Permanent pressure drop through an orifice

[DanielWW]

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!

 

[BigInch]

K doesn't care if the fluid is compressible or not. It's just a pressure drop for a given flow rate. If the pressure drop is significant and you have a gas, then you will get a change in density of the gas from what it was at the upstream pressure to whatever its new density is at the downstream pressure. If you have a slightly compressible liquid, the density will be the same for all practical purposes, providing that your downstream pressure is still above the liquid's vapor pressure.

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.

 

[DanielWW]

Thanks for the response!

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

 

[BigInch]

Both equations do equal head loss and can be equated, however it defeats the purpose of each. Each equation was designed to apply to a specific device. For an orifice (or a valve) or any pressure drop that occurs at what can be considered a single point in the system is more conveniently calculated from Cv, or Kv equation, as it is not dependent on length. fD/L is in terms of length, to meet a need to calculate a pressure drop across a long flow element. That is convenient in a piping system where one must know the pressure at many points along a pipe, since all that is required is to simply multiply by the distance to those respective points. If you are only interested in the pressure at one point along a pipe, or have a fixed length of pipe and want the pressure drop at the end, then you can use them interchangeably, but at every different length of pipe you would have to calculate a new Cv or Kv term. Equating the two seems to be unnecessary for anything except finding equivalent lengths.

As I say, the only use I can imagine for equating the two and solving for [Δ]P 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.

 

[DanielWW]

Ok, that makes sense. Let me see if I understand...

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.

 

[BigInch]

Use K values for orifice and valves, equivalent lengths for fittings, either calculated, or from tables.

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=&esrc=s&source=web&cd=1&ved=0CCgQFjAA&url=http://

http://www.ucalgary.ca/ENCH/class_n...woDwAQ&usg=AFQjCNHsZrtzJJu-0dDMEq90nl61eD8HFw

Independent events are seldomly independent.