00:01
In this problem, i started to solve problem 129 with the final bottle pressure as an input variable and put out their temperature right after charging and the temperature pressure and heat transfer it after state three is reached.
00:14
So remember this problem was we had a bottle with some air in it and we had a line with some high pressure air and we charged the bottle up to a higher pressure and then let it re -equilibrate to atmospheric temperature.
00:32
So in this problem is actually much more complicated than the previous one.
00:39
So we had the equations, i have all the equations written down here that we need.
00:44
So this is conservation of mass, conservation of energy.
00:48
Again, the ideal gas laws for the, let's see, the initial state, and then after we pressurize it, but we wind up with a higher temperature.
01:04
And then the final one that we need, when the final temperature we have is atmospheric again.
01:14
So we would figure out what the final pressure is.
01:16
So then again, the ideal gas law for the third state.
01:20
So we have three states, little low pressure, low temperature.
01:27
Then we go up to high pressure, high temperature, and then we go back down to high pressure, low temperature, and then here is, so this is the energy equation while we're filling it.
01:44
And this is the energy equation while from the time from one to three, where this is after it's cooled and that's why we have heat transfer.
01:55
We're assuming we have no heat transfer while we pressurized it.
01:59
But then once we let it settle back down, there is some heat transfer.
02:02
Again, this is the entropy equation for the entire process from pressurizing to letting it cool back down.
02:12
And then we know, again, i'm going to use constant heat capacities.
02:17
And i'm going to use these standard entropys just because i, we're going to, these two pressures, these temperatures will be kind of fixed in this so we can actually look these up.
02:29
And otherwise we could use the c -p -l -n -t ratio thing.
02:37
So here's, again, for constant heat capacity.
02:42
Again, in the problem, i think when i solved it before, i used the tables.
02:47
But if you look, if similar constant heat capacity, you get very, very similar values.
02:52
So for the enthalpy at the three different states initially, after it's pressurized, and then after it's cooled down, and then the inlet.
03:01
And then again the internal energy in the different states related to the temperature.
03:08
So we have lots of equations here and you know we have lots of unknowns.
03:16
Now what i'm going to do here is i define a whole bunch of new variables here and again i kind of work through these equations trying to get things to drop out.
03:27
So first we know k is cp over cv and then i define r hat as r over cv.
03:34
Then i defined the pressure p2 hat and p3 hat in turn of the pressure divided by the initial pressure that was in the tank.
03:47
So this is like a pressure ratio of what it was after we filled it and then what it was after we let it cooled back down to what it was initially.
03:57
And likewise i defined a pressure ratio for the inlet as this is the inlet pressure, the inlet pressure, divided by the pressure that was in the tank to begin with.
04:08
Initially, it was that atmospheric in the problem, but we could let that be a variable.
04:15
The heat transfer, again, i normalized that with the initial pressure and the volume of the tank.
04:23
The temperatures i normalized with the initial temperature in the tank.
04:30
Again, so that's the inlet temperature.
04:32
The inlet temperature is normalized by the initial temperature.
04:37
Entropies are i normalized with the specific heat capacity at constant volume.
04:45
And then the rate of entropy i normalized in this way with the temperature, the pressure, and the volume, and the initial pressure, initial temperature, and the volume of the tank.
04:57
And then i said, okay, let's just for sake, since these things, since the temperatures in state two, state one and state three are the same.
05:09
We're going to assume that's the same, so that it's initially at atmospheric temperature, and then in the end it's also back at atmospheric temperature, that we know these two things are the same.
05:21
So, if we plug everything in to this, let's see here, if we plug everything in, did i, let's see here, what did i do, the mass flow rates? did i? or the masses? did i normalize those? i think i probably did, but i didn't write it down here.
05:57
Anyway, these equations we can write in these new parameters, and they reduce to these equations here.
06:12
Let's see here.
06:13
What did i do for the masses? let's see here.
06:20
Yeah, let me actually write that in here.
06:23
Jot that in.
06:24
M -hat -1 is, let's see here.
06:34
M -1 time c -v -t times c -v divided by, t -1 times c -v divided by p -1 times v.
06:52
So then m -2 and all the other ones are likewise.
06:58
The different ones are normalized by this factor here.
07:02
So again, you notice that we're normalizing with the inlet stuff and then the volume and then one of the constants for the air.
07:11
So that's what we're using all that stuff to normalize everything.
07:14
And so those things are going to drop out of the equations.
07:18
And so notice down here we don't have any p1.
07:21
We don't have any t1.
07:22
We don't have cv and we don't have v.
07:29
Because we're scaling everything with respect to those.
07:37
Conservation of mass.
07:40
This is the energy equation for energy equation while we're filling.
07:49
Ideal gas law basically just says in the nondemential form that m -1, m -hat 1 is 1 over r -hat, so m -hat 1 is a constant.
08:02
And then ideal gas law for the after we filled and ideal gas law after we've let it cool back down we also know that that our hat is actually k minus one because of how you know that how the pressure the what do you call it the specific volumes or the heat capacities are defined right so we know that and then here's the energy equation from when we to after we from when before we pressurized it to after we let it cool down for those states and then the entropy equation which is a lengthy mess because we have all this stuff in here but um notice that we have no i think i this should be a three in here and out of four notice that you know things have dropped we don't have any in that pressure and there pressure initial temperature the the volume of the tank is dropped out and so has one of the actually the we only have one constant now for the the air because we can just use k and so we know that our hat is k minus one so we can use that and all of these so we only we only have one parameter of k for the gas so if we change the gas, we just change k in here to whatever gas it was.
09:38
So these equations look, you know, still look kind of rather ugly, but in fact we can solve them quite easily, well, for most of the stuff.
09:49
So again, m1 hat is 1 over r, one hat.
09:54
So that's just going to be a constant.
09:57
M2 hat relates the pressure ratio and the temperature ratio, r hat, t2 hat, which again, if we look back over here, it's just the ratio of the temperature after we pressurized it to the initial temperature.
10:20
Let's see here.
10:24
So i solve these.
10:26
I left what i saw for, i left what i used as parameters with the pressure, the final pressure after we let it cool down, the final temperature after we let it cool down.
10:40
That's actually going to be, again, constant, basically going to be the same as this is going to be one in most of our problem here.
10:50
I set that down to one here because we're letting it come back down to atmospheric temperature.
10:54
And so this ratio here is one.
10:58
This is the mass coming in, again normalized, you know, in this way here.
11:10
And the pressure after we've, after we've, let's see, the pressure after we've pressurized it is, you know, a function of the final pressure we want, the final temperature we want and the inlet temperature ratio.
11:27
Ratio and the ideal gas, the gas parameter...