A Carnot vapor refrigeration cycle is used to maintain a cold region at 4^∘ C when the ambient t

step 1 Hydration, heat pump and air conditioning systems. Now for this particular problem, I see, a can

A Carnot vapor refrigeration cycle is used to maintain a...

Key Concepts

Phase Change and Heat Transfer in Refrigeration

Phase changes play a central role in refrigeration cycles. The refrigerant absorbs heat during the evaporation process (turning from liquid to vapor) in the evaporator, and it releases heat during the condensation process (vapor to liquid) in the condenser. These heat transfer processes are essential for moving heat from the refrigerated space to the surroundings.

Turbine (Expander) in Refrigeration Systems

In some advanced refrigeration cycles, an expander (or turbine) is used to recover work from the refrigerant as it expands from high pressure to low pressure. This work recovery can partially offset the compressor work, improving the overall efficiency of the cycle.

Carnot Refrigeration Cycle

This is an idealized model of a refrigeration cycle that operates between two thermal reservoirs and is used to determine the maximum theoretical performance (coefficient of performance, or COP) achievable by any refrigeration system. It assumes reversible processes including isothermal heat transfer and adiabatic (isentropic) compression and expansion, making it a benchmark for comparing real systems.

Coefficient of Performance (COP)

The COP is a key performance metric in refrigeration and heat pump systems, defined as the ratio of the cooling effect (or heat removed) to the work input. It quantifies the efficiency of the refrigeration cycle, where higher values indicate more effective systems, and it is fundamentally linked to the temperature limits of operation as shown by the Carnot cycle.

Compressor in Refrigeration Systems

The compressor is a critical component that increases the pressure of the refrigerant vapor. By compressing the refrigerant, its temperature rises, making it suitable for rejecting heat in the condenser. The work done by the compressor represents a major energy input in the refrigeration cycle.

Transcript

00:01 Hydration, heat pump and air conditioning systems.

00:06 Now for this particular problem, i see, a canal paper, recognition cycle, want to determine the decompressor power, as well as the part led up by the turbine, the production of performance.

00:25 Okay? so, if we have this, see if this is...

00:50 Point one, point two, compressor, inlet, outlet, then point two to three, two, condenser inlet, three outlet, then three to four, three inlet to turbine, four to one, four to one, inlet two, evaporator and outlet, all right? so, we call here temperature, altitude degrees celsius and this is entropy at least kilo per kilogram per kilogram all right and we are to maintain four degree in the ambient 32 i'm given the evaporator sorry the condenser pressure is as now as 65 to the pascal, let's just come back to parha, which will give us 9 .65 bar.

02:17 And we also give them that the evaporator pressure is 276 kliapaska, which is 2 .7, 2 .76 bar.

02:52 All right so we also apply the principle governing the calculations of reparation cycle with respect to canal cycle and the refrigerant is arrow 184 a 184 okay given the mass flow rate to be regurrant to be equal to 0 .05 kilogram per second all right now if you look at this diagram diagram you see that s at point four is equal to s at point three as a tropic expansion okay and s at point one is equal to s at point two as a entropy compression so that we interpolate between a number and temp to get 9 .6 back you can determine to at 9 .6 bar 9 .6 bar 9 .6 bar rather you can you can determine the entropy at 2 as s2 to be equal to 0 .9947 0 .9047 0 .90 7 kilo per kilogram per kelvin.

04:55 They will also have the entropy at that point as well.

05:02 So, it would equal to, it was 27 .34.

05:10 Check all these are from using the arrow 134 table, in vision 134, 267 .34, 267 .34, for kilo this is in kilo joe this is in kilo per kilo so our enterpes and kilo per kilogram entropy kilo per entropy kilo per entropy kilo per entropy all right so if we do the same thing but this time around we're looking for the entropy's between the same pressure ring uh nine bar and ten bar is that nine point six five bar we'll have uh the uh sorry the entropy 3 equal to 2 .3774 0 .3774 0 .3774.

06:18 Logo per kilogram per kelvin per kelvin and the entire 3 at 3.

06:27 So it's equal to 103 .2 9.

06:33 1003 .2 .9.

06:39 Okay? lojo per kilogram.

06:40 All right.

06:41 Then we can also consider at the evaporator pressure okay we interpolates between uh 2 .4 and uh 2 .8 now we can get 2 .76 okay so that a 2 .76 bar right who have um um h uh at f and saturated uh liquid line okay who have the enter the gare, we need to continue that pressure.

07:29 47 .8, 5 kilo per kilogram.

07:36 And we'll have the saturated paper line for that pressure, and long as saturated with half pressure.

07:43 We'll have this to be 2426, 2 .26 .2 .28 kilo per kilogram and we will have the difference in entropy considering the the saturated before point and before line the difference okay considering that pressure saturated liquid line okay for that pressure will have to the one ninety eight point four three kilo per kilogram okay, now we can also extend the further.

08:43 We do also within the same bridge of interpolation for 2 .765, you can have, uh, essence, okay, entropy, oxygen, the saturated liquid line, you have that is 9 -1 -1.

09:02 So, 1891, all right? and we'll have all saturated before, point.

09:12 Can launch a driving bond line to that project to that point you give us 0 .9200 all right so if we look for the difference you don't have this uh 0 .9 3 0 .7 3 0 .7 3 0 .7 0 .9 now it's kilo kilo per kilogram per kevin so i said that all the entropy is like kilogee per kilogram per kelvin now haven't known on now we can just look for the adreness fraction of point one because we need to do the entropy if we use these relation with the entropy at one the attribute liquid eye okay okay and there's a entropy and entropy so it is an equal line and liquid line without pressure so if you plug in all the values you already have now for this into this relation you have this to be equal to 0 .9 .791.

10:36 Alright, so that's the quality of the refuge at that point in china's so if we now use this relation, okay, which we already know all this, we plug these values into this relation, okay? we are going to have this point one to enterp .1 .2 .2 .2 .13 .2 .443 2 .402 .4 .2 .2 .4.

11:16 2 .2 .2 .3.

11:17 3.

11:21 If we do the same thing, do the same thing, we can follow the same procedure...

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