alansimpson
Mechanical
I wish to generate a high speed air blast from a compressed air supply, 8 bar, 110 psi. Currently using a nozzle with a 2.5mm opening. Is there a nozzle shape available th achieve this?
Eng-Tips is the largest forum for Engineering Professionals on the Internet.
Members share and learn making Eng-Tips Forums the best source of engineering information on the Internet!
-
Congratulations JStephen on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Maximum air velocity from compressed air nozzle 1
- Thread starter alansimpson
- Start date
- Status
not my "ballywick" but anyways ...
bernoulli relates static pressure (110 psi) to dynamic pressure which doesn't need a nozzle area ... the nozzle area is required to determine the mass flow, and so the drop in pressure both upsteam of the nozzle and downstream ... how does the jet of compressed air react when it is exposed to the normal atmosphere ? how far away from the nozzle do you want the 110 psi ?
bernoulli relates static pressure (110 psi) to dynamic pressure which doesn't need a nozzle area ... the nozzle area is required to determine the mass flow, and so the drop in pressure both upsteam of the nozzle and downstream ... how does the jet of compressed air react when it is exposed to the normal atmosphere ? how far away from the nozzle do you want the 110 psi ?
Zekeman,
Actually the post says "Maximum air velocity from compressed air nozzle."
What part of "maximum" is confusing you? Any nozzle that exceeds MACH 1 is a "sonic nozzle". I'm going off on a limb here but I think that's why they call it a "sonic" nozzle.
You cannot exceed Mach 1 without a sonic nozzle. Sorry. The flow will be "choked".
Actually the post says "Maximum air velocity from compressed air nozzle."
What part of "maximum" is confusing you? Any nozzle that exceeds MACH 1 is a "sonic nozzle". I'm going off on a limb here but I think that's why they call it a "sonic" nozzle.
You cannot exceed Mach 1 without a sonic nozzle. Sorry. The flow will be "choked".
JkNozzle,
I think you are the one confused, First off, the guy wants the maximum blast so it DOES NOT follow that the supersonic is any more effective than the blast you get at the throat, which, as you probably know, would be at M=1,since the momentum is the thing that is effective, not the velocity. Finally, the term "sonic nozzle" is one that has the flare beyond the throat (aka Bernoulli nozzle) and not the generic which would include the truncated one, I suggested that will do the same job, probably more effectively for less cost.
I think you are the one confused, First off, the guy wants the maximum blast so it DOES NOT follow that the supersonic is any more effective than the blast you get at the throat, which, as you probably know, would be at M=1,since the momentum is the thing that is effective, not the velocity. Finally, the term "sonic nozzle" is one that has the flare beyond the throat (aka Bernoulli nozzle) and not the generic which would include the truncated one, I suggested that will do the same job, probably more effectively for less cost.
If momentum is directly related to velocity and mass how is increasing velocity at a constant mass flow not effective in increasing momentum?
He is trying to spin a small disk very fast. I agree with another poster that a wood cutting router is the best way. I disagree with you that a converging only nozzle will yield velocities in excess of Mach 1.
He is trying to spin a small disk very fast. I agree with another poster that a wood cutting router is the best way. I disagree with you that a converging only nozzle will yield velocities in excess of Mach 1.
Now for something more instructive and less smart-alecky!
This is a simple gas dynamics problem with a few twists. You are going to need to know the temperature of the air in the chamber. First the assumptions:
1) inviscid, isentropic flow
2) the pressure is constant in the chamber, that is, the mass flow out of the nozzle is constant.
outside, ambient pressure is atmospheric, 14.7 ksi==pb (back pressure)
Call the chamber pressure "pr" which stands for reservoir pressure. If the flow is 'choked', the Mach number at the throat is 1.0, the mass flow out of the nozzle is maximum.
First determine if the back pressure is low enough to choke the flow in the nozzle (meaning maximum mass flow through the nozzle). Since this is air, choked flow results if
pb <= pr*0.5283
Therefore if your chamber pressure is 110 psi, any ambient pressure less than 58.11 psi will cause the flow in the nozzle to choke--the nozzle is choked for 14.7 psi ambient. It seems odd to think that the back pressure that results in choked flow--remember one of the key assumptions-the pressure is constant in the chamber. If the nozzle area was 'large,' then it is possible the nozzle is not choked, but in that case I think you would see the pressure drop in the chamber.
This pressure 58.11 is the actual exit pressure at the nozzle exit, so call it 'pe', for exit pressure.
Because the flow is choked, the Mach number at the nozzle is 1.0 (sonic, that is). You need to know the temperature in the chamber, call that Tr (reservoir temperature). The mass flow rate through the nozzle is sometimes called "m-dot" (sorry, can't write it here, it's an 'm' with a small period centered over the top of the 'm').
m-dot=rho*A*V where 'rho' is the density of the air at a given temperature and pressure, A is the nozzle area, V is the velocity.
Mind if I switch to SI units here? They are much cleaner; therefore pr=758.21 kPa, pe=400 kPa
rho==density=pe/(R*Te), where 'pe' is the exit pressure, here 400 kPa, R is the gas constant for
air, R(air)=0.287 kJ/(kg*K), Te still needs to be calculated:
Te=0.8333*Tr--convert this temperature to Kelvin, Kelvin=Celsius+273.16
The area of your nozzle exit is Ae=(pi/4)*De^2 (you said the nozzle opening is 2.5 mm, I assume that
is the diameter De). 2.5 mm is 0.0025 meters.
m-dot=(pe/(R*Te))*Ae*Me*sqrt(gamma*R*Te)
gamma is the ratio of specific heats, in this case, 1.4 for air. The square root quantity is the speed of sound at temperature Te.
Sample calculation: I have to assume the temperature in the chamber. Say it is 100 degrees F=37.78 degrees Celsius=310.9 Kelvin. Very important to get the units right here! 1kPa=1 kN/m^2=1000 N/m^2.
m-dot=(400 kPa/(0.287 kN-m/(kg-K))*310.9 K)*(pi/4)*(0.0025 m)^2 * 1.0 *sqrt(1.4*(0.287 kN-m/(kg-K))*310.9 K*1000)
m-dot=7.78e-3 kg/s
The exit velocity from the nozzle is at most the speed sound, which is equal to sqrt(gamma*R*Te). This is choked flow. Again it seems odd, but if the back pressure pe is low enough, the exit velocity is not a function of the area of the nozzle. Assuming you have air, gamma and R are fixed, so the only way to increase the exit velocity is to boost the temperature of the chamber! Te=0.8333*Tr, increasing Tr increases Te and hence the speed of sound, which in this case is the exit velocity.
Anybody is welcome of course to check my math and my logic!
This is a simple gas dynamics problem with a few twists. You are going to need to know the temperature of the air in the chamber. First the assumptions:
1) inviscid, isentropic flow
2) the pressure is constant in the chamber, that is, the mass flow out of the nozzle is constant.
outside, ambient pressure is atmospheric, 14.7 ksi==pb (back pressure)
Call the chamber pressure "pr" which stands for reservoir pressure. If the flow is 'choked', the Mach number at the throat is 1.0, the mass flow out of the nozzle is maximum.
First determine if the back pressure is low enough to choke the flow in the nozzle (meaning maximum mass flow through the nozzle). Since this is air, choked flow results if
pb <= pr*0.5283
Therefore if your chamber pressure is 110 psi, any ambient pressure less than 58.11 psi will cause the flow in the nozzle to choke--the nozzle is choked for 14.7 psi ambient. It seems odd to think that the back pressure that results in choked flow--remember one of the key assumptions-the pressure is constant in the chamber. If the nozzle area was 'large,' then it is possible the nozzle is not choked, but in that case I think you would see the pressure drop in the chamber.
This pressure 58.11 is the actual exit pressure at the nozzle exit, so call it 'pe', for exit pressure.
Because the flow is choked, the Mach number at the nozzle is 1.0 (sonic, that is). You need to know the temperature in the chamber, call that Tr (reservoir temperature). The mass flow rate through the nozzle is sometimes called "m-dot" (sorry, can't write it here, it's an 'm' with a small period centered over the top of the 'm').
m-dot=rho*A*V where 'rho' is the density of the air at a given temperature and pressure, A is the nozzle area, V is the velocity.
Mind if I switch to SI units here? They are much cleaner; therefore pr=758.21 kPa, pe=400 kPa
rho==density=pe/(R*Te), where 'pe' is the exit pressure, here 400 kPa, R is the gas constant for
air, R(air)=0.287 kJ/(kg*K), Te still needs to be calculated:
Te=0.8333*Tr--convert this temperature to Kelvin, Kelvin=Celsius+273.16
The area of your nozzle exit is Ae=(pi/4)*De^2 (you said the nozzle opening is 2.5 mm, I assume that
is the diameter De). 2.5 mm is 0.0025 meters.
m-dot=(pe/(R*Te))*Ae*Me*sqrt(gamma*R*Te)
gamma is the ratio of specific heats, in this case, 1.4 for air. The square root quantity is the speed of sound at temperature Te.
Sample calculation: I have to assume the temperature in the chamber. Say it is 100 degrees F=37.78 degrees Celsius=310.9 Kelvin. Very important to get the units right here! 1kPa=1 kN/m^2=1000 N/m^2.
m-dot=(400 kPa/(0.287 kN-m/(kg-K))*310.9 K)*(pi/4)*(0.0025 m)^2 * 1.0 *sqrt(1.4*(0.287 kN-m/(kg-K))*310.9 K*1000)
m-dot=7.78e-3 kg/s
The exit velocity from the nozzle is at most the speed sound, which is equal to sqrt(gamma*R*Te). This is choked flow. Again it seems odd, but if the back pressure pe is low enough, the exit velocity is not a function of the area of the nozzle. Assuming you have air, gamma and R are fixed, so the only way to increase the exit velocity is to boost the temperature of the chamber! Te=0.8333*Tr, increasing Tr increases Te and hence the speed of sound, which in this case is the exit velocity.
Anybody is welcome of course to check my math and my logic!
Note alansimpson that if the nozzle is 'convergent-divergent,' meaning that the big tank empties into a channel that gradually narrows to a minimum area (called the 'throat' where the maximum Mach number is 1.0), then gets bigger ("diverges") before emptying into the atmosphere, the answer to the question "is there a nozzle shape to achieve this (that is, high speed air blast)" is very complicated because not only are the chamber conditions important, but also the area ratio of the nozzle is important (that is, the ratio of the max. area in the nozzle to the throat area). To further complicate things, for the same area ratio and different ratios of the back pressure to the chamber pressure, you can get a subsonic (Mach <1 that is) and a supersonic solution.
In a jet engine, the maximum thrust occurs when the nozzle is 'perfectly expanded.' Say you have a C-D (convergent-divergent) nozzle and your area ratio causes the pressure in the exit plane of the nozzle to be larger than the atmospheric pressure--in that case, the flow is considered 'underexpanded' and you'll see expansion waves coming out of the nozzle. If the opposite case occurs, the pressure at the exit plane drops below the atmospheric, your flow is overexpanded, and you'll see shock waves coming out of the nozzle to get the pressure up to atmospheric (pressure increases across shock waves). If the exit pressure is perfectly matched to the atmospheric, then you maximum thrust available.
if you want to talk directly about it, and some help with the calculations, write drfeaNOSPAM@yahoo.com and we'll be happy to help.
In a jet engine, the maximum thrust occurs when the nozzle is 'perfectly expanded.' Say you have a C-D (convergent-divergent) nozzle and your area ratio causes the pressure in the exit plane of the nozzle to be larger than the atmospheric pressure--in that case, the flow is considered 'underexpanded' and you'll see expansion waves coming out of the nozzle. If the opposite case occurs, the pressure at the exit plane drops below the atmospheric, your flow is overexpanded, and you'll see shock waves coming out of the nozzle to get the pressure up to atmospheric (pressure increases across shock waves). If the exit pressure is perfectly matched to the atmospheric, then you maximum thrust available.
if you want to talk directly about it, and some help with the calculations, write drfeaNOSPAM@yahoo.com and we'll be happy to help.
-
1
- #11
Or you try your luck with some free software, "GDC"
Haven't tried it, cannot tell you if it is ok.
Haven't tried it, cannot tell you if it is ok.
- Status
Similar threads
- Question
- Locked
- Question
- Locked
- Question