Warning! This post title may sound more appealing than it really is! Some of you will find this post boring, disappointing, useless and hard to swallow. So if you are not Geeks, Nerds, or simply love the subject, you should leave!

I only posted it because I’m a stubborn mule that sometimes does not follow wife’s recommendations.

This is an effort to take this subjet off US_Iceman post on “How to design refrigeration systems” which is not boring and worth reading, you can see here: http://www.refrigeration-engineer.co...ad.php?t=14013

If you think on expansion devices in general I suppose most will agree that if you subcool the liquid at the entrance of the expansion device, the compressor superheat gets lowered.

What happens with TEV?

Short version for transient conditions (not the main point of this post):

It is clear to most I suppose that during transient conditions the superheat at the expansion valve gets lower because what you really did was produce less flash gas and more refrigerant in liquid state entered the evap so the point of evaporation of the last drop should be nearer the bulb of the TEV than before. This means that superheat went down the valve opened in reaction but meanwhile the superheat at the compressor did go down.

Unfortunately I have to do some simplifying hypothesis because my wife demanded I leave partial derivatives and Taylor series expansion of two variables out of this! Therefore this is not a proof!

SYMPLIFYING HYPOTHESIS:

We should all agree that suction pressure goes up because heat exchange inside the evaporator was improved, because liquid refrigerant has much higher heat transfer coefficients than refrigerant vapour. So the suction pressure must go up.

This should also increase the mass flow of refrigerant you can clearly check this from any compressor manufacturer data.

I’ll take R717 as an example but this will be true for every refrigerant that has a dew point curve with positive slope when you draw a P-h diagram. This means that the dew point curve goes right and up when enthalpy increases (to the right).

Also I need to split the effect of the expansion valve and the suction line pressure drop so all changes in temperature are caused by the superheat of the expansion valve (suction line perfectly insulated) all pressure drop occurs between the bulb and the compressor suction.

Also that all changes are small, for if they were large it means other variables I’ll suppose constant are moving, like thermal load, etc.



Drawing from Refrigeration Utilities version 2,84 Department of Energy Engineering, Technical University of Denmark (great program TKS DTU!).

IN STEADY STATE THEN:

the superheat in point 3 is Temperature(3)-Saturation Temperature(3) i.e. horizontal line that projects to point X. Superheat at the new situation is Temperature(C)-Saturation Temperature(C) that projects to point Y.
From the drawing you can see that if the DewPoint line is tilted enough to the right the line Y-C is shorter to X-3.

With numbers:

I’m sure I don’t need to point this out but the calculations are:
Known (1)
Pressure(2)<-Pressure(1)
Temperature(2)=Saturation Temperature(1)+10c
Calculate Enthalpy(2) given Pressure(2) and Temperature(2)
Enthalpy(3)<-Enthalpy(2)
Pressure(3)=Pressure(2)-Pressure drop in suction line
Calculate Temperature(3) given Pressure(3) and Enthalpy(3)
Calculate Saturation Temperature at Pressure(3)
Subtract the last to numbers above.

The expansion valve will try to keep a 10 kelvin superheat. The pressure drop in the suction line is 0,1 bar, the saturated temperature at the evaporator outlet pressure is -20 centigrade. The evaporator saturation temperature goes up 1 kelvin.

I’ll take NIST Refprop standard database 23 version 8 numbers here. They will differ from other sources. Also I’ll use all decimals of course this is nonsense but math of very close numbers won’t be a problem.

Temp. Pressure Density Enthalpy Entropy
(°C) (bar) (kg/m³) (kJ/kg) (kJ/kg-K)
(vapor) (vapor) (vapor)
1 -20.000 1.9008 1.6033 1580.8 6.3757 (*)
2 -10.000 1.9008 1.5317 1604.7 6.4680
3 -10.430 1.8008 1.4510 1604.7 6.4937
X -21.211 1.8008 1.5240 1579.2 6.3944 (*)
Superheat 3 10.781

A -19.000 1.9867 1.6711 1582.2 6.3604 (*)
B -9.000 1.9867 1.5965 1606.1 6.4528
C -9.45 1.8867 1.5163 1606.1 6.477
Y -20.167 1.8867 1.5921 1580.6 6.3783 (*)
Superheat C 10.717

Now you see why I used all the decimals.
10.717<10.781 the original point was that if subcooling makes suction pressure go up then superheat goes down (you don’t have to tell me you can’t possibly measure this difference in the field).

The only trick here is that I used the same pressure difference in both cases and in the second case the pressure should go up if the mass flow went up. But if you cant measure this superheat I can assure you the difference in mass flow is much less. So the conclusion still stands.

CONCLUSION

Though not formally proved, at least acknowledge that If subcooling goes up in general superheat goes down if the refrigerant is well behaved. (Most refrigerants are).
In the field you cannot measure this difference. (This doesn’t mean it IS there).

Didn’t I tell you this was boring? You should see the real proof and for different refrigerants and try to calculate in what range this is valid, I can’t think of a better torture for a student, he’ll change careers immediately!