How do I find the entropies for T1 = -10 deg C and T2 = 30 deg C, respectively using the R-134a refrigerant chart (SI):

enter image description here


Instead of giving you fish I will teach you how to catch it.

There isn't a particular entropy at a given temperature. Considering the wet region there should be given a dryness fraction value for the entropy. With dryness fraction value and temperature combined, you can calculate the entropy of the refrigerant:

$$entropy (s) = sf + x * sfg$$

$\text{sf = the entropy of saturated liquid R-134a}$ $\text{x = the dryness fraction (between 0 and 1)}$ $\text{sfg = the entropy change during the evaporation process}$

Now for your question:

See the image below. The sf value can be taken from the chart. See the left value. It's between 1 and .9.

Now see the (sfg + sf) value on the right. When you equate it with a certain dryness fraction x you will get entropy at that particular point.

enter image description here

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  • $\begingroup$ thank you, good sir. but i have a question, can I use the equation S = m cp ln(t2/21) instead to get the entropy? is there a cp for this kind of refrigerant? $\endgroup$ – anakin skywalker Feb 21 '16 at 8:23
  • $\begingroup$ by that you will get change in entropy. not the "absolute" entropy $\endgroup$ – Fennekin Feb 21 '16 at 9:33

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