Wow. Thank you.

Exactly!

That's too many!!! ;)

So glad you liked it!!!

Hey, thank you very much!

Thank you!

Hi. There is no pressure drop inside the condenser because it is a heat exchanger. Heat exchangers produce a very small pressure drop so you can neglect, changes in pressure. In the condenser the refrigerant changes phase from superheated to saturated liquid. Just like water does it in a steam boiler (but gaining enthalpy instead of losing it as in this case). So short answer, because the condenser does not produce a relevant pressure drop. Same thing for the other heat exchanger, the evaporator does not produce a relevant pressure drop. Temperature does not change, pressure does not change, however, specific volume increases and enthalpy increases in this phase change. Did i help?

It is the same thing my way and your way of writing the equation.

Sorry for such a delay answer. Can you be more specific? Please use all the space you need.

Yes, using the tables: T2 is found using s2 and P2. Since we already interpolate to find h2, we can do the same thing for T2. T4 is easier. It is the saturation temperature at P1=P4.

From a superheated R-134a table at 1 MPa. What I am doing there is interpolation because I know my entropy at that point (s=0.94564) but I dont know my enthalpy which is the one that I need. So interpolating means "trapping" my unknown between two values, one less and the other greater than the value that Im looking for. The 282.74 and .9525 are the part where h and s greater than the value. Did I explain it better?

I found it in this example: Consider a refrigerator that operates on the ideal vapor compression cycle. The r-134a refrigerant enters the evaporator at 140kPa with x=30% and leaves the compressor at 60°C. The compressor consumes 450W. Find:- mass flow rate - the pressure in the condenser -COP The only thing that i can't get is how can i find h2 from T2 and s1=s2

@@zoulihr2455 I think that point 1 is P1=140 kPa and saturated steam. With that you can find enthalpy and entropy at 1. Now, at 2 you have T2 = 60°C and s1=s2. You find h2 in superheated steam tables for R-134a. Since the power of the compressor is the m*(h2-h1) = 450 W, you will be able to find mass flow (m). Hopefully this helps.

Thank you!

To tell you the truth, I don't remember. But I will say that most likely, it is the Yunus Cengel, book of Thermodynamics.

@@theclassoftorchia3856 Thank you very much, the truth is I did not find the book in pdf format unfortunately, please sir do you have time I can send you some exercises to help me or if you know where to find it in pdf format

@@wejdanlotfe4779 You can write here the problem and I will try to solve it. I think you can find the book at certain webpages related to education.

@@theclassoftorchia3856 Ex: consider Carnot cycle heat engine using steam as working fluid and having a thermal efficiency of 20% . Heat transferred ti the working fluid at 300C and during this process working fluid changes from saturated liquid to saturated vapor : find a) show this cycle on a T-s diagram that includes the sat-liquid and sat-vapor b) Calculate the quality at the beginning and end of the heat - rejection process. c) Calculate the work per kg of steam .

@@wejdanlotfe4779 Sorry for such a delayed answer. I will try to do it. And let you know about it. The thing is that the two points of the cycle touching sat liquid and sat vapor, have the same entropy as the other two points where quality can be measured. It is a square inside the saturation curve.

Thank you!