If a gas passing through a 10" pipe has a pressure of 15 psig enters a stack that is 36" in diameter. The stack is 80 ft long and is open to the atmosphere. Would the gas as it enters the stack still have a pressure? or would the pressure at the base of the stack be atmospheric.
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Pressure loss of gas from a 10" pipe to a stack
- Thread starter CHEMTT
- Start date
If there is a flow up the stack there will be a pressure drop, and therefore something more than 0 psig in the base of the stack. The magnitude of the pressure drop depends on how far along the 10" pipe you measured the 15 psig. If it was a mile away the pressure at the base of the stack will be small. It will be somewhat higher if you measured 15 psig only 1 ft away from the stack.
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
- Thread starter
- #3
The 15 psig was calculated right before the gas entered the stack. Can PV=nRT be used to calculate the pressure in the stack after it has been expanded? Also, can the transition from the pipe to the stack be treated as a sudden expansion from 10" to 36" or can it be assumed that the pressure will go to atm as it enters the stack?
PV=nRT applies to a closed system, so no, it cannot be used in your open system.
The transition from pipe to stack can be treated as a sudden expansion.
You should get about the same answers if you start at the pipe and calculate downstream (to atmosphere) or if you start at atmosphere and calculate upstream (to the pipe). Then, you will have convergence.
Good luck,
Latexman
The transition from pipe to stack can be treated as a sudden expansion.
You should get about the same answers if you start at the pipe and calculate downstream (to atmosphere) or if you start at atmosphere and calculate upstream (to the pipe). Then, you will have convergence.
Good luck,
Latexman
What happens upstream of the stack will not affect the pressure in the base of the stack. Work back from your known reference point, i.e. the discharge from the stack is at atmopheric pressure. If you know the flow then you can calculate the pressure drop up the stack. That gives you the pressure in the base of the stack.
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
Can PV=nRT be used to calculate the pressure in the stack after it has been expanded?
Since you also don't know a flowrate, you don't know how long you have before this has to happen again.
You can't do this by PV=nRT itself alone. P2, T2 and V2 suggest up to 3 unknowns and you only have that one equation. You could possibly assume T is constant and estimate a P2 by distance from the end of the stack, or by writing flow equations.
That said, assuming steady state, which may be quite an assumption, you can see that pressure in the base of the stack must be quite low, because of the sudden volume and velocity change. Assume it is 5 psig there and 0 psig at the end of the 80 ft run and, using some flowrate equation, see what flowrate you can get with that pressure drop. Put that back into the V = nRT/P and see how that looks. Keep massaging it 'till V per second equals the latest flowrate you got in the 80 ft pipe.
Let your acquaintances be many, but your advisors one in a thousand’ ... Book of Ecclesiasticus
Since you also don't know a flowrate, you don't know how long you have before this has to happen again.
You can't do this by PV=nRT itself alone. P2, T2 and V2 suggest up to 3 unknowns and you only have that one equation. You could possibly assume T is constant and estimate a P2 by distance from the end of the stack, or by writing flow equations.
That said, assuming steady state, which may be quite an assumption, you can see that pressure in the base of the stack must be quite low, because of the sudden volume and velocity change. Assume it is 5 psig there and 0 psig at the end of the 80 ft run and, using some flowrate equation, see what flowrate you can get with that pressure drop. Put that back into the V = nRT/P and see how that looks. Keep massaging it 'till V per second equals the latest flowrate you got in the 80 ft pipe.
Let your acquaintances be many, but your advisors one in a thousand’ ... Book of Ecclesiasticus
racookpe1978
Nuclear
Caution: ALL calc's will be very different if the gas coming out of the 10 inch pipe flares at the top of the stack.
If no flare, I'd make an approximation of a single sudden expansion - from 10 to 30 inch dia - plus a 90 degree elbow (as if the 10 inch pipe turned and ran vertically), and then a 30 inch pipe vertically into a open-end pipe the length of the pipe. Then just end the combined "pipe" with a second open-end expansiono into atmospheric pressure.
You originally asked if the gas in the pipe has "a pressure"? Obviously, yes it does. Even if that pressure is only 1/2 inch water above atmosphere, it has a pressure.
If no flare, I'd make an approximation of a single sudden expansion - from 10 to 30 inch dia - plus a 90 degree elbow (as if the 10 inch pipe turned and ran vertically), and then a 30 inch pipe vertically into a open-end pipe the length of the pipe. Then just end the combined "pipe" with a second open-end expansiono into atmospheric pressure.
You originally asked if the gas in the pipe has "a pressure"? Obviously, yes it does. Even if that pressure is only 1/2 inch water above atmosphere, it has a pressure.
- Thread starter
- #8
I am using the sudden expansion formula from the attached website and it is only giving me a 70% reduction in pressure due to the sudden expansion. So, at base of the stack I am getting 4.5 psig.
However, if i work back from the top of the stack it does not converge. The pressure drop across the 36" stack is only 0.2 psi, then the pressure based on that theory is 0.2 psig.
However, if i work back from the top of the stack it does not converge. The pressure drop across the 36" stack is only 0.2 psi, then the pressure based on that theory is 0.2 psig.
- Thread starter
- #11
Inlet pressure of gas is 95 psig and temperature is 1150 F.
The pipe is made up of a combination of 8" and 10" pipe. The pipe starts as 8" turns into 10" and then ties into a 36" stack.
Here are some details of the piping
Length and Fittings of 8" Pipe
Length - 55 ft
90 deg bends - 3
Valve - butterfly valve
Expansion from 8" to 10"
Expansion joint - 1
Length and fittings of 10" pipe
Length - 50 ft
90 deg bends - 2
45 deg bends - 2
Expansion from 10" to 36"
Expansion joint - 1
T-piece - 1
Based on the above info i have calculated the pressure just before it enters the stack to be ~15 psig.
The pipe is made up of a combination of 8" and 10" pipe. The pipe starts as 8" turns into 10" and then ties into a 36" stack.
Here are some details of the piping
Length and Fittings of 8" Pipe
Length - 55 ft
90 deg bends - 3
Valve - butterfly valve
Expansion from 8" to 10"
Expansion joint - 1
Length and fittings of 10" pipe
Length - 50 ft
90 deg bends - 2
45 deg bends - 2
Expansion from 10" to 36"
Expansion joint - 1
T-piece - 1
Based on the above info i have calculated the pressure just before it enters the stack to be ~15 psig.
You will have to give the flow rate and the MW of the gas for anyone to make sensible comments.
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
- Thread starter
- #13
Now that you have given more detail it becomes a rather difficult problem to solve. About the only things that we can say for sure is that PV=nRT and the standard sudden expansion formula are NOT going to give you the correct answer.
One thing that we can calculate, and which we agree on, is that with a flow of 135,000 lb/hour up the stack the base pressure will be close to 0.2 psig.
The difficult bit is to calculate what is happening in the 8" and 10" pipe sections. I would be interested to see how you calculated a pressure of 15 psig at the end of the 10" section. I agree with Latexman that you will have sonic flow at the end of the 10" section, although I get a velocity of 620 ft/second at the start of the 8" section rather than his Mach 8.42 (I suspect he typed one too many zeros into the flow in his program!).
With these very high velocities the standard ways of calculating pressure drops become inaccurate, and we can only make estimates in this case. For example, when calculating the pressure drop through fittings for incompressible flow (i.e. liquids) we do not worry about where in the line the fitting actually is because the velocity is the same all along the line. This is not true for compressible flow because the velocity increases along the line. If we accept that the Reynolds number with gas flow is sufficiently high for the fitting's K value to be the same where ever it is located in the line, then we can say that the pressure drop across the fitting is (K[ρ]v2)/2. ( [ρ] is density, v is velocity )
As the gas flows down the line [ρ] and v vary in inverse proportion to each other. But because the velocity element is squared in the pressure drop calculation, a fitting at the end of the line will have a higher pressure drop than an identical fitting at the start of the line (for the same mass flow).
In my own calculations for compressible flow I take the conservative assumption and calculate all the fittings for the density and velocity at the end of the line. This usually makes a trivial difference if we are sizing normal plant piping where the pressure drop is a small fraction of the inlet pressure.
But for your example, this assumption would lead to significant errors. Because of the lack of detail, I have assumed that the combined 8" and 10" sections can be represented by 180 ft of straight 8" pipe. This will give us an estimate of what is happening, but should not be regarded as a design-level calculation. The procedure I followed was to calculate the flow for an assumed end pressure, and then gradually lowered the assumed end pressure until I got to atmospheric pressure.
This shows that as the end pressure goes below about 40 psig there is virtually no further increase in mass flow rate. The flow rate I get is close to your 135,000 lb/hour and because of my assumptions I cannot say that you are right or wrong. But it is very difficult to fix the end pressure exactly, and as I said earlier, I would like to see how you got to 15 psig.
The result of all these calculations is that you are likely to have a sonic shock wave at the entrance to the stack, and the analysis of this is way beyond PV=nRt or the usual sudden expansion methods. If you have not already built this pipeline I would suggest looking at its design again. If this is a normal condition of flow it is going to be a very noisy piece of plant.
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
One thing that we can calculate, and which we agree on, is that with a flow of 135,000 lb/hour up the stack the base pressure will be close to 0.2 psig.
The difficult bit is to calculate what is happening in the 8" and 10" pipe sections. I would be interested to see how you calculated a pressure of 15 psig at the end of the 10" section. I agree with Latexman that you will have sonic flow at the end of the 10" section, although I get a velocity of 620 ft/second at the start of the 8" section rather than his Mach 8.42 (I suspect he typed one too many zeros into the flow in his program!).
With these very high velocities the standard ways of calculating pressure drops become inaccurate, and we can only make estimates in this case. For example, when calculating the pressure drop through fittings for incompressible flow (i.e. liquids) we do not worry about where in the line the fitting actually is because the velocity is the same all along the line. This is not true for compressible flow because the velocity increases along the line. If we accept that the Reynolds number with gas flow is sufficiently high for the fitting's K value to be the same where ever it is located in the line, then we can say that the pressure drop across the fitting is (K[ρ]v2)/2. ( [ρ] is density, v is velocity )
As the gas flows down the line [ρ] and v vary in inverse proportion to each other. But because the velocity element is squared in the pressure drop calculation, a fitting at the end of the line will have a higher pressure drop than an identical fitting at the start of the line (for the same mass flow).
In my own calculations for compressible flow I take the conservative assumption and calculate all the fittings for the density and velocity at the end of the line. This usually makes a trivial difference if we are sizing normal plant piping where the pressure drop is a small fraction of the inlet pressure.
But for your example, this assumption would lead to significant errors. Because of the lack of detail, I have assumed that the combined 8" and 10" sections can be represented by 180 ft of straight 8" pipe. This will give us an estimate of what is happening, but should not be regarded as a design-level calculation. The procedure I followed was to calculate the flow for an assumed end pressure, and then gradually lowered the assumed end pressure until I got to atmospheric pressure.
This shows that as the end pressure goes below about 40 psig there is virtually no further increase in mass flow rate. The flow rate I get is close to your 135,000 lb/hour and because of my assumptions I cannot say that you are right or wrong. But it is very difficult to fix the end pressure exactly, and as I said earlier, I would like to see how you got to 15 psig.
The result of all these calculations is that you are likely to have a sonic shock wave at the entrance to the stack, and the analysis of this is way beyond PV=nRt or the usual sudden expansion methods. If you have not already built this pipeline I would suggest looking at its design again. If this is a normal condition of flow it is going to be a very noisy piece of plant.
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
pipesnpumps
Mechanical
Is this a discharge from a pressure safety valve? If so I assume you have already considered that, if the set pressure is lower than 400 psig, that a bellows on the psv would be required due to the >10% back pressure.
Real world knowledge doesn't fall out of the sky on a parachute, but rather is gained in small increments during moments of panic or curiosity.
I either slipped in an extra 0 or forgot the MW. Probably the former. But for some funny reason, at the moment I don't care! Currently I'm sitting on the front porch lookinh due South into tha Atlantic Ocean at Oak Isalnd, NC, rocking and sipping my favorite drink. Ya'll come!
Good luck,
Latexman
Good luck,
Latexman
Hey Latexman, you are making me jealous. We are in the middle of an unusually cold winter here in Durban (8 deg C is extreme for us!) and it is a normal working day. Enjoy the holiday, and have a sip for me!
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
Katmar Software - Engineering & Risk Analysis Software
"An undefined problem has an infinite number of solutions"
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