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I'm working on designing a Cold Spray system that uses a convergent-divergent nozzle to achieve a 800 m/s exhaust velocity. I'm assuming 400 psi nitrogen at room temperature is provided at the inlet of the nozzle. The throat area is 0.0859 square inches (tiny, I know).

$$ \begin{align} \dot{m} &= \rho\ v\ A \\ \rho &= 31.920\ \frac{\text{kg}}{\text{m}^3} \text{(N}_2\text{ at 400 psi)},\\ v &= 353.575\ \frac{\text{m}}{\text{s}} \text{(speed of sound at 400 psi)},\\ A &= 5.54*10^{-5}\ \text{m}^2,\\ \Rightarrow\qquad\dot{m}&=0.625\ \frac{\text{kg}}{\text{s}} \end{align} $$

assuming $M = 1$ in the throat.

This figure seems absurdly large. Can someone shed some light on the situation?

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    $\begingroup$ Hi sjMoquin, welcome to Engineering.SE. This site supports Latex style equation editing. I've tried to edit your post for you; please double check that everything looks ok. $\endgroup$ – Chris Mueller May 29 '15 at 1:49
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    $\begingroup$ This rather horrendous page may also be of use. $\endgroup$ – Russell McMahon May 29 '15 at 12:57
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You're not that far off. In your analysis, you've just neglected the fact that the gas will accelerate as it passes through the converging nozzle. This is basically converting potential energy, in the form of pressure, into kinetic energy, velocity. As a result, the gas will no longer be at 400 psi when it reaches the throat.

The compact form to compute mass flow through a choked nozzle is

$$ \dot{m} = C_d A \sqrt{\gamma \rho P \left( \frac{2}{1+\gamma} \right)^{\frac{\gamma+1}{\gamma-1}}} $$

where $\gamma$ is the ratio of specific heats (1.4 for nitrogen), and $C_d$ is an dimensionless coefficient that depends on the efficiency of the nozzle design (let's guess 0.5 for now).

Applying the above, I got $0.18 \, kg\cdot s^{-1}$.

As a sanity check I compared this result to manufacturer reported values for flow control nozzles that I use in my lab. When I scaled up their values for a $.0252 \, in.$ nozzle at $100 \, psig$ up by area and pressure, I got $0.15 \, kg\cdot s^{-1}$. Seems reasonable.

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  • $\begingroup$ Wonderful work, and excellent explanation, Dan. Thank you. $\endgroup$ – sjMoquin May 29 '15 at 22:58

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