A frictionless piston-cylinder device, shown in Fig. P 8-118, initially contains 0.01 m^3 of ar

A frictionless piston-cylinder device, shown in Fig. P...

A frictionless piston-cylinder device, shown in Fig. $\mathrm{P} 8-118$, initially contains $0.01 \mathrm{~m}^{3}$ of argon gas at $400 \mathrm{~K}$ and $350 \mathrm{kPa}$. Heat is now transferred to the argon from a furnace at $1200 \mathrm{~K},$ and the argon expands isothermally until its volume is doubled. No heat transfer takes place between the argon and the surrounding atmospheric air, which is at $300 \mathrm{~K}$ and $100 \mathrm{kPa}$. Determine $(a)$ the useful work output, $(b)$ the exergy destroyed, and (c) the maximum work that can be produced during this process.

Key Concepts

Isothermal Process

An isothermal process is one in which the temperature remains constant throughout the process. For ideal gases, this implies that any heat added to the system is exactly balanced by the work done by the system, keeping the internal energy unchanged. This process is characterized by the relation PV = constant, and computations involving work typically lead to the logarithmic relationship, such as work being proportional to the natural logarithm of the ratio of the final volume to the initial volume.

Work Calculation in Isothermal Expansions

In an isothermal process for an ideal gas, the work output can be calculated using the formula W = nRT ln(V2/V1) or its equivalent forms involving pressures and volumes. This reflects the fact that the system does work against an external pressure during expansion while maintaining a constant temperature. Thus, knowing the volume change is key to determining the energy transferred as work.

Exergy (Availability) Analysis

Exergy, or availability, is a measure of the maximum useful work obtainable as a system comes into equilibrium with its environment. It represents the portion of a system's energy that can be converted into work under ideal, reversible conditions. Exergy analysis combines the first and second laws of thermodynamics to evaluate both the quantity and quality of energy resources, thereby indicating the potential for work extraction.

Exergy Destruction

Exergy destruction quantifies the irreversibility in a process, corresponding to the loss of potential to do work due to factors like friction, heat transfer across finite temperature differences, and other non-ideal behaviors. It is a measure of the degradation of energy quality, and minimizing exergy destruction is central to improving the efficiency of thermodynamic systems.

Maximum Work Output

The maximum work output of a process is determined by the exergy change of the system; it represents the theoretical upper limit of work achievable under reversible conditions. This concept emphasizes the difference between the actual work obtained and the ideal work that could be obtained if all irreversibilities were eliminated, serving as a benchmark for optimizing energy conversion processes.

Transcript

00:01 And this problem, we have argon in a tank with a piston that maintains it at a constant pressure, correct? let's see here.

00:14 Initially, initially has 0 .01 cubic meters of argon.

00:19 Initially, it's at 400 kelvin and 350 kilo -pacels.

00:29 And heat is transferred from a 2 ,000 kelvin.

00:33 Heat source.

00:35 So something is, you know, heating, heat source is down here, pumping heat into here.

00:42 Now, we are told the argon expands isothermally.

00:47 So somehow the heat, the temperature of the argon is kept constant.

00:52 And the volume after this process is completed is twice what it was before.

01:01 Surrounding temperature of 300 kelvin and pressure of 100 kilopascals.

01:09 So let's see here.

01:14 We know for an isothermal work problem, we can get the work out for the isothermore case.

01:22 Again, this is derived in previous chapters.

01:25 And again, we've changed from use the ideal gas law here.

01:30 Again, argonne is a very, very good ideal gas.

01:34 Approximation is very, very good for argon.

01:36 So we get a work out of 2 .43 kilojoules.

01:42 But it turns out that 1 .43 kilojoules of those was done basically pushing the atmosphere up.

01:50 So we get this work done on the surroundings, winds up being one kilojoules.

01:59 And so again, the useful work is just one, this minus one, so 1 .43.

02:06 Kilojoules.

02:07 So again, one of these, one of kilojoules used to push the surrounding atmosphere out.

02:16 Now, energy in, minus energy out, is the hinge in energy in this system.

02:23 So you have a closed system here.

02:25 The argon is not escaping or flowing.

02:28 So we have the heat transfer in minus the work out equals the change in internal energy in the argon, which is m times cv times a change in temperature.

02:40 And since it's an isotherma process, that's change in temperature is zero.

02:43 So we must have that the heat in equals the workout, and that means that the heat in is 2 .43 kilojoules.

02:52 Now, for the change in entropy, we have to consider a couple of different things.

02:59 We have to consider the heat coming in from this source and the heat going out, you know, keeping it a constant temperature.

03:09 So we have the entropy in, is the entropy out plus the entropy generated is the change in entropy on the argon where the argon remained at a fixed temperature and um so we have a uh and a um so a fixed temperature so the change in entropy um is uh is zero for the argon did it can transfer some of the whole people yeah we also had a constant pressure i think.

03:43 This maintains a constant pressure...

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