Class example from 3/13 on exergy

Exergy (Available Work)
Suppose we have a piston cylinder system shown below, and we insert a small amount of fuel
into it and ignite the fuel generating a hot gas at 2000 K, 20 bar, in a volume of 1 liter. Let’s
say the surroundings are at 300 K. We want to use this system to do work (propel our car, drive
a generator to make electricity, etc.). What is the most work we can possibly get out?
300 K, 1 bar
surroundings
2000 K
20 bar
1 liter
Our first guess might be to convert all the internal energy in the system to work, so that by the
first law:
U = W
Assuming we can’t go lower than ambient temperature, our final temperature would be 300 K.
Using the IGL, we find the mass in this system (assuming air as the gas) is m = PV/RT = .00353
kg. From the air tables, u(2000 K) = 1678.7 kJ/kg, and u(300 K) = 214.07. So the total
U = W = 5.17 kJ.
But a quick check of the 2nd law shows that for this process, S < 0, so it’s not possible.
So what is the largest possible amount of work we can extract?
300 K, 1 bar
surroundings
1 bar
W ork Out (usable)
Work on atm (not usable)
2000 K
20 bar
1 liter
Let’s first expand the system adiabatically to ambient pressure to a state 2. The farthest we can
expand it is 1 bar. In this case, Q = 0, S = 0, and W = -U = -(U2 - U1).
But we’re also doing work on the surroundings: Vsys = -Vsurr, and the surroundings are at
constant pressure, so we’ve done Psurr(V2 - V1) work on the surroundings. The extractable
amount system work is then - (U2 - U1) – Psurr(V2 - V1)
But now we still have extra heat in the system, and we can use that heat to drive a heat engine
(e.g. heat steam, run it through a turbine). We can extract this heat until the final temperature is
Tsurr, and we’ll call this final state 0. This would be the “dead state” because pressure and
temperature are the same as ambient, and so there is no further driving force from which to
extract energy.
1 bar
Heat to
Heat Engine
Heat to
surroundings
1 bar
Final system state
"Dead State"
300 K
H.E.
W ork
The heat transfer in this second step is equal to the change in internal energy of the system,
since there is no further volume change or work, so Qsys = -(Uo - U2).
So how much of this Qsys can we convert to work?
The best we can possibly do is to minimize the total entropy change (STOT = Ssys + Ssurr = 0).
We’ve decreased the entropy of the system, so we’d have to increase the entropy of the
surroundings by the same amount or:
Ssys = - Ssurr
S0 – S2 = S0 – S1= -Qsurr/Tsurr
The amount of work we can extract from Qsys is thus |Qsys| – | Qsurr| or [-(Uo-U2)] –[- Tsurr(So-S1)]
Adding the work from the two processes together, we get:
WMAX = - (U2 - U1) – Psurr(V2 - V1) - (Uo - U2) - Tsurr(So - S1)
Maximum work = “Exergy” = (U1 - U0) + Psurr(V1 - V0) – Tsurr(S1 - So)
Let’s run the numbers:
The first term is the same as our initial guess: 5.18 kJ. It represents the total internal energy of
the system.
The second term is negative, since V1 < V0. If Psurr = 1 bar, then we just need the volumes.
From the IGL, V0/V1 =(mRT0/P0)/(mRT1/P1) = (T0/T1)/(P0/P1) = (3/20)/(1/20) = 3. So the final
volume is 3 litres. The second term is thus 100000 (.001 - .003) = -200 J or -0.2 kJ.
The third term is where all the action is.
S1 = ms1 = m[so(T1) - Rln(P1/Pref)] = .00353(3.7994 - .287ln20) =.01037 kJ/K
S0 = ms0 = m[so(T0) - Rln(P0/Pref)] = .00353(1.70203 - .287ln1)=.00601 kJ/K
Tsurr(S1-So) = 1.31 kJ
Exergy = Maximum work = available work = 5.18 - .20 – 1.31 = 3.67 kJ
So the amount of usable energy (the “exergy”) is less than the internal energy of the system.