Multiple RO's in series 2

[CatBoy]

Hi!

I've been asked to check the design of a line with three restriction orifices in series. That line is connecting the discharge of a recycled H2 compressor discharge line at 2350 psia to a 14.7 psia gas seal drain pot. The line is 0.434in ID and each RO has a bore of 0.1in.

I'm a bit lost. Sonic dP for the first RO would be 1107.4psi with a Q of 480.4SCFM. I won't be able to handle 480.4SCFM in the next RO as the dP would have to be much higher than the sonic dP. So I need to do this backwards, starting with the last RO. Should I assume sonic dP for the last RO, calculate the flow and upstream pressure and use this upstream pressure as the downstream pressure for the 2nd RO... all the way to the first one and try to reach an upstream of 2350 psia?

Is that correct?

Thanks in advance!

 

[EricKvaalen]

I agree that for adiabatic flow you get critical flow easily, without a long, narrow pipe.

But achieving the isothermal case would be tricky. If you simply add the required heat before the orifice, then the temperature will increase in the part where you add the heat, but then as the gas goes through the orifice, it will nevertheless be almost adiabatic, its temperature will go down, and it will achieve sonic velocity at the narrowest point. Then, just after the orifice, its temperature will go back up as its kinetic energy is converted into enthalpy.

To achieve an isothermal critical flow situation, you'd have to add the heat "au fur et à mesure" (French for "as you go along"). In other words, you have to add the heat during the acceleration. And then take heat away during the deceleration, which would be impossible with an orifice.

http://Eric.Kvaalen.com

 

[sailoday28]

My points on the isothermal were for information.
In previous responses I noted statements in relation to adiabatic flow of a perfect gas that indicated
isothermal flow
constant stagnation temp (without considering specific heats)
const enthalpy.

Can any of the above be true for any adiabatic process? Other readers might use those type statements in other problems and get erroneous results.

For isothermal flow, I believe the heat added or taken away would be upstream between the source and the orifice. How close to reality this is will have to be determined by knowing heat transfer coef's on the inside/outside and other thermal resistance for the piping.
For adiabatic flow, the flow distribution does change just upsteam of the orifice even though one generally assumes a one dimensional analysis.

 

[sshep]

An ideal gas does not show any temp change in an adiabatic throttling process due to the assumption that enthalpy is a function of temperature only. On the otherhand I think an ideal gas assumption would be a reasonable starting point for the methodology if it can be applied.

I have been following this thread out of curiosity because it is a boundary value problem with possible discontinuities, and therefore a challenge to solve. An excel based itterative solution seems a reasonable approach, but I don't feel that the information provided in this thread has actually educated me enough to solve such a problem. I would love to see a case actually worked out with some real numbers and orifice formulas. Thanks, sshep

 

[hacksaw]

sshep, the both pressure and temperature changes are involved under the assumption of an adiabatic expansion

 

[sailoday28]

sshep (Chemical)states" I would love to see a case actually worked out with some real numbers and orifice formulas."
I suggest we responders look at the problem with 2 identical orifices, each with a Cd=1. (with the last orifice have a short distance of frictionless piping downstream of it)

I would start the problem with the second orifice choking using the formula of my original post
W/A =const. (Po/To^.5)= mass flux from second orifice.
Po the stag pressure downstream of first orifice and To for adiabatic conditions remaining constant.

Po upstream of first orifice has to be greater than Po just upstream of second orifice.
I hope there will be some responce.

 

[EricKvaalen]

First a comment to sshep--adiabatic does not mean constant enthalpy. If the gas does some work or accelerates, then its enthalpy decreases even if no heat is transferred out of the gas. (I think this is what hacksaw was saying.)

Now some comments on the calculation of a series of orifices. When an orifice is in critical flow (or "choked"), then the flow is related simply to the upstream conditions, as sailoday has stated:

W = A const. (Pb/Tb^.5)

(I use Tb and Pb for the stagnation temperature and pressure before the orifice.)

This can be solved for any unknown if the others are known.

But when the flow is not critical, the situation is more complicated. The flow is

W = A rho v = A v M Pa / (R To) = A v M Pa / (R Tb) (Pb/Pa)^(R/Cp)

(M is molecular weight, Pa is pressure after the orifice (and in the orifice), To is temperature in orifice.)
The velocity is related to the pressures by

M v^2 / 2 = Cp Tb [1 - (Pa/Pb)^(R/Cp)]

Combining the above, we have

W^2 = 2 A^2 M Pa^2 (Cp/R) [(Pb/Pa)^(R/Cp) - 1] / (R Tb)

This cannot be solved for Pa explicitly, but it can be solved for other things. For instance,

Pb = Pa [1 + W^2 R Tb (R/Cp) / (2 A^2 M Pa^2)]^(Cp/R)

In the two-orifice case, we can calculate the flow W either from P0 and P1 (that is, before and after the first orifice), or from P1 alone using the simpler equation for the second orifice (assuming it is critical). Equating these two expressions for W gives an equation for P1 in terms of P0, A, etc. This can be solved for P1, but requires an iterative method (it is not a quadratic equation).

In the case of three orifices, one can also use an iterative method, as I wrote on March 25. We know P0 and P3, but not P1 or P2. We can assume a W, find P2 (simple equation assuming third orifice is critical), then find P1 (complicated equation), then find the P0 which would give our assumed W. We compare this with the desired P0 and iterate.

If the last orifice is not critical for the assumed W, then we just have to use the complicated formula to find P2 instead of the simple formula. It doesn't change anything essential.

There are other variants, such as assuming P1, finding W from this and P0, then finding P2 from W (using choked orifice 3), then finding P1 from W and P2, and comparing with the assumed P1.

But if you do it wisely, there is only one iteration to do--no nesting or solving for more than one variable simultaneously.

http://Eric.Kvaalen.com