Question 13.9: The converging nozzle on the tank in Fig. 13–21 has a 300-mm......
The converging nozzle on the tank in Fig. 13–21 has a 300-mm exit diameter. If nitrogen within the tank has an absolute pressure of 500 kPa and an absolute temperature of 1200 K, determine the mass flow from the nozzle if the absolute pressure in the pipe at the nozzle is 300 kPa. What pressure would be needed here to create the greatest mass flow through the nozzle?
Fluid Description. We assume steady isentropic flow through the nozzle.
Analysis. The nitrogen within the tank is at rest, and so the stagnation pressure and temperature are p_0 = 500 kPa and T_0 = 1200 K. Since we know both p and p_0, then p/p_0 = 300 kPa/500 kPa = 0.6, and so we can determine M from Eq. 13–27 p_0=p(1+\frac{k-1}{2}M^2)^{k/(k-1)} or Table B–1. We get M = 0.8864.
To obtain the mass flow, \dot{m} = \rho VA, we must obtain the density and velocity at the exit. First the temperature is determined from Eq. 13–26 T_0=T(1+\frac{k-1}{2}M^2) or from Table B–1 for M = 0.8864 or p/p_0 = 0.6. We have
T = 0.8642 (1200 K) = 1037.00 K
Therefore, the exit velocity of the nitrogen is
The density can be found using the ideal gas law. The mass flow from the nozzle is therefore
\dot{m} = \rho VA = (\frac{p}{RT})VA = [\frac{300 (10^3) N/m^2}{(296.8 J/kg · K) (1037.00 K)}] (581.86 m/s)[\pi(0.15 m)^2]\dot{m} = 40.1 kg/s
For the greatest mass flow through the nozzle, it is necessary that M = 1 at the exit, and so Eq. 13–27 or Table B–1 requires
\frac{p*}{p_0} = 0.528 or p* = (500 kPa) (0.5283) = 264 kPa