Permanent pressure drop through an orifice
Permanent pressure drop through an orifice
(OP)
Hello,
I am trying to calculate the non recoverable pressure drop of a compressible fluid through an orifice. In Crane 410, it says that the pressure loss coefficient K can be calculated by K=[√(1-β^4 (1-C_d^2 ) )/(C_d β^2 )-1]^2. Can this coefficient be used for both compressible and incompressible flows? I plan on using it with an equation for isothermal compressible flow found in API 521: f*l/d=1/Ma_1^2*[1-(p_2/p_1)^2]-ln(p_1/p_2)^2. I would substitute K for the f*l/d term and solve the equation for p2 in excel using Goal Seek.
Does that procedure make sense? Is there a better way to go about it? I just started as an engineer so please forgive me if I'm way off here,
. Thanks in advance for your help!
I am trying to calculate the non recoverable pressure drop of a compressible fluid through an orifice. In Crane 410, it says that the pressure loss coefficient K can be calculated by K=[√(1-β^4 (1-C_d^2 ) )/(C_d β^2 )-1]^2. Can this coefficient be used for both compressible and incompressible flows? I plan on using it with an equation for isothermal compressible flow found in API 521: f*l/d=1/Ma_1^2*[1-(p_2/p_1)^2]-ln(p_1/p_2)^2. I would substitute K for the f*l/d term and solve the equation for p2 in excel using Goal Seek.
Does that procedure make sense? Is there a better way to go about it? I just started as an engineer so please forgive me if I'm way off here,
. Thanks in advance for your help! 




RE: Permanent pressure drop through an orifice
What doesn't make sense is setting the equation for pressure drop across an orifice = the friction pressure loss for a length of pipe, unless you want to calculate an equivalent length of pipe that would give you the same pressure drop as the orifice does. Otherwise your answer is oranges = apples.
Independent events are seldomly independent.
RE: Permanent pressure drop through an orifice
In Crane 410 it says that head loss can be calculated as K*v^2/2g. It then says that by comparing that to the Darcy equation, the hydraulic resistance of a straight pipe can be expressed in terms of a resistance coefficient by K= f*l/D. That's why I thought that the f*l/d term could simply be replaced with a resistance coefficient. I used the equation for isothermal, compressible flow mainly because it was used previously for this type of calculation by engineers at my company.
How then should I go about calculating the pressure drop using the resistance coefficient? Can I say that the headloss= K*v^2/2g? And P_delta=headloss*rho*g?
Thanks again
RE: Permanent pressure drop through an orifice
As I say, the only use I can imagine for equating the two and solving for ΔP is to find the equivalent length of an orifice, or valve, or to find an equivalent Cv, or Kv for fixed length and fixed diameter of pipe. I wouldn't be interested in recalculating Cv for every point for the next 1000 km. As I also often calculate pressure drops across valves at different %Open positions, causing Cv to become variable as well, calculating equivalent lengths from a fixed Cv isn't appealing to me either. Maybe I'm the wrong person to be answering this question.
That aside, to calculate the pressure drop along a pipe when considering compressibility, let's start upstream. Calculate the pressure drop using fL/D at the upstream density, then calculate the pressure drop again, this time using the density at downstream pressure and temperature. If the two answers are the same or close enough, stop. If there is a difference, shorten the length of pipe. Make sure that the mass is constant and that velocity remains less than sonic.
Independent events are seldomly independent.
RE: Permanent pressure drop through an orifice
Just to clarify, I'm trying to create a spreadsheet that asks the user to input information about the pipe and the pipe elements along that pipe. The goal is to calculate the pressure drop due to major losses and minor losses from valves, bends, orifices, etc. The procedure I used was to first sum the K values for all pipe elements and the fl/d value for the pipe. That value was substituted for the fl/d term in the equation for compressible,isothermal flow, and then the equation was solved for pressure drop. This procedure was based on previous work at my company.
I think that the reasoning for using the equation for compressible, isothermal flow was that the Darcy equation would not be applicable due to compressibility effects. However, after reading your last paragraph and doing some more research, it seems that as long as the pressure drop is small in comparison to the inlet pressure, the darcy equation is still applicable. (as long as you take both upstream and downstream densities into consideration, as you mentioned)
So if I'm understanding correctly, while the procedure is not technically wrong, it would probably make more sense just to use the darcy equation.
Hopefully that makes sense. Thanks for your help.
RE: Permanent pressure drop through an orifice
Ya. That's it. You never get the "right" answer, just an approximation that is good enough for your purposes, within your XL decimal setting is as close as you can get. You can use whatever pipe flow equation you like. Some for natural gas, the AGA equations for example, try to avoid compressibility and solve directly. Their accuracy is not bad within their limits, but IMO I think it is best to use Colebrook-White's equation and calculate densities via the compressibility factor using some other appropriate method, compressibility charts, the Compressed Natural Gas Association formula, Soave-Redlich-Kwong, or Benedict-Webb-Rubin-Starling, but I'm thinking about gas mixtures. CNGA is an easy one to program, but has to be used within its somewhat narrow gas specific gravity limits (dry natural gas, SG = 0.6 to 0.65 or something.
http://www.google.ae/url?sa=t&rct=j&q=&...
Independent events are seldomly independent.