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hockeyfan26
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If I'm designing an injector for a rocket engine, I have a specific mean flow rate (m/s) that must be met due to the pipe diameter and the required mass flow rate.
I've calculated all the pressure drops due to lost work from friction, valves, and orifice geometries (not from change in area, but due to the complex internal geometries that lead to losses in energy).
For calculating the necessary area of the injector plate orifices what pressure drop should I use? Should I use the pressure drop I determined to get the required flow rate or should I do this:
pressure drop over injector = (the required pressure drop to achieve the required mass flow rate ) - [(pressure drop from friction) + (pressure drop over check valves/etc)]
or this:
Pressure drop over injector = pressure drop to obtain required mean flow rate
Example: The combustion chamber is at 500 psi. I need a pressure drop of let's say 90 psi to get the required flow rate into the combustion chamber. The pressure drop from friction is 30 psi and the pressure drop from a check valve is 15 psi. For the pressure drop over the injector should I use 90psi (regulator below tank would be set to 635 psi) or 45 psi (regulator below tank would be set to 590 psi in this case)?
I guess what my question boils down to:
1) Do pressure drops due to frictional losses and losses due to flowing through valves increase the speed of the flow (or are they just considered lost energy that causes a pressure drop, but no corresponding velocity increase)?
2) As the flow of liquid/gas exits a regulator can we assume the velocity is zero at that point of our control volume, and calculate the velocity from the overall pressure drop from the regulator to the other end of the control volume?
I'm guessing that pressure drops due to friction and check valves do indeed increase flow rate since in my chemical engineering curriculum pressure gradients are generated from frictional losses and commonly used to increase volumetric flow rates in plug flow reactors. But on the other hand the increased volumetric flow rate could just be from the expansion of the reactants at the lower pressures.
I greatly appreciate any advice/help with this. Thanks!
I've calculated all the pressure drops due to lost work from friction, valves, and orifice geometries (not from change in area, but due to the complex internal geometries that lead to losses in energy).
For calculating the necessary area of the injector plate orifices what pressure drop should I use? Should I use the pressure drop I determined to get the required flow rate or should I do this:
pressure drop over injector = (the required pressure drop to achieve the required mass flow rate ) - [(pressure drop from friction) + (pressure drop over check valves/etc)]
or this:
Pressure drop over injector = pressure drop to obtain required mean flow rate
Example: The combustion chamber is at 500 psi. I need a pressure drop of let's say 90 psi to get the required flow rate into the combustion chamber. The pressure drop from friction is 30 psi and the pressure drop from a check valve is 15 psi. For the pressure drop over the injector should I use 90psi (regulator below tank would be set to 635 psi) or 45 psi (regulator below tank would be set to 590 psi in this case)?
I guess what my question boils down to:
1) Do pressure drops due to frictional losses and losses due to flowing through valves increase the speed of the flow (or are they just considered lost energy that causes a pressure drop, but no corresponding velocity increase)?
2) As the flow of liquid/gas exits a regulator can we assume the velocity is zero at that point of our control volume, and calculate the velocity from the overall pressure drop from the regulator to the other end of the control volume?
I'm guessing that pressure drops due to friction and check valves do indeed increase flow rate since in my chemical engineering curriculum pressure gradients are generated from frictional losses and commonly used to increase volumetric flow rates in plug flow reactors. But on the other hand the increased volumetric flow rate could just be from the expansion of the reactants at the lower pressures.
I greatly appreciate any advice/help with this. Thanks!
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