A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the ammonia is

A cylinder containing 1 kg of ammonia has an externally...

A cylinder containing $1 \mathrm{~kg}$ of ammonia has an externally loaded piston. Initially the ammonia is at $2 \mathrm{MPa}$ and $180^{\circ} \mathrm{C}$. It is now cooled to saturated vapor at $40^{\circ} \mathrm{C}$ and then further cooled to $20^{\circ} \mathrm{C},$ at which point the quality is $50 \% .$ Find the total work and the heat transfer for the process, assuming a piecewise linear variation of $P$ versus $V$.

Key Concepts

First Law of Thermodynamics

This principle states that energy cannot be created or destroyed, only transformed. In a closed system, the change in internal energy is equal to the heat added to the system minus the work done by the system. This concept is crucial for calculating the energy exchange during any thermodynamic process, including those that involve phase change or variable pressure and volume conditions.

Pressure-Volume Work

Work in a thermodynamic process can be determined by integrating the pressure with respect to volume (?PdV). When the process path is defined graphically, such as in a pressure-volume diagram, the area under the curve represents the work done. This concept is essential in processes where pressure varies with volume, and it forms a fundamental part of energy analysis in closed systems.

Piecewise Linear Process Analysis

In some thermodynamic problems, the process path may be approximated by piecewise linear segments. This means that the relationship between pressure and volume is assumed to be linear over specific intervals. Analyzing each segment individually allows for the calculation of work and heat transfer in processes with complex P–V behavior, facilitating the overall energy balance analysis.

Phase Change and Quality

In two-phase mixtures, the quality (or dryness fraction) indicates the proportion of the mass that is in the vapor phase. Understanding phase change—from saturated vapor to mixtures with both liquid and vapor—is crucial when analyzing processes that involve cooling or heating near the saturation line. This concept is essential for determining the internal energy, enthalpy, and other thermodynamic properties during phase transitions.

Transcript

00:01 All right, guys, we are given the following data for ammonia, that is t1 is equal to 180 degrees centigrade.

00:07 We've got t2 is equal to 40 degrees centigrade.

00:11 And we've got t3 is equal to 20 degrees centigrade and we've got p1 is equal to 2 megapascal.

00:22 Then we've got x is equal to 0 .5 and we've got m is equal to 1 kg.

00:29 Now from ammonia table b, 2 .2 corresponding to pressure p, that is 2 megapascal, and t1 is equal to 180 degrees centigrade, we can obtain initial specific volume and internal energy, that is 0 .10571 cubic meters per kg, divided by u1 is equal to 1630 .6 kilojoules per, so from ammonia table b2 .1 corresponding to temperature t2, that is 40 degrees centigrade, we can obtain stage 2 pressure, specific volume, and internal energy, that is p2 is equal to p saturated, is equal to 1554 .9 khal.

01:27 And then we've got v2 is equal to vg is equal to 0 .08 313 cubic meters.

01:36 Per kg we've got u2 is equal to u g that is equal to 1341 kilo jowls per kg so from ammonia table b2 .1 corresponding to temperature t3 that is 20 degrees centigrade we can obtain the final pressure specific volume and internal energy so p3 is equal to p saturated that is equal to 857 .5 kioscal and vf is equal to 0 .001638 cubic meters per kg.

02:19 And vfg is equal to 0 .14758 cubic meters per kg.

02:29 Now guys, we have uf that is equal to 72 .89 kilojouz per kg.

02:39 And ufg is equal to 1059.

02:43 0 .3 kilojous per kg.

02:46 Now the final specific goal volume is given as v3 is equal to vf plus x into vfg is equal to 0 .001638 plus 0 .5 multiplied by 0 .14758 and that is equal to 0 .075 428 cubic meters per k.

03:14 So the final specific internal energy is given as u3 is equal to uf plus x into ufg is equal to 272 .89 plus 0 .5 into 1059 .3 is equal to 802 .4 kilojous per kg.

03:39 Now we can calculate volume in every state that is v1 is equal to m into to v1 is equal to 0 .10571 cubic meters.

03:52 Then we've got v2 is equal to m into v2 is equal to 0 .0...

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