force of jet?

[Nabla1]

Hi, I've been wondering about this simplified problem - Say I have an airtight tank of some kind, connected to which is a compressor/impeller drawing in and compressing ambient air, to maintain a constant pressure in the tank, P_tank.

Also connected to the tank is an open pipe allowing the flow of fluid back through to the outside. Since the pressure in the tank is greater than that of the ambient air - due to the presence of the impeller - air will flow back out into the atmosphere via the pipe.

I've been thinking about how to calculate the force of the jet exiting the pipe. So far, assuming an incompressible, inviscid flow I've done this:

Let A be the cross-sectional area of the pipe opeing into the tank, and B be the cross-sectional area of the pipe opening into the atmosphere. Let P_0 be the static pressure of the ambient air.

Then the force on the fluid within the pipe will be:

F = A*P_tank - B*P_0

the acceleration of the fluid is then:

a = F/m

Where m = pV is the mass of fluid in the pipe (p=density, V=volume of pipe)

I also know that THRUST=MASS FLOW*VELOCITY, but I'm not really sure how to apply that, since the fluid is constantly accelerating.

Also, if the impeller is running at a constant power, the pressure will not necessarily be constant, since the open pipe will create a pressure drop, so there must be some kind of transient period before a steady state is reached. How do I account for this?

I'm also aware of Bernoulli's equation, but I'm not sure that's helpful here either, because of the acceleration.

If someone could help me out that would be great. Thanks alot.

 

[jmiles]

Im not an expert on this but you have amde what i believe is an incorrect assumption, the fluid is not constantly accelerating, it will have to reacha steady stae condition, your problem is F=ma it just does not apply here.

I think you were on the right track with your thrust calculation.

As well if you know what your volumetric flow is calculate the velocity based on the cross sectional outlet area of the pipe. that gives you a velocity and a mass flow and hence your thrust.

 

[Compositepro]

F=ma does apply but it must be applied correctly. Clue: the thrust of a jet is due to the rate of change in momentum of the fluid.

 

[FeX32]

Rate of change of momentum;

F=d(m*v)/dt, if you assume the mass does not change;
=m*dv/dt => m*a

=> find an expression for your velocity as is changes with time and simply re-solve the differential equation.

Also, your stated "THRUST=MASS FLOW*VELOCITY," is exactly this.




Fe

 

[MJCronin]

I didn't think that we were going to continue to do homework problems on these fora.....