subcooling and supaheat calculations??

i always forget which way around things are, so i though i would get a thread started. This way I can book mark it
is this correct.

subcooling.
:confused: we take the pressure measurement at the condenser and convert that to degrees F.
we then measure the temperature of the liquid line after the condenser again in degrees F
subcooling an example now; example (r290) 150PSIG(89F) - 82F gives 7F (what is the ideal subcooling value?)


then i would do the same for supa heat. take the pressure and convert that to degrees F with table.

:confused: is it the difference between the pressure/temperature valve and the evap temperature (or is it the) temperature 6" from the compressor. (i always get messed up with this bit)
:confused: 8F-12F is a good level of supaheat correct.

:confused: the negatives seem to stuff the answer up though.
Example: r290 at 10Hg and evap -50c and 6”from compressor is -40c.
10Hg(-60F) - -41F= -20F ??

and another system example: evap is -31c, 6”from the compressor is -20c, pressure 0spig
calculation
0psig(-43F) - -5F= -38F

or if I do it on the evap temperature.
0psig(-43F) - -25F= -18F

i remember that garry (what has happened to him, hasn’t been here for months) use to teach alot about this, he said it wasn’t about the temps its about getting these figures correct to enable the phase change system to function correctly and per long compressor life etc.
so any help i would appreciate :toast:

Subcooling is how much of a temperature drop you get after you condense your gas.

The temperature your gas condenses at is obtained from the high side pressure reading and a chart.

So for propane condensing at 160 psig, the corresponding temp is 34.5ºC
Now measure the temperature at the exit of the condensor and this will tell you how much the temp has dropped after the gas has condensed.

Say it reads 24.5ºC then the degree of subcooling is (condensing temp - condensor outlet);
34.5-24.5 = 10º of subcooling.

For superheat we want to measure how much of a temperature increase is exhibited by the evaporating gas, thus:

Take the low side pressure and look up chart for corresponding temp.

So for 0psig the evaporating temp is -42ºC (R290).

Now measure the temp of the suction line 3-6" from compressor, say it reads -27ºC.

Your degree of superheat is now (Suction Line Temp - Evaporating Temp);
-27 - (-42) = 15

If you do your calculations in ºF then you need to divide your final answer by 1.8.

Hope this helped

great info

Looking at your sums for superheat, you were subtracting the suction line temp from the evaporator temp. It should be the otherway around also use the evaporating temp from the chart at the low side pressure. So for your 2 examples;

Example: r290 at 10Hg and evap -50c and 6”from compressor is -40c.
10Hg(-60F) - -41F= -20F ??
It should be;
convert -40ºC to ºF.
-40*1.8+32 = -40ºF
Next;
-40 - -60 = +20
Next convert to ºC/Kelvin
+20/1.8 = 11.11 degrees of superheat


and another system example: evap is -31c, 6”from the compressor is -20c, pressure 0spig
calculation
0psig(-43F) - -5F= -38F
This time we can convert the chart temp to ºC first this time;
(-43 -32)/1.8 = -41.67ºC
Next subtract chart temp from suction line temp;
-20 - (-41.67) = 21.67 degrees of superheat

These systems will work fine with a superheat of 8 to 12 degF and a subcooling of 5 to 8 degF.

thanxs guys this is exactly what i want and some great examples
cheers.
so me rotary 1HP running bbq r290 runs sweet at 10Hg
and the other will need a capillary tube shortened.

chilly1 think we can get a thread stickied about this math stuff?

I is already in the stickins isn't it?

Quote:

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Originally Posted by chilly1

I is already in the stickins isn't it?

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Soooo ... Good sticky! What do ya think? :)