Repeat the calculations of Problem 10.22 for Refrigerant 134 a as the working fluid. Compare the

Repeat the calculations of Problem 10.22 for Refrigerant 134...

Key Concepts

Refrigerants and Their Thermophysical Properties

Refrigerants are working fluids used in refrigeration and air conditioning systems. Their thermophysical properties, such as boiling point, specific heat, and latent heat, determine how effectively they can transfer heat, and they are critical for designing and evaluating refrigeration cycles.

Vapor-Compression Refrigeration Cycle

This cycle is a fundamental thermodynamic process used in refrigeration systems where the refrigerant undergoes compression, condensation, expansion, and evaporation. Understanding this cycle is essential for determining system performance and calculating key performance indicators such as the coefficient of performance.

Thermodynamic Property Evaluation

Accurate analysis of refrigeration cycles involves calculating thermodynamic properties at various states of the cycle (e.g., pressures, temperatures, entropies, and enthalpies). This evaluation is crucial for performing energy balances and determining the efficiency of the system.

Performance Metrics Comparison

Comparing different working fluids involves assessing how variations in thermophysical properties affect the overall performance of the cycle. Metrics such as the efficiency or coefficient of performance (COP) help in determining the benefits or drawbacks of one refrigerant over another.

Transcript

00:01 Looking at refrigeration and air conditioning, okay? now consider these are problems here the temperature need to be kept at 20 degrees celsius.

00:15 Okay, all right.

00:41 And the outside temperature of 5 degree.

00:45 So if this is our 20 degree, and these are 5 degrees here.

01:09 And this is a evaporator temperature.

01:12 Temperature, 2 .4, sorry, minus 5 .37, actually 2 .4 bar, 2 .4 bar minus 5 .37.

01:37 Okay.

01:40 And this is 8 bar, the condenser pressure.

01:49 Condenser pressure is 8 bar.

01:51 And the temperature 31 .33 .3 .3.

02:01 All right.

02:04 So refrigerant is a r134a entropy, not joke, kilogram kelvin.

02:24 All right.

02:25 So for us to choose okay, choosing refrigerants, sorry, evaporation and condenser okay, evaporator, which is a pressure of 2 .4 bar, i've indicated for condenser, which is a pressure of 8 bar, all right? then between, so let's just do our little calculation.

03:24 So between minus 3 minus 5 to 5 degrees celsius we have 10 .37 degrees to offset so so the temperature the temperature the condenser now can just get it i just call it t -com you know give us equal to our 20 degrees we required plus 37 sorry plus 10 .37 plus 10 .37 plus 10 .37 10 .3 7 degrees celsius so it gives us equal to 30 .37 so the 10 .30 .37.

05:21 So the 30 .37.

05:22 So but from the arrow 134 -a -8 table, it by 30 .30 .37.

05:26 So but from the arrow 1 34 -a .8 table, it by 30.

05:32 31 .33 so we have taken 31 .33 there okay all right so between that point then for evaporator after ever for evaporator so let's say subna at evaporator also we have each one is of 1 to be equal to 244 24 .09 0 .9, kilo per kilogram, hf, okay, at the liquid line, saturation liquid line.

06:28 We'll have this kilojou per kilogram.

06:33 And we can also get the hfg.

06:40 We're going to give us a 201, 201 .201.

06:54 201 point, 0 .14, the photogram, then s at 1, okay? you're going to give us a 0 .9, okay, and that saturation of the ball.

07:19 We're going to have that.

07:20 So, at condenser, at condenser, we are going to have s this, the vapor line, okay? it's going to give us a adapter pressure of heat bar.

07:46 0 .966, sorry.

07:56 We can do for this as at the liquid line.

08:01 That treasure, saturation 0 .3459.

08:10 Then you can just look for the sfg, the difference between the at saturation, the entropy at saturation and equally line and vapor line for that pressure, 5, 6, 07...

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