In replacing a compressed air motor with a nozzle where does the energy go?

[bitman]

Assume a compressed air motor, whose input power 1 cubic metre/minute at 200 psi. The motor exhaust is say, 20 psi at the appropriate volume say 10 - 11 cubic metre/minute. The motor produces 1KW due to its inefficiencies.

Suppose the motor is replaced with a nozzle whose flow rate is the same as the motor. The gas is then expanded in an appropriate diameter tube such that the exhaust from the nozzle pipe combination is the same as above.

I really don't see where the energy that would be produced by the motor has gone. One would think that some sort of heating should take place but I don't see why or where. One could also argue energy lost to molecular friction but this would cause the gas to warm and the motor would suffer from molecular friction anyway.

I'm sure there must be a simple explanation, so please indulge my ignorance and curiosity.

Best Regards

bitman

 

[Simon Bridge]

Your question is a tad vague - which energy are you talking about in which process?

 

[sophiecentaur]

Yes, it is a bit vague but I think i can see where you're coming from.
If you replace the motor with a nozzle, you could choose a nozzle that would dissipate the same energy into the atmosphere that the motor would use. In this case, the turbulence (Random Kinetic Energy) in the expelled air would be dissipating the same power as the original motor. The nozzle would need to produce a reaction pressure at the end of the tube (at the place from which the motor had been connected) equal to that caused by the motor and also allow an identical flow rate of air. This would mean that the pressure times the volume flow rate was the same so the power would be the same too. The air supply would not 'know the difference'.

In most circumstances, however, the pressure or flow rate would not be the same. You might, for instance, hear a change of note from the compressor as its load changed. This is a bit like the way that a vacuum cleaner motor will race when you cover the nozzle - the appliance's motor is just not delivering the same power.

 

[bitman]

Apologies for vagueness. A compressed air motor does work, in this case 1KW worth. A nozzle just sits there and restricts the gas flow, with no apparent work being done. The energy removed from the system is the same in both the case of the nozzle and the motor. So where did the available energy go ?

Is it just a case that the molecular friction caused by the nozzle causes the gas to be heated such that an equivalent amount of energy is lost? It seems unlikely to me as the pressure at the exhaust of either system is the same. Would the nozzle get hot due to the friction ? perhaps using some of the energy.

If one were to take an electrical circuit as an analogy, with a motor being replaced with a resister, the resister would get hot and dissipate the same amount of energy.

However in this case nothing appears to get hot. The gas expands, and as a consequence its temperature falls and volume increases. In the case of a motor the expansion of the gas is converted to mechanical energy. In the case of a nozzle the gas expands to fill the space available with no work being done, so where is the energy of expansion dissipated?

Hope this clarifies things

Best Regards

bitman

 

[sophiecentaur]

High velocity air comes out of the nozzle (with K.E.) and causes more air molecules to move, turbulently and away from the nozzle. All this KE gets spread around the molecules in the vicinity until the random collisions distribute the 'directed' KE into higher KE of the atmosphere.
The reason that you aren't aware of a temperature increase is that the heat is spread over a very large volume / mass of air. Temperature and Energy are very distinct. Look at the temperature of a 1W pea bulb filament after a few seconds, compared with the temperature of a 3kW electric kettle.
If you put your system into a small sealed room you would soon feel a significant increase in temperature - just as if you had a 1kW heater running.

 

[DaveC426913]

... a vacuum cleaner motor will race when you cover the nozzle - the appliance's motor is just not delivering the same power.

[ OT ]
Is that why it does that? I've always known that if you plug a vacuum or hair dryer, the tone goes up, but never figured out why. Is it because, when the air flow is killed, the motor meets with less air resistance, so it is under less load?
[ /OT ]

 

[sophiecentaur]

The fan has 'less to push against' when the inlet or outlet are blocked. When the inlet is free, more air can pass through, using more power.

 

[DaveC426913]

The fan has 'less to push against' when the inlet or outlet are blocked. When the inlet is free, more air can pass through, using more power.

Yeah. Like lifting an outboard out of the water. Of course.