00:01
Now, in this case, we have a tank, a well -insulated, rigid tank, initially filled with nitrogen at 1 megapascal and 20 degrees c.
00:13
A valve is open, and half of the nitrogen mass is allowed to escape, and we want to determine the change in exorgy content of the tank.
00:21
Now, this is an interesting problem, and we're going to do it in a couple different ways.
00:27
The simple way first, then a more complicated way, and we're going to see that both of those give strange results.
00:36
Now, so we have our conditions here, half the mass left out.
00:42
So properties for nitrogen.
00:50
Now, these are average properties, and that's one of the reasons that we wind up with, probably wind up with a weird result here in this case.
01:00
So we can figure out the mass of nitrogen in the tank because we know the pressure, the volume, and the temperature initially, and agogas constant.
01:08
So that's 1 .150 kilograms of nitrogen.
01:16
So we know the amount that escaped is, well, the amount that's in there afterwards is also equal to the amount that escaped, and that's half of this.
01:23
So 0 .575 kilograms.
01:26
Now, energy balance says the enthalpy of the stuff going.
01:31
Out must equal the change and energy of the stuff that's in there.
01:37
So here before, um, after it's in after process, after it's let out and before.
01:46
Now, uh, this, uh, the pressure we're going to assume.
01:53
Well, the pressure, um, let's see here.
01:56
No, we're not assuming the pressure thing.
01:58
Anyway, we have the, um, this is the, uh, the mess that went out.
02:05
Times the heat capacity, constant pressure, times the temperature at the exit.
02:13
And that's another thing that we've got to be careful about.
02:19
And so then we have the heat capacity, constant volume, times the temperature at the afterwards, and then before.
02:29
And so what we do here is we know everything in here, but t2 and t -e.
02:36
So we can say, well, let's say te is the average.
02:40
So, you know, the temperature of the stuff that left is really about the average of what was in there, the temperature of what was in there to start and what was in there afterwards.
02:51
So that's an approximation.
02:54
And it may not be necessarily a good one or it may be adequate.
02:58
But that allows us to solve this equation for t2.
03:02
And that winds up being 224 .3 kelvin.
03:07
So the t2 was less, so it dropped.
03:10
So this is 293 calvin.
03:15
So then we can figure out what the pressure is afterwards.
03:19
And we wind up the pressure is 382 .8 kilopascal.
03:24
So obviously the pressure dropped too.
03:27
So the pressure and temperature drop.
03:29
And so then we can say, well, we need the going forward here to get the entropy.
03:37
We need the exit temperature.
03:39
So we're going to take the average.
03:41
You also need the exit pressure.
03:45
So we'll again take an average of p1 and p2, and that gives us 691 .4 kilopascals.
03:54
Now, so again, these are both approximations.
03:59
So now for entropy, we have the entropy in minus entropy out, plus the entropy generated is the change in entropy in the system.
04:08
The entropy out is here.
04:10
Entropy generated, then the change in entropy inside the system.
04:17
So we can see here that the entropy generated is this is m .e.
04:28
I just use that conservation of mass here, and then these two terms here.
04:34
And then we can see that i can lump these together, so i get differences in entropy here, which makes it nice to use for our ideal gas formula.
04:42
And this is where we need their pressures.
04:45
So we need the pressure at the exit because this is the entropy at the exit.
04:50
So we need these pressure at the exit.
04:57
So now if we make the assumption that it's an average, then we have all the values here.
05:04
And we can plug everything in, multiply it by t not and get the exorgy destroyed or the reversible work that we can get out.
05:13
And it turns out that that is minus 2.
05:15
0 .10 kilojoules, which should not be possible.
05:23
So something's weird here.
05:28
And there's nothing weird about the problem, right? you have nitrogen in a tank.
05:33
You let half the mass out and you should have been able to get some work out.
05:40
You shouldn't have generated entropy.
05:42
Or you shouldn't have, you know, you shouldn't have, have a, you should have a, you should have a, you should have, um, created entropy, not destroyed entropy in this process.
05:56
Um, again, you know, assuming that there's no heat transfer, you know, that's, again, you know, when there's, when you have something flowing out, you know, you can't necessarily know that, uh, you know, you're not going to have any heat transfer out of this, from the stuff that's coming out.
06:12
Um, you might have this.
06:13
Um, you might have this.
06:13
The jar or the container or the tank well insulated, but the stuff that's coming out is coming out, and it's going to have some heat transfer to the environment.
06:23
So again, there's some approximations here that the problem is, so we have, let's see here, we have, the problem is that there's very little entropy actually generated in this process.
06:42
So making assumptions can swing it from, positive to negative very easily.
06:48
Because, you know, this is pretty small.
06:50
And i didn't actually calculate this, but this is also very small.
06:54
So, you know, assumptions are going to, you know, cause it go from positive or negative fairly easily by, because it's a small value to begin with.
07:03
It should be a small positive value.
07:06
But some assumptions might have made it, you know, undershoot and that winds up being a small negative value.
07:13
So these assumptions probably aren't good.
07:16
So we can do a whole lot better.
07:22
We can do a whole lot better if we do a more thorough analysis, but it's actually a much more difficult analysis.
07:32
And i don't know if you would actually be expected to do an analysis like this in your homework.
07:39
But i can, i'll go through it and i'll try to explain it.
07:44
So what we have here is we're going to look at the the outlet of the tank in more detail.
07:56
So we're going to have a variation in gas temperature and pressure at the outlet.
08:02
So a mass balance or a conservation of mass in a differential form says that the rate of mass flow out is the time rate of change of mass, minus the time rate of change of mass in the tank.
08:19
Now, we can combine that with the first law in differential for.
08:26
And what that gives is the differential of the mass times the internal energy.
08:34
So this would be the specific internal energy.
08:37
So this is the total internal energy in the system.
08:41
The time rate of change of that is h, the enthalpy, which again is going to be a function of time as the process goes, times the rate of mass flow out of the tank or the rate of mass flow, rate of mass flow, not necessarily out, but the rate of mass flow.
09:03
Obviously, we could substitute this and this would be minus m -e dot, so it would be minus the rate of mass flow out.
09:10
Now, we can use the fact that this are of the approximation that this is an ideal gas.
09:18
And again, what was the gas? nitrogen.
09:22
Eh, that's not a great approximation.
09:24
So that's another approximation that we might not be all that great, that we have constant heat capacities.
09:32
These heat capacities way will be varying.
09:36
So using ideal gas law anyway, and that's obviously another assumption that it's an ideal gas.
09:44
We can convert this to this, which says that the heat capacity constant volume times the ideal, what ideal gas formula is not just the ideal gas law, times the volume divided by the ideal gas constant, times the time rate of change of the pressure equals the heat capacity of constant pressure times the temperature times the time rate of change of the mass.
10:13
So we just did basically some substituting with ideal gas relationships to get to here.
10:22
Now, we can do some more with ideal gas relationships.
10:28
Because, you know, we can divide through by, by t here, right? and we get b over rt, and that is m, which one we divide through, we get here.
10:44
Or is p times m, right? yeah, put it in.
10:51
So we're going to have v over rt...