Calculating Refrigerant Charge

Hello All

Im having a bit of trouble with some calculations. I would like to know how to calculate the size of charge (in grams) of R1234yf I would need to add to a hose that would result in a desired pressure at a certain temp.

Specifics:
Volume of Line: 15 cm^3
Volume of Oil: PAG 3% of volume = 0.45 cm^3
Temp: 135C
Desired pressure: 2 MPa

I tried calculating myself using ideal gas laws but I dont believe R1234yf would act ideal. My numbers seem low and I would like a math/fundamentals check.

Thanks in advance,

You are sure the temperature is 135C?
That's above what most oils can handle.
(135C=275F)

:cool:

Is the line carrying gas only - ignoring oil - two phase - liquid only?

Is the vapour pressure of the two phase mixture at 135'C equal to 2 MPa?

Viking has a valid comment about the oil at 135'C. What oil are you using? :confused:

Well considering the OP is based in the USofA, my thought is that the C is a typo as 135F = ~57.2C which makes the whole process more viable.
(And the PT charts I found for R1234yf goes up to 148F/64.4C)

:cool:

It is 135 celcius. The hose in question is a high pressure discharge line for an automotive climate system. My goal is to find the grams of R1234yf needed for an initial charge at room temperature that when heated to 135C results in a pressure of 2MPa. The oil being used is run of the mill PAG oil compatible with R1234yf.

From the PT charts I found online; at 2MPa the saturated temp for R1234yf is >148F. Assuming that my theory about the typo is correct, you are looking at a sub cooled liquid. Quick google search gives the density of R1234yf in liquid state as: 1.1g/cm3.
Your volume is: 15-0.45=14.55cm3
14.55 x 1.1 = 16.005g

Might seem low but if we compare to water (a completely useless exercise as it is not at all related) then you could fill the same volume with 14.55g of H2O.

:cool:

Could the OP give more info on how his proposed system works, in practice?

What connects to your pipe & what does the pipe connect to? In this way, folks can double-check your logic . :)

Attachment 14786
If you take a look at the R1234yf PH diagram, you see that the point with t = 135ºC and P = 2 MPa has a specific volume of ~0.013 m3/kg.
In an internal volume of 14.55 cm3 (15 cm3 - 0.45 cm3 of oil) you will need 1.12 grams of R1234yf.
Are you sure that this is what you need?