# Injector pressure drop in a liquid rocket engine

Hello everybody,

I hope you are doing fine. I'm currently designing an injector for a hydrogen peroxide - ethanol engine, and the following formula is confusing me:
Q = Cd * A * sqrt( 2 * dP / rho); where Cd is the discharge coefficient, A the total area, dP the pressure drop, and rho the fluid's density.
Supposing Cd, A, and rho are fixed once we built the injector, the only way to throttle the engine would be changing delta p (dP). What does this pressure differential represent? The difference between the entry and exit pressure of the fluid through the injection plate as a consequence of sheer, gravitational, viscous, etc forces, OR the difference between the injection pressure (taking into account the pressure drop through the plate) and the design chamber pressure? How would you change dP?

Best,

Roy S. Ramirez

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Andrew Mason
Homework Helper
Hello everybody,

I hope you are doing fine. I'm currently designing an injector for a hydrogen peroxide - ethanol engine, and the following formula is confusing me:
Q = Cd * A * sqrt( 2 * dP / rho); where Cd is the discharge coefficient, A the total area, dP the pressure drop, and rho the fluid's density.
Supposing Cd, A, and rho are fixed once we built the injector, the only way to throttle the engine would be changing delta p (dP). What does this pressure differential represent? The difference between the entry and exit pressure of the fluid through the injection plate as a consequence of sheer, gravitational, viscous, etc forces, OR the difference between the injection pressure (taking into account the pressure drop through the plate) and the design chamber pressure? How would you change dP?

Best,

Roy S. Ramirez
Hi Roy. Welcome to PF!

dP is the pressure drop across the valve or, in the case of your rocket, the injection plate. The exit or discharge pressure is determined by the ambient pressure in the exit chamber. To change dP one would have to change the pressure on the input side.

This is essentially Bernoulli's equation modified for the characteristics of a compressible fluid. Energy is conserved at all points. Energy is in the form of: internal PV energy; internal potential energy due to inter-molecular forces; gravitational potential energy (due to changes in height of the fluid); and kinetic energy of the fluid (which is a function of cross-sectional area, mass/unit volume, and flow rate (volume/unit time). If the flow through the valve is horizontal and the fluid is not compressible, the change in energy due to Δ(PV) is VΔP where ΔP is the pressure difference (intake pressure - exit pressure). That drop in pressure (which is energy/unit volume) must equal the increase in kinetic energy of the fluid per unit volume. But since the fluid is compressible that is where Cd comes in to relate actual flow rate to the theoretical flow rate for an non-compressible fluid.

AM

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