A flow of 1.8 kg / s steam at 400 kPa, 400^∘ C, is mixed with 3.2 kg / s oxygen at 400 kPa, 400

A flow of 1.8 kg / s steam at 400 kPa, 400^∘ C, is mixed...

A flow of $1.8 \mathrm{kg} / \mathrm{s}$ steam at $400 \mathrm{kPa}, 400^{\circ} \mathrm{C},$ is mixed with $3.2 \mathrm{kg} / \mathrm{s}$ oxygen at $400 \mathrm{kPa}, 400 \mathrm{K}$ in a steady flow mixing-chamber without any heat transfer. Find the exit temperature and the rate of entropy generation.

A flow of $1.8 \mathrm{kg} / \mathrm{s}$ steam at $400 \mathrm{kPa}, 400^{\circ} \mathrm{C},$ is
mixed with $3.2 \mathrm{kg} / \mathrm{s}$ oxygen at $400 \mathrm{kPa}, 400 \mathrm{K}$
in a steady flow mixing-chamber without any heat transfer. Find the exit temperature and the rate of entropy generation.

Key Concepts

Steady Flow Energy Equation

This concept involves applying the first law of thermodynamics to a control volume under steady-state conditions. In a mixing process with no work interactions or heat transfer, the sum of the enthalpy flows entering the control volume must equal the enthalpy flow leaving it. This balance allows one to compute the exit temperature or other thermodynamic state variables of the mixed fluid by linking the energy content of the individual inlet streams to that of the combined outlet stream.

Control Volume Analysis

Control volume analysis is a fundamental technique in thermodynamics where a fixed region in space is defined and the conservation laws (mass, energy, and momentum) are applied. In the context of mixing, it involves accounting for all mass and energy flows into and out of the volume. This ensures a systematic approach to solving problems governed by the conservation principles, even in cases where the process is irreversible.

Entropy Generation and Irreversibility

Entropy generation quantifies the irreversibility of a process, as dictated by the second law of thermodynamics. When two streams are mixed, the process is generally irreversible and results in an increase in the overall entropy of the system. Evaluating the rate of entropy generation involves comparing the entropy fluxes of the inlet streams with the entropy of the combined stream, thereby providing insight into the efficiency of the mixing process and the inherent losses due to irreversibility.

Transcript

00:02 Hi everyone, so for this one, you need to find the exit temperature in the weight of entropy generation, okay? so kind of looks like we have one and we have two, and it exits at three, okay? so for this, we can do the exit temperature, so we have exit temp, okay? so for that it's mh2o times h1 plus m .o2 of h2 equals mh2 of h3 plus mh2 of h3.

00:50 Okay? so you can obviously see that this is the o2 and this is the h2o.

00:58 And now we have mh2o times c .p .h .2.

01:10 Times t -p -h -2 times t1 minus the temperature of the third one, plus m -o -2, c -p -o -2 of t2 minus t3, which is going to be all this is going to go to zero.

01:25 So now we have 1 .8 times 1 .872 times 673 minus t3 plus 3.

01:38 Plus 3 .2 times 0 .922 times 400 minus t3 equals 0.

01:51 And you're just going to solve for t3, and when you do that, you're going to get 545 .6 .33 elements.

02:00 I'm going to let you solve it yourself because it should be simple algebra.

02:04 You know, you multiply this together.

02:08 Okay.

02:09 And then the next step is to then multiply out to here.

02:15 And you do the same thing here.

02:17 And then you just take the t3 on one side.

02:21 And then take the regular numbers on the other side.

02:23 Okay? i'm going to let you guys do that because that's simple, and you should be able to do that yourself without anyone instructing you.

02:32 So we've got to calculate the number of moles...