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

a. 969.1 R

b. 2237 ft/s

Step-by-step 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 &=(c_(p))/(c_(v)) \\ &=(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((P_(2))/(P_(1))\right)^((k-1) / k)\right)\right) \\ &=1400\left(1-0.88\left(1-\left((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) &+(v_(1)^(2))/(2)=h_(2)+(v_(2)^(2))/(2) \\ v_(2) &=\sqrt{2 c_(p)\left(T_(1)-T_(2)\right)} \\ &=√(2 \cdot 0.232(1400-969.2) \cdot 25037) \frac{\mathrm{ft}}{\mathrm{s}} \\ &=2237 \frac{\mathrm{ft}}{\mathrm{s}}`