Question

A mixture of 65 percent N2 and 35 percent CO2 gases (on a mass basis) enters the nozzle of a turbojet engine at 60 psia and 1400 R with a low velocity, and it expands to a pressure of 12 psia. If the isentropic efficiency of the nozzle is 88 percent, determine:
(a) the exit temperature
(b) the exit velocity of the mixture.
Assume constant specific heats at room temperature.

Answer 1

Answer:

a. 969.1 R

b. 2237 ft/s

Explanation:

First the apparent specific heats are determined from the mass fractions of the gases:

$c_{p} &=\left(\mathrm{mf} c_{p}\right)_{\mathrm{N}_{2}}+\left(\mathrm{mf} c_{p}\right) \mathrm{CO}_{2} \\ &=(0.65 \cdot 0.248+0.35 \cdot 0.203) \frac{\mathrm{Btu}}{\mathrm{lbm} \mathrm{R}} \\ &=0.232 \frac{\mathrm{Btu}}{\mathrm{lbmR}} \\ c_{v} &=\left(\mathrm{mf} c_{v}\right)_{\mathrm{N}_{2}}+\left(\mathrm{mf} c_{v}\right)_{\mathrm{CO}_{2}} \\ &=(0.65 \cdot 0.177+0.35 \cdot 0.158) \frac{\mathrm{Btu}}{\mathrm{lbmR}} \\ &=0.170 \frac{\mathrm{Btu}}{\mathrm{lbmR}}$

The isentropic coefficient then is:

$k &=\frac{c_{p}}{c_{v}} \\ &=\frac{0.232}{0.17} \\ &=1.365$

The final temperature is determined from the isentropic nozzle efficiency relation:

$T_{2} &=T_{1}-\eta_{N}\left(T_{1}-T_{2 s}\right) \\ &=T_{1}\left(1-\eta_{N}\left(1-\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}\right)\right) \\ &=1400\left(1-0.88\left(1-\left(\frac{800}{100}\right)^{(1.365-1) / 1.365}\right)\right) \mathrm{R} \\ &=969.1 \mathrm{R}$

b. The outlet velocity is determined from the energy balance:

$h_{1} &+\frac{v_{1}^{2}}{2}=h_{2}+\frac{v_{2}^{2}}{2} \\ v_{2} &=\sqrt{2 c_{p}\left(T_{1}-T_{2}\right)} \\ &=\sqrt{2 \cdot 0.232(1400-969.2) \cdot 25037} \frac{\mathrm{ft}}{\mathrm{s}} \\ &=2237 \frac{\mathrm{ft}}{\mathrm{s}}$