Assume the same setup as in Problem 3.101, but the room has a volume of 100 m^3. Show that the

step 1 All right guys, let us start with the answer of this question. We've got V1 is equal to 100 cubi

Assume the same setup as in Problem 3.101, but the room has...

Key Concepts

Use of Thermodynamic Property Data

The determination of the final state in systems involving phase changes requires referring to property tables or diagrams. These resources provide essential data—such as saturation pressure, specific volume, and latent heat—that are used to ascertain the conditions at which the liquid and vapor phases coexist.

Two?Phase Equilibrium

This concept refers to a situation where two distinct phases, typically liquid and vapor, coexist in thermodynamic equilibrium at a certain pressure and temperature. In such states, the properties of the mixture are determined by both the liquid and vapor phases, and the system's overall state can be described as a weighted average of the two, usually derived from tables or phase diagrams.

Thermodynamic Equilibrium in a Closed System

In a closed system, no mass enters or leaves, and once internal processes complete, the system reaches a state of equilibrium where intensive properties (like pressure, temperature, and specific volume) become uniform. Analyses are based on energy and mass balances which dictate the final state after a process such as expansion or mixing.

Iterative (Trial and Error) Methods

Trial and error in thermodynamic problems involves using iterative techniques to converge upon a solution that satisfies all system constraints, such as energy, mass, and volume balances. This method is often necessary when analytical solutions are not straightforward due to non-linear relationships between variables.

Transcript

00:01 All right guys, let us start with the answer of this question.

00:03 We've got v1 is equal to 100 cubic meters and t is equal to 360 degrees centigrade.

00:12 We've got p1 is equal to 20 megapascal.

00:19 And we've got p2 is equal to 200 kilopascal.

00:25 So we can consider room and reactor as one single control room.

00:32 It is insulated and evacuated.

00:34 So heat transfer and work done are zero.

00:37 So from first law of thermodynamics, we have m, u2 minus u1 is equal to q minus w is equal to 0.

00:48 So u1 is equal to u2.

00:51 So from table, water table, b1 .2 with pressure 20 megapascal, we can obtain saturation, temperature value that is t saturation is equal to 365 .81 degrees centigrade.

01:13 We can see that the saturation temperature is higher than the given temperature.

01:18 So water is in compressed liquid state from compressed liquids water table b1 .4 with temperature 360 degrees centigrade.

01:31 We can obtain specific volume and internal energy that is v1 is equal to 0 .001823 cubic meters per kg and then we've got u1 is equal to 1702 .7 .78 kilojoules per kg so water mass in a control volume is given as m is equal to v1 over small v1 so that is equal to 1 over 0 .001823 is equal to 548 .548 .54 kg.

02:20 So from water table b1 .2 pressure 200 kilopascal, we can obtain specific volume and internal energy.

02:28 That is, vf is equal to 0 .001061 cubic meters per kj.

02:37 Then we've got bg is equal to 0 .88 573 cubic meters per kg.

02:48 And then we've got uf is equal to 504 .47 kilojoules per kg.

02:58 Similarly, we've got ug is equal to 549 .49 kilojoules per kg.

03:10 So the final specific internal energy is given as u2 is equal to uf plus x into ug minus uf.

03:22 So we have x is equal to u2 minus uf divided by ug minus uf.

03:34 So putting the value is 1702 .78 minus 504 .47 divided by 2529 .49 minus 504 .47 and that is equal to 0 .57.

04:04 We can now calculate the final volume, so that is v2 is equal to vf plus x into vg minus vf and that is 0 .001601 plus 0 .59176 multiplied by 0 .88873 minus 0 .8873 minus 0 .0 .0 .0 .0 .0 .76 .5 .5 .5 .3 minus 0.

04:35 01061.

04:36 So the answer is 0 .5247.

04:43 At stage 2, we got specific internal energy the same as the initial.

04:50 So u2 is equal to 172 .1702 .8 kilojous per kg...

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