Using Bernoulli's Eq, I have attempted to calculate the exit air pressure and velocity of a system which consists of a tank (~5 gallon), 3" long 0.25" dia pipe, and an exit nozzle that is basically a cone shape (0.25" to 6" diameter). The tank will be pressurized to 10 psi. Why is my pressure at the 0.25" pipe grossly negative? I have followed a similar example that assumes free jet at exit, but I want to determine the pressure at that point if I were to place any sort of permeable material in-line. Is this calculable?
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exit tank pressure 1
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sailoday28
Mechanical
Is there inflow to the tank?
What is the temperature of air in lthe tank?
Over what length does the cone change its diameter?
What is the temperature of air in lthe tank?
Over what length does the cone change its diameter?
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1
- #3
The problem is that Bernoulli's equation is an energy equation, based on the conditions at two points, and does not take into account the conditions in between. This is it's chief advantange in many cases, but can lead to false results.
If you apply Bernoulli's equation beteen the tank (high pressure, zero velocity) and the 6" exit (atmospheric pressure), you can find a velocity and thus mass flow rate. Applying this mass flow rate at the 0.25" pipe will give some enormous velocity that requires a negative pressure to reach.
You are assuming that the sum of the kinetic energy (velocity squared) and potential energy (pressure) are constant. On paper, either can be increased without limit by making the other smaller or even negative. But of course in real life, you'll get imaginary velocities or negative absolute pressures if you run it too far, and that indicates that your conditions are not in line with the theory. With fluids, you'll get cavitation effects that you might not expect.
You can get a similar effect by analyzing the flow of water in a vertical pipe open at the bottom; you can calculate negative absolute pressures and enormous velocities.
Suppose you replaced that 0.25" pipe with a valve. Applying Bernoulli's equation between the tank and the 6" exit will give exactly the same flow rate regardless of whether that valve is open, partially closed, or even fully closed.
Right offhand, I don't recall the assumptions behind Bernoulli's equation, but it may be based on incompressible flow- look it up in a fluids book and see. With 10 PSI and atmospheric, you'll have a noticable variation in density, and if the equation is set up for liquid flow, it will need some adjustment for that.
An additional concern is compressibile fluid flow in general. Anytime your calculated velocity anywhere in the system exceeds about Mach 0.3 or so, compressibility and sonic effects begin to become important, and I suppose this would be the case in your problem. Once again, check up on supersonic/ sonic flow through nozzles in a fluids book.
If you apply Bernoulli's equation beteen the tank (high pressure, zero velocity) and the 6" exit (atmospheric pressure), you can find a velocity and thus mass flow rate. Applying this mass flow rate at the 0.25" pipe will give some enormous velocity that requires a negative pressure to reach.
You are assuming that the sum of the kinetic energy (velocity squared) and potential energy (pressure) are constant. On paper, either can be increased without limit by making the other smaller or even negative. But of course in real life, you'll get imaginary velocities or negative absolute pressures if you run it too far, and that indicates that your conditions are not in line with the theory. With fluids, you'll get cavitation effects that you might not expect.
You can get a similar effect by analyzing the flow of water in a vertical pipe open at the bottom; you can calculate negative absolute pressures and enormous velocities.
Suppose you replaced that 0.25" pipe with a valve. Applying Bernoulli's equation between the tank and the 6" exit will give exactly the same flow rate regardless of whether that valve is open, partially closed, or even fully closed.
Right offhand, I don't recall the assumptions behind Bernoulli's equation, but it may be based on incompressible flow- look it up in a fluids book and see. With 10 PSI and atmospheric, you'll have a noticable variation in density, and if the equation is set up for liquid flow, it will need some adjustment for that.
An additional concern is compressibile fluid flow in general. Anytime your calculated velocity anywhere in the system exceeds about Mach 0.3 or so, compressibility and sonic effects begin to become important, and I suppose this would be the case in your problem. Once again, check up on supersonic/ sonic flow through nozzles in a fluids book.
sailoday28
Mechanical
Bernoulli's equation (the original) is based on non friction flow. It is applicable to both liquids and gases.
The engineering Bernoulli's equation includes friction losses.
Yes, low and even negative pressures can be achieved if the upstream conditions of velocity (KE) are higher than the exity KE. Older washing machines used this principal to to dump the water from the tub by a liquid ejector.
The engineering Bernoulli's equation includes friction losses.
Yes, low and even negative pressures can be achieved if the upstream conditions of velocity (KE) are higher than the exity KE. Older washing machines used this principal to to dump the water from the tub by a liquid ejector.
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- #6
Thanks for the help folks.
sailoday28, I plan to pressure the tank to 10 psi and close it off. Temp is ~70F and the cone changes from .25" to 6" over 2.5". I was hoping to neglect friction for ease of calculation. After referring to my fluids book, maybe that's not the case.
jstephen, I had looked into the Mach 0.3 velocity but was unsure due to the wild numbers I was getting for pressure. Make's sense, I have roughly Mach 58 at the .25" pipe!! Hence the -33202 psi at the same point.
So if I were to close off the nozzle with my permeable material (possibly creating another cavity depending on the through-flow of the material), I would basically see no flow (or some small flow based on mat'l characteristics) and close to a balanced pressure in the system. Right? I understand there would be a minute pressure drop at the exit of the .25" pipe based on out flow.
Do you guys know if there is already a test rig out there that will accomplish air flow through a material for measurement of permeability?? (Shot in the dark!)
sailoday28, I plan to pressure the tank to 10 psi and close it off. Temp is ~70F and the cone changes from .25" to 6" over 2.5". I was hoping to neglect friction for ease of calculation. After referring to my fluids book, maybe that's not the case.
jstephen, I had looked into the Mach 0.3 velocity but was unsure due to the wild numbers I was getting for pressure. Make's sense, I have roughly Mach 58 at the .25" pipe!! Hence the -33202 psi at the same point.
So if I were to close off the nozzle with my permeable material (possibly creating another cavity depending on the through-flow of the material), I would basically see no flow (or some small flow based on mat'l characteristics) and close to a balanced pressure in the system. Right? I understand there would be a minute pressure drop at the exit of the .25" pipe based on out flow.
Do you guys know if there is already a test rig out there that will accomplish air flow through a material for measurement of permeability?? (Shot in the dark!)
I didn't realize the cone was that abrupt. You could probably make all calculations just neglecting the cone and working with the orifice. Flow at the cone outlet would no longer be axial flow.
If the material you're testing is mostly impermeable, and IF you can get it sealed to the cone all around, then flow pressure drop should be mostly in the material. IE, 10 PSI at one side of the material, atmospheric at the other side.
If the material is very permeable, you might do better using a fan, duct, and orifice plate to measure flow and manometer across the material tested to measure pressure drop.
If the material you're testing is mostly impermeable, and IF you can get it sealed to the cone all around, then flow pressure drop should be mostly in the material. IE, 10 PSI at one side of the material, atmospheric at the other side.
If the material is very permeable, you might do better using a fan, duct, and orifice plate to measure flow and manometer across the material tested to measure pressure drop.
sailoday28
Mechanical
JStephen (Mechanical) Dec 21, 2004
sailoday, I'm referring to absolute pressure, which can be negative in the equations, but not in real fluids. And this is the case regardless of whether friction is considered.
Liquid water is a real fluid and it can flow at pressures less than atmospheric as long as the temperature is less than the corresponding the vapor pressure. For example consider saturated liquid water at 10psia and 193.19F
Liquid water can flow at 10 psia as long as the temp is not greater than 193.19.
sailoday, I'm referring to absolute pressure, which can be negative in the equations, but not in real fluids. And this is the case regardless of whether friction is considered.
Liquid water is a real fluid and it can flow at pressures less than atmospheric as long as the temperature is less than the corresponding the vapor pressure. For example consider saturated liquid water at 10psia and 193.19F
Liquid water can flow at 10 psia as long as the temp is not greater than 193.19.
sailoday28
Mechanical
JStephen (Mechanical) Dec 21, 2004
sailoday, I'm referring to absolute pressure, which can be negative in the equations, but not in real fluids. And this is the case regardless of whether friction is considered.
My last response was meant to refer to fluids in an environment of less than atmospheric pressure.
However, real fluids can have negative pressures (fluid in tension). This condition of metastable equilibrium, while not stable, has been tabulated in steam tables. See for instance Keenan, Keyes, Hill and Moore, Thermo Proper of Water...... John Wiley and Sons.
sailoday, I'm referring to absolute pressure, which can be negative in the equations, but not in real fluids. And this is the case regardless of whether friction is considered.
My last response was meant to refer to fluids in an environment of less than atmospheric pressure.
However, real fluids can have negative pressures (fluid in tension). This condition of metastable equilibrium, while not stable, has been tabulated in steam tables. See for instance Keenan, Keyes, Hill and Moore, Thermo Proper of Water...... John Wiley and Sons.
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- #11
sailoday28
Mechanical
Sorry, but I don't follow where this permeable material is to be placed.
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- #13
At the 6" opening of the nozzle. It will be clamped into place. The whole system is built just to measure the pressure drop over time through a material.
There are ASTM stds and test equipment, I'm taking a more applicable and economical approach.
There are ASTM stds and test equipment, I'm taking a more applicable and economical approach.
sailoday28
Mechanical
Won't you require a valve between the tank and cone?
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- #15
sailoday28
Mechanical
What you are after is not clear. If the process is slow and the system insulated a quasi-steady analysis could be performed to a limited degree. For ex. Temp or press in tank vs time could yield mass flow thru system and the permeable material.
If you want pressure just upstream of the material, which is a function of time, then additional system information must be provided.
If you want pressure just upstream of the material, which is a function of time, then additional system information must be provided.
I have seen this situation come up a number of times.
Maybe there should be a name for it, maybe the "Bernoulli
paradox". Basically the other commentators have it correct in that the usual textbook Bernoulli equation does not allow for energy lost to friction which in the real world one must nearly always account for it.
In this case the problem can be immediately noticed in the statement of the problem. An air stream originates at 10 psig
goes through a narrow pipe and discharges to the atmosphere.
If we assume the velocity is zero in the tank and is zero at some point in the atmosphere then energy is lost making the use of the "textbook" Bernoulli equation impossible without adding a friction term.
My advice is to go the CRANE handbook and use the equations for compressible flow.
Maybe there should be a name for it, maybe the "Bernoulli
paradox". Basically the other commentators have it correct in that the usual textbook Bernoulli equation does not allow for energy lost to friction which in the real world one must nearly always account for it.
In this case the problem can be immediately noticed in the statement of the problem. An air stream originates at 10 psig
goes through a narrow pipe and discharges to the atmosphere.
If we assume the velocity is zero in the tank and is zero at some point in the atmosphere then energy is lost making the use of the "textbook" Bernoulli equation impossible without adding a friction term.
My advice is to go the CRANE handbook and use the equations for compressible flow.
rb, friction is not the only problem there. It's a more general problem with energy-type solutions. You can have similar issues in dynamics problems, among others.
Suppose you have a level surface with a 1-meter-high hill. You roll a ball at the hill. How fast must you roll it to get it to stop on the hill? It's a simple problem to solve using energy methods.
But suppose you put a 2-meter-high hill between you and the 1-meter hill and ask the same question. The energy solution is exactly the same. But there is no real answer- any velocity that gets it over the 2 meter hill will be too high to stop on the 1-meter hill. What happens is that if you use the energy solution, and then check the ball at the 2-meter hill, it will have negative kinetic energy and therefore imaginary velocity. Works fine in the equations, but not in real life. And this same type of thing comes up in some fluids problems.
Suppose you have a level surface with a 1-meter-high hill. You roll a ball at the hill. How fast must you roll it to get it to stop on the hill? It's a simple problem to solve using energy methods.
But suppose you put a 2-meter-high hill between you and the 1-meter hill and ask the same question. The energy solution is exactly the same. But there is no real answer- any velocity that gets it over the 2 meter hill will be too high to stop on the 1-meter hill. What happens is that if you use the energy solution, and then check the ball at the 2-meter hill, it will have negative kinetic energy and therefore imaginary velocity. Works fine in the equations, but not in real life. And this same type of thing comes up in some fluids problems.
sailoday28
Mechanical
RB
If we assume the velocity is zero in the tank and is zero at some point in the atmosphere then energy is lost making the use of the "textbook" Bernoulli equation impossible without adding a friction term.
I agree with the above except----for low Mach Nos. assume atmos pressure at exit AND velocity at exit. OR add a sudden expansion loss between exit and atmosphere (where zero velocity is used)
Consider a systme with just static head and atmos pressure on inlet and exit. Bernoulli's equation will allow calc of exit velocity with or without friction.
If we assume the velocity is zero in the tank and is zero at some point in the atmosphere then energy is lost making the use of the "textbook" Bernoulli equation impossible without adding a friction term.
I agree with the above except----for low Mach Nos. assume atmos pressure at exit AND velocity at exit. OR add a sudden expansion loss between exit and atmosphere (where zero velocity is used)
Consider a systme with just static head and atmos pressure on inlet and exit. Bernoulli's equation will allow calc of exit velocity with or without friction.
How were the numbers of Mach 58 and -33202 psi pressure calculated? Bernoulli's equation would only be able to give you one of those values. You have to pick a point at the exit where you know the pressure or velocity and then you can use the equation to calculate the other. You can't use the equation to find the value for two unknown variables.
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