calculating pressures in split flow from a compressor

[mattc80]

Hi,

I have an NPK09 24v DC air compressor attached to which I have 4mm inner diameter tubing split into 3 then these tubes are split into 3 again leaving 9 4mm inner diameter outlet tubes (see diagram). I wish to calculate the theoretical mass flow rates, exit pressures and hence force exerted by the air stream at these outlets as I will eventually be using each of these nine outlets to displace a 1 gram mass. What I ultimately want to achieve is to determine the minimum volumetric flow rate and pressure delivered from the pump that will still allow me to vertically displace this mass when placed on one of these outlets.

So far I have calculated that, at a delivery of 15 l/min and hence pressure of 1 atm (see attached data sheet for the npk 09 pump);
volumetric flow rate = 2.5x10^-4 m^3s^-1
air velocity from pump = 4.97ms^-1
mach number - 0.0145 << 1 hence assume incompressible flow
I am assuming dry, inviscid and newtonian 20 degrees c air

Can I use the bernoulli equation to calculate Pe, the pressure at the outlet tubes?

Am I right in saying the sum of the mass flow rates through the outlet tubes = the mass flow rate in?

Once I have pressure and mass flow rate can I use this equation to find the force at the outlet tubes - F = mdot * ve + (Pe - Pat)*A, where Pat is atmospheric pressure, A is the CSA of the tube, mdot is mass flow rate and ve is exit velocity.

What other assumptions must I make?

Cheers for any help.

 

[BigInch]

I think you're correct, theoretically. The problem will come as you attempt to keep the pipe outflows balanced. I think the weights will not displace equally and possibly have an effect on outlet pressures of each discharge pipe, which will have an effect on the inlet pressures and flows going into each outlet pipe. I don't have a good feeling about trying to maintain all 9 outlets at equal flows and pressures with weights. It's hard enough, if not nearly impossible, to do it with needle valves.

Independent events are seldomly independent.