Steam Flow Through Gate Valve 1

You are dealing with relatively low pressures, 3.4 atm, compared to the critical pressure of water, 218 atm. The reduced pressure is therefore 3.4/218 = 0.016. The reduced temperature under these conditions is 412K/647K = 0.64. From the generalized compressibility charts, the compressibility factor for water (steam) at these conditions is very close to 1. In other words, steam does behave very much as an ideal gas under these circumstances.

Does that help, or have I misundersatood your question?

If you really have 50 psia on the upstream side and atmospheric pressure on the downstream side of this gate valve, you will have chocked flow.

I suggest getting/borrowing a copy of Crane's technical paper 410. It will have the equations in there for compressible flow. How big is this valve? If you really have (about) 35 psi dP across just an open gate valve, that steam is really going to move.

If you have a size, I can give you a rough estimate but Crane would be your best bet.

Thank you both for your help! I guess I was wrong on both of my previous statements. I do have portions of the Crane paper and have found it pretty useful (now that I have looked at it).

TD2K:
I am curious about the Darcy equation (Crane equation 1-11). I take it that the 0.525 value is a conversion factor. I am calculating the conversion factor to be 0.473 ... (1/144)*(32.2*144)^0.5 ... any insight?

Also, for a 4" line (50 psia inlet) with a 4" gate valve (wide open, discharge to atm) I calculate approx. 24,000 to 27,000 #/hr (depending on the conversion factor used). Is this in the ballpark?

Thanks for the help, I appreciate it!!

jproj

I get about 31,000 lb/hr using 8 for the L/D of a gate valve and K = 1.0 for the exit. Looks like you are in the ballpark.

I'll have to take a look at the question you raised on the Darcy equation in a bit.

The conversion factor between equation 1-4 and 1-11 has actually quite a few numbers in there to make the conversion, it's quite a bit more complicated that what you have. Equation 1-4 has a v^2 in it, you have to replace that with w^2/(a^2*rho^2) to wind up with a w for equation 1-11 (a is the pipe area). And watch the units, you can get tripped up there quite easily. After that, it's just rearrangement of the terms to match equation 1-11.

When I worked through it starting out from equation 1-4, I got the same 0.525 as Crane does for equation 1-11. If you still have problems (or care ;-)), drop me an email at testdog2000@yahoo.com and I'll forward the steps to you.

I also did a hand check of my spreadsheet that I used to come up with the 31,000 lb/hr and got essentially the same answer. Are you including any 4" pipe in determining the total K for the system that could account for your lower flow rate?