}}{2}=h_{2}+\frac{v_{2}^{2}}{2} \\ v_{2} &=\sqrt{2 c_{p}\left(T_{1}-T_{2}\right)} \\ &=\sqrt{2 \cdot 0.232(1400-969.2) \cdot 25037} \frac{\mathrm{ft}}{\mathrm{s}} \\ &=2237 \frac{\mathrm{ft}}{\mathrm{s}}[/tex]
