# Determine supersonic flow velocity, tank to regulator topipe

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1. Jan 7, 2016

### Michu

Hello All,

I am trying to figure out the solution to a problem I am trying to do for fun. I've been trying this for a while and has annoyed me so any help is greatly appreciated! I attached a picture of the problem/ control area to help.

[PLAIN]http://postimg.org/image/5pytdzie3/[URL]http://postimg.org/image/5pytdzie3/[/URL] [Broken]

The situation is that there is a tank filled at a certain known pressure P1 and has known area A1. This is connected to a regulator that lets out a certain pressure in a smaller pipe. Inside the pipe P2 is known since it can be set by the regulator and A2 is also known. Typically I would try to use the Bernoulli equation but my aim is to work with compressible air so I can't do that. Working backwards I was thinking using the continuity equation:

(d/dt)( ∫_cv of ρ d∀ ) + (∫_cs of ρVdA) = 0 which then goes to just ρ_2*V_2*A_2=0 so that doesn't work either.

I was also thinking maybe using the Isentropic Equations such as (P_o/P) = [1+ ((γ -1)/2)*M^2] ^(λ/(λ-1)) but that is giving me that I need 5 atm in the tank and 1 atm in the pipe and gives 1.7 Mach which is way to high just by reasoning.

Any advice or help is greatly appreciated, this problems is just annoying me and I cant stop thinking about it.
Also I am new to the forums so if I did something wrong sorry!

Thank you,
~Michu

Last edited by a moderator: May 7, 2017
2. Jan 7, 2016

### SteamKing

Staff Emeritus
You can't use Bernoulli and continuity with compressible fluids and expect to get reasonable answers.

In situations of this type, the analysis is done assuming either isothermal (constant temperature) conditions or adiabatic conditions (no heat exchange with surroundings) and accounting for friction losses in the pipe outlet. If the ratio P1/P2 is large enough, somewhere in the outlet pipe, the velocity of the escaping air will go to Mach 1, at which point the pipe, being of constant cross-sectional area, cannot flow any more air, and the flow is said to be choked.

The article at the link gives a quick intro to compressible pipe flow:

http://www.aft.com/documents/AFT-CE-Gasflow-Reprint.pdf

Last edited by a moderator: May 7, 2017
3. Jan 7, 2016

For the record, the continuity equation is valid in compressible flow. Mass is still conserved after all.

However, it doesn't make a lot of sense to me to try and apply any of these equations across a regulator in the first place (at least in this manner), compressible or not. Typically, if ou want the flow rate in the outlet pipe, you need to know its exit pressure and you can do your calculations from there (since you already set the upstream pressure with the regulator).

4. Jan 23, 2016

### Michu

Hello,

Thank you guys for responding. I do not see how this can be isothermal since the pressure differential between the stagnation air in the tank and the pressure in the pipe will be very different and result in different temperatures. Adiabatic could work, but I am not sure where to start with that. However, boneh3ad's response made me think that I am doing this wrong. Since there is a regulator, I think it would automatically set what would be considered the back pressure. With this in mind I tried to start the problem with the pipe at the certain pressure and then it releases the moving air into the atmosphere. Would it make sense in this sense to then use the Energy Equation and the Isentropic Equations?

To demonstrate what I mean:
Energy Equation: q + h + (V^2)/2 + gz =Constant
h=(Cp)(T)
Isentropic Equations: (P2/P1) = (T2/T1)^ [gamma/(gamma-1)]
In the problem no heat is added q=0 and no change in elevation z1=z2 and we have a constant Cp

Then (P2/P1) = (T2/T1)^ [gamma/(gamma-1)] becomes T1= (T2/ (P2/P1)^ [(gamma-1)/gamma] and T1 can be found
then Energy Equation simplifies to CpT1 + (V1^2)/2 = CpT2 + (V2^2)/2 but the 2nd location is atmsphere so V2=0 then:
V1 = sqrt{2*Cp*(T2-T1)}

I think this would work since we know P1 (pressure in the pipe set by the regulator), P2 and T2 are set by the atmosphere.