Conservation of Energy OR Momentum?

  • Thread starter 246ohms
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  • #1
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Main Question or Discussion Point

I have a closed system (a wind tunnel with entrained air) where a high speed air flow sucks up ambient air to provide a combined flow of the 2 air streams over the model. The resulant flow has a mass flow rate equal to the combination of the 2 air streams.

However in calculating the velocity of the combined flow there seems to be 2 answers.

If I use momentum then m1*v1 + m2*v2 = (m1+m2)*v3

but if I use energy then m1*v1^2 + m2*v2^2 = (m1+m2)*v3^2

m1 and v1 being the high speed air mass rate and velocity and m2 and v2 are the entrained air stream mass rate and velocity. v3 is the needed velocity after mixing.

Both equations are Laws of conservation but in this instance which one is correct and why. There seems a lot of history about which is correct but no real conclusion is drawn.

Many thanks for any guidance.

ohms
 

Answers and Replies

  • #2
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I would use momentum conservation. Whenever you have two masses stick together, usually the energy is not conserved (some escape to to heat mostly), but momentum is always conserved.
 
  • #3
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In collisions momentum is always conserved. You must be very careful when you say energy is conserved....it is ....but it is not KINETIC ENERGY that is always conserved.
Usually some kinetic energy is 'lost' by being converted to heat as a result of friction, or sound.
 
  • #4
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I would use momentum conservation. Whenever you have two masses stick together, usually the energy is not conserved (some escape to to heat mostly), but momentum is always conserved.
In collisions momentum is always conserved. You must be very careful when you say energy is conserved....it is ....but it is not KINETIC ENERGY that is always conserved.
Usually some kinetic energy is 'lost' by being converted to heat as a result of friction, or sound.
The masses lose both energy and momentum to friction. It's just that this momentum loss is vanishingly small (i.e. [itex]p=E/c[/itex]), even more than that really because the friction emits photons in a pattern that is nearly random from the point of view of the COM frame of the collision. But yet, for the purposes of these kinds of problems, assuming conservation of momentum is definitely the way to go.
 
  • #5
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Many thanks for your comments. In the meantime further studies have resulted in no firm resolution of which model to use.

There have been many similar post on the momentum versus energy discussion and many interesting answers but non seem to catogorically put one model as the preferred solution over the other except by use of innate understanding of the problem.

From a work point of view it appears that the primary stream has 'work' available and is stored as kinetic energy, using this to do work on the entrained flow would mean using the energy equation. But this negates the momentum equation. To do this I have assumed ideal conditions with no turbulence or shear forces to produce heat. Even if you use P*A (pressure times area) in the energy equation the balance still fails.

The most interesting anomaly comes when using h+0.5*c^2 as the systems energy at each of the control volume boundary and equating them. Using momentum to solve for the values of c (speed) and hence h and then Htotal adds energy to the system.

Any more ideas.

Thanks

ohms
 
  • #6
cjl
Science Advisor
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In general, the correct equation for fluid flows is actually continuity. There can be pressure forces that are difficult to account for, and assuming your air velocity is less than about a hundred meters per second or so, you can just assume it to be incompressible. This leads to the simple relation where A1v1 + A2v2 = A3v3 (where A is the cross sectional area of each region).
 
  • #7
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This is a supersonic case where the wind tunnel is driven by a supersonic flow and mixes with the entrained flow to get medium flow with more volume. Continuity is considered and essential to help solve the combined flow.

I just hope when it is built and tested it meets the results from the momentum equation or else the various tunnel sections will be a long way off.

Thanks
ohms
 

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