I'm trying to calculate the pressure drop in a PSV discharge line. The set pressure is low so the pressure drop allowable is low to meet back pressure limitations. For the pipe exit to atmosphere, crane quotes a k-value of 1.0. For a 10" pipe this adds a lot to the equivalent length (~14m) versus a straight pipe length of only ~3m. This takes me over the allowable presure drop. Should I be including this K-value in the calc, or is the frictional pressure drop recovered by kinetic energy gain across the fitting, and thus the pressures are the same? Would appreciate any insight into this problem.
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PSV Pipeline Exit Pressure Drop 1
- Thread starter spudton
- Start date
spudton,
1) The velocity head (K=1) is not recovered at the pipe exit.
2) Crane calls it an exit loss. But remember, this pressure gets converted to velocity at the inlet, and is only "lost" at the exit. You can design an exit that recover most of the velocity head.
3) I would not include this exit loss (K=1) in the RV back pressure calcs. My justification is:
(a) on inlet losses we only use non-recoverable losses,
(b) the pipe exit is at atmospheric (0 guage), thus the pressure at the RV discharge is 0 plus frictional loss.
1) The velocity head (K=1) is not recovered at the pipe exit.
2) Crane calls it an exit loss. But remember, this pressure gets converted to velocity at the inlet, and is only "lost" at the exit. You can design an exit that recover most of the velocity head.
3) I would not include this exit loss (K=1) in the RV back pressure calcs. My justification is:
(a) on inlet losses we only use non-recoverable losses,
(b) the pipe exit is at atmospheric (0 guage), thus the pressure at the RV discharge is 0 plus frictional loss.
- Thread starter
- #4
Thanks for both your responses.
To clarify, are you saying that the velocity head loss across the pipe exit is lost as energy (heat) to the atmosphere, rather than a frictional pressure drop across the fitting?
Also (to help get this clear in my head), if there was a vessel at the exit of the pipe, rather than dischrge to atmosphere, the velocity head would be seen as a pressure drop across the fitting?
In the Crane manual example 4-22, the K=1.0 is included in the head loss calc. Doesn't this contradict what CJKruger says above?
Thanks again
To clarify, are you saying that the velocity head loss across the pipe exit is lost as energy (heat) to the atmosphere, rather than a frictional pressure drop across the fitting?
Also (to help get this clear in my head), if there was a vessel at the exit of the pipe, rather than dischrge to atmosphere, the velocity head would be seen as a pressure drop across the fitting?
In the Crane manual example 4-22, the K=1.0 is included in the head loss calc. Doesn't this contradict what CJKruger says above?
Thanks again
spudton,
No, this doesn’t contradict what CJKruger said. It depends on how you do your calculations. I always start with Bernoulli’s theorem so I can keep track of pressure, velocity, and elevation changes. I bet CJKruger does that part in his head on simple safety valve inlet and outlet line pressure drop calculations, like the majority of engineers do. Let me explain.
Start with Bernoulli’s theorem:
Z1 + 144P1/[ρ]1 + v12 /2g = Z2 + 144P2/[ρ]2 + v2 2/2g + hL
Station 1 is at the exit inside the pipe and Station 2 is outside the pipe.
Z1 = Z2, v2 = 0, and rearrange:
144P1/[ρ]1 - 144P2/[ρ]2 = hL - v12/2g
[Δ]P = P1 - P2 and you’ll see shortly that [ρ]1 = [ρ]2
144[Δ]P/[ρ] = hL - v1 2
hL = K1v12/2g
K1 = 1 so hL = v12/2g
Therefore:
144[Δ]P/[ρ] = v12/2g - v12/2g
144[Δ]P/[ρ] = 0
[Δ]P = 0 (this is why [ρ]1 = [ρ]2)
You see, I have to include the exit loss for things to work out because of the way I do my calculations. CJKruger and I are both right, once you know we go about our calculations differently.
In fact, Bernoulli’s theorem was used to derive Kexit loss = 1.
Sorry for confusing everyone, but I think I’ve explained it now.
Good luck,
Latexman
No, this doesn’t contradict what CJKruger said. It depends on how you do your calculations. I always start with Bernoulli’s theorem so I can keep track of pressure, velocity, and elevation changes. I bet CJKruger does that part in his head on simple safety valve inlet and outlet line pressure drop calculations, like the majority of engineers do. Let me explain.
Start with Bernoulli’s theorem:
Z1 + 144P1/[ρ]1 + v12 /2g = Z2 + 144P2/[ρ]2 + v2 2/2g + hL
Station 1 is at the exit inside the pipe and Station 2 is outside the pipe.
Z1 = Z2, v2 = 0, and rearrange:
144P1/[ρ]1 - 144P2/[ρ]2 = hL - v12/2g
[Δ]P = P1 - P2 and you’ll see shortly that [ρ]1 = [ρ]2
144[Δ]P/[ρ] = hL - v1 2
hL = K1v12/2g
K1 = 1 so hL = v12/2g
Therefore:
144[Δ]P/[ρ] = v12/2g - v12/2g
144[Δ]P/[ρ] = 0
[Δ]P = 0 (this is why [ρ]1 = [ρ]2)
You see, I have to include the exit loss for things to work out because of the way I do my calculations. CJKruger and I are both right, once you know we go about our calculations differently.
In fact, Bernoulli’s theorem was used to derive Kexit loss = 1.
Sorry for confusing everyone, but I think I’ve explained it now.
Good luck,
Latexman
-
1
- #7
I suspect that this thread has got very confused for two reasons. But before I explain any further let me state that I am assuming that your fluid is a gas or vapor. If it is not then stop reading now and disregard the rest of this post.
Firstly, the form of the Bernoulli equation that CJKruger and Latexman are referencing is specifically for incompressible fluids, and if you are using a gas it will just mislead and confuse you.
Secondly, it is difficult to look at just the discharge pipe. Of course, it is not impossible to analyze the discharge pipe alone, but it is much easier to understand conceptually if you consider the entire system from the pressure vessel through to the atmosphere.
OK, so I suppose I should justify these statements.
Latexman has set out the Bernoulli Equation in detail. This form has the units of length for each term, and this "length" is usually called the head. However, the head is in terms of the flowing fluid and with a gas the density is constantly changing along the length of the pipe so a foot of head at the start of the pipe is not the same as a foot of head at the discharge. With an incompressible liquid the density does not change and you can add the various terms involving head, but with a vapor you are adding apples and oranges. If you multiply each term of the equation by (Density x Gravity) then you convert each term to pressure units and then you can add them because you will have applied different densities at each point.
Also, I don't agree with Latexman's statement hL = V12/(2g). When we calculate the pressure drops in pipes we often take the inlet and outlet losses into account this way by lumping them in with the friction calc (i.e. Darcy-Weisbach), but what we are really doing is applying a partial form of Bernoulli. In my understanding hL is specifically the frictional resistance and the V12 term is taken care of separately in Bernoulli. In my opinion, Latexman is taking the V12 term into account twice, plus he is getting the signs confused and therefore they seem to cancel each other out. I would not do it that way.
CJKruger makes a very important poing by saying "But remember, this pressure gets converted to velocity at the inlet, and is only "lost" at the exit". This is true of liquids, but does not tell the whole story for gases. Naturally a portion of the pressure energy is converted to velocity at the inlet when gases are involved. But as the gas flows along the pipe and the pressure decreases the gas expands and the velocity continues to increase and this consumes energy. This requires a modified version of Bernoulli and as the math is a bit hairy for an old man like me I will leave it to you to check in your favorite text book! The outcome of this modified treatment is an equation like 1-6 in Crane TP410 (which is expressed in terms of pressure and not head for the reason I explained above).
Now let us consider the overall system in order to put the discharge pipe into context. If we assume the pressure vessel is large enough that we can regard the gas velocity in the vessel as zero, then we can regard the pressure energy in the vessel alone as supplying the energy for the following aspects
a) forcing the gas into the outlet nozzle (i.e. the inlet loss)
b) overcoming friction in the pipe between the vessel and the valve
c) overcoming the pressure drop through the valve
d) overcoming friction in the pipe between the valve and the atmosphere
e) accelerating the gas from zero to its final velocity at the exit. (i.e the exit loss)
You really need to look at this overall system in checking your valve. If you want to look at the discharge pipe alone you would have to take the velocity at the start of the discharge pipe into account, because the gas contains both kinetic and pressure energy at this point. Your acceleration term would then take into account only the increase in velocity from the start to the end of the discharge pipe.
If you have read this long story all the way through, then I guess you deserve an answer to your original question. Yes, you absolutely cannot disregard the gas velocity in sizing PSV piping. But if you are looking specifically at the discharge piping remember to take only the velocity difference from the valve's outlet flange to the end of the pipe into account. The losses in terms of changes of direction, acceleration, etc inside the valve will be taken care of in the Cv of the valve.
Katmar Software
Engineering & Risk Analysis Software
Firstly, the form of the Bernoulli equation that CJKruger and Latexman are referencing is specifically for incompressible fluids, and if you are using a gas it will just mislead and confuse you.
Secondly, it is difficult to look at just the discharge pipe. Of course, it is not impossible to analyze the discharge pipe alone, but it is much easier to understand conceptually if you consider the entire system from the pressure vessel through to the atmosphere.
OK, so I suppose I should justify these statements.
Latexman has set out the Bernoulli Equation in detail. This form has the units of length for each term, and this "length" is usually called the head. However, the head is in terms of the flowing fluid and with a gas the density is constantly changing along the length of the pipe so a foot of head at the start of the pipe is not the same as a foot of head at the discharge. With an incompressible liquid the density does not change and you can add the various terms involving head, but with a vapor you are adding apples and oranges. If you multiply each term of the equation by (Density x Gravity) then you convert each term to pressure units and then you can add them because you will have applied different densities at each point.
Also, I don't agree with Latexman's statement hL = V12/(2g). When we calculate the pressure drops in pipes we often take the inlet and outlet losses into account this way by lumping them in with the friction calc (i.e. Darcy-Weisbach), but what we are really doing is applying a partial form of Bernoulli. In my understanding hL is specifically the frictional resistance and the V12 term is taken care of separately in Bernoulli. In my opinion, Latexman is taking the V12 term into account twice, plus he is getting the signs confused and therefore they seem to cancel each other out. I would not do it that way.
CJKruger makes a very important poing by saying "But remember, this pressure gets converted to velocity at the inlet, and is only "lost" at the exit". This is true of liquids, but does not tell the whole story for gases. Naturally a portion of the pressure energy is converted to velocity at the inlet when gases are involved. But as the gas flows along the pipe and the pressure decreases the gas expands and the velocity continues to increase and this consumes energy. This requires a modified version of Bernoulli and as the math is a bit hairy for an old man like me I will leave it to you to check in your favorite text book! The outcome of this modified treatment is an equation like 1-6 in Crane TP410 (which is expressed in terms of pressure and not head for the reason I explained above).
Now let us consider the overall system in order to put the discharge pipe into context. If we assume the pressure vessel is large enough that we can regard the gas velocity in the vessel as zero, then we can regard the pressure energy in the vessel alone as supplying the energy for the following aspects
a) forcing the gas into the outlet nozzle (i.e. the inlet loss)
b) overcoming friction in the pipe between the vessel and the valve
c) overcoming the pressure drop through the valve
d) overcoming friction in the pipe between the valve and the atmosphere
e) accelerating the gas from zero to its final velocity at the exit. (i.e the exit loss)
You really need to look at this overall system in checking your valve. If you want to look at the discharge pipe alone you would have to take the velocity at the start of the discharge pipe into account, because the gas contains both kinetic and pressure energy at this point. Your acceleration term would then take into account only the increase in velocity from the start to the end of the discharge pipe.
If you have read this long story all the way through, then I guess you deserve an answer to your original question. Yes, you absolutely cannot disregard the gas velocity in sizing PSV piping. But if you are looking specifically at the discharge piping remember to take only the velocity difference from the valve's outlet flange to the end of the pipe into account. The losses in terms of changes of direction, acceleration, etc inside the valve will be taken care of in the Cv of the valve.
Katmar Software
Engineering & Risk Analysis Software
Harvey, Good explaination but you are thinking in terms of a system containing a control valve. The original post is for a PSV tail pipe so there is no Cv to consider. I think you will agree that taking the calculation from the PSV outlet flange to final discharge is the correct approach.
Isn't a PSV just a quick opening control valve? ;-)
Katmar Software
Engineering & Risk Analysis Software
Katmar Software
Engineering & Risk Analysis Software
Not really. A control valve is treated thermodynamically as an adiabatic reduction but a PSV is treated as a nozzle that follows the path of an isentropic expansion. We do not consider Cv in the sizing of a PSV nor can you find Cv's listed with any PSV that I know of.
By the way, how do you input that smiling face?
By the way, how do you input that smiling face?
Probably Cv was the wrong term to use, but whether you use an orifice coefficient, or K value, or even a graphical method for the PSV pressure drop the priciple remains the same i.e. the published data for the PSV should not assume that there is any pressure recovery at the outlet.
For the smiley face - I just typed in the traditional semi-colon, dash, right bracket and the Eng-Tips software converted it into the graphic. When you are in the "Preview Post" mode there is a link to how to do all the smileys (or emoticons, to use the posh word!).
Harvey
For the smiley face - I just typed in the traditional semi-colon, dash, right bracket and the Eng-Tips software converted it into the graphic. When you are in the "Preview Post" mode there is a link to how to do all the smileys (or emoticons, to use the posh word!).
Harvey
Harvey,
In PSV system analysis and calculations, one does not consider a pressure drop through a PSV, it's just not treated in the same way as other valves. Once the theoretical orifice flow area is obtained, it is multiplied (or divided if you will) by the certifed coefficient of discharge which is measured in a lab for that particular model valve. This coefficient adjusts the valve size for the true nature of flow since the PSV is not an ideal nozzle. This coefficient should not be mistaken as a coefficient that can be used in a hydraulic calculation because it has nothing really to do with pressure drop.
The two pressures of concern are the pressure at the inlet to the PSV and the back pressure against the PSV tailpipe flange. Yes, this is a "pressure drop" but they are determined from inlet piping and outlet piping hydraulics, not what happens inside the PSV itself. The inlet pressure loss is based only on non-recoverable losses (if a liquid, static head is not even considered) and the base pressure used in the calculation is the PSV set pressure, not even the actual relieving pressure, which is a percentage higher. Therefore, you don't even calculate a real pressure drop accross these valves, and it is not necessary to do so.
The Coefficient of Discharge is reported as an average of several "runs" and is not manipulated, so whatever it turns out to be, it is. If there are pressure recovery effects within the nozzle and valve body, or not, it will be reflected in this value. The PSV body is extremely complicated especially compared to other types of valves.
Thanks for the smiley face tip.
In PSV system analysis and calculations, one does not consider a pressure drop through a PSV, it's just not treated in the same way as other valves. Once the theoretical orifice flow area is obtained, it is multiplied (or divided if you will) by the certifed coefficient of discharge which is measured in a lab for that particular model valve. This coefficient adjusts the valve size for the true nature of flow since the PSV is not an ideal nozzle. This coefficient should not be mistaken as a coefficient that can be used in a hydraulic calculation because it has nothing really to do with pressure drop.
The two pressures of concern are the pressure at the inlet to the PSV and the back pressure against the PSV tailpipe flange. Yes, this is a "pressure drop" but they are determined from inlet piping and outlet piping hydraulics, not what happens inside the PSV itself. The inlet pressure loss is based only on non-recoverable losses (if a liquid, static head is not even considered) and the base pressure used in the calculation is the PSV set pressure, not even the actual relieving pressure, which is a percentage higher. Therefore, you don't even calculate a real pressure drop accross these valves, and it is not necessary to do so.
The Coefficient of Discharge is reported as an average of several "runs" and is not manipulated, so whatever it turns out to be, it is. If there are pressure recovery effects within the nozzle and valve body, or not, it will be reflected in this value. The PSV body is extremely complicated especially compared to other types of valves.
Thanks for the smiley face tip.
sailoday28
Mechanical
A portion of this thread relates to the exit loss "K".
For subsonic flow, the pressure at the exit of an expansion is that of the surrounding ambient.
Consider a sudden expansion for a smaller to a larger diameter. The K for the friction losses is simply derived by taking into account the momentum fo the fluid.
For expansion to the atmosphere, the ratio of areas is either zero or infinite, depending on how one relates velocity ratios.
For a simple explantion see
Friction Losses
Regards
For subsonic flow, the pressure at the exit of an expansion is that of the surrounding ambient.
Consider a sudden expansion for a smaller to a larger diameter. The K for the friction losses is simply derived by taking into account the momentum fo the fluid.
For expansion to the atmosphere, the ratio of areas is either zero or infinite, depending on how one relates velocity ratios.
For a simple explantion see
Friction Losses
Regards
sailoday28
Mechanical
Latexman (Chemical)Your derivations was for incompressible flow. My statement includes compressible fluids.
Regards
Regards
Latexman's "hint" made me realize that if I apply my earlier recommendation of only taking into account the change in velocity through the discharge piping it is close to nothing when the valve set pressure is low and the pipe is short (spudton mentioned 3m).
Also, the exit loss for a PSV with no discharge piping is much the same as for one with discharge piping when the set pressure is low. The discharge piping has little or no effect on the exit loss and it only adds frictional loss to the situation of having the valve alone.
For a low set pressure PSV I would therefore agree with Latexman and CJKruger that you can disregard the exit loss.
Katmar Software
Engineering & Risk Analysis Software
Also, the exit loss for a PSV with no discharge piping is much the same as for one with discharge piping when the set pressure is low. The discharge piping has little or no effect on the exit loss and it only adds frictional loss to the situation of having the valve alone.
For a low set pressure PSV I would therefore agree with Latexman and CJKruger that you can disregard the exit loss.
Katmar Software
Engineering & Risk Analysis Software
- Thread starter
- #18
Sorry to labor this point, but I do not believe you are correct to use pressure recovery according to the Bernoulli principle as the reason why you can ignore the gas velocity head. In the specific instance you have, with a low set pressure you can ignore it, but if you assumed that the Bernoulli principle was the reason why you might be tempted to ignore it at higher pressures too. And this could lead to very wrong results.
The real reason why you can ignore the velocity head for low set pressures is that the velocity head of the gas entering the discharge pipe is virtually the same as the velocity head at the exit to the atmosphere. So there is no net change to the velocity head component caused by the discharge pipe. The discharge pipe only introduces additional frictional losses and with low set pressures these can be calculated sufficiently accurately using incompressible fluid assumptions.
Have a look at
for an example of the calculation for the discharge piping for a PSV with a set pressure 150 psi. This example compares the short-cut method where incompressible flow is assumed, with a full-blown compressible fluid analysis similar to the Crane TP410 Eq 1-6 procedure that I referred to earlier. The difference between the two results is dramatic to say the least.
If you apply the rigorous procedure to an instance where you could have got away with the short-cut method you will have done a bit of extra work, but you will still get the right answer. On the other hand, if you apply the short-cut method where it does not apply you would get a very wrong answer. Engineering experience is all about knowing when you can take short cuts.
I agree with Latexman - post your actual conditions and we can see how much difference it makes.
Katmar Software
Engineering & Risk Analysis Software
The real reason why you can ignore the velocity head for low set pressures is that the velocity head of the gas entering the discharge pipe is virtually the same as the velocity head at the exit to the atmosphere. So there is no net change to the velocity head component caused by the discharge pipe. The discharge pipe only introduces additional frictional losses and with low set pressures these can be calculated sufficiently accurately using incompressible fluid assumptions.
Have a look at
for an example of the calculation for the discharge piping for a PSV with a set pressure 150 psi. This example compares the short-cut method where incompressible flow is assumed, with a full-blown compressible fluid analysis similar to the Crane TP410 Eq 1-6 procedure that I referred to earlier. The difference between the two results is dramatic to say the least.
If you apply the rigorous procedure to an instance where you could have got away with the short-cut method you will have done a bit of extra work, but you will still get the right answer. On the other hand, if you apply the short-cut method where it does not apply you would get a very wrong answer. Engineering experience is all about knowing when you can take short cuts.
I agree with Latexman - post your actual conditions and we can see how much difference it makes.
Katmar Software
Engineering & Risk Analysis Software
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