Blower fitted with De Laval Nozzle

[boneh3ad]

The statement you responded to (which I agreed with) was: "...no further increase in speed occurs despite increase in input pressure..." It doesn't mention temperature, so it is reasonable to assume the intent is for temperature to remain constant. It is unhelpful and argumentative to go beyond the bounds of what was stated (without at least dealing with the statement as given). For example, a more reasonable response would have been: "That's correct as stated, but be careful; there are other things not mentioned, such as temperature, that could be changed that you didn't mention, that could allow the speed to change."

Agreed. So do you think it is reasonable to assume the OP has a device that has a label that says "compressor" on it, but he called it a "blower" instead?

It would be more helpful if you tried to answer the OP's questions in a way that makes a reasonable effort to interpret the question in the way the OP intended...and for that matter, apply that to other posts/posters as well. Otherwise, you are going off on tangents that distract from the question being asked.

Do you mean like the 5 post discussion I had with the OP before you even joined the thread or the several in depth posts I provided after that walking him/her through nozzle calculations?

If you want to nitpick about the topic of the original post, the. Go read it yourself. OP specifically asks about achieving a certain flow velocity, which is intimately and undeniably related to the flow temperature. The fact that OP did not mention temperature does not somehow make it irrelevant. Based on the context of this actual post, it is more likely that OP did not know it was important, making it still more important to discuss.

You offered a bit of advice that was factually inaccurate and potentially misleading with regards to what it means for a nozzle to be choked and in the context of this thread. For example, in reality, OP wants a certain outlet velocity (400 m/s). If instead they wound up with 350 m/s, my post and reality suggest OP could add some heat to the flow to raise that speed even though the nozzle is choked. Your post suggests that is not a viable route to take.

 

[T C]

So far, all the discussions are based on pressurised air releasing through a nozzle while what I was asked about a blower fitted with a c/d nozzle. When compressed gas is released through a nozzle, it has higher pressure but zero initial velocity while in my case the the fluid enters with nozzle with a specific velocity.

As you make the nozzle smaller, the airflow profile will ride the blower curve until it reaches a maximum pressure at which point you've reached your maximum velocity...and it may stall and drop the pressure/velocity

What can be that maximum velocity?

 

[jack action]

So far, all the discussions are based on pressurised air releasing through a nozzle while what I was asked about a blower fitted with a c/d nozzle. When compressed gas is released through a nozzle, it has higher pressure but zero initial velocity while in my case the the fluid enters with nozzle with a specific velocity.

You have to convert to stagnation (or total) properties: stagnation temperature, stagnation pressure. These properties are usually denoted with a zero subscript ($T_0$, $p_0$). This will represent the equivalent "reservoir" (with zero initial velocity) that can be compared to your compressed gas with initial velocity.

 

[T C]

My gas isn't compressed. It enters the nozzle with a specific velocity.

 

[jack action]

My gas isn't compressed. It enters the nozzle with a specific velocity.

It still has an absolute pressure, even if it is atmospheric pressure. With this absolute pressure $p$ and the Mach number $M$, you can find the total pressure $p_0$ (see the links in my previous post).

 

[boneh3ad]

My gas isn't compressed. It enters the nozzle with a specific velocity.

Whether you know it or not, it is compressed. Your blower, however it operates, is compressing the air entering the inlet and exhausting it with a higher total pressure and likely a slightly higher temperature. This total pressure and temperature are what will determine what you can and cannot achieve through a de Laval nozzle.

 

[T C]

By absolute pressure, do you want to mean a summation of static and dynamic pressure?

 

[boneh3ad]

Absolute pressure means pressure referenced to zero (as opposed to gauge pressure, referenced to atmosphere).

Total pressure is the pressure that occurs if a flow is isentropically slowed to zero velocity (sometimes called stagnation pressure). For an incompressible flow, it equals the sum of static and dynamic pressure. For a compressible flow, it does not. Your flow here is compressible.

 

[T C]

Actually there is a basic difference when compressed air is used and when a blower is used. In case of compressed air, we can assume that it has been compressed adiabatically and then left to cool down so that the temperature can come to ambient level. That means a big part of the energy spent in compressing the gas is lost as heat. And also when this gas is released, then it expands adiabatically and thus its temperature falls.
But in case of a blower, the case would be like a compressor where the gas is compressed adiabatically and then released instantly without being cooled. The formula that Jack Action has mentioned has the temperature parameter but in the formula, it isn't mentioned whether the temperature is before being released or after being released.

 

[cjl]

What bonehead is trying to tell you though is that it doesn't matter. You say

In case of compressed air, we can assume that it has been compressed adiabatically and then left to cool down so that the temperature can come to ambient level. That means a big part of the energy spent in compressing the gas is lost as heat. And also when this gas is released, then it expands adiabatically and thus its temperature falls.

but we can suppose whatever reservoir conditions we would like. In Jack's formula, the temperature is the temperature just upstream of the nozzle (technically, the stagnation temperature just upstream of the nozzle, but as long as we suppose the exit to your blower is well below mach 1, this will basically be the exhaust temperature of the blower). There's no fundamental difference between your blower and a reservoir with the same temperature and pressure located upstream.

 

[jack action]

Actually there is a basic difference when compressed air is used and when a blower is used.

There are absolutely no differences.

In case of compressed air, we can assume that it has been compressed adiabatically and then left to cool down so that the temperature can come to ambient level.

The cool down period you refer to, is actually a decompression.

The formula that Jack Action has mentioned has the temperature parameter but in the formula, it isn't mentioned whether the temperature is before being released or after being released.

That is what I've been trying to tell you all along. All equations you will find for a nozzle (or any turbomachinery for that matter) refer to the inlet and outlet conditions of that nozzle.

What are the pressure, temperature and Mach number of the fluid a the inlet? That is all you need to know. How you got there doesn't matter. Whether you compress the gas with an axial turbine followed by a long cooling journey through a pipe OR that your nozzle is facing a strong headwind caused by a tornado OR that you connect the nozzle to a heated box filled with your gas:

If the result inlet gas conditions (pressure, temperature, Mach number) are the same - and assuming the nozzle discharges in the same environment - the result will be the same.

 

[JBA]

Apart from all of the above, since he apparently has the blower in question at hand, whatever it is, all that is necessary to determine the stagnation pressure is a ram tube (funnel) connected to a pressure gauge to determine the stagnation pressure at his blower outlet and a thermometer to determine the discharge temperature.

 

[boneh3ad]

Apart from all of the above, since he apparently has the blower in question at hand, whatever it is, all that is necessary to determine the stagnation pressure is a ram tube (funnel) connected to a pressure gauge to determine the stagnation pressure at his blower outlet and a thermometer to determine the discharge temperature.

Maybe I am missing something here, but this may not be enough. The shape of the CD nozzle may alter the output of the blower as well if it can't keep up with the mass flow rate at the given pressure required to start the nozzle.

 

[JBA]

This is only a first step to determine if the blower discharge pressure at its current mass flow is capable of a pressure sufficient to support a sonic flow in a nozzle throat. If so, then the appropriate nozzle throat size would be selected that would provide a sonic nozzle throat velocity without creating any additional backpressure on the blower. Since the blower discharge velocity will not be changed from its current condition there will be no risk of creating sonic flow at the matching nozzle entry connection; and, then the sizing of the diverging nozzle discharge would be based upon providing the desired increase in final discharge velocity.

This all assumes that the current blower stagnation pressure is sufficient to providing a nozzle throat pressure high enough to create a pressure at the nozzle discharge, after expansion, high enough to exceed the downstream back pressure of the system and attain the desired supersonic discharge velocity.

The target is to be able to increase a velocity at the C/D nozzle discharge velocity greater than that currently provided by the blower at its current design capacity.

 

[russ_watters]

My gas isn't compressed. It enters the nozzle with a specific velocity.

Others have responded that your gas is compressed, but that is unlikely to be true in a realistic sense. The only thing you have told us about your blower is that it can generate 100m/s flow. That's actually right on the edge of what is typically considered incompressible. Strictly speaking, any airflow requires compression, but it is generally ignored up to 100m/s and considered incompressible to make the calculations easier.

Your blower - again, since you haven't told us what it is, I'm assuming - likely generates 100m/s at the outlet, where the static pressure is zero, so it can be considered incompressible.

Ironically, though, while you think this helps you, it doesn't. As we keep trying to tell you, over and over and over and over and over and over again, generating supersonic flow requires compression. The fact that you aren't doing it (or, rather, are doing very little), is the reason why you can't achieve supersonic flow.

 

[russ_watters]

Here is a selection for the highest pressure blower you're likely to find:

180 inches of water gauge is 6.5 psi or an available pressure ratio of 0.69. That's a lot better than I would have expected, but still quite a bit above the maximum required for choked flow of <0.5. And it requires 170 horsepower for just 200SCFM of airflow. That's like a midsized car engine at full throttle to run a bathroom fan.

 

[jack action]

Here is a selection for the highest pressure blower you're likely to find:
View attachment 224961
180 inches of water gauge is 6.5 psi or an available pressure ratio of 0.69. That's a lot better than I would have expected, but still quite a bit above the maximum required for choked flow of <0.5. And it requires 170 horsepower for just 200SCFM of airflow. That's like a midsized car engine at full throttle to run a bathroom fan.

Oh! Is that what HVAC guys call a "blower"?

Try a Roots-type blower, you can achieve 2:1 pressure ratio easily:

Funny how we see different things from the same statement.

 

[russ_watters]

Oh! Is that what HVAC guys call a "blower"?

Try a Roots-type blower, you can achieve 2:1 pressure ratio easily:

View attachment 224973​

Funny how we see different things from the same statement.

Wow, thanks, I hadn't heard of that. And it fits nicely into the nomenclature overlap, being referred to as a blower, a pump and a compressor in the same article! I would say that's confusing, but it really is in an overlap range: the flow is definitely compressible even though the pressure is below what is common for "compressors".

 

[boneh3ad]

I'd like to point out that 100 m/s could be considered compressible depending on the local speed of sound (which depends on temperature). That's unlikely here since the air is likely warm. The flow through the blower is very possibly compressible, though.

Also, the requirements to choke a CD nozzle are typically greater than the common 0.528 pressure ratio ($p_b/p_0$). Running it without any shocks so the exit velocity is supersonic requires a lower ratio. The 0.528 number is for converging nozzles only.

 

[jack action]

Wow, thanks, I hadn't heard of that. And it fits nicely into the nomenclature overlap, being referred to as a blower, a pump and a compressor in the same article! I would say that's confusing, but it really is in an overlap range: the flow is definitely compressible even though the pressure is below what is common for "compressors".

Actually, it is a simple "air displacer". The air gets into a chamber and the chamber rotates towards the outlet. If you put it in front a restrictor (ex.: an engine with a smaller defined volumetric flow), then the pressure will accumulate between the two, i.e. a compression. So it is in fact a «blower».

There is also a screw type compressor which looks a lot like a roots type, but it actually compresses the air within (the chamber decreases in size as it moves), so it is a compressor. That is why they are more efficient than Roots type.

Here's the cute history of its origin:

The Roots device is a positive displacement pump, used as an air pump or fluid pump. It moves air, but does not compress. The basic design was patented in 1860 by Philander Higley and Francis Marion Roots (hence the name) of Connersville, Indiana, as an air pump for mines, grain elevators, blast furnaces, and other industrial applications. As an air pump, it pulls air by trapping it with meshing lobes against the outer side of the blower case and pushes air from the intake side to the exhaust side. In an engine application, when pumped into a confined space such as an intake manifold, positive pressure is developed, along with an increase in temperature of the pressurized air (not a good thing). An intercooler can be used to help cool the charged air.

 

[JBA]

While we are listing the available types of compressors it would be unfair not to include the sliding vane type as well.

 

[cjl]

Here is a selection for the highest pressure blower you're likely to find:
View attachment 224961
180 inches of water gauge is 6.5 psi or an available pressure ratio of 0.69. That's a lot better than I would have expected, but still quite a bit above the maximum required for choked flow of <0.5. And it requires 170 horsepower for just 200SCFM of airflow. That's like a midsized car engine at full throttle to run a bathroom fan.

A centrifugal compressor can push a pressure ratio of 4:1 or more per stage, with very high airflow. You could easily size a blower that would start a C-D nozzle, and maintain the flow rate to keep it operating, although the power rating would necessarily be extremely high. As an extreme example, multistage axial jet engine compressors can flow hundreds of kg/s of air with a pressure ratio of 40 or 50 to 1, but they require powers on the order of tens of megawatts to drive.

 

[T C]

One question just comes to my mind against boneh3ad's post no. 19. If a turbogenerator is used to produce power by using the velocity of the released air/fluid, does that alter the temperature or remain the same? Suppose that the turbine blades can withstand the blow of liquid oxygen droplets.

 

[russ_watters]

One question just comes to my mind against boneh3ad's post no. 19. If a turbogenerator is used to produce power by using the velocity of the released air/fluid, does that alter the temperature or remain the same? Suppose that the turbine blades can withstand the blow of liquid oxygen droplets.

When a gas flows through a turbine, it does work against the turbine, expands and cools.

 

[T C]

I know that and that's why I want to know whether in such case the fall in temperature will be same or more.

 

[russ_watters]

I know that and that's why I want to know whether in such case the fall in temperature will be same or more.

Than without the turbine? The temperature drop should generally be more with the turbine I believe...but you are also reducing the exit speed, so I'm not sure how this fits with what you were trying to do before.

 

[T C]

This is a new question. As the thread is still alive; that's why I have posted that in this thread for quick response. This has nothing to do with the question with which I have started this thread.

 

[T C]

Another question just comes to mind against post 19 here. In case of an ideal/reversible process, can we say that if we have air/gas at 108 k temp and 614.8 m speed and that has been passed through a divergent nozzle, then when the speed is zero then the pressure will be 33.3 bar?

 

[jack action]

Another question just comes to mind against post 19 here. In case of an ideal/reversible process, can we say that if we have air/gas at 108 k temp and 614.8 m speed and that has been passed through a divergent nozzle, then when the speed is zero then the pressure will be 33.3 bar?

Yes, as long as you consider the ideal process. If the speed is reduced by going through a normal shock (usual method in CD nozzle), then there will be a lost that will be reflected in the total pressure (i.e. the pressure when the fluid will be brought to rest).

Finally, the ratio of stagnation pressures is shown below where

is the upstream stagnation pressure and

occurs after the normal shock. The ratio of stagnation temperatures remains constant across a normal shock since the process is adiabatic.

 

[cjl]

Just to clarify, it is not usual for there to be a normal shock in a C-D nozzle unless it is poorly designed or operating off of the expected design conditions. Usually, you would expect it to be fairly close to ideal, so yes, it is reversible. It is quite difficult to slow down the flow without a shock though - much more difficult than accelerating isentropically, so you would be hard pressed to actually demonstrate that in experiment.

 

[T C]

That means if an air/gas stream has 614.8 m/s speed and at 1 barA pressure, then we can raise the pressure to 33.3 barA by using a c/d or divergent nozzle?

 

[jack action]

In theory and ideal conditions, yes.

 

[boneh3ad]

True, a normal shock should not exist inside of a C-D nozzle as long as the pressure ratio is sufficient for the nozzle design. That said, if you want to then slow the resulting supersonic flow down, you almost always have a normal shock somewhere. If you're extremely clever you can maybe avoid that with a borderline magical diffuser. Otherwise, it is nearly impossible to isentropically slow a supersonic flow down to subsonic speeds without a shock. Compression waves simply coalesce too readily into shock waves to make that practical in most cases.

Also, if your gas already had 614 m/s velocity and a pressure of 1 barA, you wouldn't need a C-D nozzle to increase the pressure. You would just need some means of slowing the flow down to zero velocity isentropically. In theory, you could use a C-D diffuser to do it, but in practice, that is wishful thinking and a shock will still likely form.

 

[T C]

Also, if your gas already had 614 m/s velocity and a pressure of 1 barA, you wouldn't need a C-D nozzle to increase the pressure. You would just need some means of slowing the flow down to zero velocity isentropically. In theory, you could use a C-D diffuser to do it, but in practice, that is wishful thinking and a shock will still likely form.

Is there any other way to do the job?

 

[cjl]

Is there any other way to do the job?

There are quite a few ways, but as boneh3ad said, it's almost impossible to slow a flow down from supersonic isentropically. You can do better than just a normal shock - most high speed aircraft use a series of oblique shocks for their engine intakes, as the loss in an oblique shock is lower, but you'll pretty much always end up at a lower pressure and higher temperature than the isentropic relation would indicate, due to the losses inherent in the shocks.