# Mass air flow through a nozzle at different upstream pressures

by InquisitiveOne
Tags: flow, mass, nozzle, pressures, upstream
 PF Gold P: 1,491 So basically, you can get the mass flux (rate of mass flow per unit area passing through some imaginary plane in the nozzle) at any give point in a nozzle pretty simply. It is $$\dfrac{\dot{m}}{A} = \rho u$$ where $\dot{m}$ is the mass flow rate, $A$ is the cross-sectional area of the aforementioned plane, $\rho$ is the density and $u$. It is common to denote the the sonic flow conditions at the throat with a star, so from conservation of mass, you can also say $$\rho u A = \rho^* u^* A^*.$$ So, $$\dfrac{\dot{m}}{A} = \rho^* u^* \dfrac{A^*}{A}.$$ From there, we know that from the ideal gas law, $p^* = \rho^*RT^*$ and, since the flow is choked and therefore sonic at the throat, $u^* = a^* = \sqrt{\gamma R T^*}$. Here, $R$ is the specific gas constant, $\gamma$ is the ratio of specific heats (1.4 for air), and $T$ is temperature. These can be combined with the previous equation to get $$\dfrac{\dot{m}}{A} = \sqrt{\dfrac{\gamma}{R}} \dfrac{p^*}{\sqrt{T^*}} \dfrac{A^*}{A}.$$ Using the various isentropic relations for compressible flow, this can be expanded to $$\dfrac{\dot{m}}{A} = \dfrac{p_{01}}{\sqrt{T_{01}}}\sqrt{\dfrac{\gamma}{R}}\left( \dfrac{2}{\gamma + 1} \right)^{\frac{\gamma+1}{2(\gamma-1)}}\dfrac{A^*}{A}.$$ You'll note there that the term $A^*/A$ is always less than or equal to 1, and it is equal to 1 at the throat, so for constant parameters, the maximum mass flux is at the throat. Anyway, that obviously gets a little simpler, $$\boxed{\dfrac{\dot{m}}{A^*} = \dfrac{p_{01}}{\sqrt{T_{01}}}\sqrt{\dfrac{\gamma}{R}}\left( \dfrac{2}{\gamma + 1} \right)^{\frac{\gamma+1}{2(\gamma-1)}}.}$$ So, that equation gives the mass flux at the throat, $\dot{m}/A^*$, as a function of the total pressure, $p_{01}$, and total temperature, $T_{01}$, which, for an isentropic flow are the same as your reservoir temperature and pressure. The rest of the terms are known constants. In nearly all situations, you can treat that portion of the flow between the reservoir and the throat as isentropic. If you want the actual mass flow rate, just move the area term over to the right side of the equation. This only works if the flow is choked, though. If not, then the throat is not sonic and you can't use this, and whether or not your flow is choked depends on the upstream and downstream pressures and the geometry.