A steady flow of 0.01 kmol / s of 50 % carbon dioxide and 50 % water on a mole basis at 1200 K a

A steady flow of 0.01 kmol / s of 50 % carbon dioxide and 50...

A steady flow of $0.01 \mathrm{kmol} / \mathrm{s}$ of $50 \%$ carbon dioxide and $50 \%$ water on a mole basis at 1200 $\mathrm{K}$ and $200 \mathrm{kPa}$ is used in a heat exchanger where $300 \mathrm{kW}$ is extracted from the flow. Find the flow exit temperature and the rate of change of entropy using Table A.9.

Key Concepts

Steady-State Energy Balance

This concept involves applying the first law of thermodynamics to an open system, ensuring that the energy entering a control volume, minus the energy leaving (via work, heat, and carried energy), equals the accumulation of energy. In the context of a heat exchanger, it allows one to relate the heat extracted from the flow to the change in enthalpy (and hence temperature) of the fluid stream.

Entropy Balance in Open Systems

The second law of thermodynamics governs the entropy balance in control volumes, accounting not only for the entropy carried by incoming and outgoing mass but also for the entropy generation due to irreversibilities. This is essential in evaluating the rate of change of entropy in a process such as heat transfer, where heat extraction can affect the entropy distribution in the system.

Thermodynamic Property Tables

Thermodynamic property tables provide critical data, such as enthalpy and entropy values at specific temperatures and pressures, which are necessary for analyzing thermodynamic processes. These tables are used to look up and interpolate properties of substances (or mixtures) that help in calculating changes in energy and entropy during processes like heating, cooling, or phase change.

Flow Mixture Analysis

When dealing with a mixture, the properties of the overall flow are determined by the contributions of each individual component based on their mole, mass, or volume fractions. Understanding how to apply the principle of mixtures is crucial for correctly computing the average properties needed in energy and entropy balance calculations, especially in systems where the composition affects thermal behavior.

Transcript

00:01 Hi everybody, so we are asked, find the flow temp, exosur, and the rate change entropy of table a9, okay? so, for this, we have your cue is mh2 minus h1, which is going to be in.

00:24 So what we're going to have is eventually h2 equals q over, um, n plus h1 because this is going to equal to n h2 minus h and now this has been equal to negative 300 divided by 0 .01 times 0 .5 times 4473 plus 0 .5 times 345 times 3456.

01:06 And this thing, this beast equals 9489 .5 joules per curmole.

01:17 Okay.

01:19 And so now at 500, okay, we have, we got up on h2 at 500.

01:28 Okay, so it's going to be at 0 .5 times 8305 plus 6 .6.

01:36 692 and so here we get 6 -7613 give that 7 looking nice 0 .5 kilojoules per k no okay and now we have it at 600k and with h2 equals 0 .5 times 1 2906 plus 104 499 equals 11702 .2 .5 kilo, kilo, 4 kilo mole.

02:19 Okay.

02:20 And so your temperature two is actually going to be from 500 plus 600 minus 500 times the 948.

02:34 Point five minus 76135 and we have divided by one one one seven seven oh two point five so we end up getting five four five point eight seven nine kelvin's okay and when i'm done we need to find the entropy.

03:09 So we have entropy change as m equals st2 minus st1 minus r log p2 divided by p1 and from this we have 0 .01 times 0 .5 times 238.

03:38 0 .757.

03:40 I mean, about three looking nice.

03:43 7 minus 279 .3 .9...

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