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5.6

## University of Liverpool

Skip to 0 minutes and 4 secondsOK, so we're going to take a look now at a worked example for a compressed air powered car. And we're going to make some assumptions, which I'm basing on a prototype car that Peugeot showed at the 2014 Paris Auto Show. And also a company called MDI, who are actually producing an air-powered car. So I'm going to assume that we've got a tank of compressed air, and that it's got a volume of 100 litres, and that we've compressed it a lot, and so it's compressed to 248 bars. And I'm going to allow my air to escape, and so it's going to escape through here.

Skip to 0 minutes and 59 secondsAnd so I'm going to talk about state 1, where the air is compressed, and state 2, where the air has expanded into atmospheric conditions. And I'm going to use as my system the air itself. So I'm going to draw my system boundary around my storage tank and around the air that has escaped. So this system boundary here is the air that we're using to power the car. And so inside the car, this air will escape into some sort of turbine and then from the turbine out into the atmosphere. So I'm just ignoring what's happening in the turbine for the moment. And so we're going to apply the first law of thermodynamics initially. So apply the first law of thermodynamics.

Skip to 1 minute and 58 secondsAnd we'll apply it to the air. And we're going to do it before-- so that's what we're going to call 1, and after, which we're going to call 2, escape from the tank.

Skip to 2 minutes and 27 secondsAnd so we can write the first law in the normal way. So we're going to have a heat transfer term. So I'm going to use one and two this time instead of the usual in and out. And this is going to be heat transfer in state - sorry, work transfer in state 1 and 2, plus the effects of any energy that flows in or out.

Skip to 3 minutes and 1 secondAnd then that'll be equal to the change in the internal energy plus the change in the kinetic energy plus the change in the potential energy.

Skip to 3 minutes and 16 secondsAnd if we make the assumption that the air doesn't escape from our system, then actually this term here will go to zero. And if we assume that our vehicle is moving along a level road at a constant speed, then if the road is level, we can set this term here equal to zero. And if we're moving at a constant speed, we can set this term here equal to zero. And that simplifies things significantly. And so we're left with these terms here for us to deal with. If we assume that the air expands from this tank out into this area here gradually, and so there's no change in its energy content.

Skip to 4 minutes and 13 secondsIn other words, it expands isothermally, then we can also say that this goes to zero as well. So perhaps we should note that. We are assuming that isothermal expansion of the air with a vehicle at a constant speed on level road. So they were the assumptions we used to allow us to cancel these terms down. So now we're left with Q2 minus Q1 plus W2 minus W1 equal to zero. So a much simpler expression. So now we want to access this work. How much work is being done by the air expanding? And so if we assume an ideal gas, so assume air is an ideal gas, that means just that pV is equal to mRT.

Skip to 5 minutes and 49 secondsAnd we've got our change in heat equal to our change in work, and so that actually allows us to say-- so let's talk about dQ rather than writing out Q2 minus Q1 each time-- is equal to the change in work. And that's going to be equal to pdV if we differentiate pressure times volume for work. And so we can substitute this expression into that, and we end up with mRTdV divided by V. OK, so what I've done here is equate heat and work from this expression here, said that my work is pressure times volume, so a change in work will be at constant pressure times the change in volume.

Skip to 6 minutes and 39 secondsAnd then I've substituted from the ideal gas equation here to end up with a change in heat is equal to mRTdV upon V. And now I want to use the second law of thermodynamics. So now applying the second law of thermodynamics.

Skip to 7 minutes and 14 secondsAnd we'll say for an ideal reversible system.

Skip to 7 minutes and 27 secondsAnd so that means we can say that the change in entropy - S for entropy - will be equal, because we're saying it's a reversible system, to dQ upon T, that's Clausius' expression. And we can also say that by definition, Gibbs energy is-- G for Gibbs energy-- H minus Ts. And H, the enthalpy is equal to the internal energy plus the product of pressure times volume. And so Gibbs energy is going to be internal energy plus the pressure times volume product minus the product of the temperature times the entropy. And so I want to hang on to this equation here.

Skip to 8 minutes and 28 secondsLet's call it equation (II) I also want to hang on to this one over here, equation (I) and use those two in the next stage. But let me just wipe the board now so I can move on. So this is the expression that I got a moment ago. And what we're now interested in is the change in Gibbs energy. And so that's going to be dG, and we're going to need to differentiate each of these terms here. So we're going to have the change in internal energy plus the change in the pressure volume product minus the change in the temperature entropy product. And for isothermal conditions, so isothermal expansion, there will be no change in the internal energy.

Skip to 9 minutes and 27 secondsSo we can set that equal to zero. And we can deal with this term here. We can say the change in pressure times volume will be equal to - if we substitute the ideal gas equation, it's going to be the change in the mass times the gas constant times T. And if it's isothermal, this piece here will be zero. And so this will go to zero, because dT is equal to zero. So that just leaves us with this term here to consider. And if we differentiate this by parts, we're going to have a dT term here again, which will go to zero. So we'll just be left with a dS term.

Skip to 10 minutes and 20 secondsAnd so hence we can write that dG is equal to minus Tds.

Skip to 10 minutes and 33 secondsSo if we now substitute into this expression the ideal gas equation we have up here, and also from the second law of thermodynamics, then we can rewrite this as dG equals mRT over V, dV.

Skip to 11 minutes and 0 secondsAnd now we need to integrate over the whole process of the air going from one here through to 2. And so we're going to end up with the integral between Gibbs energy at 1 to Gibbs energy at 2 dG equal to integral, and we're going to do it with respect to volume, so volume at 1, volume at 4, of mRT over v, dV. So these subscripts here refer to the states over here. And so that if we do that, that leads us to equal to mRT ln - natural logarithm - of V2 over V1. We can substitute again for the ideal gas equation here. And so that leaves us with p1V1 ln V2 over V1.

Skip to 12 minutes and 11 secondsAnd the final step now is to put some numbers into this, and we know what the initial volume is, it's 100 litres. And we know the pressure that we start with is 248 bars. It escapes to the atmosphere, so our pressure when it escapes will be atmospheric pressure. What we don't know is what happens in terms of the volume when it escapes. And so we have to convert this ratio here into a pressure ratio, because we know the pressure at the beginning, and we know the pressure at the end. And so we can write that V2 over V1, the volume ratio is equal to p1 over p2, the pressure ratio to the one upon gamma.

Skip to 13 minutes and 1 secondAnd so that will be equal to 248 over 1. And the normal value for the gamma-- for air-- is 1.4. And so that comes out at 51. So that gives us the volume ratio that we can now substitute into that expression. And so we'll end up with delta G is going to be equal to our initial pressure, which is 248 bar. So that's 10 to the three times our initial volume-- which is 0.1 of a cubic metres is 100 litres, times the log of 51. And that will finally come out to be 9.75 times 10 to the 6 joules, or about 10 megajoules.

Skip to 14 minutes and 5 secondsAnd that's equivalent to about the equivalent energy from a third of a litre of gasoline in terms of available energy. Now, MDI, who already build a compressed air powered car, claim about 90% efficiency, for their air motors, which is much, much better than we can get from an internal combustion engine. And so they claim that this 10 megajoules that we can get out of one tank of air gives their cars a range of about 220 kilometres with three people in it on one tank of compressed air. And that seems a bit ambitious, perhaps, based on this value of available energy that we've got, based on changing Gibbs energy when air escapes.

Skip to 14 minutes and 48 secondsBut we've made some idealizations here in doing a calculation, and we've also assume that we just satisfy the second law. So maybe these numbers are a little bit off because of those assumptions, but it's in the right sort of ballpark.

# Air powered car (worked example)

Calculate the available energy (change in Gibbs energy) in a 100 litre tank of compressed air at 248bars if it is released isothermally through an air-motor to power a vehicle.

This worked example correlates very closely to the real world as engineers working it was used by engineers working on the Peugeot 208 Air Hybrid that was demonstrated at the 2014 Paris Auto Show and the air-powered car produced by MDI.

Once you have worked through this challenge, watch the video or check your findings in the solution PDF at the bottom of the page.

NB. The question above states that the expansion of the air is isothermal, i.e. no change in temperature (T1=T2), but in the video and original solution pdf I have solved it assuming adiabatic expansion, i.e. no heat transfer and T1 does not equal T2. I have provided a corrected solution pdf in which I have assumed isothermal conditions. My apologies for the confusion caused.