Question 12.6: Airflow through a Converging–Diverging Nozzle Air enters a c...

Air flows through a converging–diverging nozzle. The throat and the exit conditions and the mass flow rate are to be determined.
Assumptions   1  Air is an ideal gas with constant specific heats at room temperature. 2  Flow through the nozzle is steady, one-dimensional, and isentropic.
Properties   The specific heat ratio of air is given to be k = 1.4. The gas constant of air is 0.287 kJ/kg⋅K.
Analysis   The exit Mach number is given to be 2. Therefore, the flow must be sonic at the throat and supersonic in the diverging section of the nozzle. Since the inlet velocity is negligible, the stagnation pressure and stagnation temperature are the same as the inlet temperature and pressure, P_0 = 1.0  MPa and T_0 = 800  K. Assuming ideal-gas behavior, the stagnation density is

\rho_0=\frac{P_0}{R T_0}=\frac{1000  kPa }{\left(0.287  kPa \cdot m ^3 / kg \cdot K \right)(800  K )}=4.355  kg / m ^3

(a) At the throat of the nozzle Ma = 1, and from Table A–13 we read

\frac{P^*}{P_0}=0.5283 \quad \frac{T^*}{T_0}=0.8333 \quad \frac{\rho^*}{\rho_0}=0.6339

Thus,

P^* = 0.5283P_0 = (0.5283)(1.0  MPa) = 0.5283  MPa
T^* = 0.8333T_0 = (0.8333)(800  K) = 666.6  K
\rho* = 0.6339\rho_0 = (0.6339)(4.355  kg/m³) = 2.761  kg/m³

Also,

V^*=c^*=\sqrt{k R T^*}=\sqrt{(1.4)(0.287  kJ / kg \cdot K )(666.6  K )\left(\frac{1000  m ^2 / s ^2}{1  kJ / kg }\right)}

= 517.5 m/s

(b) Since the flow is isentropic, the properties at the exit plane can also be calculated by using data from Table A–13. For Ma = 2 we read

\frac{P_e}{P_0}=0.1278 \frac{T_e}{T_0}=0.5556 \frac{\rho_e}{\rho_0}=0.2300   Ma _e^*=1.6330   \frac{A_e}{A^*}=1.6875

Thus,

P_e = 0.1278P_0 = (0.1278)(1.0  MPa) = 0.1278  MPa
T_e = 0.5556T_0 = (0.5556)(800  K) = 444.5  K
\rho_e = 0.2300\rho_0 = (0.2300)(4.355  kg/m^3) = 1.002  kg/m^3
A_e = 1.6875A^* = (1.6875)(20  cm^2) = 33.75  cm^2

and

V_e = Ma_e^*c^* = (1.6330)(517.5  m/s) = 845.1  m/s

The nozzle exit velocity could also be determined from V_e = Ma_ec_e, where c_e is the speed of sound at the exit conditions:

V_e= Ma _e c_e= Ma _e \sqrt{k R T_e}=2 \sqrt{(1.4)(0.287  kJ / kg \cdot K )(444.5  K )\left(\frac{1000  m ^2 / s ^2}{1  kJ / kg }\right)}

= 845.2 m/s

(c) Since the flow is steady, the mass flow rate of the fluid is the same at all sections of the nozzle. Thus it may be calculated by using properties at any cross section of the nozzle. Using the properties at the throat, we find that the mass flow rate is

\dot{m}=\rho^* A^* V^*=\left(2.761  kg / m ^3\right)\left(20 \times 10^{-4}  m ^2\right)(517.5  m / s )=2.86  kg / s

Discussion   Note that this is the highest possible mass flow rate that can flow through this nozzle for the specified inlet conditions.