Does the Joule-Thomson effect explain how a Thermal Expansion Valve works for refrigeration systems? 3

[timfinnegan]

As the title suggests, can the JT effect be used to explain why the pressure drops across a thermal expansion valve for a typical refrigeration system (e.g. air conditioner or heat pump)?

From what I understand, the liquid refrigerant dropping in pressure as it goes through a narrower passage can be more or less approximated by Bernoulli's equation as it speeds up, however, by the time the fluid exits the TXV, Bernoulli's equation no longer applies because the incompressibility assumption no longer holds true.

The problem is that, at least from what I've been reading online, the JT effect doesn't seem to account for phase change?

Also, does the fluid become a saturated mixture while it's in the TXV or right after it leaves the TXV?

And as a final question, for the venturi effect, why does the pressure drop in the narrow passage and then recover (as per Bernoulli's equation) yet the JT effect does not have this recovery effect? The pressure seems to drop permanently even after leaving the narrow passage into a larger space. What is the physics underlying these two different effects?

 

[LittleInch]

I don't beleive so no.

JT is for when a gas is reduced in pressure without heat entering - I can never remember which that is called, but its the effect that is important here.

A valve in a refrigeration unit usually flashes a high pressure liquid into a low pressure gas, thus requiring energy for the heat of vaporisation. The only way it can get that heat is from the fluid itself so it reduces in temperature.

In my simple mind JT doesn't recover because the pressure doesn't recover.

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[timfinnegan]

The system I had in mind is just a simple refrigeration system that you might find in your home or in a commercial space. The expansion is considered to be essentially adiabatic because the refrigerant passes through the valve quickly and has very little surface area or time to perform any heat transfer with surroundings.

I believe there are some CO2 systems that use the transcritical cycle which uses the flashing effect as you described but this is not what I had in mind when asking the question.

 

[LittleInch]

Maybe try this then

https://en.wikipedia.org/wiki/Heat_pump_and_refrigeration_cycle

"but this is not what I had in mind when asking the question." so what was?

The flashing is exactly what happens to the liquid coming out of the condensor.

It isn't J-T though.

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[timfinnegan]

When it leaves the TXV, it has flash gas because we dip into the saturated region on a PH diagram, however, this is an adiabatic process so this statement does not hold true: "A valve in a refrigeration unit usually flashes a high pressure liquid into a low pressure gas, thus requiring energy for the heat of vaporisation".

Heat transfer DOES occur in the evaporator and that one can be explained quite easily. What I don't understand fully is how a TXV manages to take 100% liquid refrigerant and drop its pressure so much that it leaves the TXV as a saturated mixture (roughly 20% mass fraction vapor) with negligible heat transfer to the surroundings.

I was looking for a physical phenomenon that explains this and the best thing I could find was the JT effect.

 

[GBTorpenhow]

Heat transfer can occur internally to the fluid system in an adiabatic process, if that is what is hanging you up. Adiabatic means no heat transfer with the environment.

An adiabatic pressure drop through a valve is isenthalpic, i.e. a straight vertical drop on a PH diagram. Temperature goes down when this occurs. JT effect strictly doesn't involve a phase change, and does not always reduce temperature with reducing pressure.

 

[timfinnegan]

The part that is hanging me up is this: I don't understand why the pressure of the fluid is lower after the valve.

What principle causes the pressure to drop? Bernoulli's equation can describe why the pressure is lower INSIDE the valve (the fluid is forced through a narrower passage which accelerates it and thus the pressure drops) but when it exits the valve, the size of the tube is the same as the inlet of the valve.

In the above image (venturi effect), the pressure drops in the narrow passage, HOWEVER, it is recovered when it enters the larger section again.

This is NOT the case for expansion valves. Below is an image of a throttling valve which has a similar geometry to a TXV:

Something about the sharp edges of this geometry prevents us from recovering the pressure even after the narrow passage. What physical process can explain this? What exactly are the molecules of the fluid doing differently between a venturi and a valve that allows it to recover pressure in one but not the other?

 

[GBTorpenhow]

The recoverable part is hydrodynamic, ie the Bernoulli part. The nonrecoverable part is essentially friction, no different from any other head loss.

Edit to add: In real life venturis, not all of the pressure is recovered.

 

[timfinnegan]

I see, does this mean that the heat generated by the friction is contained in the fluid since this is a near adiabatic process?

And thank you, it is making more sense now.

 

[Snickster]

As the title suggests, can the JT effect be used to explain why the pressure drops across a thermal expansion valve for a typical refrigeration system (e.g. air conditioner or heat pump)?

As previosly indicated the JT relates only to a drop in pressure of a gas in which temperature drops at constant enthalpy. Pressue drops across a TXV becuse the pressure is maintained in the evaporator at a lower pressure by the compessor removing evaporated gas from the evaporator. For a given evaporator pressure the compressor is pumping at a given rate, if compessor starts pumping out vapor faster the pressure will drop and if pumps out slower pressure will rise, until vaporization rate and compressor flow again reaches equillibrium. In this manner evaporator temperature of evaporation is determined by the compressor capacity versus heat input rate. In fact refrigerant compressors are equiped with low suction pressure shut down to prevent the evaporator temperature from droping so much that the evaporator freezes, this could be during low heat input where the compressor even at lowest speed is pumping faster than the fluid is vaporizing so the temperature of evaporation reaches below 32 F. This why when your filter gets real dirty and blower air flow is reduced so much there is not enough air flowing over the evaporator coils to input any significant heat into the evaporator and the coil freezes up.

From what I understand, the liquid refrigerant dropping in pressure as it goes through a narrower passage can be more or less approximated by Bernoulli's equation as it speeds up, however, by the time the fluid exits the TXV, Bernoulli's equation no longer applies because the incompressibility assumption no longer holds true.

Correct, as the fluid speeds up in the valve the velocity increases and pressure drops per the bernoulli equation where pressure head is converted into velocity head, and also there is irrecoverable friction loss. After it exits the valve and into the evaporator the pressure in the evaporator is determined as indicated above.

The problem is that, at least from what I've been reading online, the JT effect doesn't seem to account for phase change?

Correct JT only explains why temperature change occurs at constant enthalpy for a gas when pressure changes.

Also, does the fluid become a saturated mixture while it's in the TXV or right after it leaves the TXV?

Yes as soon as the refrigerant enters the lower pressure evaporator it immediately reaches saturated liquid state due to auto refrigeration (uses available heat contained within itself above that contained above the saturation point to adiabatically boil itself off forcing itself down to saturation point) and then boils as a saturated liquid until all has turned into vapor at constant saturation temperature corresponding to the evaporator pressure.

And as a final question, for the venturi effect, why does the pressure drop in the narrow passage and then recover (as per Bernoulli's equation) yet the JT effect does not have this recovery effect? The pressure seems to drop permanently even after leaving the narrow passage into a larger space. What is the physics underlying these two different effects?

Because JT effect only happens with a gas so there is no JT.

 

[Snickster]

The recoverable part is hydrodynamic, ie the Bernoulli part. The nonrecoverable part is essentially friction, no different from any other head loss.

Edit to add: In real life venturis, not all of the pressure is recovered.

In a venturi pressure head/energy is converted to velocity head/KE with no enegy loss and then converted back to pressure head without loss, assuming perfect venturi with no friction loss. For a throttling valve pressure drop is cased by friction loss which is a force times distance type work the fluid need to expend to get through the valve. This is a permanent energy loss and that goes back into the fluid as heat (which is true for both liquid and gas flow) so that the enthalpy of the fluid remains constant although pressure permanently drops.

 

[LittleInch]

The bernoulli equation / theorem is based on frictionless fluids. As noted above in reality venturis, even small smooth ones have some level of friction losses and hence pressure on the d/s side is always lower than the upstream side.

In a control valve scenario the friction losses are much much bigger. As noted above by others, if you maintain the fluid as say a liquid at all times ( i.e. at no point does the liquid go below bubble point) then the liquid will increase slightly in temperature due to the frictional losses as it reduces in pressure.

In a refrigerant liquid, this heat gain is much smaller than the heat loss due to the change of state from liquid to gas and hence the gas cools down. Maintenance of the pressure difference and low pressure in the evaporator is undertaken by the compressor.

It's still nothing really to do with the JT effect which is where you drop pressure across a control valve or restriction and any heat gain from the friction is small compared to the J-T cooling. Different gases behave differently at different pressure, so e.g. Natural gas in the 20- 70 bar range drop temperature approx 1 Deg C per 2 bar pressure drop. Hydrogen on the other hand increases very slightly in temperature for the same pressure drop.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[FacEngrPE]

If you add in the enthalpy term to the Bernoulli's equation it will closely model the behavior of a refrigerant in A TXV.

 

[MortenA]

@OP in my opinion yes, because its a throttling process with a (almost) constant enthropy - but as other has mentioned in a regular refrigeration system you would get condensated refrigerant after the valve flooding the area that you want cooled (may be a "radiator" might be pipes with a cooling medium e.g. a brine or glycol/water mix. The advantage of this is that you get a big cold reservoir in form of your boiling liquid where - as long as pressure remains constant then you will have a constant temperature in the boiling liquid.

--- Best regards, Morten Andersen