Air Flow Question.

V2 = V1 A1/A2.

Edit: Oops, true for incompressible fluids.

Hi Tornado, you answer yourself right.

In this case , it is air.

I think that you are right because, when the system becomes stable, i.e. the flow has been established, the same mass of air coming in will have to come out even for a compressible media.

Please read my reply in #5.

If you said same mass of air coming in and going out , and also , there is increase of speed, it means there is gain in energy (1/2 mv^2), BUT there is no input of energy !!

I agree with Pritam. Until the air exits from the nozzle, the upstream pressure and density will likely not vary by much, and the given formula will still be pretty close.

If this is correct, it means the amount of air Q before the tube is moving at V1, the same amount or air is moving at higher speed V2, meaning there is a gain of energy (1/2 mv^2) !!

But , there is no input of energy. Therefore your statement is wrong

You are right, since no energy input is present on the track total energy is constant,i.e.

p+ ρ*v²/2= constant.

If fluid is non viscous and non compressible there is a mass flow continuity and this is an other story.

Figure 1 shows such a flow, simulation is based on a velocity potential which is 100 at the bottom and 0 at the top. It is an axis-symmetric nozzle. Figure 2 shows for same simulation the pressure distribution. Pressures are negative since for the simulation the initial pressure was considered as = 0. One sees the "brake effect" of the nozzle and the increase of pressure at the cone back.

Hi Nick name,

thx for the nice simulation result, but I can't understand it. Good if you can explain.

p+ ρ*v²/2= constant.

Yes ,the above applies to non compressible fluid only.

In this case, air is compressible.

So, can any one show the analytical way , to get V2?

I thought there should be some simple formula for this ?

You should shift to "mass flow" and postulate that it is constant. Then consider that the total energy is a constant too. The transformation can be considered as either isotherm if changes are small or isentropic (adiabatic) if changes are more important. You could consider only the truncated cone and assume that the cylinder is of no loss.

Yes the simple formula is the same: p+ ρ*v²/2= constant.

I will elaborate it a little. The above formula can very well be applied for air, even though it is compressible, because there is no appreciable change in total pressure. Even for fans this formula is used where there is little change in pressure. Above formula can also be written as:

p1+ ρ1*v1²/2= p2+ ρ2*v2²/2= constant. OR

p1+ ρ1*v1²/2= p2+ ρ2*v2²/2 + delta p to acount for losses in pipe flow

The term "p+ ρ*v²/2" is called total pressure which is sum of static pressure (p) and velocity pressure(ρ*v²/2). So, at reduced cross section of pipe when velocity is increased to v2, it does not mean that total pressure is increased. In fact the static pressure component is decreased, so that the total pressure is almost constant. Hope its clear bravo88.

In order to give the best explanation please formulate your questions as clear as you can and I promise you get all explanations you need. As a start it is a finite elements simulation of the flow of a non viscous incompressible flow. I considered a cylinder-cone nozzle so that the program computed only half of it (till centre line). I took a large space around because you wrote about wind so that the computing space had to be a lot bigger in all directions so that the effect of the nozzle was not under the space limits influences. For your case - a compressible fluid- this simulation ONLY gives an image, a hint, but is not at all quantitatively correct.

"p+ ρ*v²/2= constant. Yes ,the above applies to non compressible fluid only."

I believe the equation applies to all fluids - compressible or not!

Most engineering solutions will consider the above model as isothermal and therefore the only change to cosnider is pressure. Many books can help you find the pressure drop in the conical section. So,

P2= P1- deltaP

ρ2= ρ1*p1/p1 (pressure in abs units)

So we now know p1, V1, ρ1, p2,ρ2, and therefore V2 can be calculated with the above equaition.

I am sure it does not have to be more complex than that.

Do you really know all the parameters you wrote ?

All you know is that a wind "w1" blows in a tube and you have to define the outlet air velocity.

Which are the values with index #2 ? How did you find them ? May be you are right, please elaborate and explain how you obtained the values.

You insists! yes,there is input of energy!.-

Yes, I think you are right. There is input of energy. There is definitely no input of energy by the pipe, but by the wind blowing towards the pipe.

That solves my 'mystery' that there is gain of kinetic energy at the nozzle.

I feel you understood the case.Just remember add what others said here:you may try air as a liquid in most common cases even in fans one said and i add even in speedometers for airplanes;that is not after your understanding,is an empiric result,not "cellestial inpiration";2)When is not possible hold that hypo,next to,is consider ρ•A•v=constant (which is neither Bernouilli) in a stationary flow, if not the case will come even harder! As an exercise lets think there is not a kinetic change→v1=v2 and ρ2=(A1/A2)•ρ1 ,this density change could becomes in differents ways (severals) but you should consider at least two extreme ones:adiabatic and isothermal ways...Bernouilli equation we use commonly is not complete but works fine in most of technical cases.-

The air speed V2= A1*V1/A2

You are repeating the answer in #1

Consider mass flow through the two control surfaces A1 and A2 is equal and continuous. That is true whether or not the fluid is compressible. The flow through nozzles is isentropic. The speeds through the control surfaces are inversely proportional to the control surface areas, with high speed through low area. So consider the ratio of speeds V2/V1 is equal to the inverse of the ratio of areas:

V2/V1 = A1/A2 ==> V2 = (A1/A2)*V1

Yes, Exactly this will give you speed the others give you volume and flow.

Do you assume that the specific volume of gas is constant ? This assumption allows you to write that velocity is proportional to the areas ratio. If the mass flow is constant and the flow is isentropic then you should not assume that the specific volume is constant. There is no reason for it to be.

There are two equations to solve the problem:

1- mass flow is constant thus d(M')/dx= 0

2- total energy is constant i.e. E total = p+rho*V²/2+ the work done to compress the gas = constant (Rho = 1/v) which can be also put in the form dE/dx=0.

I wrote "compress" because to pass same mass at higher velocity some work has to be done.

One MUST have two equations since there are 2 unknowns: p and v both being function of the axial coordinate and of the cross area change rate dA/dx. All other solutions are based on assumptions which have no base. In the case of an incompressible fluid the 3rd term disappears and you find the basic equation which unfortunately is not valid for this case.

Sorry, but there is a change in kinetic energy,when inside the nozzle you may suppose remains the same quantity of energy, from inlet to outlet there is a difference of kinetic energy. A•V=constant, even if true, is not Bernouilli equation.Think better in a liquid:it will need some work of the pressure passing thru a small hole, this work is per second (P1-P2)•(A1•v1)=.5•ρ•(v2**2-v1**2)•(A1•v1);(A1•v1)=(A2•v2).Left side says work done per second is pressure difference by volume per second passed thru the hole,on the wright side of equation says kinetic energy per unit of volume difference by volume per second.Course you may put it into a simpler way ordering like ΔP=-.5•ρ•Δ(v**2) what is the Bernouili formula.-

See answer #3.

Sundog

Do you want an answer or the correct answer ?

If you want an answer then of course you can use the ratio of sections as in the case of an in-compressible fluid.

If you want a correct answer then the equation is a LOT more complex even in a simplified form.

I try to form the right equation and I shall send it when it is ready.

thx, appreciate it !

You said that the simulation pictures were not clear to you. Here is one to be easier to understand.

The nozzle presence perturbs the flow and has as direct follow up a velocity reduction in front of nozzle. In the simulation flow is from down to up. As you can notice flow has a constant velocity till it reaches the conical section. Here it is accelerated and after the nozzle there is a zone where the speed is still higher than the one of outer flow. The way as velocity grows shows that its velocity change is related to the dA/dx value i.e. the area change rate in the cone. It is also possible to notice the nozzle "shadow" effect which manifests itself by the low velocity zones on the cone outer surface. Although air is a compressible stuff if velocity is small it can be accepted as "almost" in-compressible. The problem is that the speed value at the cylindrical part entry (V1) is NOT known so that to apply the several times mentioned formula is not possible without making an error since one takes as reference an unknown value ! This is the problem I am still confronted to.

Great, I gave you GA .