A piston/cylinder compressor takes R-410 a as saturated vapor 500 kPa and compresses it in a rev

A piston/cylinder compressor takes R-410 a as saturated...

Key Concepts

Specific Compression Work

Specific compression work is the work required to compress a unit mass of a working fluid. In thermodynamic processes, especially in compressors, it is calculated from the integral of pressure with respect to specific volume changes during the process. This quantity is crucial for analyzing the energy efficiency and performance of compression equipment.

Thermodynamic Property Relations

Thermodynamic property relations involve the equations and correlations that relate variables such as pressure, temperature, specific volume, and entropy. These relations, often provided by property tables or equations of state, are essential for accurately determining the conditions at various points in a process, particularly when dealing with real fluids like refrigerants.

Reversible Adiabatic Process

A reversible adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings and the process is carried out in such a way that the system is always in thermodynamic equilibrium. Because there is no heat transfer and the process is reversible, the process is isentropic, meaning the entropy remains constant throughout.

Saturated Vapor

Saturated vapor refers to a state where a fluid is at the temperature and pressure at which it changes phase from liquid to vapor. It is at the boundary of phase change, meaning its properties, such as temperature, pressure, and specific volume, are well-defined by saturation conditions. This concept is essential when starting a thermodynamic process from a state of phase equilibrium.

Isentropic Process

An isentropic process is one where the entropy of the system remains constant. This concept is a direct consequence of a reversible adiabatic process, and it provides the framework to relate the state variables (pressure, temperature, volume, entropy) along the compression or expansion path. It is widely used to simplify the analysis of idealized thermodynamic systems.

Transcript

00:01 In this question it is given that a refrigerant r410a is compressed through a piston cylinder arrangement.

00:10 This refrigerant was in the state of saturated vapor.

00:15 As a result of compression, the pressure increases from 500 kioskal to 3 ,000 kioskol.

00:28 We are required to evaluate the final temperature and the specific compression work.

00:34 So let's see how to solve this question.

00:37 Refer table b .4 .2 and from there, the internal energy corresponding to pressure 500kcal is equals to 248 .29 kilojoules per kilogram and entropy corresponding to this pressure is s1 is equals to 1 .0647.

01:07 Kilojoule per kilogram kelvin.

01:11 Now since it is given that the process is reversible and adiabetic hence entropy at state 1 will be equal to entropy at state 2 so we can say entropy s 2 will be equals to s 1 .06 47 kilojoules per kilogram kelvin.

01:36 Now again refer the table b .4 so the entropy s at 60 degree centigrade corresponding to pressure 3 ,000 kioskal is equals to 0 .9933 kilojoules per kilogram kelvin and entropy s at 80 degree centigrade corresponding to same pressure is equals to 1 .0762 kilojoules per kilogram per kilogram.

02:10 Kelvin similarly now let's find the internal energy you at 60 degree centigrade from that table and corresponding to 3 ,000 kiosk pressure u at 60 degree centigrade is equal to 274 .96 kilojoules per kilogram and you at 80 degree centigrade corresponding to same pressure is equal to 298 .38 .38 0 .38 kilojoule per kilogram.

02:46 Now let's apply the interpolation method.

02:50 So we can write t2 minus 60 degree centigrade divided by 80 degree centigrade minus 60 degree centigrade and this will be equals to s2 minus s60 divided by s at 80 degree minus entropy s at 60 degrees...

Please give Ace some feedback