Register to reply

At what volumetric flow rate will air fill a space vacuum at Earth's STP?

by rcoopster
Tags: earth, flow, rate, space, vacuum, volumetric
Share this thread:
rcoopster
#1
Jan18-12, 03:09 PM
P: 12
At atmospheric pressure of 14.2 psi, if I had a 1 ft square hole through a 5 foot thick wall into an endless space vacuum, I am trying to find out the the final maximum volumetric flow rate that our atmospheric pressure would push air through the hole.

I tried using Poiseuille's Equation, but I feel like the number I got was way too high, and I have been unable to find any other equation or method of solution which fits the variable of the question.

Assiatance would be greatly appreciated.
Phys.Org News Partner Physics news on Phys.org
Mapping the optimal route between two quantum states
Spin-based electronics: New material successfully tested
Verifying the future of quantum computing
technician
#2
Jan18-12, 03:42 PM
P: 1,506
You could try using change of momentum.
If you know the pressure then you know the force on a given hole area.
This force pushes air out of the hole and changes its momentum
Force = rate of change of momentum = A x ρ x v where A = area of hole, ρ=air density and v is the speed you are trying to calculate.
I don't think the thickness of the wall is relevant in this technique.
russ_watters
#3
Jan18-12, 04:02 PM
Mentor
P: 22,243
I'd start with Bernoulli's equation and work in some complexities like an exit loss coefficient and compressibility.

technician
#4
Jan18-12, 04:26 PM
P: 1,506
At what volumetric flow rate will air fill a space vacuum at Earth's STP?

Must correct my first response.... rate of change of momentum = ρAv^2
Not ρAv
Using your data (converted to metric!!!!!!)
I got a velocity of about 380m/s
obafgkmrns
#5
Jan18-12, 04:48 PM
P: 126
This question is more complex than it may seem: momentum equation & Bernoulli's equation won't suffice. Technician's figure of 380 m/s shows that the flow will undoubtedly choke, and the sharp edge of the opening will lead to separation. (On a brighter not, the flow will be laminar, so you can dispense with turbulence.) A believable answer is going to require CFD.
Curl
#6
Jan18-12, 11:50 PM
P: 757
What if you take the molecular approach? Calculate all the molecules that would have hit that section of the wall in unit time - those molecules will go through. Thus you have found the mass flow rate.
bbbeard
#7
Jan19-12, 12:24 AM
P: 192
Quote Quote by rcoopster View Post
At atmospheric pressure of 14.2 psi, if I had a 1 ft square hole through a 5 foot thick wall into an endless space vacuum, I am trying to find out the the final maximum volumetric flow rate that our atmospheric pressure would push air through the hole.

I tried using Poiseuille's Equation, but I feel like the number I got was way too high, and I have been unable to find any other equation or method of solution which fits the variable of the question.

Assiatance would be greatly appreciated.
Well, first, standard atmospheric pressure is 14.696 psia. Do you have an application for which 14.2 psia is the correct figure?

Second, ducting from atmospheric pressure to a vacuum will necessarily choke the flow. That is, at some point in the duct, the Mach number will reach 1 and at that point you will have reached the maximum flow rate. However, experience shows that flows like this will exhibit a vena contracta and the effective flow area will be less than the geometric flow. For a sharp-edged orifice the contraction ratio is somewhere around 0.6 to 0.65, but for a long channel it could be considerably higher. Your mileage may vary.

Third, because the Mach number is considerable, compressibility effects are important. The density drops as Mach 1 is approached. Therefore the volumetric flow rate is not conserved in the duct. The dodge that many engineers accept is to speak in terms of "standard" cubic feet per second (or per minute), that is, the computed mass flow rate (which is conserved in the duct) is expressed as a theoretical volume flow rate at standard temperature and pressure.

Fourth, the standardized volumetric flow rate depends on the upstream temperature. Are you assuming the upstream temperature is the standard atmospheric temperature (15 C = 59 F = 518.67 R = 288.15 K)?

Anyway, with all these caveats, for 100% contraction coefficient (no vena contracta), I get a mass flow rate of 49.4 lbm/s for a 1 square foot flow area. The standard density of air is 0.076478 lbm/ft3, so this mass flow corresponds to 646 SCFS or 38,760 SCFM.

BBB
rcoopster
#8
Jan20-12, 10:16 AM
P: 12
BBB,

You seem brilliant and well educated. So here is my next question in sequence of design.

How much power can be produced by the dynamic pressure of a 49.4 lbm/sec mass or 646 SCFS flow of air?

For instance, If I had a 1 sq ft tube with the above stated air flow flowing through it (with the flow in this case be produced mechanically), and I held it a couple feet in front of a 1 sq ft opening in a large expandable container in standard atmospheric conditions, can I use the dynamic pressure of the air flow to perform work or produce power?

For example, I calculate the air flow to produce a direct dynamic pressure of 3.45 psi of 496 psf, and I am assuming this dynamic pressure to be in addition to the 14.7 standard atmospheric pressure, implying that inside my container I would have the ability to compress air to a guage pressure of 3.45 psi or a total pressure of 18.15 psi. Is this incorrect?

Using P1*VI/T1 = P2*V2/T2 (and not really accounting for the complexities of temperate change) with the given volumetric flow rate of 646 SCFS, would I be able to produce 523.2 ft^3/s at 496 pfs which equals 258,984 ft*lb/sec or 470 hp (without any real world losses taken into account)?

Can you see any major flaws with my logic of solution?
bbbeard
#9
Jan20-12, 11:09 PM
P: 192
Quote Quote by rcoopster View Post
BBB,

You seem brilliant and well educated. So here is my next question in sequence of design.

How much power can be produced by the dynamic pressure of a 49.4 lbm/sec mass or 646 SCFS flow of air?

For instance, If I had a 1 sq ft tube with the above stated air flow flowing through it (with the flow in this case be produced mechanically), and I held it a couple feet in front of a 1 sq ft opening in a large expandable container in standard atmospheric conditions, can I use the dynamic pressure of the air flow to perform work or produce power?

For example, I calculate the air flow to produce a direct dynamic pressure of 3.45 psi of 496 psf, and I am assuming this dynamic pressure to be in addition to the 14.7 standard atmospheric pressure, implying that inside my container I would have the ability to compress air to a guage pressure of 3.45 psi or a total pressure of 18.15 psi. Is this incorrect?

Using P1*VI/T1 = P2*V2/T2 (and not really accounting for the complexities of temperate change) with the given volumetric flow rate of 646 SCFS, would I be able to produce 523.2 ft^3/s at 496 pfs which equals 258,984 ft*lb/sec or 470 hp (without any real world losses taken into account)?

Can you see any major flaws with my logic of solution?
I can't picture what setup you are envisioning, but the maximum power that you can extract from airflow at a given total temperature is [itex]\dot{m}c_{p}T_{t}[/itex]. That's assuming you run it through an ideal turbine that extracts all the energy down to absolute zero, and assuming air is an ideal gas down to absolute zero. In fact the air starts to liquefy well before that, so that's a pretty bad assumption. With non-zero exit temperature, try

[itex]Pwr=\dot{m}c_{p}(T_{ti}-T_{te})[/itex]

with specified inlet and exit temperatures. E.g. the critical temperature of oxygen is 154.6 K = 278.2 R, so maybe you want to stay above that. That means the maximum power you could extract would be

(49.4 lbm/s)(186.7 ft lbf/lbm R)(518.67 - 278.2 R) = 2.2E6 ft lbf/s = 4000 HP

The mass flow rate was for a choked duct with no contraction, not for a choked turbine, so there's more fuzz there. But this is probably the right order of magnitude. Now all you need is a permanent "source" of vacuum next to atmospheric pressure. ;-)
rcoopster
#10
Jan21-12, 06:40 PM
P: 12
BBB,

Thank you for your response!

In order to give you a picture of the set up I am envisioning, I have included one. I call it a Vacuum Wheel.

According to your calculations, through a 1 sq ft hole into a vacuum, air would flow through the hole with a velocity of 646 SCFS. So, in theory anyways, I would be able to assimilate a vacuum if I expanded a container's volume at a slightly higher rate like 650 CFS. True?

Well I am trying to do something of this sort with my vacuum wheel, but instead of continuously expanding the volume of a container, I am trying to continuously re-create a vacuum using the centripetal force created by the angular acceleration of the air as it travels around the wheel.

So here is one of my next questions, given the flow rate or angular velocity of the wheel at 650 ft/sec how tight would I need to make the radius of the wheel to create enough centripetal force to create a vacuum as shown by the design?

But moving forward, according to the design of my wheel, it is my understanding that air flows are created by the pushing expansive force of the molecules under higher pressures and not caused necessarily by the so called "sucking force" of low pressure. To me this implies that velocity of the airflow up to the red arrow on the diagram is actually produced and input into the system by the atmospheric pressure force and not necessariy by the wheel itself. Up to this point the air entering the system is being pushed, not pulled. It is not until the wheel fin moves into the flow and begins pushing the flow forward that the wheel become responsible for any increases or maintenances of the velocity of the air flow.

It is also my theoretical understanding that the dynamic pressure and momentum of the air flow as it exits into high pressures will in essence "protect or hide" the fin of the wheel in the airflow from the higher than atmospheric pressure on the exiting side of the wheel which leads me to my next question.

How much power to you think that it would take to rotate the wheel with an exterior angular velocity of 650 ft/sec? (RPMs determined obviously by the radius figured by my earlier question.) Or, what pressure force do you think the air flow would exert on the forward face of the wheel fin submerged in the airflow?

This power value will obviously need to be compared then to the power which is able to be generated by the air flow, whether the this power be generated on the entry or exiting side of the wheel, I don't know if one would be more efficient than the other.

I know that in all case energy in = energy out.

But I guess that the real and final question that I have for you is this:

In this application does energy in = energy input by Earth's atmospheric pressure + energy input mechanically to rotate the wheel, or does energy in = only energy input mechanically to rotate the wheel.

I pray for your integrity.
Attached Thumbnails
Vacuum Wheel.png  
rcoopster
#11
Jan24-12, 09:37 AM
P: 12
Still looking for answers, anyone?
bbbeard
#12
Jan25-12, 10:47 AM
P: 192
Well, the first question you asked was about the flow rate through a duct that extends through a wall. The second question was about extracting power from flow, which generally requires some sort of turbine (although one can also extract work from a flow using a nozzle), which is a different flow situation, but since turbines choke, it's not that unreasonable to compare the flow rates and the power extraction. Your latest question, though, is about a pump, which is a different thermodynamic device. In a turbine, one extracts shaft power: flow enters the turbine at a higher energy state and leaves at a lower energy state. In a pump, one adds shaft power. The flow enters at a low energy state and leaves at a higher energy state.

The pump in your diagram is a vane pump. It is certainly theoretically possible to arrange a vane pump so that there is a pretty good vacuum at certain locations in the pump, although I don't think the flowfield would look quite like what you have drawn. The presence of a vacuum does not make the pump more efficient, though, and certainly does not reverse the flow of work into device. I'm sort of at a loss of for what else to say, because the assumptions underlying your calculations are inconsistent, and it's not clear how to "fix" your drawing so that it makes some kind of mechanical sense. For example, one might consider a centrifugal pump instead of a vane pump, since you get better single-stage pressure ratios out of a centrifugal pump. But the pump you've shown essentially takes air at atmospheric pressure, whirls it around, and exhausts to atmospheric pressure. So it really just dissipates the shaft power internally and exhausts a bunch of hot air.

Or one could rearrange the layout of the device. Imagine the device were a box that one kept at low pressure. At one end of the box, a turbine operating at 100% efficiency extracts power from air that flows through the turbine into the box. At the other end is a pump that operates at 100% efficiency that keeps the box evacuated by removing exactly the amount of air that enters by pumping it back up to atmospheric pressure and sending it back into the atmosphere. If both devices are 100% efficient, then both processes are vertical lines on a T-s diagram and the power required by the pump is exactly what is output from the turbine -- and there is no net output of work. If the devices are not 100% efficient, then the entropy increases during both the expansion and compression phases, and the process lines drift to the right on at T-s diagram. In that case the pump requires more power than the turbine will supply. The net work is into the device. These considerations are true regardless of how many stages the turbine and pump have, and how low a pressure one takes the box to.
rcoopster
#13
Jan25-12, 02:09 PM
P: 12
BBB,

Once again, thank you for your response.

I don't want to drag it out, so I will keep it as simple as possible. The entire premise of the idea basically rides completely upon this one question. Is the atmospheric pressure of the air in any way shape or form responsible for any of the mass acceleration of the airflow?

Obviously, without the spinning wheel, there wouldn't be any airflow at all, but due to the creation of low pressure by the wheel, does the atmospheric pressure actually accelerate the air molecules into the low pressure cavity?

If it doesn't, I have not further arguments.

Thanks,

rcoopster
bbbeard
#14
Jan25-12, 04:08 PM
P: 192
Quote Quote by rcoopster View Post
BBB,

Once again, thank you for your response.

I don't want to drag it out, so I will keep it as simple as possible. The entire premise of the idea basically rides completely upon this one question. Is the atmospheric pressure of the air in any way shape or form responsible for any of the mass acceleration of the airflow?

Obviously, without the spinning wheel, there wouldn't be any airflow at all, but due to the creation of low pressure by the wheel, does the atmospheric pressure actually accelerate the air molecules into the low pressure cavity?

If it doesn't, I have not further arguments.

Thanks,

rcoopster
Not to engage in sophistry here, but I usually discourage people from thinking about fluid flow in terms of cause and effect unless they have a really firm grasp of the physical principles underlying the flow. The example I always give is fluid flow in a pipe with friction. Consider a length of pipe with incompressible liquid flowing through it. Since there is friction (i.e. viscosity), the pressure at the entrance to the pipe is higher than the pressure at the exit.

Now, is the pressure drop "caused by" the flow in the pipe, or is the flow in the pipe "caused by" the pressure difference between the entrance and exit? One can argue either end of this, say, by positing a positive displacement pump at the entrance to the pipe which forces a fixed flow rate through the pipe, thereby making the pressure drop the "effect" of the flow rate "cause", or conversely, by positing a pump that provides the pressure difference that makes the flow the "effect" of the pressure difference "cause". I encourage people to think in terms of a flow that accommodates itself to whatever boundary conditions exist, rather than thinking in terms of general flow features "causing" other flow features.

So, when you ask "Is the atmospheric pressure of the air in any way shape or form responsible for any of the mass acceleration of the airflow?" I would say on general grounds pressure is not the kind of thing to which one can assign "responsibility". So what are you really asking? Can a pressure boundary condition be used to determine a flow rate? Yes. Can a downstream pressure be the result of a specified flow rate and geometry? Also yes. Help me understand what it is you're looking for here.
bbbeard
#15
Jan25-12, 06:28 PM
P: 192
On further reflection, perhaps I am being too obtuse.

The straightforward answer to your question is that the negative of the gradient of pressure ([itex]-\nabla P[/itex]) is the net pressure force per unit volume acting on the fluid. Another formulation is that [itex]-\nabla P/\rho[/itex] is the force per unit mass due to the pressure gradient. (If you're not familiar with the gradient, it is simply the rate at which pressure increases, i.e. the derivative of pressure with respect to distance, formulated as a vector which points in the direction in which the pressure increase is the greatest.) So if the pressure at point A is higher than the pressure at point B, that pressure force accelerates fluid particles from A to B (though it may not be the only force!)
rcoopster
#16
Jan26-12, 10:42 AM
P: 12
BBB,

Let me see if I can explain what I'm looking for.

I understand that if I slowly pump air out of one end of a concealed box, and then at the other end of the box I have a turbine that intakes air, the exposed pressure differences on both the turbine and the pump with be relatively the same and with same volume of flow passing through them, with friction losses obviously the turbine will produce less energy that it takes to turn the pump. This is very clear to me, but I do believe the design of my idea to have some differentiating factors.

According to your calculations, through a 1 sq ft hole to a vacuum, Earth's atmospheric pressure would create a 49.4 lbm/sec flow rate into the vacuum, and so now just assume with me for a second, in my vacuum wheel design, that the energy that it takes to create that mass flow into the vacuum wheel comes completely from the same source that it would if we had an actual vacuum on the other side of a five ft thick wall, just like in my initial question. The air flow is coming from the same source, the only difference is the source of vacuum. But without knowing a better way to define it, I referred to this source of energy which creates the mass flow as the Earth’s atmospheric pressure force. So now, just like in your box example, what I believe you are going to say is that yes, the atmospheric pressure will accelerate air to fill the vacuum, but it is going to take just as much if not more energy to “create” the vacuum than you will be able to “extract” from the air filling the vacuum. This would be the logic with your box idea, saying that it takes just as much energy to create the low pressure as you can extract when using the low pressure to produce power. Am I incorrect?

And so now we have some givens, the rate at which air will fill a vacuum is determined by the atmospheric pressure, and therefore, so is the amount of power that we can theoretically extract from that airflow. But to me, what is yet to be determined in my design it the amount of power that it will take to create the vacuum at the corresponding rate. The rotational velocity of the wheel is determined by the potential flow rate of how fast air can fill a vacuum, but the real question that I asked previously, is what will the pressure differential be on the wheel fin submerged in the airflow. By the time the wheel fin enters into the airflow, the flow already has substantial velocity and forward momentum, which is created by what I am deeming to be called “atmospheric pressure, and I believe that this forward velocity and momentum, where the air flow is already moving with a velocity which is close to the wheel fin, greatly reduces the net pressure difference exerted on the wheel fin to a pressure well below atmospheric pressures difference compared to a vacuum. For example what would the net pressure differential on a 1 sq ft plate moving 650 ft/sec in the same direction as a 645 ft/sec air flow? Would it be similar to a 5 ft/sec flow? I know my situation is different, but I believe that the pressure differential acting on the wheel in the counter axis of rotation once the air flow begins drafting through the wheel will be substantially less than atmospheric pressure, or even less than the difference between the total pressure of the flow and the atmospheric pressure which I calculated to be around 3.45 psi.
I would still like this number from you. Assume a 4 ft outside radius, with around a 1800 rpm rotation which gives the wheel fin about a 660 ft/sec rotational velocity. Do you believe that the airflow is going to exert around a 400 lb/ft^2 counter rotational force on the wheel fin? The thing is, is that it is only the high rate of flow and momentum which creates a centrifugal force which allows the wheel in my design to move any air at all. With a slow rotation, as far as air moving through the wheel from the entry to the exit, no work will be performed at all, and I can’t reason where that much resistance to rotation would come from, but maybe you can enlighten me.

I am sorry for the confusion, but the diagram that I included only really included one part of the complete two part system. The vacuum wheel is the pump, but I did not include the turbine or method of energy generation.

I predicted that I would be able to produce about 500 hp by compressing the air on the exiting side of the wheel to about 18.5 psi of total pressure, and using a large surface area, slow rotation volume displacement turbine, the number that you came back at me with was much higher. The turbine used to do that was not shown in my diagram, and for theoretically purposed we could simplify it to the power produced when expanding a large piston. I know like to look at the two systems as complete separate.

How much power would it take to rotate the wheel at 1800 rpms?

How much power can be produced by the airflow which will naturally flow through the wheel spinning at 1800 rpms?

Maybe the two independent calculations will be exactly the same, I don’t know, but I bet that if you honestly and independently calculated the two numbers the later would be higher than the first.

From an energy, and I could be completely wrong, when below atmospheric pressures are used to generate flow velocity, as with my vacuum wheel, I believe that power or energy is being input into the system by the energy in the atmosphere itself. Obviously, it still requires a power input source to turn the vacuum pump, or vacuum wheel, but when the total available power output of the system is measured, it will equal the power input into the system by the atmosphere and by the power plant rotating the wheel. A basic 1 + 1 = 2 formula where 2 can then turn around and be used to power 1 and where the other 1 is coming from an independent renewable power source.

You seem to me to have a better grasp on things than 99% of the other professors that I have worked with over the years, and I have a great deal of respect for you willingness to entertain my questions.

I would love to see your numerical calculations for the two proposed power input and power output values of the machine.

Thank you very much.

rcoopster
bbbeard
#17
Jan27-12, 09:32 PM
P: 192
Quote Quote by rcoopster View Post
The rotational velocity of the wheel is determined by the potential flow rate of how fast air can fill a vacuum, but the real question that I asked previously, is what will the pressure differential be on the wheel fin submerged in the airflow. By the time the wheel fin enters into the airflow, the flow already has substantial velocity and forward momentum, which is created by what I am deeming to be called “atmospheric pressure, and I believe that this forward velocity and momentum, where the air flow is already moving with a velocity which is close to the wheel fin, greatly reduces the net pressure difference exerted on the wheel fin to a pressure well below atmospheric pressures difference compared to a vacuum.
Okay, let's see if I can answer some of these questions. Let's start with the misconceptions inherent in your statement above.

The first obstacle is that the vacuum wheel as you have drawn it cannot work. Consider the air flowing into the entrance. The way you have drawn it, the flow approaches a sharp corner with vacuum on one side, and then continues merrily in the direction it was flowing, ignoring the vacuum next to it, and I'm guessing you think that somehow this momentum has been "earned" and that the flow won't significantly change direction. Compressible flows absolutely do not behave like that. In a convergent passage like you have drawn, with anything less than half an atmosphere of pressure downstream, the flow will accelerate to Mach 1. When that flow exhausts into a vacuum, it expands around the corner in what is called a "Prandtl-Meyer fan". The Mach number continues to rise as the flow turns the corner. Going from M=1 to M=infinity, the flow goes from axial and turns 130 degrees, i.e. it flows back upstream into the vacuum channel.

Now, the way you have depicted the wheel working, that accelerated flow runs straight into the vane coming out of the vacuum channel. This hypersonic flow rams into the vane, and what you would see would be a complicated set of shocks, reflecting and intersecting and moving with the vane, with a subsonic layer next to the vane material and the hub, because these strong shocks would decelerate the hypersonic flow back to a subsonic condition, generating a lot of entropy in the process. It's very difficult to compute the result of this moving set of shocks and expansions.

As the vane passes by the entrance channel, the flow field gets more complicated. Part of the flow slams into the vane from behind, generating another set of compression shocks and sneaking flow upstream into the vacuum channel again. It's hard to tell, exactly, but I think it is likely the flow will reach the next vane after shocking off the outer curved walls and expanding around the inner curved walls before that vane has a chance to move appreciably. The other part of the flow expands around the vane and generates a set of expansions and shocks. Meanwhile, the curved main channel that you intended the flow to go down will also throw off a set of shocks because supersonic flow doesn't just change direction without shocking (there is one exception, for highly uniform flow at a fixed Mach number you can design a curved compression surface that will isentropically compress the flow, but this surface (a) is not circular, and (b) only works for one particular Mach number).

So for these reasons the wheel will not provide anything approaching a steady vacuum to draw flow into the device. It will stall and sputter as the flow intermittently fills the vacuum channel as the vanes pass the entrance.

That's why it's practically impossible to provide a reasonable answer to how much power the device will demand. But let's push on some of your other questions. For example you ask

Quote Quote by rcoopster View Post
For example what would the net pressure differential on a 1 sq ft plate moving 650 ft/sec in the same direction as a 645 ft/sec air flow? Would it be similar to a 5 ft/sec flow?
Again, compressible flows don't behave like this. Even in a one-dimensional setting, if the vane is moving at 650 ft/s, the flow behind and ahead of the vane will be moving at 650 ft/s. So your question doesn't make any sense. It's not like a vane or propeller in water, where there is a possibility of cavitation. The flow will accelerate or decelerate to match the boundary conditions. And the boundary conditions are of primary importance. You cannot calculate the pressure on a surface just knowing the speed of the surface.


Quote Quote by rcoopster View Post
How much power would it take to rotate the wheel at 1800 rpms?

How much power can be produced by the airflow which will naturally flow through the wheel spinning at 1800 rpms?

Maybe the two independent calculations will be exactly the same, I don’t know, but I bet that if you honestly and independently calculated the two numbers the later would be higher than the first.
No, it won't. And in fact once you take into account all the entropy-generating shocks in the system, it takes considerably more power to pump the flow than you can extract from it.
rcoopster
#18
Jan30-12, 12:42 PM
P: 12
BBB,

Thank you for your response and candid insight.

From the way you described the complexity of the airflow, and how quickly the flow is able to accelerate to rates above Mach 1, and how at the proposed rotational velocity the pump would sputter and stall, it seems to me like the pump is running way to slow, and that it would run more efficiently at speeds above Mach 1, since it is really only the momentum and velocity of the flow which allows the pump to work at all.

When I originally began working with the idea, I proposed rotational velocity between 1200 to 1400 ft/sec, but I didn't know if the air would flow that fast, hence my original question. Another thing that I would probably do is add more vanes, like go from three to six. This would have a vane passing the entrance to the vacuum wheel about every .003 sec, which I think would keep to much air from reversing and traveling back up into the vacuum channel, but I could be wrong?

I am intrigued by the ability you describe of the air change direction so quickly, but you did also state that hypersonic flows will not change direction without a compression shock.

I have another question which will greatly help my understanding. I have attached a small diagram of an airflow. Assume a hypersonic airflow being pumped into the entry, with standard atmospheric pressure outside Exit B and a vacuum outside exit A, at Mach 1 what percentage of the airflow do you think would exit through exit B?

Thank you for you thoughts,

rcoopster
Attached Images
File Type: bmp Air Flow Diagram.bmp (246.1 KB, 9 views)


Register to reply

Related Discussions
Volumetric flow rate (simple?) Engineering, Comp Sci, & Technology Homework 1
Thermodynamics (Mass flow rate & Volume flow rate) Engineering, Comp Sci, & Technology Homework 4
Volumetric flow rate Engineering Systems & Design 7
Space as a vacuum and earth's atmosphere Astronomy & Astrophysics 9