WEBVTT 00:00:00.000 --> 00:00:02.430 align:middle line:90% [SQUEAKING] 00:00:02.430 --> 00:00:03.402 align:middle line:90% [RUSTLING] 00:00:03.402 --> 00:00:05.346 align:middle line:90% [CLICKING] 00:00:05.346 --> 00:00:10.510 align:middle line:90% 00:00:10.510 --> 00:00:14.100 align:middle line:84% GIAN PAOLO BERETTA: OK, good morning. 00:00:14.100 --> 00:00:18.040 align:middle line:84% So we started talking about systems with chemical reactions. 00:00:18.040 --> 00:00:20.760 align:middle line:90% 00:00:20.760 --> 00:00:31.740 align:middle line:84% And we-- to just go quickly over what we have done so far, 00:00:31.740 --> 00:00:38.070 align:middle line:84% we considered the two most representative setups, 00:00:38.070 --> 00:00:46.440 align:middle line:84% the closed system up here and the open system setup, 00:00:46.440 --> 00:00:49.530 align:middle line:84% for which, of course, when we analyze, 00:00:49.530 --> 00:00:54.030 align:middle line:84% we have to make at least the energy and the entropy balance, 00:00:54.030 --> 00:00:56.040 align:middle line:90% and sometimes also-- 00:00:56.040 --> 00:00:58.085 align:middle line:90% here also the volume balance. 00:00:58.085 --> 00:01:01.890 align:middle line:90% 00:01:01.890 --> 00:01:05.040 align:middle line:84% And these balances take these forms, 00:01:05.040 --> 00:01:09.720 align:middle line:84% where these differences between properties of the products 00:01:09.720 --> 00:01:12.760 align:middle line:84% and the reactants, energy, entropy-- 00:01:12.760 --> 00:01:16.620 align:middle line:84% here, the enthalpies because we have bulk flow interactions-- 00:01:16.620 --> 00:01:20.230 align:middle line:90% are important. 00:01:20.230 --> 00:01:26.895 align:middle line:84% And they have the feature that we need to-- 00:01:26.895 --> 00:01:30.360 align:middle line:90% 00:01:30.360 --> 00:01:34.860 align:middle line:84% we need to deal with the fact that the composition changes now 00:01:34.860 --> 00:01:39.330 align:middle line:84% because as the process proceeds, we have 00:01:39.330 --> 00:01:41.450 align:middle line:90% a chemical reaction in place-- 00:01:41.450 --> 00:01:45.130 align:middle line:84% one or more chemical reactions in place. 00:01:45.130 --> 00:01:49.300 align:middle line:84% We introduced the notation for stoichiometry 00:01:49.300 --> 00:01:53.030 align:middle line:84% and the proportionality relations. 00:01:53.030 --> 00:01:56.090 align:middle line:84% And we defined the properties of reaction as a way 00:01:56.090 --> 00:02:02.150 align:middle line:84% to concentrate the part of the change 00:02:02.150 --> 00:02:05.330 align:middle line:84% in the properties that has to do with the change in composition 00:02:05.330 --> 00:02:12.530 align:middle line:84% in this last term, where we have the product of the reaction 00:02:12.530 --> 00:02:17.670 align:middle line:84% coordinate and what we called the property of reaction. 00:02:17.670 --> 00:02:19.650 align:middle line:84% For example, this is the energy of reaction, 00:02:19.650 --> 00:02:21.060 align:middle line:90% the enthalpy of reaction. 00:02:21.060 --> 00:02:24.170 align:middle line:84% But the same way we can define also 00:02:24.170 --> 00:02:25.730 align:middle line:84% the entropy, the Gibbs free energy, 00:02:25.730 --> 00:02:29.120 align:middle line:90% and the volume of the reaction. 00:02:29.120 --> 00:02:30.920 align:middle line:90% And remember that if-- 00:02:30.920 --> 00:02:37.650 align:middle line:84% the 0 here means standard pressure, so 1 atmosphere. 00:02:37.650 --> 00:02:41.630 align:middle line:84% And the 0 here without the temperature dependence 00:02:41.630 --> 00:02:44.400 align:middle line:90% means also standard temperature. 00:02:44.400 --> 00:02:50.000 align:middle line:84% So STP means-- is standard temperature, 25 degrees C, 00:02:50.000 --> 00:02:54.440 align:middle line:90% and 1 atmosphere. 00:02:54.440 --> 00:03:00.900 align:middle line:84% It may seem that we have to measure properties of reaction 00:03:00.900 --> 00:03:03.360 align:middle line:84% for every reaction that we have to deal with. 00:03:03.360 --> 00:03:06.000 align:middle line:84% But fortunately, that's not necessary. 00:03:06.000 --> 00:03:09.650 align:middle line:84% All we need to do is to measure the properties of reaction 00:03:09.650 --> 00:03:18.640 align:middle line:84% for what we will define as the reactions of formation. 00:03:18.640 --> 00:03:21.130 align:middle line:84% The properties of reaction-- for example, this 00:03:21.130 --> 00:03:25.240 align:middle line:84% is the reaction of formation from molecular hydrogen, which 00:03:25.240 --> 00:03:31.990 align:middle line:84% is the most stable species for hydrogen at STP. 00:03:31.990 --> 00:03:39.580 align:middle line:84% And it's the formation from that to atomic hydrogen, which 00:03:39.580 --> 00:03:43.760 align:middle line:84% is not-- which is also an elemental species, 00:03:43.760 --> 00:03:46.780 align:middle line:90% but it's not the most stable. 00:03:46.780 --> 00:03:50.360 align:middle line:84% And in order to go from here to there, 00:03:50.360 --> 00:03:54.710 align:middle line:84% we have to break this hydrogen bond. 00:03:54.710 --> 00:03:57.840 align:middle line:90% And that requires energy. 00:03:57.840 --> 00:04:03.500 align:middle line:84% And this is why the curve that represents the equilibrium 00:04:03.500 --> 00:04:07.310 align:middle line:84% state for hydrogen, even at 0 temperature, 00:04:07.310 --> 00:04:09.950 align:middle line:84% is above the curve that represents 00:04:09.950 --> 00:04:15.050 align:middle line:90% the molecular hydrogen states. 00:04:15.050 --> 00:04:17.329 align:middle line:84% And at 0 temperature, this difference 00:04:17.329 --> 00:04:21.925 align:middle line:84% is precisely what we call the bond energy. 00:04:21.925 --> 00:04:25.000 align:middle line:90% 00:04:25.000 --> 00:04:28.420 align:middle line:84% Instead, what we call the property of reaction 00:04:28.420 --> 00:04:30.320 align:middle line:90% doesn't refer to 0 temperature. 00:04:30.320 --> 00:04:35.160 align:middle line:84% It refers to standard temperature, T0. 00:04:35.160 --> 00:04:40.870 align:middle line:84% So it is the difference between that property-- in this case, 00:04:40.870 --> 00:04:45.940 align:middle line:84% the enthalpy, if we go vertical-wise, or the entropy, 00:04:45.940 --> 00:04:49.020 align:middle line:84% if you go horizontal-wise, from-- 00:04:49.020 --> 00:04:52.530 align:middle line:84% the difference in the enthalpy or entropy 00:04:52.530 --> 00:04:56.650 align:middle line:84% between this state and that state. 00:04:56.650 --> 00:05:00.610 align:middle line:84% OK, so these represent graphically the enthalpy 00:05:00.610 --> 00:05:03.105 align:middle line:84% of reaction and the entropy of reaction. 00:05:03.105 --> 00:05:09.530 align:middle line:90% 00:05:09.530 --> 00:05:14.900 align:middle line:84% We ended the last lecture with this graph 00:05:14.900 --> 00:05:20.930 align:middle line:84% that wants to point out that even the bond energies can 00:05:20.930 --> 00:05:26.880 align:middle line:84% be described somehow by means of the Lennard-Jones potential. 00:05:26.880 --> 00:05:31.770 align:middle line:84% That is a possible, reasonable model-- 00:05:31.770 --> 00:05:36.710 align:middle line:84% perhaps not the best, but it is simple enough 00:05:36.710 --> 00:05:39.380 align:middle line:90% that one can use it. 00:05:39.380 --> 00:05:46.600 align:middle line:84% And the values of the parameters, the Lennard-Jones 00:05:46.600 --> 00:05:56.270 align:middle line:84% parameters, for atomic bonds, bond energies, bond energies 00:05:56.270 --> 00:06:00.140 align:middle line:84% are at least three orders of magnitude 00:06:00.140 --> 00:06:03.760 align:middle line:84% stronger than the van der Waals forces. 00:06:03.760 --> 00:06:06.090 align:middle line:84% They are all electrostatic forces, 00:06:06.090 --> 00:06:13.760 align:middle line:84% but the ones that you need in order to form this bond 00:06:13.760 --> 00:06:15.680 align:middle line:90% are much stronger. 00:06:15.680 --> 00:06:21.340 align:middle line:84% Not only they occur at closer distances-- 00:06:21.340 --> 00:06:27.610 align:middle line:84% you see, this is at 0.7 Angstrom with respect 00:06:27.610 --> 00:06:33.400 align:middle line:84% to the bottom of the Van der Waals 00:06:33.400 --> 00:06:37.210 align:middle line:84% forces, which is about 3 point something 00:06:37.210 --> 00:06:41.620 align:middle line:90% Angstrom for the liquid-- 00:06:41.620 --> 00:06:46.340 align:middle line:84% for example, this is liquid water. 00:06:46.340 --> 00:06:49.170 align:middle line:84% And also notice that, yes, I have plotted this, 00:06:49.170 --> 00:06:52.670 align:middle line:84% but in order to make it visible, I had to multiply it by a factor 00:06:52.670 --> 00:06:54.200 align:middle line:90% of 1,000. 00:06:54.200 --> 00:06:57.800 align:middle line:84% Otherwise, it would be almost flat as 00:06:57.800 --> 00:07:05.290 align:middle line:84% compared to the strength of the atomic bonds, covalent bonds. 00:07:05.290 --> 00:07:08.300 align:middle line:90% 00:07:08.300 --> 00:07:16.560 align:middle line:84% OK, now, the properties of reaction, 00:07:16.560 --> 00:07:19.680 align:middle line:84% as we have seen, at 0 temperature, 00:07:19.680 --> 00:07:27.670 align:middle line:84% are related to the bond energies. 00:07:27.670 --> 00:07:29.560 align:middle line:90% So, for example, if I take-- 00:07:29.560 --> 00:07:35.510 align:middle line:84% this is the reaction of formation of water, 00:07:35.510 --> 00:07:42.210 align:middle line:84% of two molecules of water, from two molecules of hydrogen, 00:07:42.210 --> 00:07:46.790 align:middle line:84% of molecular hydrogen, and one molecule of oxygen. 00:07:46.790 --> 00:07:49.610 align:middle line:84% In order to go from this configuration 00:07:49.610 --> 00:07:55.510 align:middle line:84% to this configuration, you have to break these two hydrogen 00:07:55.510 --> 00:07:57.880 align:middle line:84% bonds and the oxygen bond and then 00:07:57.880 --> 00:08:01.030 align:middle line:90% reassemble in this other way. 00:08:01.030 --> 00:08:04.880 align:middle line:84% At the end, in this other way, in two water molecules, 00:08:04.880 --> 00:08:14.060 align:middle line:84% you have four times an oxygen-hydrogen bond, OK? 00:08:14.060 --> 00:08:14.875 align:middle line:90% Four times. 00:08:14.875 --> 00:08:18.590 align:middle line:90% 00:08:18.590 --> 00:08:22.170 align:middle line:84% The strength of these bonds are listed-- 00:08:22.170 --> 00:08:24.080 align:middle line:90% are plotted, are measured. 00:08:24.080 --> 00:08:28.640 align:middle line:84% You find tables that provide this strength 00:08:28.640 --> 00:08:34.070 align:middle line:84% and also the typical distance between atoms, 00:08:34.070 --> 00:08:37.630 align:middle line:84% between the nuclei in these bonds. 00:08:37.630 --> 00:08:42.549 align:middle line:84% For example, the blue one is the hydrogen. It's the closest. 00:08:42.549 --> 00:08:53.310 align:middle line:84% It's at 74 picometers-- that's 0.74 Angstrom. 00:08:53.310 --> 00:08:57.070 align:middle line:90% The oxygen is here. 00:08:57.070 --> 00:09:01.680 align:middle line:84% And in the middle, we have the hydrogen and oxygen bond. 00:09:01.680 --> 00:09:04.250 align:middle line:90% 00:09:04.250 --> 00:09:08.810 align:middle line:84% And, in fact, if you take the depth-- 00:09:08.810 --> 00:09:12.800 align:middle line:84% so the strength of these potentials-- 00:09:12.800 --> 00:09:16.760 align:middle line:84% so the epsilons for the Lennard-Jones for these bonds, 00:09:16.760 --> 00:09:21.080 align:middle line:84% and you do the simple counting of what's 00:09:21.080 --> 00:09:25.010 align:middle line:84% the difference between having four of these ones, four 00:09:25.010 --> 00:09:32.060 align:middle line:84% of these ones that are worth 459, as compared 00:09:32.060 --> 00:09:38.690 align:middle line:84% to two of these ones, the hydrogen that are 432 00:09:38.690 --> 00:09:50.260 align:middle line:84% and one oxygen, 495, you find that the difference is in favor 00:09:50.260 --> 00:09:51.380 align:middle line:90% of this one. 00:09:51.380 --> 00:09:52.700 align:middle line:90% So these are more stable. 00:09:52.700 --> 00:10:03.960 align:middle line:84% They are less-- they are more strongly bound. 00:10:03.960 --> 00:10:12.710 align:middle line:84% And this number, which is by the accounting of the bonds, 00:10:12.710 --> 00:10:18.860 align:middle line:84% is very close to the enthalpy of formation of 2 moles of water. 00:10:18.860 --> 00:10:24.880 align:middle line:84% The difference stays mostly in the fact 00:10:24.880 --> 00:10:32.590 align:middle line:84% that whereas the bonds are computed at 0 kelvin, 00:10:32.590 --> 00:10:36.820 align:middle line:84% the enthalpies of formation are listed at standard temperature 00:10:36.820 --> 00:10:37.700 align:middle line:90% and pressure. 00:10:37.700 --> 00:10:42.310 align:middle line:84% So you have to correct in going from here to here 00:10:42.310 --> 00:10:50.440 align:middle line:84% for the specific heats from 0 temperature to 298 K. 00:10:50.440 --> 00:10:55.200 align:middle line:84% And if you do that correction, you get closer to this number. 00:10:55.200 --> 00:10:58.470 align:middle line:84% So in a sense, this is to say that the Lennard-Jones 00:10:58.470 --> 00:11:06.600 align:middle line:84% representation is not bad for giving a first look at what 00:11:06.600 --> 00:11:09.810 align:middle line:84% these properties of formation, and at least the enthalpy 00:11:09.810 --> 00:11:12.665 align:middle line:90% of formation, comes from. 00:11:12.665 --> 00:11:15.210 align:middle line:90% 00:11:15.210 --> 00:11:23.890 align:middle line:84% Now, this slide wants to describe 00:11:23.890 --> 00:11:28.300 align:middle line:84% what's behind the Hess relation and these properties 00:11:28.300 --> 00:11:30.890 align:middle line:90% of formation. 00:11:30.890 --> 00:11:34.720 align:middle line:84% So it turns out that we don't need 00:11:34.720 --> 00:11:38.210 align:middle line:84% to measure all the properties of reactions, 00:11:38.210 --> 00:11:44.470 align:middle line:84% but only those of a formation from a set which is complete 00:11:44.470 --> 00:11:45.680 align:middle line:90% enough that-- 00:11:45.680 --> 00:11:51.430 align:middle line:84% so we define a complete set of independent chemical species 00:11:51.430 --> 00:11:56.890 align:middle line:84% from which you can construct any other molecule. 00:11:56.890 --> 00:12:01.150 align:middle line:84% So these are like the building blocks of chemical compounds. 00:12:01.150 --> 00:12:07.883 align:middle line:84% And typically, they are called the elemental species. 00:12:07.883 --> 00:12:12.360 align:middle line:90% 00:12:12.360 --> 00:12:18.790 align:middle line:84% In order to define this set, you need to request two conditions. 00:12:18.790 --> 00:12:27.870 align:middle line:84% The first is that, yes, every molecule must be-- 00:12:27.870 --> 00:12:32.040 align:middle line:84% you construct any chemical species, 00:12:32.040 --> 00:12:34.520 align:middle line:84% and you have to be able to construct it 00:12:34.520 --> 00:12:37.700 align:middle line:84% from these elemental species, according 00:12:37.700 --> 00:12:42.540 align:middle line:84% to a chemical reaction of formation, which is this one. 00:12:42.540 --> 00:12:48.590 align:middle line:84% Now, in our notation, where we put everything 00:12:48.590 --> 00:12:51.890 align:middle line:84% on the left-hand side with the plus stoichiometric coefficient 00:12:51.890 --> 00:12:57.410 align:middle line:84% for the products and with the negative stoichiometric 00:12:57.410 --> 00:13:00.740 align:middle line:84% coefficient for the reactants, this 00:13:00.740 --> 00:13:06.240 align:middle line:84% is the way we would write our reaction. 00:13:06.240 --> 00:13:09.350 align:middle line:84% But you could also write it this other way. 00:13:09.350 --> 00:13:13.190 align:middle line:84% And formally, these will be useful in one step below. 00:13:13.190 --> 00:13:15.710 align:middle line:84% The other condition that you want to impose 00:13:15.710 --> 00:13:22.070 align:middle line:84% is that the set of these species be independent from one another 00:13:22.070 --> 00:13:26.180 align:middle line:84% so that it is impossible to convert one 00:13:26.180 --> 00:13:29.450 align:middle line:84% into the other or several into the other. 00:13:29.450 --> 00:13:32.840 align:middle line:84% That condition is written by saying 00:13:32.840 --> 00:13:38.750 align:middle line:84% that if you ever write a chemical reaction that involves 00:13:38.750 --> 00:13:43.170 align:middle line:84% only elemental species, that reaction has 00:13:43.170 --> 00:13:46.450 align:middle line:90% to be identically trivial-- 00:13:46.450 --> 00:13:47.160 align:middle line:90% so 0. 00:13:47.160 --> 00:13:49.560 align:middle line:84% All the stoichiometric coefficients must be 0. 00:13:49.560 --> 00:13:53.290 align:middle line:84% Otherwise, it would be possible to transform gold-- 00:13:53.290 --> 00:13:57.150 align:middle line:84% I mean, lead in gold, like-- instead of chemistry, you 00:13:57.150 --> 00:13:58.040 align:middle line:90% do alchemy. 00:13:58.040 --> 00:14:02.340 align:middle line:90% 00:14:02.340 --> 00:14:07.500 align:middle line:84% So the practical choice is to select-- 00:14:07.500 --> 00:14:09.390 align:middle line:84% the practical-- the conventional choice 00:14:09.390 --> 00:14:12.990 align:middle line:84% is to select as elemental the species that 00:14:12.990 --> 00:14:16.380 align:middle line:90% have only one type of nucleus. 00:14:16.380 --> 00:14:19.600 align:middle line:84% For example, the oxygen molecule has two nuclei, 00:14:19.600 --> 00:14:23.320 align:middle line:84% but they are of the same kind, so only one kind. 00:14:23.320 --> 00:14:27.740 align:middle line:90% And plus, since you know that-- 00:14:27.740 --> 00:14:30.700 align:middle line:84% for example, oxygen, you can find it 00:14:30.700 --> 00:14:33.490 align:middle line:84% in several molecular configuration, 00:14:33.490 --> 00:14:43.660 align:middle line:84% like atomic oxygen, the ion, or ozone, O3, and O2-- 00:14:43.660 --> 00:14:52.300 align:middle line:84% you pick the one that is most stable at standard condition. 00:14:52.300 --> 00:14:57.360 align:middle line:84% So in this case, for oxygen, you take O2. 00:14:57.360 --> 00:15:01.560 align:middle line:84% And it turns out that you have to select 00:15:01.560 --> 00:15:06.100 align:middle line:84% one species for each atomic number in the periodic table. 00:15:06.100 --> 00:15:09.150 align:middle line:84% I'm sure you remember the periodic table. 00:15:09.150 --> 00:15:15.880 align:middle line:84% And Z is the number of protons in the nucleus of the atom. 00:15:15.880 --> 00:15:23.980 align:middle line:84% And the actual-- the isotopes, which has to do with the number 00:15:23.980 --> 00:15:28.660 align:middle line:84% of neutrons that are in the nucleus of the atom, 00:15:28.660 --> 00:15:35.710 align:middle line:84% don't count for chemistry, because chemistry is mostly-- 00:15:35.710 --> 00:15:41.040 align:middle line:84% what's responsible for chemical reactions is the electrical-- 00:15:41.040 --> 00:15:46.660 align:middle line:84% the electrons and the protons and not the weight 00:15:46.660 --> 00:15:54.340 align:middle line:84% of the nucleus, which is instead important for nuclear reactions. 00:15:54.340 --> 00:15:58.130 align:middle line:90% 00:15:58.130 --> 00:16:04.730 align:middle line:84% And, of course, I think you can proceed in a similar way 00:16:04.730 --> 00:16:08.040 align:middle line:84% also if you are interested in nuclear reactions. 00:16:08.040 --> 00:16:12.260 align:middle line:84% So you want to apply this same kind of reasoning 00:16:12.260 --> 00:16:16.350 align:middle line:84% to the properties of formation of nuclei. 00:16:16.350 --> 00:16:22.470 align:middle line:84% You would have to define a set of complete and independent 00:16:22.470 --> 00:16:24.760 align:middle line:90% nucleons. 00:16:24.760 --> 00:16:28.210 align:middle line:84% And typically, you would choose the proton, the neutron, 00:16:28.210 --> 00:16:30.820 align:middle line:90% and the electron, and-- 00:16:30.820 --> 00:16:31.560 align:middle line:90% what else? 00:16:31.560 --> 00:16:35.460 align:middle line:90% 00:16:35.460 --> 00:16:36.030 align:middle line:90% I don't know. 00:16:36.030 --> 00:16:39.770 align:middle line:84% Maybe quarks are the different level. 00:16:39.770 --> 00:16:46.100 align:middle line:84% Anyway, so then, for this reaction of formation, 00:16:46.100 --> 00:16:48.860 align:middle line:84% you select your compound, and you 00:16:48.860 --> 00:16:50.870 align:middle line:84% find the stoichiometric coefficients 00:16:50.870 --> 00:16:55.160 align:middle line:84% in terms that are needed in order to construct your compound 00:16:55.160 --> 00:16:56.710 align:middle line:90% from the elemental species. 00:16:56.710 --> 00:16:58.070 align:middle line:90% These are the alphas. 00:16:58.070 --> 00:17:01.940 align:middle line:84% And then you write down according 00:17:01.940 --> 00:17:03.530 align:middle line:84% to our rules-- so the summation of 00:17:03.530 --> 00:17:09.020 align:middle line:84% the stoichiometric coefficients times the properties 00:17:09.020 --> 00:17:14.390 align:middle line:84% of the pure components that appear in the reaction at STP. 00:17:14.390 --> 00:17:17.730 align:middle line:84% And this gives you the property of reaction. 00:17:17.730 --> 00:17:20.880 align:middle line:84% And since it's the reaction of formation, 00:17:20.880 --> 00:17:25.099 align:middle line:84% we call it the property of formation of your compound i. 00:17:25.099 --> 00:17:27.770 align:middle line:90% 00:17:27.770 --> 00:17:31.160 align:middle line:90% Now, it turns out-- 00:17:31.160 --> 00:17:35.036 align:middle line:90% and this is the good-- 00:17:35.036 --> 00:17:38.870 align:middle line:84% the practical aspect of this choice-- 00:17:38.870 --> 00:17:46.890 align:middle line:84% it turns out that you can safely arbitrarily select 00:17:46.890 --> 00:17:53.230 align:middle line:84% the reference value for the property at STP 00:17:53.230 --> 00:17:56.650 align:middle line:90% of the elemental species. 00:17:56.650 --> 00:18:00.880 align:middle line:84% And since it's arbitrary-- in other words, it will not-- 00:18:00.880 --> 00:18:05.670 align:middle line:84% it will cancel out your calculations in chemistry-- 00:18:05.670 --> 00:18:08.760 align:middle line:84% you can set you can select any number. 00:18:08.760 --> 00:18:11.915 align:middle line:84% So the simplest is to set it equal to 0. 00:18:11.915 --> 00:18:15.280 align:middle line:90% 00:18:15.280 --> 00:18:17.150 align:middle line:90% Why so? 00:18:17.150 --> 00:18:23.500 align:middle line:84% Well, because in your chemical problems, 00:18:23.500 --> 00:18:26.490 align:middle line:84% you will have to deal with chemical reactions that have 00:18:26.490 --> 00:18:30.390 align:middle line:90% this generic form equal to 0. 00:18:30.390 --> 00:18:32.500 align:middle line:84% So 0 equals the stoichiometric coefficients 00:18:32.500 --> 00:18:35.820 align:middle line:90% times the chemical compounds. 00:18:35.820 --> 00:18:39.450 align:middle line:84% Now, each chemical compound, like here, 00:18:39.450 --> 00:18:41.740 align:middle line:84% can be obtained from the elemental species. 00:18:41.740 --> 00:18:44.790 align:middle line:84% So we can formally substitute this formula here 00:18:44.790 --> 00:18:47.490 align:middle line:90% for the compound in here. 00:18:47.490 --> 00:18:53.940 align:middle line:84% And we get the chemical reaction that involves 00:18:53.940 --> 00:18:56.900 align:middle line:90% only the elemental species. 00:18:56.900 --> 00:18:58.880 align:middle line:84% By the condition of independence that 00:18:58.880 --> 00:19:04.790 align:middle line:84% requires that this set of stoichiometric coefficients 00:19:04.790 --> 00:19:09.240 align:middle line:90% be all identically equal to 0. 00:19:09.240 --> 00:19:17.090 align:middle line:84% This is more or less the formal expression 00:19:17.090 --> 00:19:22.850 align:middle line:84% of what you do when you fix the stoichiometric coefficients 00:19:22.850 --> 00:19:24.930 align:middle line:90% in a reaction. 00:19:24.930 --> 00:19:27.120 align:middle line:84% And you balance the reaction by putting 00:19:27.120 --> 00:19:29.640 align:middle line:84% the proper stoichiometric coefficients. 00:19:29.640 --> 00:19:35.010 align:middle line:84% That's what you're searching for, 00:19:35.010 --> 00:19:38.850 align:middle line:90% the nu sub i's that satisfy-- 00:19:38.850 --> 00:19:40.360 align:middle line:90% that solve this equation. 00:19:40.360 --> 00:19:47.854 align:middle line:90% 00:19:47.854 --> 00:19:51.200 align:middle line:84% So now, since we are interested in general 00:19:51.200 --> 00:19:54.970 align:middle line:84% in this kind of reaction, the property 00:19:54.970 --> 00:19:58.150 align:middle line:84% of that reaction-- for example, the energy of reaction 00:19:58.150 --> 00:20:00.945 align:middle line:84% for that reaction-- is given, is defined this way. 00:20:00.945 --> 00:20:05.320 align:middle line:90% 00:20:05.320 --> 00:20:07.540 align:middle line:84% If you substitute this expression here-- 00:20:07.540 --> 00:20:15.730 align:middle line:84% because, you see, here we have these pure component energies 00:20:15.730 --> 00:20:22.610 align:middle line:84% at STP for your substances that appear in your reaction. 00:20:22.610 --> 00:20:27.760 align:middle line:84% And you can write them in terms of the properties of formation 00:20:27.760 --> 00:20:31.550 align:middle line:84% and these other summation that you take on the other side. 00:20:31.550 --> 00:20:33.140 align:middle line:90% So this is the substitution. 00:20:33.140 --> 00:20:35.900 align:middle line:90% 00:20:35.900 --> 00:20:38.360 align:middle line:84% And then you bring the summation in, 00:20:38.360 --> 00:20:46.620 align:middle line:84% and you find that this second contribution has coefficients 00:20:46.620 --> 00:20:48.840 align:middle line:84% that are identically 0 for the reason 00:20:48.840 --> 00:20:50.890 align:middle line:90% that we just mentioned before. 00:20:50.890 --> 00:20:54.450 align:middle line:90% So this summation cancels away. 00:20:54.450 --> 00:20:58.830 align:middle line:84% And therefore, you find that the property of your reaction 00:20:58.830 --> 00:21:01.710 align:middle line:84% is equal to the weighted sum on the stoichiometric coefficients 00:21:01.710 --> 00:21:06.831 align:middle line:84% of your reaction of the properties of formation 00:21:06.831 --> 00:21:10.720 align:middle line:84% of the various components that appear in your reaction. 00:21:10.720 --> 00:21:12.575 align:middle line:84% This is called the Hess relation. 00:21:12.575 --> 00:21:17.690 align:middle line:90% 00:21:17.690 --> 00:21:20.930 align:middle line:84% And the fact that these things are multiplied by 0 00:21:20.930 --> 00:21:23.060 align:middle line:90% allows you to select any-- 00:21:23.060 --> 00:21:28.781 align:middle line:90% their value being not important. 00:21:28.781 --> 00:21:33.330 align:middle line:84% So you conventionally select those equal to 0. 00:21:33.330 --> 00:21:39.410 align:middle line:84% And also, you could safely select 00:21:39.410 --> 00:21:44.380 align:middle line:84% that the STP value for the pure components 00:21:44.380 --> 00:21:48.290 align:middle line:84% are equal to their properties of formation, 00:21:48.290 --> 00:21:51.365 align:middle line:84% like we do here, setting these equal to 0. 00:21:51.365 --> 00:21:54.090 align:middle line:90% 00:21:54.090 --> 00:22:00.790 align:middle line:84% There is another minor but dangerous complication-- 00:22:00.790 --> 00:22:04.290 align:middle line:84% dangerous if you forget about it in your calculations-- 00:22:04.290 --> 00:22:08.910 align:middle line:90% that most of the tables have-- 00:22:08.910 --> 00:22:12.310 align:middle line:84% since much of this has been developed, 00:22:12.310 --> 00:22:16.530 align:middle line:84% when people dealt with combustion problems, 00:22:16.530 --> 00:22:23.410 align:middle line:84% and the products of combustion are in the gaseous phase, 00:22:23.410 --> 00:22:30.550 align:middle line:84% in practical applications, people 00:22:30.550 --> 00:22:36.800 align:middle line:84% are listing the values of formation of the species 00:22:36.800 --> 00:22:40.230 align:middle line:84% when they are in the gaseous phase. 00:22:40.230 --> 00:22:44.370 align:middle line:84% But many species-- for example, water-- 00:22:44.370 --> 00:22:47.850 align:middle line:84% at standard conditions is not in the gaseous phase 00:22:47.850 --> 00:22:51.990 align:middle line:84% because at 25 degrees C and 1 atmosphere, water is liquid. 00:22:51.990 --> 00:22:55.950 align:middle line:84% Yet in the table, you find the property 00:22:55.950 --> 00:23:01.340 align:middle line:84% of formation of water in the fictitious gaseous state. 00:23:01.340 --> 00:23:02.970 align:middle line:90% So what's that? 00:23:02.970 --> 00:23:06.990 align:middle line:84% OK, that is sort of an extrapolation of water 00:23:06.990 --> 00:23:10.830 align:middle line:84% vapor from the range of states in which it behaves 00:23:10.830 --> 00:23:13.840 align:middle line:84% as an ideal gas for which you know 00:23:13.840 --> 00:23:16.080 align:middle line:84% the properties-- for example, the specific 00:23:16.080 --> 00:23:20.100 align:middle line:90% heats at constant pressure. 00:23:20.100 --> 00:23:22.650 align:middle line:84% And you know them as a function of temperature. 00:23:22.650 --> 00:23:24.100 align:middle line:90% They are given by correlations. 00:23:24.100 --> 00:23:25.800 align:middle line:90% We'll see some in a moment. 00:23:25.800 --> 00:23:31.620 align:middle line:84% And what you do is to extrapolate those properties 00:23:31.620 --> 00:23:36.390 align:middle line:84% down to standard conditions as if that substance kept 00:23:36.390 --> 00:23:41.190 align:middle line:84% behaving as an ideal gas down to those STP. 00:23:41.190 --> 00:23:47.160 align:middle line:84% And that's how you compute this, from this temperature 00:23:47.160 --> 00:23:54.780 align:middle line:84% T where it behaves as an ideal gas down to here. 00:23:54.780 --> 00:23:59.700 align:middle line:84% And also, for the entropy, you just apply the laws of ideal gas 00:23:59.700 --> 00:24:01.930 align:middle line:90% and extrapolate down to STP. 00:24:01.930 --> 00:24:05.040 align:middle line:90% 00:24:05.040 --> 00:24:09.950 align:middle line:90% OK, so these are details. 00:24:09.950 --> 00:24:12.710 align:middle line:90% And when you compute-- 00:24:12.710 --> 00:24:18.360 align:middle line:84% yeah, these tables give values for the enthalpies of formation 00:24:18.360 --> 00:24:23.970 align:middle line:84% of a number of compounds, from which you can cook up 00:24:23.970 --> 00:24:26.590 align:middle line:90% all sorts of chemical reactions. 00:24:26.590 --> 00:24:29.150 align:middle line:90% 00:24:29.150 --> 00:24:34.640 align:middle line:84% For example, we are interested in combustion of methane 00:24:34.640 --> 00:24:39.780 align:middle line:90% or the combustion of isooctane. 00:24:39.780 --> 00:24:44.580 align:middle line:84% Isooctane is close to gasoline, and methane 00:24:44.580 --> 00:24:48.990 align:middle line:84% is natural gas, the main component in natural gas. 00:24:48.990 --> 00:24:54.690 align:middle line:84% So these are important for energy applications, 00:24:54.690 --> 00:24:56.580 align:middle line:90% as you know. 00:24:56.580 --> 00:25:01.190 align:middle line:90% So in order to write this-- 00:25:01.190 --> 00:25:07.880 align:middle line:84% this is the representation of the complete combustion, 00:25:07.880 --> 00:25:11.250 align:middle line:84% so the simplest model for oxidation of the fuel, 00:25:11.250 --> 00:25:14.330 align:middle line:90% complete oxidation of the fuel. 00:25:14.330 --> 00:25:18.350 align:middle line:84% And so the stoichiometric coefficients are minus 1, 00:25:18.350 --> 00:25:21.060 align:middle line:90% minus 2, plus 1, plus 2. 00:25:21.060 --> 00:25:24.080 align:middle line:84% Then you pick from this table the values 00:25:24.080 --> 00:25:27.460 align:middle line:90% of the enthalpies of formation. 00:25:27.460 --> 00:25:30.815 align:middle line:84% So, for example, you look for CH4, which is methane. 00:25:30.815 --> 00:25:31.880 align:middle line:90% It's up here. 00:25:31.880 --> 00:25:35.150 align:middle line:90% It's minus 74.9. 00:25:35.150 --> 00:25:39.650 align:middle line:84% So that's where you get this number. 00:25:39.650 --> 00:25:46.880 align:middle line:84% The elemental species have no value 00:25:46.880 --> 00:25:48.390 align:middle line:90% for the property of formation. 00:25:48.390 --> 00:25:50.210 align:middle line:90% It's 0 for them. 00:25:50.210 --> 00:25:52.450 align:middle line:90% So you can recognize them here. 00:25:52.450 --> 00:25:55.680 align:middle line:84% These are the elemental species-- 00:25:55.680 --> 00:26:02.310 align:middle line:84% hydrogen, nitrogen, molecular oxygen. So you get 0. 00:26:02.310 --> 00:26:04.020 align:middle line:90% Then you have CO2. 00:26:04.020 --> 00:26:12.860 align:middle line:84% So you look for CO2, which is here, minus 393.8. 00:26:12.860 --> 00:26:18.780 align:middle line:84% And then you get water, which is here at the bottom, 242. 00:26:18.780 --> 00:26:22.950 align:middle line:90% Then you can calculate this. 00:26:22.950 --> 00:26:27.160 align:middle line:84% These numbers are in megajoules per kilomole-- 00:26:27.160 --> 00:26:30.180 align:middle line:90% or, say, kilojoules per mole. 00:26:30.180 --> 00:26:36.760 align:middle line:84% And if you divide by the molecular weight for methane, 00:26:36.760 --> 00:26:43.900 align:middle line:84% 16, you get this number, negative 50 megajoules 00:26:43.900 --> 00:26:44.745 align:middle line:90% per kilogram. 00:26:44.745 --> 00:26:47.830 align:middle line:90% 00:26:47.830 --> 00:26:53.100 align:middle line:84% For isooctane, it's a much higher molecular weight. 00:26:53.100 --> 00:27:04.350 align:middle line:84% So mole-wise, it's a much higher enthalpy of reaction. 00:27:04.350 --> 00:27:07.290 align:middle line:84% And that's because you have many more 00:27:07.290 --> 00:27:10.810 align:middle line:90% bonds to break per molecule. 00:27:10.810 --> 00:27:13.520 align:middle line:84% But if you divide by the molecular weight, 00:27:13.520 --> 00:27:16.150 align:middle line:84% you get a number which is of the same-- 00:27:16.150 --> 00:27:20.180 align:middle line:84% very close, same order of magnitude as this one. 00:27:20.180 --> 00:27:24.370 align:middle line:84% And actually, this turns out to be true for all hydrocarbons. 00:27:24.370 --> 00:27:26.120 align:middle line:84% I thought I had it in the next slide. 00:27:26.120 --> 00:27:26.635 align:middle line:90% We don't. 00:27:26.635 --> 00:27:28.650 align:middle line:90% It will come up in a moment. 00:27:28.650 --> 00:27:32.403 align:middle line:90% 00:27:32.403 --> 00:27:32.903 align:middle line:90% OK. 00:27:32.903 --> 00:27:36.074 align:middle line:90% 00:27:36.074 --> 00:27:40.980 align:middle line:84% So negative means that the reaction is exothermic. 00:27:40.980 --> 00:27:46.790 align:middle line:84% And we will see now what the sign of these properties 00:27:46.790 --> 00:27:52.490 align:middle line:84% imply in determining the direction 00:27:52.490 --> 00:27:56.455 align:middle line:84% of spontaneous evolution of the reaction. 00:27:56.455 --> 00:28:00.000 align:middle line:90% 00:28:00.000 --> 00:28:06.620 align:middle line:84% So let's take the steady-state reactor setup. 00:28:06.620 --> 00:28:15.420 align:middle line:84% And let's assume that we want the reaction to occur 00:28:15.420 --> 00:28:18.910 align:middle line:84% at constant temperature and pressure, 00:28:18.910 --> 00:28:24.330 align:middle line:84% meaning that the inlet flows and the outlet flows 00:28:24.330 --> 00:28:29.070 align:middle line:84% are going to be kept at the same temperature and pressure. 00:28:29.070 --> 00:28:35.110 align:middle line:84% And this is going to be done with a heat interaction 00:28:35.110 --> 00:28:38.370 align:middle line:90% at some temperature TQ. 00:28:38.370 --> 00:28:42.350 align:middle line:84% For this moment, for this slide, this temperature 00:28:42.350 --> 00:28:44.450 align:middle line:84% of the reservoir is not necessarily 00:28:44.450 --> 00:28:47.810 align:middle line:84% the same temperature of the inlet 00:28:47.810 --> 00:28:51.090 align:middle line:84% at which you want the reactor to operate. 00:28:51.090 --> 00:28:54.090 align:middle line:90% 00:28:54.090 --> 00:28:59.990 align:middle line:84% We do our energy balance and our entropy balance. 00:28:59.990 --> 00:29:08.230 align:middle line:84% We combine them by eliminating Q dot. 00:29:08.230 --> 00:29:09.920 align:middle line:90% And here's the outcome. 00:29:09.920 --> 00:29:15.370 align:middle line:84% It gives the entropy production by irreversibility multiplied 00:29:15.370 --> 00:29:19.090 align:middle line:84% by TQ, the temperature of the heat source-- 00:29:19.090 --> 00:29:23.240 align:middle line:84% or, say, heat bath or heat reservoir, thermal reservoir, 00:29:23.240 --> 00:29:31.120 align:middle line:84% whatever you want to call it, is equal to minus epsilon dot, 00:29:31.120 --> 00:29:37.010 align:middle line:84% which is the reaction rate, the rate of the reaction coordinate, 00:29:37.010 --> 00:29:42.130 align:middle line:84% times this combination of the enthalpy 00:29:42.130 --> 00:29:46.070 align:middle line:84% and the entropy of reaction, which 00:29:46.070 --> 00:29:49.880 align:middle line:84% looks like the same kind of combination 00:29:49.880 --> 00:29:56.960 align:middle line:84% that we have in available energy, or, actually, exergy. 00:29:56.960 --> 00:30:01.520 align:middle line:84% If you look back into when we did a review of exergies 00:30:01.520 --> 00:30:05.240 align:middle line:84% in bulk flow, the difference in enthalpy 00:30:05.240 --> 00:30:07.790 align:middle line:84% minus the temperature of the reservoir times the difference 00:30:07.790 --> 00:30:11.520 align:middle line:84% in the entropies between inlet and outlet, 00:30:11.520 --> 00:30:14.870 align:middle line:84% that is the exergy of your flows. 00:30:14.870 --> 00:30:20.180 align:middle line:84% So this is the exergy related to-- 00:30:20.180 --> 00:30:23.780 align:middle line:84% the exergy difference between the products 00:30:23.780 --> 00:30:30.055 align:middle line:84% and the reactants with respect to that reservoir at temperature 00:30:30.055 --> 00:30:31.940 align:middle line:90% TQ. 00:30:31.940 --> 00:30:36.370 align:middle line:84% Now, when epsilon dot is positive, 00:30:36.370 --> 00:30:39.440 align:middle line:90% the reaction is forward. 00:30:39.440 --> 00:30:47.560 align:middle line:84% It means that it's proceeding into producing 00:30:47.560 --> 00:30:49.976 align:middle line:90% products, consuming reactants. 00:30:49.976 --> 00:30:54.325 align:middle line:90% 00:30:54.325 --> 00:31:01.010 align:middle line:84% So the fact that, of course, entropy cannot be destroyed-- 00:31:01.010 --> 00:31:05.620 align:middle line:84% so that this has to have always a positive sign 00:31:05.620 --> 00:31:09.170 align:middle line:90% if it has to really occur-- 00:31:09.170 --> 00:31:13.430 align:middle line:84% gives you a criterion to decide whether the reaction will 00:31:13.430 --> 00:31:16.700 align:middle line:84% be forward or backwards at that temperature, 00:31:16.700 --> 00:31:19.280 align:middle line:84% because if you want it forward, you 00:31:19.280 --> 00:31:22.340 align:middle line:90% need epsilon dot to be positive. 00:31:22.340 --> 00:31:26.940 align:middle line:84% So if that's positive, in order for this to be positive, 00:31:26.940 --> 00:31:31.260 align:middle line:84% you need this linear combination to be negative. 00:31:31.260 --> 00:31:34.810 align:middle line:90% 00:31:34.810 --> 00:31:38.330 align:middle line:84% So in this table, you can summarize the possibilities. 00:31:38.330 --> 00:31:43.940 align:middle line:84% So if you want that to be positive, 00:31:43.940 --> 00:31:50.420 align:middle line:84% and you want it to be positive at all combinations, 00:31:50.420 --> 00:31:53.990 align:middle line:84% you could make sure that the enthalpy of reaction is 00:31:53.990 --> 00:31:58.880 align:middle line:84% negative-- so that is an exothermic reaction-- 00:31:58.880 --> 00:32:04.280 align:middle line:84% and also that this term doesn't influence 00:32:04.280 --> 00:32:07.260 align:middle line:84% the sign of the overall linear combination. 00:32:07.260 --> 00:32:13.800 align:middle line:84% So certainly, if the entropy of reaction is positive, 00:32:13.800 --> 00:32:17.780 align:middle line:84% you're safe because this also gives a negative contribution. 00:32:17.780 --> 00:32:24.810 align:middle line:84% You're also safe if it is negative, but not too negative, 00:32:24.810 --> 00:32:27.555 align:middle line:90% because this will-- 00:32:27.555 --> 00:32:30.550 align:middle line:84% with the minus sign in front, if it is negative, 00:32:30.550 --> 00:32:31.750 align:middle line:90% it will be positive. 00:32:31.750 --> 00:32:37.590 align:middle line:84% It eventually could cancel out the negativeness 00:32:37.590 --> 00:32:39.300 align:middle line:90% of the enthalpy. 00:32:39.300 --> 00:32:45.210 align:middle line:84% And so you could reverse the fact, which happens 00:32:45.210 --> 00:32:49.350 align:middle line:90% if this is not satisfied-- 00:32:49.350 --> 00:32:53.150 align:middle line:84% or, actually, if this is satisfied. 00:32:53.150 --> 00:32:56.390 align:middle line:84% OK, so, you see, these properties of reaction 00:32:56.390 --> 00:33:00.890 align:middle line:84% give you the key to understand whether the reaction will 00:33:00.890 --> 00:33:07.080 align:middle line:84% go in the direction you want, because, typically, you 00:33:07.080 --> 00:33:09.990 align:middle line:84% bother constructing reactors because you 00:33:09.990 --> 00:33:16.110 align:middle line:84% want to consume some products in order to-- some reactants 00:33:16.110 --> 00:33:18.110 align:middle line:84% in order to produce some products that you want. 00:33:18.110 --> 00:33:23.770 align:middle line:90% 00:33:23.770 --> 00:33:35.910 align:middle line:84% Now, if we repeat the same, but with the same temperature as-- 00:33:35.910 --> 00:33:39.330 align:middle line:84% so the heat bath or the thermal reservoir 00:33:39.330 --> 00:33:42.970 align:middle line:84% has the same temperature at which the reaction occurs. 00:33:42.970 --> 00:33:46.510 align:middle line:84% So when you have your reaction, for example, 00:33:46.510 --> 00:33:52.970 align:middle line:84% placed in a calorimeter or in a thermostat-- 00:33:52.970 --> 00:33:59.400 align:middle line:84% so it's a thermostatted reaction at constant pressure, again-- 00:33:59.400 --> 00:34:03.200 align:middle line:84% and this pressure is the atmospheric pressure-- 00:34:03.200 --> 00:34:06.630 align:middle line:84% again, your balance equations is exactly like before, 00:34:06.630 --> 00:34:09.420 align:middle line:84% only the TQ now is becoming equal to T. 00:34:09.420 --> 00:34:12.750 align:middle line:90% And then this makes-- 00:34:12.750 --> 00:34:19.179 align:middle line:84% this linear combination is the Gibbs free energy of reaction 00:34:19.179 --> 00:34:24.050 align:middle line:84% because h minus Ts is the Gibbs free energy. 00:34:24.050 --> 00:34:28.239 align:middle line:84% So this is why they typically say 00:34:28.239 --> 00:34:35.770 align:middle line:84% that the reaction occurs in the direction of negative Gibbs 00:34:35.770 --> 00:34:37.210 align:middle line:90% free energy. 00:34:37.210 --> 00:34:40.360 align:middle line:84% Yes, you need the Gibbs free energy 00:34:40.360 --> 00:34:43.389 align:middle line:84% of reaction to be negative in order to the epsilon 00:34:43.389 --> 00:34:45.306 align:middle line:90% dot to be positive. 00:34:45.306 --> 00:34:50.480 align:middle line:90% 00:34:50.480 --> 00:34:55.699 align:middle line:84% However, this depends on temperature 00:34:55.699 --> 00:34:58.520 align:middle line:84% because the Gibbs free energy is a function of temperature. 00:34:58.520 --> 00:35:02.010 align:middle line:90% 00:35:02.010 --> 00:35:05.370 align:middle line:84% And actually, in the next slide, we 00:35:05.370 --> 00:35:08.190 align:middle line:84% will prove this relation here, that the way 00:35:08.190 --> 00:35:12.330 align:middle line:84% the Gibbs free energy changes in temperature is this way, 00:35:12.330 --> 00:35:16.870 align:middle line:84% is related to the enthalpy of the reaction. 00:35:16.870 --> 00:35:21.690 align:middle line:84% So the rate of change of the Gibbs free energy divided 00:35:21.690 --> 00:35:29.005 align:middle line:84% by RT with respect to temperature 00:35:29.005 --> 00:35:32.630 align:middle line:84% is equal to the enthalpy of reaction divided by RT squared. 00:35:32.630 --> 00:35:36.500 align:middle line:90% 00:35:36.500 --> 00:35:41.150 align:middle line:90% This is that relation. 00:35:41.150 --> 00:35:42.870 align:middle line:90% And that is the proof. 00:35:42.870 --> 00:35:47.700 align:middle line:84% The proof comes from an old friend of ours, 00:35:47.700 --> 00:35:49.970 align:middle line:84% which is this relation that gives 00:35:49.970 --> 00:35:52.430 align:middle line:84% the enthalpy in terms of the chemical potential 00:35:52.430 --> 00:35:54.180 align:middle line:90% for the pure substance. 00:35:54.180 --> 00:35:56.930 align:middle line:84% The chemical potential is the Gibbs free energy. 00:35:56.930 --> 00:36:00.360 align:middle line:84% If you go back in your notes, you find this relation. 00:36:00.360 --> 00:36:08.190 align:middle line:84% And if you put it inside here, out comes this thing. 00:36:08.190 --> 00:36:16.860 align:middle line:90% And this plot has here-- 00:36:16.860 --> 00:36:20.700 align:middle line:84% it plots versus-- this is not temperature. 00:36:20.700 --> 00:36:23.240 align:middle line:84% It's 1 over T. It's a 1 over T scale. 00:36:23.240 --> 00:36:27.800 align:middle line:84% So actually, the scale is here is 10 to the 4 over T. 00:36:27.800 --> 00:36:31.304 align:middle line:84% So it's a 1 over T scale, which is precisely 00:36:31.304 --> 00:36:36.000 align:middle line:84% the thing with respect to which we're doing the derivative. 00:36:36.000 --> 00:36:42.540 align:middle line:84% And this axis instead shows delta g over RT. 00:36:42.540 --> 00:36:47.790 align:middle line:90% 00:36:47.790 --> 00:36:52.930 align:middle line:84% Yeah, there is the logarithm of 10 because this is done on-- 00:36:52.930 --> 00:36:55.822 align:middle line:90% 00:36:55.822 --> 00:36:59.040 align:middle line:90% this is the log of-- 00:36:59.040 --> 00:37:02.920 align:middle line:84% the Napierian log instead of log of 10. 00:37:02.920 --> 00:37:10.320 align:middle line:84% But apart from this factor, this shows the dependence 00:37:10.320 --> 00:37:17.310 align:middle line:84% of delta g over RT versus 1 over T. It's almost linear. 00:37:17.310 --> 00:37:26.580 align:middle line:84% And these slopes, the slopes are represented by the negative 00:37:26.580 --> 00:37:36.045 align:middle line:90% of the enthalpy of reaction. 00:37:36.045 --> 00:37:38.910 align:middle line:90% 00:37:38.910 --> 00:37:41.600 align:middle line:84% So for the exothermic reaction, this is positive. 00:37:41.600 --> 00:37:46.790 align:middle line:84% So you see these ones that go up are the exothermic reaction. 00:37:46.790 --> 00:37:53.090 align:middle line:84% For example, CO plus 1/2 O2 is CO2, or water is here, 00:37:53.090 --> 00:37:59.140 align:middle line:84% H2 plus 1/2 O2 is H2O, and so on. 00:37:59.140 --> 00:38:01.600 align:middle line:90% And you see that-- 00:38:01.600 --> 00:38:07.980 align:middle line:84% so if this is positive, because the relation 00:38:07.980 --> 00:38:10.080 align:middle line:84% that we proved in the previous slide 00:38:10.080 --> 00:38:13.530 align:middle line:84% was that the entropy produced by reversibility 00:38:13.530 --> 00:38:21.240 align:middle line:90% divided by the reaction rate-- 00:38:21.240 --> 00:38:24.010 align:middle line:84% so if this is positive, that is also positive. 00:38:24.010 --> 00:38:25.990 align:middle line:84% You need this number to be positive. 00:38:25.990 --> 00:38:36.670 align:middle line:84% So from here up, you have all product-favored reactions. 00:38:36.670 --> 00:38:44.140 align:middle line:84% But for each reaction, for even that of formation of water, 00:38:44.140 --> 00:38:48.250 align:middle line:84% if you go sufficiently high in temperature-- 00:38:48.250 --> 00:38:50.200 align:middle line:90% and as I go down-- 00:38:50.200 --> 00:38:53.270 align:middle line:84% high in temperature means 1 over T goes down. 00:38:53.270 --> 00:38:56.200 align:middle line:84% So if I go down here, eventually, I 00:38:56.200 --> 00:38:58.890 align:middle line:90% will find a temperature-- 00:38:58.890 --> 00:39:08.010 align:middle line:84% for water is of the order of, what, maybe 5,000 K. 00:39:08.010 --> 00:39:10.350 align:middle line:90% It becomes negative. 00:39:10.350 --> 00:39:14.590 align:middle line:90% So it becomes reactant-favored. 00:39:14.590 --> 00:39:18.520 align:middle line:84% So water dissociates instead of-- 00:39:18.520 --> 00:39:22.005 align:middle line:84% And the same is for true for CO2. 00:39:22.005 --> 00:39:24.560 align:middle line:90% 00:39:24.560 --> 00:39:28.920 align:middle line:84% CO2 also dissociates at higher temperatures-- 00:39:28.920 --> 00:39:30.180 align:middle line:90% high temperatures. 00:39:30.180 --> 00:39:34.100 align:middle line:90% 00:39:34.100 --> 00:39:39.830 align:middle line:84% This, of course, is important because in combustion, it's 00:39:39.830 --> 00:39:45.770 align:middle line:84% true that the oxidation of a fuel produces CO2 and water. 00:39:45.770 --> 00:39:49.200 align:middle line:84% But if you are very high in temperature, 00:39:49.200 --> 00:39:52.610 align:middle line:84% like in an adiabatic flame setup like the one 00:39:52.610 --> 00:39:56.720 align:middle line:84% we will consider in a moment, the temperatures 00:39:56.720 --> 00:40:02.590 align:middle line:90% are so high that the reaction-- 00:40:02.590 --> 00:40:07.570 align:middle line:84% that CO2 and even water, a little part dissociates. 00:40:07.570 --> 00:40:10.530 align:middle line:90% 00:40:10.530 --> 00:40:12.630 align:middle line:84% And of course, if it dissociates, 00:40:12.630 --> 00:40:20.660 align:middle line:84% it consumes some energy in order to break those bonds that 00:40:20.660 --> 00:40:22.255 align:middle line:90% are particularly stable. 00:40:22.255 --> 00:40:27.670 align:middle line:90% 00:40:27.670 --> 00:40:32.200 align:middle line:84% Another setup that is important is precisely 00:40:32.200 --> 00:40:36.890 align:middle line:84% the adiabatic setup in which we have 00:40:36.890 --> 00:40:41.000 align:middle line:90% a reactor which is insulated-- 00:40:41.000 --> 00:40:45.420 align:middle line:90% so no heat, no work interaction. 00:40:45.420 --> 00:40:51.890 align:middle line:84% Just the reaction itself makes things happen. 00:40:51.890 --> 00:40:57.980 align:middle line:84% And this time, of course, you cannot have the same temperature 00:40:57.980 --> 00:40:59.390 align:middle line:90% of the outlet. 00:40:59.390 --> 00:41:01.580 align:middle line:84% So in the outlet, you have a different temperature 00:41:01.580 --> 00:41:02.310 align:middle line:90% of the products. 00:41:02.310 --> 00:41:07.170 align:middle line:84% And we call it T ad for adiabatic. 00:41:07.170 --> 00:41:14.430 align:middle line:84% Of course, you have to write an energy balance. 00:41:14.430 --> 00:41:16.710 align:middle line:84% And the energy balance this time will be simply 00:41:16.710 --> 00:41:23.140 align:middle line:84% that the net of the enthalpy coming in with this bulk flow 00:41:23.140 --> 00:41:28.210 align:middle line:84% must be equal to the enthalpy going out with this outlet bulk 00:41:28.210 --> 00:41:30.280 align:middle line:90% flow. 00:41:30.280 --> 00:41:32.260 align:middle line:90% So these must be equal, too. 00:41:32.260 --> 00:41:37.110 align:middle line:84% So this gives you one relation that 00:41:37.110 --> 00:41:40.990 align:middle line:84% allows you to solve for the adiabatic temperature-- 00:41:40.990 --> 00:41:44.160 align:middle line:90% adiabatic product temperature. 00:41:44.160 --> 00:41:48.665 align:middle line:84% And once you solve for that, you go into the entropy balance. 00:41:48.665 --> 00:41:51.290 align:middle line:90% 00:41:51.290 --> 00:41:55.536 align:middle line:84% And that allows you to compute this difference. 00:41:55.536 --> 00:41:59.360 align:middle line:90% 00:41:59.360 --> 00:42:03.470 align:middle line:84% And you can write this S dot irreversible-- 00:42:03.470 --> 00:42:09.290 align:middle line:90% one way is to write it this way. 00:42:09.290 --> 00:42:13.190 align:middle line:90% 00:42:13.190 --> 00:42:14.185 align:middle line:90% Yes, here it is. 00:42:14.185 --> 00:42:17.690 align:middle line:90% 00:42:17.690 --> 00:42:23.450 align:middle line:90% Yeah, you essentially add a 0. 00:42:23.450 --> 00:42:25.700 align:middle line:90% You add the entropy balance-- 00:42:25.700 --> 00:42:27.810 align:middle line:84% i'm sorry, the energy balance, which is 0. 00:42:27.810 --> 00:42:28.560 align:middle line:90% So you can add it. 00:42:28.560 --> 00:42:30.160 align:middle line:84% It doesn't change the entropy balance. 00:42:30.160 --> 00:42:34.260 align:middle line:90% 00:42:34.260 --> 00:42:38.160 align:middle line:84% And you multiply the entropy balance by T 00:42:38.160 --> 00:42:43.760 align:middle line:84% so that it becomes dimensionally an energy. 00:42:43.760 --> 00:42:50.010 align:middle line:84% And then what you get, you can split the outcome as two parts. 00:42:50.010 --> 00:42:53.330 align:middle line:84% One is our familiar Gibbs free energy 00:42:53.330 --> 00:42:56.850 align:middle line:84% of reaction term multiplied by the epsilon dot. 00:42:56.850 --> 00:42:59.520 align:middle line:90% 00:42:59.520 --> 00:43:03.720 align:middle line:84% And these other terms are essentially-- 00:43:03.720 --> 00:43:06.090 align:middle line:90% as they're written here-- 00:43:06.090 --> 00:43:11.550 align:middle line:84% are essentially the difference in the exergy of your products 00:43:11.550 --> 00:43:16.640 align:middle line:84% at this temperature from the reactants. 00:43:16.640 --> 00:43:18.230 align:middle line:90% I'm sorry. 00:43:18.230 --> 00:43:20.870 align:middle line:90% I take it back. 00:43:20.870 --> 00:43:24.350 align:middle line:84% It's the difference between the exergy 00:43:24.350 --> 00:43:30.500 align:middle line:84% that the products have at this temperature-- 00:43:30.500 --> 00:43:31.620 align:middle line:90% you see, this is still P-- 00:43:31.620 --> 00:43:37.870 align:middle line:90% 00:43:37.870 --> 00:43:42.520 align:middle line:84% and that they have a temperature of the reactants. 00:43:42.520 --> 00:43:44.710 align:middle line:84% So it's the difference, between Tad and T, 00:43:44.710 --> 00:43:50.430 align:middle line:90% of the exergy of your products. 00:43:50.430 --> 00:43:56.730 align:middle line:90% 00:43:56.730 --> 00:44:01.990 align:middle line:84% And so you see if you want the reaction to occur 00:44:01.990 --> 00:44:08.930 align:middle line:84% in that direction, you need delta g to be negative. 00:44:08.930 --> 00:44:14.043 align:middle line:84% But it must be negative enough to overcome this exergy here. 00:44:14.043 --> 00:44:18.210 align:middle line:90% 00:44:18.210 --> 00:44:23.200 align:middle line:84% And this exergy is what will be-- 00:44:23.200 --> 00:44:27.290 align:middle line:84% you could deploy in applications. 00:44:27.290 --> 00:44:31.250 align:middle line:84% For example, if this is high temperature, 00:44:31.250 --> 00:44:38.560 align:middle line:84% you could use it to run the heat exchanger of a thermal cycle 00:44:38.560 --> 00:44:40.460 align:middle line:84% to produce electricity and so on. 00:44:40.460 --> 00:44:43.610 align:middle line:90% 00:44:43.610 --> 00:44:46.245 align:middle line:84% We will do the calculation for fuels. 00:44:46.245 --> 00:44:52.410 align:middle line:90% 00:44:52.410 --> 00:44:59.360 align:middle line:84% Again, another setup is the exergy kind of discussion. 00:44:59.360 --> 00:45:06.920 align:middle line:84% Here we use TQ as a different temperature, 00:45:06.920 --> 00:45:12.350 align:middle line:84% as the temperature at which you want your reaction to occur. 00:45:12.350 --> 00:45:14.210 align:middle line:90% And you assume also that this-- 00:45:14.210 --> 00:45:16.860 align:middle line:84% I shouldn't really call it a reactor, 00:45:16.860 --> 00:45:18.590 align:middle line:84% but it's a reactor that is going to be 00:45:18.590 --> 00:45:23.550 align:middle line:84% part of some more complex system that contains also machinery, 00:45:23.550 --> 00:45:25.580 align:middle line:84% thermal cycles, whatever you want, 00:45:25.580 --> 00:45:27.870 align:middle line:90% in order to produce some work. 00:45:27.870 --> 00:45:29.960 align:middle line:84% So this is like a box, which could 00:45:29.960 --> 00:45:36.620 align:middle line:84% be a power plant that takes in reactants, 00:45:36.620 --> 00:45:38.470 align:middle line:90% produces your products. 00:45:38.470 --> 00:45:41.060 align:middle line:84% It could be a materials processing plant 00:45:41.060 --> 00:45:49.090 align:middle line:84% that takes in raw materials and produces your material, what 00:45:49.090 --> 00:45:55.426 align:middle line:90% you desire to manufacture. 00:45:55.426 --> 00:45:57.070 align:middle line:90% Now, in some cases-- 00:45:57.070 --> 00:46:00.060 align:middle line:84% you write, as this is ?] usual, the energy and entropy balance, 00:46:00.060 --> 00:46:03.830 align:middle line:90% and you compute now the work. 00:46:03.830 --> 00:46:09.130 align:middle line:84% So the work, in some cases, depending 00:46:09.130 --> 00:46:13.540 align:middle line:84% on the conditions and the values of these properties of reaction, 00:46:13.540 --> 00:46:16.650 align:middle line:90% sometimes, you can do work. 00:46:16.650 --> 00:46:20.390 align:middle line:84% And the maximum work you can do is given by this. 00:46:20.390 --> 00:46:24.020 align:middle line:84% And if you have entropy produced by irreversibility, 00:46:24.020 --> 00:46:26.870 align:middle line:84% that will detract from the maximum you can do. 00:46:26.870 --> 00:46:30.260 align:middle line:84% So your actual work will be less than the maximum 00:46:30.260 --> 00:46:33.080 align:middle line:84% because there is a minus sign here. 00:46:33.080 --> 00:46:37.880 align:middle line:84% Sometimes, you cannot extract work, 00:46:37.880 --> 00:46:42.910 align:middle line:84% but you actually need to spend work in order to make that 00:46:42.910 --> 00:46:45.310 align:middle line:90% occur. 00:46:45.310 --> 00:46:48.910 align:middle line:84% This is the minimum work you need to spend. 00:46:48.910 --> 00:46:52.800 align:middle line:90% 00:46:52.800 --> 00:46:56.160 align:middle line:84% And if you have irreversibility, that will detract. 00:46:56.160 --> 00:46:59.280 align:middle line:84% And now the sign becomes plus, and it 00:46:59.280 --> 00:47:02.320 align:middle line:90% will add up to the minimum. 00:47:02.320 --> 00:47:06.210 align:middle line:84% So in practice you'll spend more work 00:47:06.210 --> 00:47:08.080 align:middle line:90% than the minimum requirement. 00:47:08.080 --> 00:47:18.568 align:middle line:90% 00:47:18.568 --> 00:47:25.940 align:middle line:84% So let's do something more specific about the oxidation 00:47:25.940 --> 00:47:35.630 align:middle line:84% of hydrocarbons in this very simplified view 00:47:35.630 --> 00:47:40.400 align:middle line:84% in which we consider the overall basic reaction 00:47:40.400 --> 00:47:43.930 align:middle line:84% mechanism of complete oxidation of a hydrocarbon 00:47:43.930 --> 00:47:51.050 align:middle line:84% as if there were only that one, which is not true. 00:47:51.050 --> 00:47:54.560 align:middle line:84% But in this simplified view, you already 00:47:54.560 --> 00:47:59.346 align:middle line:84% can do a lot of interesting considerations and calculations. 00:47:59.346 --> 00:48:01.850 align:middle line:90% 00:48:01.850 --> 00:48:07.310 align:middle line:84% First of all, you notice that the mechanism 00:48:07.310 --> 00:48:11.150 align:middle line:90% involves a fuel, a hydrocarbon. 00:48:11.150 --> 00:48:17.210 align:middle line:84% k is the number of carbon nuclei in the molecule, 00:48:17.210 --> 00:48:24.520 align:middle line:84% and l is the number of hydrogen nuclei in the molecule. 00:48:24.520 --> 00:48:32.220 align:middle line:84% You need oxygen. And the two main products are CO2 and water. 00:48:32.220 --> 00:48:38.820 align:middle line:84% And that's the balanced stoichiometric coefficients. 00:48:38.820 --> 00:48:43.660 align:middle line:84% Typically, when you burn a fuel like this, 00:48:43.660 --> 00:48:46.050 align:middle line:84% you don't do it in pure oxygen, although you 00:48:46.050 --> 00:48:48.150 align:middle line:90% could do oxy-combustion. 00:48:48.150 --> 00:48:50.060 align:middle line:90% But you do it in-- 00:48:50.060 --> 00:48:53.550 align:middle line:90% you use air as an oxidant. 00:48:53.550 --> 00:49:01.080 align:middle line:84% And so oxygen is going to be only one 00:49:01.080 --> 00:49:07.300 align:middle line:84% of the components in the mixture of your oxidant. 00:49:07.300 --> 00:49:10.433 align:middle line:84% The mole fraction of oxygen in air is about 21%. 00:49:10.433 --> 00:49:13.690 align:middle line:90% 00:49:13.690 --> 00:49:21.460 align:middle line:84% So if you want 1 mole of oxygen to be available for your fuel, 00:49:21.460 --> 00:49:26.710 align:middle line:90% you need this divided by-- 00:49:26.710 --> 00:49:32.450 align:middle line:84% so 1 over this is, like, 4.77 moles of air. 00:49:32.450 --> 00:49:37.550 align:middle line:84% And therefore, the stoichiometric amount 00:49:37.550 --> 00:49:40.820 align:middle line:84% that you need, since for 1 mole of the fuel, 00:49:40.820 --> 00:49:47.030 align:middle line:84% you need k plus l over 4 molecules of oxygen, that 00:49:47.030 --> 00:49:53.000 align:middle line:90% means 4.77 times this of air. 00:49:53.000 --> 00:49:59.660 align:middle line:84% So this gives the stoichiometric air/fuel ratio. 00:49:59.660 --> 00:50:03.260 align:middle line:90% The s stands for stoichiometric. 00:50:03.260 --> 00:50:10.360 align:middle line:84% If you are more interested in weight fractions, 00:50:10.360 --> 00:50:15.180 align:middle line:84% this is the air-to-fuel ratio mass-wise. 00:50:15.180 --> 00:50:18.000 align:middle line:84% You have to multiply by the molecular weight of air 00:50:18.000 --> 00:50:20.435 align:middle line:84% and the divide by the molecular weight of the fuel. 00:50:20.435 --> 00:50:27.760 align:middle line:90% 00:50:27.760 --> 00:50:33.470 align:middle line:84% Also, if you need to completely burn your fuel, 00:50:33.470 --> 00:50:36.640 align:middle line:84% you want to make sure that if you have a combustion 00:50:36.640 --> 00:50:49.760 align:middle line:84% application, you don't leave unburned hydrocarbons around. 00:50:49.760 --> 00:50:53.630 align:middle line:84% So, for example, if you think of a combustion chamber 00:50:53.630 --> 00:50:56.750 align:middle line:84% in the internal combustion engine, 00:50:56.750 --> 00:50:59.450 align:middle line:90% yeah, it's loaded with fuel. 00:50:59.450 --> 00:51:02.720 align:middle line:84% And you have the molecules of fuel spread around 00:51:02.720 --> 00:51:04.260 align:middle line:90% in the combustion chamber. 00:51:04.260 --> 00:51:09.680 align:middle line:84% And if your air is just the stoichiometric amount, 00:51:09.680 --> 00:51:12.830 align:middle line:84% chances are that the last molecule of fuel 00:51:12.830 --> 00:51:15.200 align:middle line:84% will be on one side and the last molecule of oxygen 00:51:15.200 --> 00:51:18.180 align:middle line:84% on the other side, and they will not encounter. 00:51:18.180 --> 00:51:21.440 align:middle line:90% So you miss that. 00:51:21.440 --> 00:51:24.060 align:middle line:84% And so you're going to have some unburned fuel. 00:51:24.060 --> 00:51:29.160 align:middle line:84% So for this reason, you provide extra air, extra oxygen. 00:51:29.160 --> 00:51:31.730 align:middle line:90% You provide some excess air. 00:51:31.730 --> 00:51:36.050 align:middle line:84% So the actual fuel/air ratio that you use 00:51:36.050 --> 00:51:39.590 align:middle line:84% is more than the stoichiometric amount. 00:51:39.590 --> 00:51:41.420 align:middle line:84% And the ratio of what you actually 00:51:41.420 --> 00:51:45.610 align:middle line:84% put over the stoichiometric is called equivalence ratio. 00:51:45.610 --> 00:51:56.140 align:middle line:90% 00:51:56.140 --> 00:51:59.680 align:middle line:84% We already talked about the heating values of fuels 00:51:59.680 --> 00:52:01.420 align:middle line:90% being in the range-- 00:52:01.420 --> 00:52:04.670 align:middle line:90% more or less in the range. 00:52:04.670 --> 00:52:10.770 align:middle line:84% And now, hopefully-- OK, here is the list 00:52:10.770 --> 00:52:18.470 align:middle line:84% for the various hydrocarbons with molecular weights that 00:52:18.470 --> 00:52:29.990 align:middle line:84% range from 2 of hydrogen to 16 of methane, 114 of isooctane, 00:52:29.990 --> 00:52:33.050 align:middle line:90% and up. 00:52:33.050 --> 00:52:37.470 align:middle line:84% And you see that the enthalpy of reaction, 00:52:37.470 --> 00:52:40.080 align:middle line:84% except for hydrogen, which is much higher, 00:52:40.080 --> 00:52:43.880 align:middle line:84% or for carbon, which is lower, in terms of megajoules 00:52:43.880 --> 00:52:48.480 align:middle line:84% per kilogram, the others are all in the range between 40-- 00:52:48.480 --> 00:52:53.040 align:middle line:84% this one-- but between 44 and 48, more or less-- 00:52:53.040 --> 00:52:53.905 align:middle line:90% 50 for methane. 00:52:53.905 --> 00:52:56.680 align:middle line:90% 00:52:56.680 --> 00:53:00.960 align:middle line:84% And also, you could compute the entropies. 00:53:00.960 --> 00:53:06.270 align:middle line:84% Notice that the entropy of reaction-- 00:53:06.270 --> 00:53:11.070 align:middle line:84% of combustion does not necessarily 00:53:11.070 --> 00:53:13.890 align:middle line:90% have to have a positive sign. 00:53:13.890 --> 00:53:15.790 align:middle line:90% For some fuels, it's negative. 00:53:15.790 --> 00:53:18.930 align:middle line:90% 00:53:18.930 --> 00:53:20.880 align:middle line:84% What's important for the application 00:53:20.880 --> 00:53:23.740 align:middle line:84% that you have already used in your homework 00:53:23.740 --> 00:53:29.480 align:middle line:84% is that the difference between the enthalpy and the Gibbs 00:53:29.480 --> 00:53:36.580 align:middle line:84% free energy is small, percentage-wise 00:53:36.580 --> 00:53:40.630 align:middle line:84% in terms of the Gibbs free energy, 00:53:40.630 --> 00:53:45.090 align:middle line:84% which is related to the exergy of the fuel. 00:53:45.090 --> 00:53:48.080 align:middle line:90% 00:53:48.080 --> 00:53:54.527 align:middle line:84% You see that this ratio is between plus and minus 2.4%. 00:53:54.527 --> 00:53:57.560 align:middle line:90% 00:53:57.560 --> 00:54:02.720 align:middle line:84% So this allows you for simple calculations 00:54:02.720 --> 00:54:07.318 align:middle line:84% to neglect the difference and say that the exergy of the fuel, 00:54:07.318 --> 00:54:09.860 align:middle line:84% which is the Gibbs free energy-- minus the Gibbs free energy, 00:54:09.860 --> 00:54:16.460 align:middle line:84% is almost equal to the lower heating value of the fuel, which 00:54:16.460 --> 00:54:17.320 align:middle line:90% is the enthalpy. 00:54:17.320 --> 00:54:20.030 align:middle line:90% 00:54:20.030 --> 00:54:26.020 align:middle line:84% This other table provides more specific values 00:54:26.020 --> 00:54:36.030 align:middle line:84% for heating values, and also the air/fuel stoichiometric ratios 00:54:36.030 --> 00:54:46.250 align:middle line:84% for some practical fuels, and introduces to the idea 00:54:46.250 --> 00:54:50.300 align:middle line:90% that heating values-- 00:54:50.300 --> 00:54:56.790 align:middle line:84% as you know, if you buy a burner for your house, 00:54:56.790 --> 00:55:01.700 align:middle line:84% you have to deal with the salesperson that 00:55:01.700 --> 00:55:04.276 align:middle line:90% gives you efficiencies. 00:55:04.276 --> 00:55:06.890 align:middle line:90% 00:55:06.890 --> 00:55:10.010 align:middle line:84% And sometimes-- most of the times, the efficiencies 00:55:10.010 --> 00:55:14.470 align:middle line:84% are based on the lower heating value. 00:55:14.470 --> 00:55:21.040 align:middle line:84% But nowadays, they're selling a lot of condensing burners 00:55:21.040 --> 00:55:28.930 align:middle line:84% where they exploit also the heat of condensation 00:55:28.930 --> 00:55:33.600 align:middle line:84% of the water that is in the products of combustion. 00:55:33.600 --> 00:55:35.340 align:middle line:84% And therefore, the proper reference 00:55:35.340 --> 00:55:39.810 align:middle line:84% for computing efficiencies there would be the higher heating 00:55:39.810 --> 00:55:40.680 align:middle line:90% value. 00:55:40.680 --> 00:55:42.450 align:middle line:84% Sometimes, they play with this difference 00:55:42.450 --> 00:55:44.650 align:middle line:84% because that difference can be large. 00:55:44.650 --> 00:55:50.810 align:middle line:84% For example, for natural gas, it's about 10%, 00:55:50.810 --> 00:55:55.880 align:middle line:84% which is not an unimportant amount. 00:55:55.880 --> 00:56:04.290 align:middle line:84% So where does that difference come from? 00:56:04.290 --> 00:56:06.960 align:middle line:84% So we have to distinguish between lower and higher heating 00:56:06.960 --> 00:56:07.910 align:middle line:90% values. 00:56:07.910 --> 00:56:14.910 align:middle line:84% And we have to therefore consider how much of the water 00:56:14.910 --> 00:56:23.776 align:middle line:84% that is present in the products of combustion condenses. 00:56:23.776 --> 00:56:33.480 align:middle line:90% 00:56:33.480 --> 00:56:42.080 align:middle line:84% Now, the setup here is the setup of Raoult's law. 00:56:42.080 --> 00:56:44.570 align:middle line:90% We are at pressure p. 00:56:44.570 --> 00:56:47.090 align:middle line:90% 00:56:47.090 --> 00:56:53.150 align:middle line:90% We have water vapor plus-- 00:56:53.150 --> 00:56:57.350 align:middle line:90% let's call it 2-- 00:56:57.350 --> 00:56:59.810 align:middle line:84% the other products of combustion. 00:56:59.810 --> 00:57:03.320 align:middle line:84% And water vapor is the one that can condense-- 00:57:03.320 --> 00:57:13.270 align:middle line:84% that condenses, at least part of it, at a given temperature. 00:57:13.270 --> 00:57:18.410 align:middle line:84% And, of course, we have dealt with this problem, 00:57:18.410 --> 00:57:24.580 align:middle line:84% and we proved the Raoult's law, that the Raoult law tells 00:57:24.580 --> 00:57:30.820 align:middle line:84% you what is the amount of water vapor 00:57:30.820 --> 00:57:35.590 align:middle line:84% that is in the gaseous phase when there is equilibrium 00:57:35.590 --> 00:57:38.330 align:middle line:90% with the liquid phase. 00:57:38.330 --> 00:57:44.640 align:middle line:84% And there cannot be more water vapor here. 00:57:44.640 --> 00:57:48.530 align:middle line:84% So that amount given by the Raoult's law 00:57:48.530 --> 00:57:58.470 align:middle line:84% is the maximum that can be dissolved in the gaseous phase. 00:57:58.470 --> 00:58:02.350 align:middle line:84% If you have instead less than that amount, 00:58:02.350 --> 00:58:03.940 align:middle line:90% then there is no condensation. 00:58:03.940 --> 00:58:09.370 align:middle line:90% 00:58:09.370 --> 00:58:12.660 align:middle line:90% It's like for today. 00:58:12.660 --> 00:58:14.800 align:middle line:90% Today, we have rain. 00:58:14.800 --> 00:58:19.480 align:middle line:84% So that means that moist air has more water 00:58:19.480 --> 00:58:23.270 align:middle line:90% than it can stay there. 00:58:23.270 --> 00:58:26.270 align:middle line:84% And therefore, it drops, condensates, 00:58:26.270 --> 00:58:28.870 align:middle line:90% and then we get rain. 00:58:28.870 --> 00:58:34.510 align:middle line:84% On a drier day, there is less water vapor, 00:58:34.510 --> 00:58:38.780 align:middle line:84% and it's below the threshold for condensation. 00:58:38.780 --> 00:58:43.030 align:middle line:84% So this is all in this expression here. 00:58:43.030 --> 00:58:47.660 align:middle line:84% But, of course, it's Raoult's law with the mole fraction 00:58:47.660 --> 00:58:51.590 align:middle line:84% in the liquid phase equal to 1, because we 00:58:51.590 --> 00:58:53.915 align:middle line:84% assume that this liquid phase is just pure water. 00:58:53.915 --> 00:58:56.290 align:middle line:84% There's no salt dissolved or other things. 00:58:56.290 --> 00:58:59.990 align:middle line:90% 00:58:59.990 --> 00:59:03.150 align:middle line:84% All right, so Raoult's law gives us the maximum. 00:59:03.150 --> 00:59:07.310 align:middle line:84% So the actual amount is called the relative-- 00:59:07.310 --> 00:59:09.530 align:middle line:84% the actual amount divided by the maximum 00:59:09.530 --> 00:59:14.680 align:middle line:84% is called the relative humidity, relative to the maximum 00:59:14.680 --> 00:59:17.690 align:middle line:90% that you can dissolve. 00:59:17.690 --> 00:59:21.890 align:middle line:84% And the saturation pressure is that of pure water. 00:59:21.890 --> 00:59:25.520 align:middle line:84% And at standard temperature, 25 C, 00:59:25.520 --> 00:59:28.970 align:middle line:84% that pressure is about 3.17 atmospheres 00:59:28.970 --> 00:59:34.320 align:middle line:84% (slide was wrong, I meant kiloPascals). 00:59:34.320 --> 00:59:40.760 align:middle line:84% So you see that by taking the pressure down here-- 00:59:40.760 --> 00:59:41.690 align:middle line:90% so you can-- 00:59:41.690 --> 00:59:44.470 align:middle line:84% this is about 3 atmospheres (read kiloPascals) 00:59:44.470 --> 00:59:49.010 align:middle line:84% divided by 101, which is atmospheric pressure. 00:59:49.010 --> 00:59:51.140 align:middle line:84% This ratio is equal to the maximum. 00:59:51.140 --> 00:59:57.850 align:middle line:84% So the maximum-- instead of y1s, this should be y1gmax-- 00:59:57.850 --> 01:00:00.243 align:middle line:90% is 3%. 01:00:00.243 --> 01:00:06.780 align:middle line:90% 01:00:06.780 --> 01:00:12.390 align:middle line:84% So out of 100 moles of stuff in the products of combustion, 01:00:12.390 --> 01:00:16.650 align:middle line:84% at most 3 can be water molecules. 01:00:16.650 --> 01:00:20.280 align:middle line:84% If you have more, your combustion reaction 01:00:20.280 --> 01:00:27.470 align:middle line:84% produces more water, then the extra will condense. 01:00:27.470 --> 01:00:31.700 align:middle line:84% So if you do the inventory for water, 01:00:31.700 --> 01:00:37.490 align:middle line:84% this is the amount produced by your reaction. 01:00:37.490 --> 01:00:40.580 align:middle line:84% So in the products, that's the amount. 01:00:40.580 --> 01:00:43.410 align:middle line:84% Some of it remains in the gaseous phase. 01:00:43.410 --> 01:00:46.350 align:middle line:84% Some of it goes into the liquid phase. 01:00:46.350 --> 01:00:49.430 align:middle line:90% It condenses. 01:00:49.430 --> 01:00:54.600 align:middle line:84% And the stoichiometry tells you that per unit amount of fuel, 01:00:54.600 --> 01:00:58.060 align:middle line:84% the amount of water produced is l over 2. 01:00:58.060 --> 01:01:04.420 align:middle line:84% So this l over 2 will be split into partly condensed and partly 01:01:04.420 --> 01:01:05.295 align:middle line:90% water vapor. 01:01:05.295 --> 01:01:13.060 align:middle line:90% 01:01:13.060 --> 01:01:17.910 align:middle line:84% The mole fraction in the water vapor 01:01:17.910 --> 01:01:21.180 align:middle line:84% is defined as the amount of water 01:01:21.180 --> 01:01:28.680 align:middle line:84% that remained in the vapor phase divided by the overall amount 01:01:28.680 --> 01:01:30.070 align:middle line:90% in the vapor phase. 01:01:30.070 --> 01:01:34.050 align:middle line:84% And so the overall amount is the overall products 01:01:34.050 --> 01:01:38.730 align:middle line:84% of combustion minus the water that has condensed. 01:01:38.730 --> 01:01:41.070 align:middle line:84% So it went away from the gaseous phase. 01:01:41.070 --> 01:01:42.620 align:middle line:90% So this is the mole fraction. 01:01:42.620 --> 01:01:47.160 align:middle line:84% And the total amount is given by this stoichiometry. 01:01:47.160 --> 01:01:53.320 align:middle line:84% So if you combine all that, you get 01:01:53.320 --> 01:01:57.460 align:middle line:84% that the maximum amount of water that 01:01:57.460 --> 01:02:01.390 align:middle line:84% can stay in the gaseous phase is limited 01:02:01.390 --> 01:02:05.410 align:middle line:84% by this combination of the l and k's that describe how many 01:02:05.410 --> 01:02:10.660 align:middle line:84% carbons and hydrogens you have in the fuel molecule. 01:02:10.660 --> 01:02:14.560 align:middle line:84% And when you have condensation, which happens when l over k 01:02:14.560 --> 01:02:20.480 align:middle line:84% is greater than 0.4, this is the amount that condenses. 01:02:20.480 --> 01:02:23.348 align:middle line:90% 01:02:23.348 --> 01:02:28.780 align:middle line:84% OK, so if that amount condenses, it 01:02:28.780 --> 01:02:37.190 align:middle line:84% releases the enthalpy of vaporization of water. 01:02:37.190 --> 01:02:41.395 align:middle line:84% And that enters into your balance equation. 01:02:41.395 --> 01:02:47.130 align:middle line:90% 01:02:47.130 --> 01:02:52.923 align:middle line:84% And so the heat that you need to-- 01:02:52.923 --> 01:02:57.060 align:middle line:90% 01:02:57.060 --> 01:03:02.170 align:middle line:84% when you compute the heating value at standard temperature, 01:03:02.170 --> 01:03:06.660 align:middle line:84% you're actually making an experiment in which you 01:03:06.660 --> 01:03:17.550 align:middle line:84% have your reactants come in at STP, produce your products, 01:03:17.550 --> 01:03:26.380 align:middle line:84% and you need to put some heat out at temperature T 01:03:26.380 --> 01:03:31.820 align:middle line:84% so as to cool the products down again to this temperature here. 01:03:31.820 --> 01:03:34.930 align:middle line:84% So that's the setup in which you measure this value. 01:03:34.930 --> 01:03:38.880 align:middle line:90% 01:03:38.880 --> 01:03:41.100 align:middle line:84% And this is the entropy balance-- 01:03:41.100 --> 01:03:42.690 align:middle line:90% I'm sorry, the energy balance. 01:03:42.690 --> 01:03:45.870 align:middle line:84% That defines per unit-- so the amount 01:03:45.870 --> 01:03:49.080 align:middle line:84% of heat that you have to extract per unit of amount of fuel 01:03:49.080 --> 01:03:55.260 align:middle line:84% that you have burned in that experimental setup 01:03:55.260 --> 01:03:59.970 align:middle line:90% is the actual heating value. 01:03:59.970 --> 01:04:02.340 align:middle line:84% And part of it is equal to the negative 01:04:02.340 --> 01:04:05.100 align:middle line:90% of the enthalpy of reaction. 01:04:05.100 --> 01:04:14.023 align:middle line:84% And part of it is due to the condensation enthalpy. 01:04:14.023 --> 01:04:18.945 align:middle line:90% 01:04:18.945 --> 01:04:23.060 align:middle line:84% So if you do the calculation, for example, for methane, 01:04:23.060 --> 01:04:30.050 align:middle line:84% the value is, in megajoules per kilogram, 54.7. 01:04:30.050 --> 01:04:38.530 align:middle line:84% If you neglect this condensation, it's only 50.1. 01:04:38.530 --> 01:04:41.410 align:middle line:84% And the other extreme is if you want 01:04:41.410 --> 01:04:43.960 align:middle line:84% to have an even quicker calculation 01:04:43.960 --> 01:04:48.720 align:middle line:84% and forget about the amount of water that 01:04:48.720 --> 01:04:52.290 align:middle line:84% remains in the gaseous phase and assume 01:04:52.290 --> 01:04:55.860 align:middle line:84% that everything, every single molecule of water that 01:04:55.860 --> 01:05:00.690 align:middle line:84% has been produced, condenses, then you put l over 2 here. 01:05:00.690 --> 01:05:05.430 align:middle line:84% And you get what is called the higher heating value. 01:05:05.430 --> 01:05:09.370 align:middle line:90% It's 55.5. 01:05:09.370 --> 01:05:12.290 align:middle line:84% So there is heating value and heating value. 01:05:12.290 --> 01:05:15.760 align:middle line:84% So the actual-- the lower if you neglect condensation, 01:05:15.760 --> 01:05:19.940 align:middle line:84% and the higher is you a bit overestimate condensation. 01:05:19.940 --> 01:05:30.590 align:middle line:90% 01:05:30.590 --> 01:05:35.220 align:middle line:84% So this reactor that we just considered, 01:05:35.220 --> 01:05:38.450 align:middle line:90% it's represented again here. 01:05:38.450 --> 01:05:45.860 align:middle line:84% And in this setup, with this setup, 01:05:45.860 --> 01:05:50.010 align:middle line:84% we can consider three possible cases. 01:05:50.010 --> 01:05:55.840 align:middle line:84% The one is the heating value, the one that we just did, 01:05:55.840 --> 01:06:03.470 align:middle line:84% in which there is no work, and you set the outlet temperature 01:06:03.470 --> 01:06:10.400 align:middle line:84% of the products to be standard, like the inlet 01:06:10.400 --> 01:06:17.150 align:middle line:84% in this setup that defines the combination from which 01:06:17.150 --> 01:06:20.540 align:middle line:90% you measure the heating value. 01:06:20.540 --> 01:06:27.270 align:middle line:84% The exergy setup, the exergy of the fuel, is one in which, 01:06:27.270 --> 01:06:39.890 align:middle line:84% again, you have the products out at atmospheric or standard 01:06:39.890 --> 01:06:41.890 align:middle line:90% temperature. 01:06:41.890 --> 01:06:46.870 align:middle line:84% But your reactor has some system inside that 01:06:46.870 --> 01:06:52.460 align:middle line:84% allows you to extract work, and actually does it reversibly. 01:06:52.460 --> 01:06:59.240 align:middle line:84% So if you do it reversibly-- so in order 01:06:59.240 --> 01:07:01.170 align:middle line:84% to obtain the maximum work, you set 01:07:01.170 --> 01:07:04.100 align:middle line:84% the entropy produced by irreversibility equal to 0 01:07:04.100 --> 01:07:06.590 align:middle line:90% in your entropy balance. 01:07:06.590 --> 01:07:09.410 align:middle line:90% And the outcome is this one. 01:07:09.410 --> 01:07:13.660 align:middle line:84% The exergy of the fuel turns out to be 01:07:13.660 --> 01:07:18.930 align:middle line:84% equal to the negative of the Gibbs free energy at STP. 01:07:18.930 --> 01:07:21.940 align:middle line:90% 01:07:21.940 --> 01:07:26.800 align:middle line:84% And notice that in this calculation, see, 01:07:26.800 --> 01:07:31.150 align:middle line:84% while here you only have the enthalpy difference-- 01:07:31.150 --> 01:07:34.230 align:middle line:84% and in the enthalpy difference, then you need to account 01:07:34.230 --> 01:07:37.620 align:middle line:90% for condensation-- 01:07:37.620 --> 01:07:40.252 align:middle line:84% here you have the enthalpy difference, but also 01:07:40.252 --> 01:07:42.210 align:middle line:84% the entropy difference, and they are multiplied 01:07:42.210 --> 01:07:44.160 align:middle line:90% by the temperature T0. 01:07:44.160 --> 01:07:48.850 align:middle line:84% So one brings in the enthalpy of vaporization. 01:07:48.850 --> 01:07:51.850 align:middle line:84% The other brings in the entropy of vaporization. 01:07:51.850 --> 01:07:53.615 align:middle line:84% And they are related to one another. 01:07:53.615 --> 01:07:57.790 align:middle line:90% 01:07:57.790 --> 01:08:00.820 align:middle line:84% For a pure substance, you have this identity 01:08:00.820 --> 01:08:08.410 align:middle line:84% that h fg is equal to T s fg at every temperature because 01:08:08.410 --> 01:08:12.100 align:middle line:84% of the condition of mutual equilibrium between the liquid 01:08:12.100 --> 01:08:13.220 align:middle line:90% and the vapor. 01:08:13.220 --> 01:08:16.830 align:middle line:84% So these two contributions cancel out. 01:08:16.830 --> 01:08:21.390 align:middle line:84% So this delta g negative does not 01:08:21.390 --> 01:08:25.720 align:middle line:84% have contributions from condensation. 01:08:25.720 --> 01:08:29.569 align:middle line:90% 01:08:29.569 --> 01:08:33.354 align:middle line:84% The condensation doesn't play a role in the exergy of the fuel. 01:08:33.354 --> 01:08:39.520 align:middle line:90% 01:08:39.520 --> 01:08:46.529 align:middle line:84% If you want, what I'm neglecting here 01:08:46.529 --> 01:08:56.069 align:middle line:84% is the mixing entropy terms that are here, depending 01:08:56.069 --> 01:08:59.149 align:middle line:90% on how you put in your fuels. 01:08:59.149 --> 01:09:05.340 align:middle line:84% Certainly, the air is a mixture of oxygen and nitrogen. 01:09:05.340 --> 01:09:09.250 align:middle line:84% Sometimes, you put in your fuel premixed. 01:09:09.250 --> 01:09:13.620 align:middle line:84% So you mix the fuel with the air. 01:09:13.620 --> 01:09:14.859 align:middle line:90% So you have a mixture. 01:09:14.859 --> 01:09:17.100 align:middle line:84% And in the entropy balance, you should 01:09:17.100 --> 01:09:19.800 align:middle line:90% consider the mixing terms. 01:09:19.800 --> 01:09:23.920 align:middle line:84% And then also the products of combustion is a mixture. 01:09:23.920 --> 01:09:26.290 align:middle line:90% And so you have an entropy term. 01:09:26.290 --> 01:09:30.899 align:middle line:84% So somehow, these are approximations in which we are 01:09:30.899 --> 01:09:35.649 align:middle line:90% neglecting mixing entropies. 01:09:35.649 --> 01:09:39.520 align:middle line:84% The final setup of this is one in which 01:09:39.520 --> 01:09:43.819 align:middle line:84% you compute the adiabatic flame temperature-- 01:09:43.819 --> 01:09:49.520 align:middle line:84% no heat, no works, just an adiabatic insulated reactor 01:09:49.520 --> 01:09:53.575 align:middle line:84% in which you just have this combustion reaction occur. 01:09:53.575 --> 01:09:57.450 align:middle line:90% 01:09:57.450 --> 01:10:01.140 align:middle line:90% The energy balance is this one. 01:10:01.140 --> 01:10:03.840 align:middle line:84% Now the enthalpy of the products has 01:10:03.840 --> 01:10:10.720 align:middle line:84% to be computed at temperature T ad, T adiabatic. 01:10:10.720 --> 01:10:16.230 align:middle line:84% We can add and subtract the enthalpy of the products 01:10:16.230 --> 01:10:20.610 align:middle line:84% at standard temperature so that this difference here 01:10:20.610 --> 01:10:25.580 align:middle line:84% is the same that appears here in this other setup. 01:10:25.580 --> 01:10:27.080 align:middle line:90% And we've already computed it. 01:10:27.080 --> 01:10:29.740 align:middle line:84% It's the heating value times epsilon dot. 01:10:29.740 --> 01:10:34.150 align:middle line:84% And this is just the sensible heat that is-- 01:10:34.150 --> 01:10:36.370 align:middle line:84% sensible, yeah, heat that is required 01:10:36.370 --> 01:10:45.210 align:middle line:84% to heat up the products from standard 01:10:45.210 --> 01:10:46.780 align:middle line:90% to adiabatic flame temperature. 01:10:46.780 --> 01:10:49.110 align:middle line:84% And you can compute that by simply using 01:10:49.110 --> 01:10:52.830 align:middle line:90% the properties of an ideal gas. 01:10:52.830 --> 01:11:00.160 align:middle line:84% You can actually start even from standard temperature, 01:11:00.160 --> 01:11:05.160 align:middle line:84% even for water, because in the heating value, which 01:11:05.160 --> 01:11:07.890 align:middle line:84% is on that part, you assume that water 01:11:07.890 --> 01:11:12.460 align:middle line:84% is in the fictitious gaseous state. 01:11:12.460 --> 01:11:15.960 align:middle line:84% So you can use those correlations for specific heats 01:11:15.960 --> 01:11:18.760 align:middle line:90% and integrate. 01:11:18.760 --> 01:11:24.070 align:middle line:84% Of course, this gives you an equation in which T adiabatic 01:11:24.070 --> 01:11:26.350 align:middle line:90% is the only unknown. 01:11:26.350 --> 01:11:29.320 align:middle line:90% And you can solve for it. 01:11:29.320 --> 01:11:34.450 align:middle line:84% Once you find it, you can put it in the entropy balance. 01:11:34.450 --> 01:11:36.100 align:middle line:84% Also, the entropy balance, which is 01:11:36.100 --> 01:11:39.910 align:middle line:84% this one, entropy of the outlet, is 01:11:39.910 --> 01:11:41.960 align:middle line:90% greater than that of the inlet. 01:11:41.960 --> 01:11:43.840 align:middle line:84% And the difference is the entropy 01:11:43.840 --> 01:11:45.770 align:middle line:90% produced by irreversibility. 01:11:45.770 --> 01:11:50.260 align:middle line:84% You can add and subtract the entropy of the products 01:11:50.260 --> 01:11:51.990 align:middle line:90% at standard conditions. 01:11:51.990 --> 01:11:55.930 align:middle line:90% 01:11:55.930 --> 01:12:03.490 align:middle line:84% And so this last difference is equal to the entropy 01:12:03.490 --> 01:12:08.250 align:middle line:84% of the reaction times the reaction coordinate. 01:12:08.250 --> 01:12:12.390 align:middle line:84% And this initial difference you can 01:12:12.390 --> 01:12:15.825 align:middle line:84% obtain by integrating the formulas for an ideal gas 01:12:15.825 --> 01:12:16.325 align:middle line:90% mixture. 01:12:16.325 --> 01:12:21.070 align:middle line:90% 01:12:21.070 --> 01:12:22.940 align:middle line:84% When you do these kind of calculations-- 01:12:22.940 --> 01:12:28.180 align:middle line:84% and we have a couple of tables that show these numbers later 01:12:28.180 --> 01:12:30.120 align:middle line:90% on-- 01:12:30.120 --> 01:12:32.520 align:middle line:90% you compute the entropy of-- 01:12:32.520 --> 01:12:38.500 align:middle line:84% these typical values are in the range between 26% 01:12:38.500 --> 01:12:44.980 align:middle line:84% and 30% of the negative of the Gibbs free energy of-- 01:12:44.980 --> 01:12:49.480 align:middle line:84% so if you take T0 and put it on the other side, 01:12:49.480 --> 01:12:54.010 align:middle line:84% T0 times S irreversible is the waste of-- 01:12:54.010 --> 01:12:58.630 align:middle line:84% the exergy destruction-- is the source 01:12:58.630 --> 01:13:02.510 align:middle line:90% of inefficiency of your system. 01:13:02.510 --> 01:13:11.540 align:middle line:84% And that is equal to about 26%, 30% of the exergy of the fuel. 01:13:11.540 --> 01:13:15.820 align:middle line:84% So that means that by burning your fuel 01:13:15.820 --> 01:13:21.250 align:middle line:84% in a steady-- in a reactor like this, 01:13:21.250 --> 01:13:28.000 align:middle line:84% you're wasting about 25% to 30% of the availability of the fuel 01:13:28.000 --> 01:13:35.760 align:middle line:84% without having consumed one single unit of energy. 01:13:35.760 --> 01:13:41.670 align:middle line:84% And this is, if you remember, the reason why I told you 01:13:41.670 --> 01:13:46.510 align:middle line:84% that if we ever will follow history 01:13:46.510 --> 01:13:52.120 align:middle line:84% and we'll have power plants or a production of electricity that 01:13:52.120 --> 01:13:59.620 align:middle line:84% will exceed our current 65% of the best available technology, 01:13:59.620 --> 01:14:02.230 align:middle line:84% we will have to find alternatives 01:14:02.230 --> 01:14:09.570 align:middle line:84% to burning fuels this way, in adiabatic flames, 01:14:09.570 --> 01:14:17.440 align:middle line:84% because we need to overcome this 25% loss of efficiency-- 01:14:17.440 --> 01:14:19.190 align:middle line:90% of potential efficiency. 01:14:19.190 --> 01:14:24.420 align:middle line:90% 01:14:24.420 --> 01:14:27.080 align:middle line:84% Now, if you've seen in the previous slide, 01:14:27.080 --> 01:14:29.530 align:middle line:84% we need to do those integrals to compute 01:14:29.530 --> 01:14:37.900 align:middle line:84% the enthalpy and the entropy, the difference for the products 01:14:37.900 --> 01:14:41.830 align:middle line:84% of combustion between the standard 01:14:41.830 --> 01:14:45.580 align:middle line:90% and adiabatic flame temperature. 01:14:45.580 --> 01:14:49.140 align:middle line:90% One way to do that calculation-- 01:14:49.140 --> 01:14:53.340 align:middle line:84% this is just a detail, but you can find correlations 01:14:53.340 --> 01:14:56.400 align:middle line:84% for the specific heats of the various substances. 01:14:56.400 --> 01:15:02.730 align:middle line:84% And one way-- one correlation that is given in the book 01:15:02.730 --> 01:15:08.850 align:middle line:84% and from which I have extracted some of these numbers, 01:15:08.850 --> 01:15:15.690 align:middle line:84% this kind of correlation is able to cover with a single 01:15:15.690 --> 01:15:23.660 align:middle line:84% correlation the range from standard to 5,000 K with very 01:15:23.660 --> 01:15:25.560 align:middle line:90% good approximation. 01:15:25.560 --> 01:15:31.290 align:middle line:84% Otherwise, in typical software they use, 01:15:31.290 --> 01:15:39.810 align:middle line:84% they split the range with a correlation that holds between 01:15:39.810 --> 01:15:44.430 align:middle line:84% standard, say, 25 C and, say, 1,000 K, 01:15:44.430 --> 01:15:47.230 align:middle line:84% and another correlation from 1,000. 01:15:47.230 --> 01:15:49.400 align:middle line:84% So you have to split the integral into two parts. 01:15:49.400 --> 01:15:54.000 align:middle line:90% 01:15:54.000 --> 01:15:58.410 align:middle line:84% The thing that was into these integrals here, you see, 01:15:58.410 --> 01:16:00.990 align:middle line:84% it's the summation of the specific heats 01:16:00.990 --> 01:16:06.380 align:middle line:84% of the various products weighted on their mole fractions 01:16:06.380 --> 01:16:10.530 align:middle line:90% in the product stream. 01:16:10.530 --> 01:16:14.770 align:middle line:84% And also, for the entropy, it's the same thing, only divided 01:16:14.770 --> 01:16:16.010 align:middle line:90% by T. 01:16:16.010 --> 01:16:18.750 align:middle line:90% So you need to compute these. 01:16:18.750 --> 01:16:20.680 align:middle line:84% And with that kind of correlation, 01:16:20.680 --> 01:16:25.110 align:middle line:84% you can do it simply by using a correlation 01:16:25.110 --> 01:16:30.600 align:middle line:84% with similar fractional powers of the temperature 01:16:30.600 --> 01:16:32.670 align:middle line:84% with the coefficients that are weighted sum 01:16:32.670 --> 01:16:35.880 align:middle line:84% of the individual coefficients of the various molecules which 01:16:35.880 --> 01:16:37.410 align:middle line:90% you take from this. 01:16:37.410 --> 01:16:41.520 align:middle line:84% And then also the integral of these fractional powers 01:16:41.520 --> 01:16:45.180 align:middle line:84% of temperature can be done explicitly. 01:16:45.180 --> 01:16:47.520 align:middle line:84% So here's the result of this integral. 01:16:47.520 --> 01:16:49.590 align:middle line:84% That's the result of that integral. 01:16:49.590 --> 01:16:52.590 align:middle line:84% And so that is the actual expression 01:16:52.590 --> 01:16:57.060 align:middle line:84% that you will have to solve numerically in order 01:16:57.060 --> 01:17:01.185 align:middle line:84% to obtain the adiabatic flame temperature from the energy 01:17:01.185 --> 01:17:01.685 align:middle line:90% balance. 01:17:01.685 --> 01:17:04.410 align:middle line:90% 01:17:04.410 --> 01:17:06.870 align:middle line:84% I am mentioning these details because I'm not 01:17:06.870 --> 01:17:12.180 align:middle line:84% going to go over other details, but in the past, 01:17:12.180 --> 01:17:14.820 align:middle line:84% I have developed this Excel file that I 01:17:14.820 --> 01:17:20.610 align:middle line:84% will share with you on Canvas, just 01:17:20.610 --> 01:17:24.360 align:middle line:84% in case in the future you have to maybe teach 01:17:24.360 --> 01:17:25.960 align:middle line:90% a course like this. 01:17:25.960 --> 01:17:30.760 align:middle line:84% Then you may have your students play with these equations. 01:17:30.760 --> 01:17:33.220 align:middle line:90% This is several sheets. 01:17:33.220 --> 01:17:34.590 align:middle line:90% It allows you to compute-- 01:17:34.590 --> 01:17:37.840 align:middle line:84% to use all the equations that we have seen. 01:17:37.840 --> 01:17:40.810 align:middle line:84% It allows you to compute, for example, 01:17:40.810 --> 01:17:45.810 align:middle line:84% the enthalpy versus entropy diagram 01:17:45.810 --> 01:17:51.930 align:middle line:90% for reactants and products. 01:17:51.930 --> 01:17:55.556 align:middle line:90% It can allow you to identify-- 01:17:55.556 --> 01:17:58.830 align:middle line:90% 01:17:58.830 --> 01:18:00.910 align:middle line:84% for example, in this case, the temperature. 01:18:00.910 --> 01:18:03.870 align:middle line:84% I assume the reactants are at 700 K-- 01:18:03.870 --> 01:18:08.100 align:middle line:84% and to compute the adiabatic flame temperature, so 01:18:08.100 --> 01:18:12.000 align:middle line:84% the temperature of this other state for the products. 01:18:12.000 --> 01:18:14.135 align:middle line:84% So you can play with this if you like. 01:18:14.135 --> 01:18:18.240 align:middle line:90% 01:18:18.240 --> 01:18:27.980 align:middle line:84% Of course, we left it alone, but we are not always 01:18:27.980 --> 01:18:34.900 align:middle line:84% working with steady-state flow reactors. 01:18:34.900 --> 01:18:39.970 align:middle line:84% Often, like in an internal combustion engine or in bombs-- 01:18:39.970 --> 01:18:42.620 align:middle line:90% not necessarily used for wars. 01:18:42.620 --> 01:18:46.720 align:middle line:84% You use bombs also to measure in the lab, 01:18:46.720 --> 01:18:56.950 align:middle line:84% like, the heating power of fuels or the flame propagation rates. 01:18:56.950 --> 01:19:02.570 align:middle line:84% So you have a closed system set up. 01:19:02.570 --> 01:19:08.900 align:middle line:84% And you can perform exactly the same reasoning that we've done. 01:19:08.900 --> 01:19:10.400 align:middle line:84% The only difference is that you have 01:19:10.400 --> 01:19:12.055 align:middle line:90% to perform a volume balance. 01:19:12.055 --> 01:19:14.580 align:middle line:90% 01:19:14.580 --> 01:19:20.300 align:middle line:84% And if you can assume that both the initial and final states 01:19:20.300 --> 01:19:23.900 align:middle line:84% are in the gaseous phase, the ideal gas law 01:19:23.900 --> 01:19:26.120 align:middle line:90% gives you the volume. 01:19:26.120 --> 01:19:30.410 align:middle line:84% And then, if the volume remains constant, like VP equals VR, 01:19:30.410 --> 01:19:34.160 align:middle line:84% then you find that the ratio of the final temperature 01:19:34.160 --> 01:19:41.390 align:middle line:84% over the final pressure is equal to the initial ratio. 01:19:41.390 --> 01:19:44.390 align:middle line:84% But there is also the ratio of nP over nR. 01:19:44.390 --> 01:19:50.100 align:middle line:90% 01:19:50.100 --> 01:19:52.120 align:middle line:84% So, for example, for the heating value, 01:19:52.120 --> 01:19:56.080 align:middle line:84% you could measure a heating value at constant volume. 01:19:56.080 --> 01:19:58.430 align:middle line:90% You would have no heat exchange. 01:19:58.430 --> 01:20:01.840 align:middle line:84% You would have-- I'm sorry, no work exchange. 01:20:01.840 --> 01:20:05.320 align:middle line:84% And you would have heat exchange enough 01:20:05.320 --> 01:20:10.540 align:middle line:84% to bring your products down to the environmental temperature. 01:20:10.540 --> 01:20:14.350 align:middle line:84% But you have to accept that since it's a closed volume 01:20:14.350 --> 01:20:23.210 align:middle line:84% setup, the final pressure will not go down to atmospheric. 01:20:23.210 --> 01:20:27.590 align:middle line:84% And it will be a pressure p that-- 01:20:27.590 --> 01:20:31.020 align:middle line:84% determined by the ideal gas law is determined by this thing. 01:20:31.020 --> 01:20:37.824 align:middle line:90% 01:20:37.824 --> 01:20:44.660 align:middle line:84% OK, so if you do the energy balance, 01:20:44.660 --> 01:20:47.780 align:middle line:90% you get the heating value. 01:20:47.780 --> 01:20:51.930 align:middle line:84% The heating value this time, in this setup, 01:20:51.930 --> 01:20:54.960 align:middle line:84% is equal not to the enthalpy, but the internal energy 01:20:54.960 --> 01:21:05.850 align:middle line:84% of combustion plus the amount of energy that is involved 01:21:05.850 --> 01:21:06.970 align:middle line:90% in condensation. 01:21:06.970 --> 01:21:10.920 align:middle line:84% And I presume that here I should change this 01:21:10.920 --> 01:21:20.578 align:middle line:84% into a U. That's the internal energy of condensation and not 01:21:20.578 --> 01:21:21.120 align:middle line:90% the enthalpy. 01:21:21.120 --> 01:21:26.410 align:middle line:90% 01:21:26.410 --> 01:21:30.730 align:middle line:84% Or if you consider an adiabatic setup, 01:21:30.730 --> 01:21:39.180 align:middle line:84% like a true bomb setup, where there is no work, no heat, 01:21:39.180 --> 01:21:43.020 align:middle line:84% this is the final pressure in terms of the final temperature. 01:21:43.020 --> 01:21:46.350 align:middle line:84% And you compute the final temperature 01:21:46.350 --> 01:21:51.480 align:middle line:90% by, again, the energy balance. 01:21:51.480 --> 01:21:56.430 align:middle line:84% The only difference is that instead of the specific heats 01:21:56.430 --> 01:22:01.200 align:middle line:84% at constant pressure, since the energy for a gas 01:22:01.200 --> 01:22:04.360 align:middle line:84% depends on the specific heats at constant volume, 01:22:04.360 --> 01:22:06.755 align:middle line:84% it's the cv that you have to integrate here. 01:22:06.755 --> 01:22:09.840 align:middle line:90% 01:22:09.840 --> 01:22:13.930 align:middle line:84% And you can then compute the pressure. 01:22:13.930 --> 01:22:16.500 align:middle line:84% Once you have the temperature, you can compute the pressure 01:22:16.500 --> 01:22:17.370 align:middle line:90% from here. 01:22:17.370 --> 01:22:19.230 align:middle line:84% And then you can compute the entropy 01:22:19.230 --> 01:22:22.150 align:middle line:84% produced by irreversibility with the entropy balance. 01:22:22.150 --> 01:22:26.655 align:middle line:84% And you find that this time the loss of-- 01:22:26.655 --> 01:22:29.400 align:middle line:90% 01:22:29.400 --> 01:22:33.330 align:middle line:84% the exergy destruction-- so the loss of efficiency-- 01:22:33.330 --> 01:22:35.700 align:middle line:90% is a bit smaller. 01:22:35.700 --> 01:22:38.030 align:middle line:90% Well, why? 01:22:38.030 --> 01:22:42.310 align:middle line:84% Because you are left with products that are in a container 01:22:42.310 --> 01:22:44.560 align:middle line:90% and at very high pressure. 01:22:44.560 --> 01:22:47.990 align:middle line:84% And that high pressure has some availability. 01:22:47.990 --> 01:22:50.900 align:middle line:84% You could use it to expand and, therefore, to make work. 01:22:50.900 --> 01:22:57.850 align:middle line:84% So that is has transformed part of the exergy of the fuel 01:22:57.850 --> 01:23:00.170 align:middle line:90% into that. 01:23:00.170 --> 01:23:02.440 align:middle line:90% And that's why-- 01:23:02.440 --> 01:23:04.000 align:middle line:90% These are the numbers. 01:23:04.000 --> 01:23:08.910 align:middle line:84% So here are numbers for a closed combustion chamber. 01:23:08.910 --> 01:23:14.940 align:middle line:90% 01:23:14.940 --> 01:23:21.480 align:middle line:84% Temperatures are very high, and pressures of the order of 10 01:23:21.480 --> 01:23:22.480 align:middle line:90% atmospheres. 01:23:22.480 --> 01:23:26.190 align:middle line:90% 01:23:26.190 --> 01:23:30.750 align:middle line:84% And here that ratio is indeed in the range that I said. 01:23:30.750 --> 01:23:36.480 align:middle line:84% For the steady-state flow burner, 01:23:36.480 --> 01:23:40.530 align:middle line:84% the adiabatic flame temperatures are lower. 01:23:40.530 --> 01:23:44.405 align:middle line:84% And the irreversibility is higher. 01:23:44.405 --> 01:23:52.670 align:middle line:90% 01:23:52.670 --> 01:24:03.620 align:middle line:84% Actually, these calculations are done 01:24:03.620 --> 01:24:12.020 align:middle line:84% not using simply the reaction of oxidation of the fuel, 01:24:12.020 --> 01:24:17.600 align:middle line:84% but also using the other two reactions, that of dissociation 01:24:17.600 --> 01:24:25.420 align:middle line:84% of CO2 and of H2O, because those two reactions are 01:24:25.420 --> 01:24:30.100 align:middle line:84% important at high temperatures and are also 01:24:30.100 --> 01:24:35.440 align:middle line:84% energy-intensive in the sense that they consume up 01:24:35.440 --> 01:24:40.300 align:middle line:84% some of the energy released by the combustion reaction. 01:24:40.300 --> 01:24:44.780 align:middle line:84% And therefore, they are important. 01:24:44.780 --> 01:24:47.440 align:middle line:84% Otherwise, if you neglect those, you 01:24:47.440 --> 01:24:51.465 align:middle line:84% would get even higher values for the temperatures here. 01:24:51.465 --> 01:24:58.560 align:middle line:90% 01:24:58.560 --> 01:25:03.610 align:middle line:84% And in the Excel file, you find that in some of the sheets 01:25:03.610 --> 01:25:09.990 align:middle line:84% you are neglecting this dissociation of CO2 and H2O. 01:25:09.990 --> 01:25:12.960 align:middle line:84% And some other sheets, you find them. 01:25:12.960 --> 01:25:20.340 align:middle line:84% Those are the ones where you also compute the equilibrium 01:25:20.340 --> 01:25:21.180 align:middle line:90% because-- 01:25:21.180 --> 01:25:24.030 align:middle line:84% at the moment we haven't done yet. 01:25:24.030 --> 01:25:27.280 align:middle line:84% Next time, we'll talk about chemical equilibrium. 01:25:27.280 --> 01:25:32.250 align:middle line:84% And those reactions are assumed to be in chemical equilibrium 01:25:32.250 --> 01:25:33.850 align:middle line:90% at these temperatures. 01:25:33.850 --> 01:25:38.900 align:middle line:90% 01:25:38.900 --> 01:25:40.546 align:middle line:90% So let's summarize. 01:25:40.546 --> 01:25:46.660 align:middle line:90% 01:25:46.660 --> 01:25:53.010 align:middle line:84% For the closed system setup, what 01:25:53.010 --> 01:26:03.990 align:middle line:84% we have seen in this energy versus entropy diagram, 01:26:03.990 --> 01:26:07.170 align:middle line:84% this is the curve that represents 01:26:07.170 --> 01:26:10.140 align:middle line:84% the stable equilibrium states of the reactants. 01:26:10.140 --> 01:26:12.690 align:middle line:84% And this is the one that represents 01:26:12.690 --> 01:26:14.355 align:middle line:90% those of the products. 01:26:14.355 --> 01:26:17.592 align:middle line:90% 01:26:17.592 --> 01:26:23.450 align:middle line:84% T0 is this slope here, which touches the reactants 01:26:23.450 --> 01:26:24.450 align:middle line:90% at this point. 01:26:24.450 --> 01:26:31.080 align:middle line:84% So these are the reactant mixture. 01:26:31.080 --> 01:26:35.970 align:middle line:84% So it would be your fuel, including 01:26:35.970 --> 01:26:37.860 align:middle line:84% the air that you're using to oxidize 01:26:37.860 --> 01:26:45.340 align:middle line:84% the fuel at 25 C, 1 atmosphere, so in this state here. 01:26:45.340 --> 01:26:51.270 align:middle line:84% And these are the products of reaction 01:26:51.270 --> 01:26:56.870 align:middle line:84% after you cool them down at standard conditions. 01:26:56.870 --> 01:27:03.770 align:middle line:84% So the difference in energy between this and that 01:27:03.770 --> 01:27:06.990 align:middle line:90% is these blue arrows. 01:27:06.990 --> 01:27:12.570 align:middle line:90% That's the energy of reaction. 01:27:12.570 --> 01:27:15.570 align:middle line:84% And the difference in the entropies-- 01:27:15.570 --> 01:27:17.730 align:middle line:90% these black arrows here-- 01:27:17.730 --> 01:27:21.970 align:middle line:90% is the entropy of the reaction. 01:27:21.970 --> 01:27:26.490 align:middle line:84% And, again, sometimes, this point 01:27:26.490 --> 01:27:28.600 align:middle line:90% will be to the left of that one. 01:27:28.600 --> 01:27:32.170 align:middle line:84% Sometimes, for other fuels, it will be on the right. 01:27:32.170 --> 01:27:36.380 align:middle line:84% There is no fundamental reason why it should be left or right. 01:27:36.380 --> 01:27:41.596 align:middle line:90% 01:27:41.596 --> 01:27:44.060 align:middle line:90% We start with the products-- 01:27:44.060 --> 01:27:51.950 align:middle line:84% I mean, with the reactants at STP. 01:27:51.950 --> 01:28:05.290 align:middle line:84% And we run our adiabatic constant energy combustion-- 01:28:05.290 --> 01:28:07.940 align:middle line:84% adiabatic flame temperature setup. 01:28:07.940 --> 01:28:10.130 align:middle line:90% So the energy remains constant. 01:28:10.130 --> 01:28:14.816 align:middle line:84% And in the final state, we end up here, 01:28:14.816 --> 01:28:18.090 align:middle line:84% with this state that has a temperature, which 01:28:18.090 --> 01:28:20.240 align:middle line:84% is represented by this slope, which is higher. 01:28:20.240 --> 01:28:27.550 align:middle line:90% 01:28:27.550 --> 01:28:29.540 align:middle line:90% Now, talking about-- 01:28:29.540 --> 01:28:32.460 align:middle line:90% I mean, yes, available energies. 01:28:32.460 --> 01:28:35.940 align:middle line:84% So let me call omega, as we did in the initial part 01:28:35.940 --> 01:28:36.700 align:middle line:90% of the course-- 01:28:36.700 --> 01:28:40.350 align:middle line:84% omega, the available energy, with respect to a reservoir. 01:28:40.350 --> 01:28:42.330 align:middle line:84% So the 0 up there means with respect 01:28:42.330 --> 01:28:44.620 align:middle line:84% to a reservoir at standard temperature. 01:28:44.620 --> 01:28:47.810 align:middle line:90% 01:28:47.810 --> 01:28:51.230 align:middle line:84% So in any place of this diagram, which 01:28:51.230 --> 01:28:56.260 align:middle line:84% you remember, at any point, we can compute 01:28:56.260 --> 01:28:58.480 align:middle line:90% the energy and the entropy-- 01:28:58.480 --> 01:29:08.230 align:middle line:84% this is the expression for the available energy-- 01:29:08.230 --> 01:29:13.510 align:middle line:84% energy minus the energy when the system is in mutual equilibrium 01:29:13.510 --> 01:29:14.450 align:middle line:90% with the reservoir. 01:29:14.450 --> 01:29:16.960 align:middle line:90% 01:29:16.960 --> 01:29:20.680 align:middle line:84% But now we have two options here. 01:29:20.680 --> 01:29:24.260 align:middle line:90% One is with the off case-- 01:29:24.260 --> 01:29:26.530 align:middle line:90% no reactions. 01:29:26.530 --> 01:29:31.480 align:middle line:84% So for no reactions, the composition cannot change. 01:29:31.480 --> 01:29:35.980 align:middle line:90% And so at our reference state-- 01:29:35.980 --> 01:29:42.430 align:middle line:84% the reactants are in mutual equilibrium with our reservoir 01:29:42.430 --> 01:29:46.060 align:middle line:90% at this state here. 01:29:46.060 --> 01:29:55.080 align:middle line:84% So this is the state with energy U0R and S0R, so it's here. 01:29:55.080 --> 01:30:00.480 align:middle line:84% If, instead, we allow the chemical reactions, 01:30:00.480 --> 01:30:09.210 align:middle line:84% then the mutual equilibrium with the reservoir will be this one. 01:30:09.210 --> 01:30:16.420 align:middle line:84% The most stable is when the products have occurred, 01:30:16.420 --> 01:30:22.800 align:middle line:84% the reactions have allowed to go to a more stable 01:30:22.800 --> 01:30:27.170 align:middle line:84% mutual equilibrium, which is this one, the state at T0 01:30:27.170 --> 01:30:28.380 align:middle line:90% of the products. 01:30:28.380 --> 01:30:29.395 align:middle line:90% So I can compute-- 01:30:29.395 --> 01:30:31.980 align:middle line:90% 01:30:31.980 --> 01:30:37.010 align:middle line:84% of course, this state has zero availability with respect 01:30:37.010 --> 01:30:39.050 align:middle line:90% to the off situation. 01:30:39.050 --> 01:30:41.970 align:middle line:84% But with respect to the on situation, 01:30:41.970 --> 01:30:45.240 align:middle line:84% so with the reactions, it has an availability 01:30:45.240 --> 01:30:48.480 align:middle line:84% which is given by this blue line here. 01:30:48.480 --> 01:30:55.040 align:middle line:84% So that's the available energy of the inlet state with respect 01:30:55.040 --> 01:30:58.510 align:middle line:90% to the outlet state. 01:30:58.510 --> 01:31:02.740 align:middle line:90% 01:31:02.740 --> 01:31:07.370 align:middle line:84% When I do the flame, the adiabatic combustion, 01:31:07.370 --> 01:31:12.660 align:middle line:84% I end up with this state here and its availability, 01:31:12.660 --> 01:31:14.580 align:middle line:90% its available energy-- 01:31:14.580 --> 01:31:16.140 align:middle line:84% you remember the construction, you 01:31:16.140 --> 01:31:20.880 align:middle line:84% have to go down here geometrically until you reach 01:31:20.880 --> 01:31:24.150 align:middle line:84% the line that has the slope of the temperature 01:31:24.150 --> 01:31:25.210 align:middle line:90% of the reservoir. 01:31:25.210 --> 01:31:26.650 align:middle line:90% So this is the blue-- 01:31:26.650 --> 01:31:29.290 align:middle line:90% this is the available energy. 01:31:29.290 --> 01:31:31.850 align:middle line:90% It's less than the initial. 01:31:31.850 --> 01:31:40.980 align:middle line:84% And what's missing is related to the amount of entropy 01:31:40.980 --> 01:31:43.080 align:middle line:84% that you generated by irreversibility. 01:31:43.080 --> 01:31:50.950 align:middle line:84% If you look at this triangle, this tangent is T0. 01:31:50.950 --> 01:31:55.960 align:middle line:84% And so this side of the triangle is 01:31:55.960 --> 01:32:00.550 align:middle line:84% equal to T0 times this length, which is S irreversible. 01:32:00.550 --> 01:32:06.520 align:middle line:90% 01:32:06.520 --> 01:32:10.300 align:middle line:84% Now, sometimes-- well, first observation 01:32:10.300 --> 01:32:17.170 align:middle line:84% here is that these two curves approach each other 01:32:17.170 --> 01:32:21.390 align:middle line:90% as you go high in temperature. 01:32:21.390 --> 01:32:27.270 align:middle line:84% So if we could have the combustion occur by preheating 01:32:27.270 --> 01:32:32.080 align:middle line:84% our reactants at a higher temperature-- for example, 01:32:32.080 --> 01:32:35.470 align:middle line:90% T1R here-- 01:32:35.470 --> 01:32:40.690 align:middle line:84% when I do the oxidation, the adiabatic flame, 01:32:40.690 --> 01:32:43.670 align:middle line:84% I produce less entropy by irreversibility, 01:32:43.670 --> 01:32:48.200 align:middle line:84% and therefore, I consume less exergy. 01:32:48.200 --> 01:32:51.210 align:middle line:90% That increases the efficiency. 01:32:51.210 --> 01:32:57.290 align:middle line:84% So preheating the reactants is a way to improve the efficiency. 01:32:57.290 --> 01:33:01.350 align:middle line:84% Of course, preheating requires some availability, 01:33:01.350 --> 01:33:04.490 align:middle line:84% some available energy, this much. 01:33:04.490 --> 01:33:08.570 align:middle line:84% And this is the available energy of the heated reactants 01:33:08.570 --> 01:33:12.140 align:middle line:84% with respect to the reactants themselves. 01:33:12.140 --> 01:33:18.360 align:middle line:84% So I'm using the off expression for the available energy. 01:33:18.360 --> 01:33:24.910 align:middle line:84% And then, when I have obtained these products, of course, 01:33:24.910 --> 01:33:30.290 align:middle line:84% at a higher temperature, see, T1 prime, 01:33:30.290 --> 01:33:35.030 align:middle line:84% they have this much available energy. 01:33:35.030 --> 01:33:39.770 align:middle line:84% And even if I pay off what I spent in order 01:33:39.770 --> 01:33:44.720 align:middle line:84% to do the preheating, I'm still gaining in terms of the fact 01:33:44.720 --> 01:33:50.150 align:middle line:84% that, of course, if I do things reversibly, 01:33:50.150 --> 01:33:54.440 align:middle line:90% I have less energy waste than-- 01:33:54.440 --> 01:33:58.320 align:middle line:84% exergy waste than in the previous case. 01:33:58.320 --> 01:34:05.980 align:middle line:84% And the last observation is the following, that, OK, well, 01:34:05.980 --> 01:34:11.860 align:middle line:84% could we keep preheating and so get closer and closer? 01:34:11.860 --> 01:34:15.670 align:middle line:90% This concept has been-- 01:34:15.670 --> 01:34:25.190 align:middle line:84% so the question is, is it possible to go to an extreme 01:34:25.190 --> 01:34:28.300 align:middle line:84% to where these two curves, one of the reactants 01:34:28.300 --> 01:34:33.520 align:middle line:84% and one of the products, touch each other? 01:34:33.520 --> 01:34:36.970 align:middle line:84% And they do, at a certain temperature, 01:34:36.970 --> 01:34:41.940 align:middle line:90% provided you do what-- 01:34:41.940 --> 01:34:47.926 align:middle line:84% you have to mix some of the products with your reactants. 01:34:47.926 --> 01:34:51.990 align:middle line:84% In fact, this is also done in internal combustion engines. 01:34:51.990 --> 01:34:56.360 align:middle line:84% It's called exhaust gas recirculation. 01:34:56.360 --> 01:35:00.860 align:middle line:84% You take some of the gases and mix them 01:35:00.860 --> 01:35:03.680 align:middle line:90% with the reactant mixture. 01:35:03.680 --> 01:35:07.700 align:middle line:84% That makes these curves get closer to one another 01:35:07.700 --> 01:35:10.520 align:middle line:84% and, therefore, reduces the irreversibility 01:35:10.520 --> 01:35:13.490 align:middle line:90% and increases the efficiency. 01:35:13.490 --> 01:35:19.450 align:middle line:84% Of course, if you do too much, this mixing 01:35:19.450 --> 01:35:22.630 align:middle line:90% involves an entropy of mixing. 01:35:22.630 --> 01:35:27.170 align:middle line:84% So what you gain by getting these two curves close, 01:35:27.170 --> 01:35:32.870 align:middle line:90% you lose in the mixing entropy. 01:35:32.870 --> 01:35:36.020 align:middle line:90% And that's also irreversible. 01:35:36.020 --> 01:35:38.540 align:middle line:90% 01:35:38.540 --> 01:35:42.038 align:middle line:84% These concepts are very theoretical, of course. 01:35:42.038 --> 01:35:53.400 align:middle line:84% You know Keenan has brought up this question in 1941. 01:35:53.400 --> 01:36:00.620 align:middle line:84% Also Obert, 1948 and then also 1973. 01:36:00.620 --> 01:36:10.210 align:middle line:84% And later on, every 10 years or so, people go back to this one. 01:36:10.210 --> 01:36:11.710 align:middle line:84% I haven't been following, so I don't 01:36:11.710 --> 01:36:15.290 align:middle line:84% know if the frequency of 10 years is still respected. 01:36:15.290 --> 01:36:19.070 align:middle line:84% But at least we are talking about it now. 01:36:19.070 --> 01:36:22.280 align:middle line:90% 01:36:22.280 --> 01:36:29.420 align:middle line:84% We ourselves-- you can go to this reference-- 01:36:29.420 --> 01:36:32.990 align:middle line:90% we worked on some calculation. 01:36:32.990 --> 01:36:37.680 align:middle line:84% And we computed the amount of water 01:36:37.680 --> 01:36:42.070 align:middle line:90% that you have to dissolve in-- 01:36:42.070 --> 01:36:46.060 align:middle line:84% this was the combustion of hydrogen. 01:36:46.060 --> 01:36:51.280 align:middle line:84% And we wanted to look at what dilution you should dilute 01:36:51.280 --> 01:36:55.030 align:middle line:84% your products of combustion, which in this case 01:36:55.030 --> 01:37:00.070 align:middle line:84% is just water, in order for the temperature at which they touch 01:37:00.070 --> 01:37:04.420 align:middle line:84% not to be too high, because if it is too high, 01:37:04.420 --> 01:37:07.810 align:middle line:84% the problem here is that you want to do the preheating. 01:37:07.810 --> 01:37:12.790 align:middle line:84% And while you're preheating, you should be 01:37:12.790 --> 01:37:14.840 align:middle line:90% able to contain the reaction. 01:37:14.840 --> 01:37:20.030 align:middle line:84% You should freeze the reaction and allow the reaction-- only, 01:37:20.030 --> 01:37:23.600 align:middle line:84% so to speak, turn it on only when you are here. 01:37:23.600 --> 01:37:27.310 align:middle line:84% Then you turn it on, and nothing happens 01:37:27.310 --> 01:37:31.450 align:middle line:84% because the reactants and products are in equilibrium. 01:37:31.450 --> 01:37:34.910 align:middle line:84% But then, as you cool down the products, 01:37:34.910 --> 01:37:36.740 align:middle line:90% you go down this curve. 01:37:36.740 --> 01:37:41.200 align:middle line:90% So you obtain more than-- 01:37:41.200 --> 01:37:43.270 align:middle line:84% you obtain the exergy, essentially, 01:37:43.270 --> 01:37:46.870 align:middle line:84% of the fuel with respect to this one. 01:37:46.870 --> 01:37:50.440 align:middle line:84% So in the preheating, you spend this exergy. 01:37:50.440 --> 01:37:54.710 align:middle line:84% And in the cooling down of the products, 01:37:54.710 --> 01:37:56.540 align:middle line:90% you obtain this energy. 01:37:56.540 --> 01:37:59.400 align:middle line:84% And the difference is the exergy of the fuel. 01:37:59.400 --> 01:38:02.500 align:middle line:90% 01:38:02.500 --> 01:38:04.780 align:middle line:84% But the problem in practice is that as you 01:38:04.780 --> 01:38:09.280 align:middle line:84% go above a certain temperature, any mixture 01:38:09.280 --> 01:38:13.180 align:middle line:84% will have a self-combustion-- an auto-combustion limit. 01:38:13.180 --> 01:38:17.050 align:middle line:90% And it will burn by itself. 01:38:17.050 --> 01:38:20.150 align:middle line:84% And that's why we were looking for a way 01:38:20.150 --> 01:38:23.900 align:middle line:84% to dilute enough so by diluting you 01:38:23.900 --> 01:38:27.390 align:middle line:84% reduce the temperature at which this happens. 01:38:27.390 --> 01:38:35.120 align:middle line:84% And therefore, you have to go below the self-combustion 01:38:35.120 --> 01:38:35.850 align:middle line:90% temperature. 01:38:35.850 --> 01:38:37.490 align:middle line:90% And here is this. 01:38:37.490 --> 01:38:42.170 align:middle line:84% So, for example, we could find that-- 01:38:42.170 --> 01:38:47.070 align:middle line:84% I think we did, like, 1,500 or maybe 1,000. 01:38:47.070 --> 01:38:50.120 align:middle line:84% This is the amount of water molecules 01:38:50.120 --> 01:38:51.390 align:middle line:90% that you have to put in. 01:38:51.390 --> 01:38:54.520 align:middle line:90% 01:38:54.520 --> 01:39:00.360 align:middle line:84% I don't think anybody has ever done this kind of thing 01:39:00.360 --> 01:39:01.980 align:middle line:90% practically. 01:39:01.980 --> 01:39:08.490 align:middle line:84% But conceptually, it's something to be kept in mind. 01:39:08.490 --> 01:39:12.100 align:middle line:84% So I think, yes, we can stop here today. 01:39:12.100 --> 01:39:17.460 align:middle line:84% And next time, we'll do chemical equilibrium. 01:39:17.460 --> 01:39:19.010 align:middle line:90% OK. 01:39:19.010 --> 01:39:27.000 align:middle line:90%