Review (distorted) on heat pumps and refrigerators machinery

[bolt]

goodnight Capt_Maloche

Do you know specific examples of flow rates in existing regulators:

for example compressor heat pump 3 kw COP = 4 = useful heating 12 kw with such gas: what precise flow can there be in the regulator (capillary or not)? and which functional delta P (pressure drop)?

just to realize the reality

I neglect the accessories (eg exchangers fans)
in general, how much does the power of the accessories represent in% tage of the compressor?

bolt

[PITMIX]

We agree that the energy recovered can never run the compressor. But it's true that the idea of ​​a bolt is great.

[Targol]

I think that ends this debate.

For those who have understood what was written above this sentence, surely. For balls like me, it closes nothing at all.

Because knowing nothing about it does not prevent bad faith:

Yes, but if there is a spatio-temporal fault that opens in your compressor and aliens take advantage of it to swing lots of energy into the system, huh, what's going on, huh? Is the debate still closed?

.... Okay, I quit smoking glue

[Capt_Maloche]

What do you want, it's a job

[Other]

Hello

Do your job to live whole
it tells you something before landing! the recitation of the pilots.

we are far from freon 22 or 410 a little joke in passing ...

Andre

[Philippe Schutt]

But often the COP of PACs with water withdrawal does not take into account the kw which are used to convey water to the PAC

If you re-inject the water at the same depth as you pumped it, and your pipes are large enough not to have too much pressure drop, the consumption of your pump should not be very bad ...

For your regulator system, Guy Nègre also put a regulator on his compressed air cars. I always wondered why, since it is a loss of energy when he has a big problem of autonomy ...

[PITMIX]

Yes, but in Guy Nègre's car the air is compressed in a tank in order to obtain better autonomy precisely. Then it is necessary to relax the air in order to use it at the adequate pressure.
The regulator will create a pressure drop but no autonomy problem.

[bolt]

If you re-inject the water at the same depth as you pumped it, and your pipes are large enough not to have too much pressure drop, the consumption of your pump should not be very bad ...

Agree for dynamic pumping levels (water level reached during pumping given useful for the heat pump) very little different from static levels (tablecloth at rest)

This involves drilling (one to sample, and one to reject) with very good communication (very good flow potential: level moving very little during pumping or rejection)

knowing that a submersible (centrifugal) pump of 3 m3 / ha has a hydraulic efficiency of less than 60%, to be multiplied by the efficiency of the electric motor (eg 57% motive energy / energy consumed for a 370w mono motor ; 64% for a 370w tri at full load) and it is like diesel engines, if they are used unless the full load their efficiency decreases again

the 1,1kw mono turns at 72.5% and in tri 73 to 74%

Note that there are special engines with more efficient performance, (a little) but the price is not the "a little" more efficient, it is rather the double.

On the other hand, if you have a static level at -60 m, which goes down to -90 m in dynamic level at 3 m3 / h, you need a pump that makes 9 bar to raise the water -1 bar maximum for l in the 2nd drilling + pressure drops to push the water into the pipes

In the first case (niv dyn 10 m.) 370w is more than enough: can do 3 m3 / h at 3,5 bar (I do not know a weaker engine in references

In the second case, the minimum 1,1kw is required for the 3 m3 / h at 8.5 bar

See therefore, what heat pump capacity can we rotate with eg 3 m3 / h at 11 ° C start
or 3 m3 / h at 17 ° C flow

what do you think Capt_Maloche?

For your regulator system, Guy Nègre also put a regulator on his compressed air cars. I always wondered why, since it is a loss of energy when he has a big problem of autonomy ...

The compressed air reserve has too much pressure when it is full "to bursting" to be used without a regulator
the engine would not hold up (or so it would start "on the wheel caps, and finish the reserve normally)

Energy loss is difficult to understand for compressed air motors.

It is mainly done during the compression of the air to charge the reserve, but if it is a wind energy compressor we don't care

Arrived in the engine, if the air entered directly, by over-relaxing it might cool down too much (in addition to the mechanical forces not foreseen)

Energetically, and especially functionally, it is best to relax the excessively compressed air, and as this relaxation cools it, pass it through an exchanger to redilate it (recover more useful volume by passing ) (and have air not too cold for the engine)
See even, if it is a mixed engine (compressed air in addition to a normal engine: take calories (lost) in the exhaust (like the pantone) to increase even more the effective useful volume of the air reserve compressed.

bolt

[Philippe Schutt]

On the other hand, if you have a static level at -60 m, which goes down to -90 m in dynamic level at 3 m3 / h, you need a pump that makes 9 bar to raise the water -1 bar maximum for l in the 2nd drilling + pressure drops to push the water into the pipes

if the level drops by 30m, we can admit that on the other side it goes up by the same amount? would still remain 30m, or 3 bars.
do you have to look for water at 60m?

Energetically, and especially functionally, it is best to relax the excessively compressed air, and as this relaxation cools it, pass it through an exchanger to redilate it (recover more useful volume by passing ) (and have air not too cold for the engine)

yes, it does. when the air is at a more manageable pressure it passes it through successive engine stages with intermediate heaters. But the first pressure drop, the largest, is wasted with this system. The diameter of the 1st piston could be very small, like a diesel injection pump to limit the torque. it does not seem to me infeasible, and given the dead end in which it is, I am puzzled ...

[bolt]

-1 bar maximum for suction in the 2nd drilling

if the level drops by 30m, we can admit that on the other side it goes up by the same amount? would still remain 30m, or 3 bars.

Even if the water from the 2nd borehole is 30 m deep, the water returned to it will not be sucked more than 10,33 meters

to "empty" does not exist "empty" and a half and even less empty X 3 (30 meters.

do you have to look for water at 60m?

here it is for the example, but I have already seen a static level of 15 m. descend more than 100 m. for less than 3 m3 / h

bolt