Refrigerant 22 enters the compressor of an ideal vaporcompression refrigeration system as satura

step 1 We'll be looking at refrigeration and air conditioning. I will have this problem of a of an idea

Refrigerant 22 enters the compressor of an ideal...

Key Concepts

Vapor Compression Refrigeration Cycle

This cycle involves the compression, condensation, throttling, and evaporation of a refrigerant to provide cooling. It is the fundamental process used in refrigeration and air-conditioning systems, involving phase changes of the refrigerant to absorb heat from a space and reject it to the surroundings.

Compressor Work and Isentropic Compression

In an ideal vapor compression system, the compressor operates under the assumption of an isentropic (constant entropy) process. The work required by the compressor is determined by the mass flow rate and the specific enthalpy difference between the suction and discharge states. This concept is crucial for calculating the power input needed for the cycle.

Mass Flow Rate and Volumetric Flow Rate Relation

The conversion between volumetric flow rate and mass flow rate is based on the specific volume of the refrigerant at the inlet conditions. Accurate determination of the mass flow rate is essential because it directly influences the calculations of both the compressor work and the cooling capacity of the system.

Refrigerating Capacity and Ton of Refrigeration

Refrigerating capacity is the rate at which a refrigeration system can remove heat from a space, typically measured in kilowatts. One ton of refrigeration is a standard unit of cooling capacity defined as 3.517 kW. Understanding this concept allows for the conversion between energy units and the practical sizing of refrigeration equipment.

Coefficient of Performance (COP)

The coefficient of performance is a measure of a refrigeration system's efficiency, defined as the ratio of the heat removed in the evaporator (the refrigerating effect) to the input work done by the compressor. COP provides a quantitative assessment of performance, indicating how effectively the work input is used to produce cooling.

Transcript

00:01 We'll be looking at refrigeration and air conditioning.

00:02 I will have this problem of a of an ideal vapor compression, a refutation cycle.

00:22 So we have this is 0 .2 prime, 0 .1, 0 .3, 0 .4.

00:38 All right.

00:39 And the temperatures are at the evaporator.

00:43 We have at this minus 40 degree.

00:47 Celsius and we have the condenser to be celsius and we have the condenser to be 32 degrees okay 32 degrees okay we are given the flow the volumetric flow okay and now to we want to want to determine the compressor okay compressor power we want to determine the the revision capacity in terms as well like the performance.

01:31 To do this, we apply the principles governing the calculations of reflation and air conditioning.

01:37 And we'll look at point one, at point one.

01:46 Point one.

01:48 Of course, legends fixed this.

01:49 This is in kilo, the unit.

01:52 So wherever you see, we using, wherever you see, entropy, other things, you as we are walking along just know that it is in kilo per kilogram okay so we have that's that's sorry just know that it's in kilo per kilogram caravan okay so at point one we have a s surface 1 to be equal to what we have at okay so if you check it from the steam table you'll have it to be 0 .05 i've already told you the unit so the unit stands for entropy.

02:42 Okay then you also have the entropy there would be equal to 233, 33, 47.

02:53 So this unit also for entropy stands for all that we'll be doing to get along.

03:01 All right, so that's at that point two.

03:05 We're going to have this point two.

03:11 We're going to have to check it from the steam sorry from your table of refrigerator 22 or 22 you're going to have that to check it from there you are going to have this to be 0 .0 because at that point at point sorry this is point 2 okay let me write it is along the liquid sorry the before cycle paper line zero point so have s2 you got the 0 .842 the unit stands for the unit then at that point also can also look for each point so see along the saturated for line 1 3 2 over kilogram okay then we can also appoint 2 a point 2 prime point 2 prime of course it's going to have the same entropy value as what we have in point there point 1 so 0 .2 prime at point 2 prime s 2 prime so as that's that i will know the value already all right then at point three three you're gonna have from the steam table from the sorry from the arrow 22 table you have all those values okay we have this to be at sf will give you 0 .3101 please all the units stand here measurement stands so just use them then out there for this f there.

05:57 They will have it to be equal to 84.

06:02 We is simply as s3, okay? so to make it, let's just put it as s3.

06:08 So we know that, okay, and s3, h3 is going to give us 84 .14.

06:20 All right.

06:24 Then we're going to point four.

06:25 So if we go for that at point four, at point four, we are going to have hf...

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