Question

Air enters an adiabatic nozzle at 500 kPa and a temperature of 200 °C with a velocity of 100 m/s. It exits the nozzle at a pressure of 100 kPa. Assuming that the expansion through the nozzle occurs reversibly, determine (a) the exit temperature and (b) the exit velocity of the air. The specific heats of air can be assumed to be constant with Cv = 0.742 kJ/kg oC and Cp = 1.029 kJ/kg oC.

Answer 1

Since the process is adiabatic, we have :

$PV^{\gamma}=\text{constant} \Rightarrow P_1V_1^{\gamma}=P_2V_2^{\gamma} \\\text{or,} P_1^{1-\gamma}T_1^{\gamma}=P_2^{1-\gamma}T_2^{\gamma} \\ \therefore T_2 = \bigg(\frac{P_1}{P_2}\bigg)^{\gamma -1}T_1 \hspace{0.5cm}(\because \text{ PV=RT and} \gamma= C_p/C_v) \\T_2= 608.49 K= 335.49 ^{\circ}C$

This is the exit temperature after substituting all the values.

(b) Since the mass flow rate is constant, so :

$\rho_1Av_1=\rho_2Av_2 \text(\text{where, A is the area of cross-section of the nozzle and}\hspace{0.1cm} v_2 \hspace{0.1cm}\text{is the exit velocity})$

$\therefore \frac{P_1}{RT_1}v_1=\frac{P_2}{RT_2}v_2 \\ v_2 = \frac{P_1}{P_2}\times \frac{T_2}{T_1}\times v_1=\frac{500}{100}\times\frac{608.49}{473}\times 100=643.22 \text{m/s}$

This is the exit velocity.