Actual Vapour Compression Cycle-1

[@Hum4292]

Excellent Lecture Sir...... Explanation was super clear......... Thankyou.

[@ZEESHANUSTAD]

Cleared all my PH and TS diagram concepts

[@edaurelio2]

Good lecture, however the Refrigerating Effect should be "ha-hd", since cooling process happens only at the evaporator side. (22:29)

[@priyanshubharti5680]

He wrote that because hc=hd as it is a constant enthalpy expansion process

[@Sk_entertainment-9]

rocking

[@nitingaikwad8756]

Superb lecture sir

[@saikumar-tn1ku]

Thank you a lot sir

[@amirzaib3017]

thanks sir

[@jaganmodi5026]

How can the sub cooling from saturated liquid state to sub cooled state is achieved at constant pressure??. I think it follows x=0 line; that is saturated liquid line only.

[@edaurelio2]

Subcooling can be achieved by bringing the refrigerant from the condenser outlet (saturated liquid at this point, hence quality is zero) to an intermediate heat exchanger, which also has the refrigerant from the evaporator outlet flowing inside of it on the opposite direction. Since the saturated refrigerant gives off heat to the refrigerant at evaporator outlet, and since it is a constant pressure process, the net effect to the saturated refrigerant is a lower temperature at exit of intermediate heat exchanger. Therefore, it becomes subcooled in the process.

[@Sk_entertainment-9]

ONE AND ONLY

[@pavan2909]

great

[@jitendrapatidar1539]

everything clear

[@kanigirijagadeesh6710]

sir enthalpy of R-134a at -20 0 C is 238.41kJ/kg.....

[@Sk_entertainment-9]

#LOCA RELEASED

[@sinkilmansinghchandraul9308]

why we are desaturating that vapor after the point 2 , what is the need.?

[@edaurelio2]

By saying "desaturating", I assume you meant to refer to the condensation process or heat rejection process as the superheated refrigerant lowers its temperature from state 2 to state 3 in a constant pressure process inside the heat exchanger. The condensation of refrigerant is required in order to reject an amount of heat, and therefore not violate the Kelvin-Plank Statement and the 2nd Law of Thermodynamics. We can apply the Kelvin-Plank to a refrigeration cycle such that it is impossible for a system to operate in a thermodynamic cycle such that the sole effect is the net transfer of work into the system while transferring a net amount of energy to/from a single reservoir. That is, we need at least two thermal reservoirs (a high and a low temperature reservoir). Now, for the 2nd Law of Thermodynamics, a "cycle" without condensation process but with only evaporation process will result in a negative value for entropy generation, which would be a violation of the 2nd Law.

[@Sk_entertainment-9]

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[@omprakashmourya3595]

RAC sir ji

[@Sk_entertainment-9]

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[@Sk_entertainment-9]

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[@Sk_entertainment-9]

#yoyohoneysingh

[@rohansingh1163]

Is vapour compression cycle reversible cycle

[@pranavshinde500]

No

[@edaurelio2]

A vapor compression cycle is not a reversible cycle because in its simplest form or design, the expansion process happens in throttling valve which is highly irreversible. The carnot cycle is the reversible cycle.

[@fathullahism]

how to get value Tb ?

[@edaurelio2]

Since the table presented in the lecture does not have property values at Superheated state, the professor used 1st Law of Thermodynamics for Control Volume at the superheated region of the condenser. And because it is assumed to be a constant pressure process (no pressure drops across the tubes), we can make use of the definition Cp = dh/dT to compute for the change in heat energy.

[@Sk_entertainment-9]

MAKHNA

[@omprakashmourya3595]

Hindi me MCQ

[@Sk_entertainment-9]

o mere makhna

[@Sk_entertainment-9]

mere makhna

[@Sk_entertainment-9]

makhna

[@prathameshgawade536]

Ganja fook ke padhai mat Kiya karo bhai