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0:00
<b>Give me one afternoon today and I'll save you 20 hours later.</b> <b>This is every core concept of engineering thermodynamics, tightly edited,</b> <b>zero fluff designed for three people.</b> <b>The student previewing the course, the student panicking right before</b> <b>finals and the engineer prepping for the FE exam.</b>
0:24
<b>So I spent seven years operating nuclear reactors on Navy submarines before</b> <b>getting my PhD at Duke.</b> <b>Now I head the mechanical engineering department at High Point University.</b> <b>I'm joined as always by TA Serenity and TA Indiana who just dove off the</b> <b>table there. They're cats, but still more reliable than most human TAs.</b> <b>The professor will cover the theory, the demos, and the math.</b> <b>Just leave the cute cat humor to us professionals.</b>
0:54
<b>So grab a notebook. Don't write down everything.</b> <b>Just the aha moments.</b> <b>And we're starting at the end, the power plant.</b>
1:09
<b>Before we go deep on any single concept,</b> <b>I want you to see the whole machine first.</b> <b>So this is a steam power plant.</b> <b>Most of the world's electricity is produced using this four step process,</b> <b>the Rankine cycle.</b>
1:27
<b>So in the real world, a power plant looks like a massive maze of pipes and</b> <b>cooling towers and soaring structures.</b> <b>But thermodynamically, it boils down to just four essential components.</b> <b>So let's track the working fluid through this cycle.</b>
1:49
<b>First, we have the pump from state one to state two.</b> <b>It takes cold liquid water at low pressure and pressurizes it to cold liquid</b> <b>water at high pressure, which significantly increases power plant efficiency.</b> <b>Easily justifies the relatively small energy costs to run the pump.</b> <b>Next, from state two to state three, high pressure water enters the boiler.</b> <b>It doesn't matter if your boiler is coal, natural gas, nuclear, geothermal, solar.</b>
2:25
<b>Those are all just different ways of producing massive amounts of heat to boil</b> <b>the water to high pressure, high temperature steam full of energy.</b>
2:39
<b>Then from state three to state four, the moneymaker, the turbine.</b> <b>It's a really fancy windmill that high energy steam expands rapidly through the</b> <b>turbine blades, spinning a shaft connected to a generator producing electricity.</b> <b>Finally, the low energy leftover steam enters the condenser.</b> <b>It takes less energy to pressurize water than steam.</b> <b>So we reject heat to the environment, fully condensing all of the steam</b> <b>back to water at state one.</b>
3:19
<b>Now black smoke from a thin chimney, that's the pollution from coal in the</b> <b>boiler, but the big clean white fluffy clouds from fat concrete cooling towers.</b> <b>That's just water vapor carrying rejected heat from the condenser.</b> <b>It's totally safe.</b> <b>It's a perfectly closed loop.</b> <b>Fluid goes in a circle, only work and heat cross the boundary.</b> <b>And that's the core math of this course,</b> <b>calculating work, heat, and thermal efficiency.</b>
3:57
<b>So for the rest of the video, this cycle is your anchor.</b> <b>But to analyze the macro, we have to master the micro.</b> <b>So let's build the foundation from the ground up.</b> <b>Then I get to knock it down.</b> <b>That's literally the opposite of what we're trying to achieve.</b> <b>To analyze the Rankine cycle we just saw, we need a common language.</b> <b>In thermo, everything starts with the system.</b> <b>A system is simply the mass or location we've chosen to study.</b>
4:32
<b>Look at your figure.</b> <b>Draw a dashed box.</b> <b>Everything inside the box is your system.</b> <b>Everything outside the box is surroundings.</b> <b>The box itself is the boundary.</b> <b>There are two main types of systems you'll see on exams.</b> <b>An open system where mass and energy are both flowing.</b> <b>Think of a water pipe or a turbine or a nozzle.</b> <b>In these cases, we focused on a fixed region in space.</b> <b>We usually call this a control volume.</b>
5:12
<b>Your TA Serenity is a control volume.</b> <b>Heat enters when she basks in the sun.</b> <b>Mass enters when she eats snacks.</b> <b>And mass exits when she...</b> <b>gross.</b> <b>I prefer to be called an open system</b> <b>because my stomach is always open for more snacks.</b> <b>For a closed system, no mass can cross the boundary, only energy.</b> <b>If I have gas trapped in a piston cylinder, the mass stays put.</b> <b>But energy, in the form of heat or work, can still go in and out.</b>
5:49
<b>So TA Indiana resting happily inside his box?</b> <b>That's a closed system.</b> <b>Forget what Schrodinger said about cats and boxes.</b> <b>It's perfectly safe in here.</b> <b>You might also see isolated systems.</b> <b>These are a special type of closed system.</b> <b>Nothing crosses the boundary, not mass, not energy.</b> <b>We'll usually indicate this by calling the</b> <b>boundary rigid, meaning no work is performed,</b> <b>and insulated, that no heat crosses the boundary.</b>
6:22
<b>Rigid and insulated are keywords that evil professors use to trap students.</b> <b>So keep your eye out for them.</b>
6:32
<b>Now, sometimes you have to choose your system.</b>
6:36
<b>If I draw a dashed box around the turbine, I'm</b> <b>studying an open system, a control volume, right?</b> <b>Just the turbine.</b> <b>If I draw my box around the entire power plant, now I have a closed system since</b> <b>all of the mass stays inside the box I drew.</b>
6:57
<b>Once you define your system, you have to describe it.</b> <b>We do this using properties.</b> <b>The big three, you already know.</b> <b>You use these every day.</b> <b>Pressure temperature volume.</b>
7:11
<b>In fluid mechanics, pressure is usually</b> <b>gauge pressure, just relative to the atmosphere.</b> <b>In thermo, we always use absolute pressure.</b> <b>For temperature, thermo uses Kelvin and Rankine, not Celsius or Fahrenheit.</b> <b>Small exception when you're calculating a delta T.</b> <b>In that case, you can use any temperature scale you want.</b> <b>Just think of me.</b> <b>I'm an absolute unit and when under pressure, I remain cool as a cucumber.</b>
7:44
<b>For volume, actually, you'll use specific volume</b> <b>a lot more often, which is volume divided by mass.</b> <b>And that's just the inverse of density.</b> <b>Same idea just flipped.</b> <b>So high density, low specific volume.</b>
8:03
<b>The next three properties, you probably haven't seen these before.</b> <b>Internal energy, enthalpy and entropy.</b> <b>Something to look forward to in chapters two, three and seven.</b> <b>So when a system is uniform and unchanging, we say it's in a state of equilibrium.</b> <b>Thermo problems are almost always in equilibrium.</b> <b>But look out for two other similar terms.</b> <b>Steady state, meaning not changing in time and transient the opposite.</b> <b>It is changing in time.</b>
8:41
<b>A common mistake is to use steady state and equilibrium interchangeably.</b> <b>But these are not the same thing.</b> <b>And your evil professor will take points off on your test for mixing them up.</b> <b>So here's an example to help you keep them separate.</b> <b>When the temperature stopped changing, this bar is in steady state with high</b> <b>temperatures on one side and cold on the other.</b> <b>But since temperature isn't uniform, the rod is not in equilibrium.</b>
9:14
<b>And transient can sometimes be in equilibrium,</b> <b>which kind of breaks your brain a little bit.</b> <b>So consider a nail in your car tire, a very slow leak.</b> <b>We call this quasi equilibrium, where at each instant</b> <b>in time, you can treat it as if it's in equilibrium.</b> <b>But eventually your tire does go flat.</b> <b>That's the transient.</b>
9:41
<b>A state is basically a snapshot of a system.</b> <b>Like what are pressure, temperature and volume right now?</b> <b>Those values define the current state.</b> <b>So numbers one through four in the Rankine cycle figure, each of these are a state.</b>
10:04
<b>For properties, two is the magic number.</b>
10:08
<b>If you know two independent, intensive properties like pressure and temperature,</b> <b>you have fixed the state, which means you can</b> <b>look up all the rest of your properties in tables.</b> <b>Ooh, foreshadowing, spooky, be afraid, TA Serenity.</b> <b>She's very scared.</b>
10:30
<b>If a system moves from one equilibrium state to another, that's called a process.</b>
10:38
<b>These lines connecting the four points, these are each processes.</b>
10:44
<b>Now, if a system goes through a series of processes and ends up exactly where it</b> <b>started, like our Rankine cycle, that's called a cycle.</b>
10:55
<b>In a cycle, your net change in internal energy is zero because no matter what path</b> <b>you took, you ended up back where you started.</b> <b>Empty bowl gets filled with food.</b> <b>I eat it until it's empty again.</b> <b>That's the dinner cycle.</b> <b>Who do I need to talk to you about the process of filling it up more promptly?</b> <b>So let's get to the first law of thermodynamics,</b> <b>which states you do not talk about thermodynamics.</b>
11:25
<b>I don't think students understand Fight Club references.</b>
11:30
<b>Before the first law, there's actually a zeroth law.</b> <b>And yes, they named it the zeroth law because they forgot about it the first time.</b> <b>So if system A is in thermal equilibrium with system</b> <b>B, and system B is in thermal equilibrium with system C,</b> <b>then system A is in thermal equilibrium with system C.</b> <b>Same as the transitive property in math.</b> <b>The thermometer is system B.</b>
12:01
<b>You use it to measure the two objects A and C, and if you get the same thermometer</b> <b>reading, then A and C are the same temperature.</b> <b>So in our house, if TA Serenity steals the warm spot on the couch where I was just</b> <b>sitting, then she is leveraging the zeroth law.</b> <b>She is matching my thermal state.</b> <b>Hey, in this house, I'm the thermal frame of reference.</b> <b>But now, the heavyweight champion, the first law of thermodynamics.</b>
12:34
<b>At its core, it's simply the principle of conservation of energy.</b> <b>Energy cannot be created or destroyed.</b> <b>It can only change forms.</b> <b>Before we look at the thermodynamics side, let's quickly review the mechanical</b> <b>energy terms you brought with you, like from physics and dynamics.</b>
12:58
<b>So first, kinetic energy.</b> <b>This comes from speed.</b> <b>Second, potential energy.</b> <b>This comes from height due to gravity.</b>
13:08
<b>Third, mechanical work.</b> <b>The simplest version of this was just force times distance.</b> <b>But in this course, we unlock a massive hidden vault of energy, internal energy.</b> <b>We use the capital letter U or lowercase U for internal energy per unit mass.</b>
13:32
<b>Now, to be formal, internal energy is the microscopic energy stored within the</b> <b>molecular structure of our working fluid.</b> <b>It's the sum of all the microscopic, kinetic and potential energies of the atoms.</b> <b>So they're spinning, vibrating, the chemical bonds.</b> <b>So when you heat them up, you are cranking up its internal energy.</b> <b>Napping in the sun isn't just napping.</b> <b>I'm adding internal energy to my system.</b> <b>Now, don't write this on a test, but</b>
14:14
<b>internal energy is energy stored as temperature.</b>
14:19
<b>There are counter examples to this.</b> <b>It's not a true statement.</b> <b>But as a memory aid, if you were to think of internal energy as temperature energy,</b> <b>then for me, you're at least pretty close.</b> <b>And hey, good news.</b> <b>For most thermo problems, kinetic and potential energy aren't changing.</b> <b>So we'll be able to ignore them.</b> <b>In this course, we'll usually only care about internal energy.</b>
14:52
<b>So how do we change the internal energy of a system?</b> <b>If mass isn't crossing the boundary, then energy</b> <b>can only cross in two distinct forms, heat and work.</b>
15:05
<b>Heat, Q, is energy transferred across the system</b> <b>boundary solely due to temperature difference, hot to cold.</b> <b>If there's no delta T, there's zero heat transfer.</b>
15:20
<b>Then work, W, it's energy transferred across the</b> <b>system boundary for any reason that's not temperature.</b> <b>Think of a spinning shaft, an electrical wire or a piston expanding or compressing.</b> <b>Now, we also deeply care about the rate of energy transfer.</b> <b>We call the time derivative of Q, Q dot.</b> <b>It's the rate of heat transfer.</b> <b>If Q is energy in kilojoules, then Q dot is power in kilojoules per second, which</b>
15:54
<b>is kilowatts in SI units, BTUs per hour in English units.</b>
16:02
<b>The rate of work is power, W dot.</b> <b>It shares the exact same units, kilowatts or actually horsepower in English units.</b> <b>Pray for SI units on your test.</b> <b>BTU and horsepower are awful units to work with.</b> <b>I mostly pray for tuna.</b> <b>Results have been mixed.</b> <b>Let's talk about the single biggest trap on thermodynamics exams, sign conventions.</b> <b>But you'll never get the signs wrong because you understand power plants.</b>
16:36
<b>That's why we started this video with power plants first.</b>
16:42
<b>The boiler and the turbine are the most</b> <b>important pieces, so both of their terms are positive.</b> <b>The boiler adds energy to the power plant as</b> <b>heat, so heat is positive when it enters the system.</b> <b>The turbine removes energy from the system to produce</b> <b>electricity, so work is positive when it leaves the system.</b> <b>When the system does work to the surroundings, that's positive.</b> <b>When you calculate the efficiency of your power plant, net</b>
17:18
<b>work divided by heat input, both of those terms are positive.</b> <b>I eat snacks, that's heat in.</b> <b>I kick my rocket, that's work out.</b> <b>No wonder I'm always out in energy deficit.</b> <b>This open system demands a snack refill.</b>
17:37
<b>Alright, one more vocabulary word before the equation.</b> <b>Adiabatic.</b>
17:43
<b>An adiabatic process is one where there is zero heat transfer across the boundary.</b> <b>This happens for two reasons.</b> <b>Either the system is perfectly insulated, which this is not, or the process happens</b> <b>so incredibly fast that heat simply doesn't have time to flow.</b> <b>Let's look at a classic demo, the compression igniter.</b> <b>I place a tiny scrap of cotton inside this glass cylinder.</b> <b>If I push the piston down slowly, heat transfers out to the room.</b>
18:19
<b>It's not adiabatic.</b> <b>But watch what happens if I slam the piston down violently.</b> <b>That instantaneous compression is adiabatic.</b> <b>Cool.</b> <b>Because the work was done so quickly, the heat didn't have time to escape.</b> <b>The massive spike in internal energy drove the</b> <b>temperature past the auto ignition point of the cotton.</b> <b>That flame is basically me when somebody turns on the vacuum.</b> <b>Adiabatic.</b> <b>Instantaneous.</b> <b>Panic.</b>
18:55
<b>No time for heat to escape.</b> <b>You want an adiabatic process?</b> <b>You can try yourself right now.</b> <b>Smash the like button.</b> <b>Seriously, it takes a small amount of work, but there's no heat transfer.</b> <b>That makes it adiabatic.</b> <b>Go ahead.</b> <b>Smash the like button.</b> <b>You'll see.</b>
19:13
<b>All right, here is the equation we have been</b> <b>building toward, the energy balance equation.</b> <b>Don't write this first equation down yet.</b> <b>I'm doing a small derivation.</b> <b>For a closed system, since no mass is crossing</b> <b>the boundary, the first law is this equation.</b> <b>Initial energy plus energy added or removed equals final energy.</b> <b>In thermo, we usually care about change in energy.</b> <b>Final minus initial.</b>
19:46
<b>So we can rewrite this with change in energy.</b> <b>Then substitute all our different forms of energy.</b>
19:56
<b>And in most textbooks and FE exam problems, the system is stationary, meaning</b> <b>changes in macro, kinetic and potential energy are completely negligible.</b> <b>So finally, this is the one you'll use.</b> <b>Box this equation.</b> <b>We've got some math problems in this video, too.</b> <b>It's not all lecture and explosions, and we don't even have to solve them quietly</b> <b>because your TA Serenity can sleep through anything.</b>
20:23
<b>So the piston cylinder, that's a closed system, no mass in or out.</b> <b>So I start with the closed system energy equation, delta U equals Q minus W.</b>
20:36
<b>The system performs boundary work on the surroundings, so the 50 is positive work.</b>
20:43
<b>Internal energy is our final answer.</b> <b>So how do we find Q?</b>
20:49
<b>If you ever find yourself with too many unknowns</b> <b>in your equation, go back to the problem statement.</b> <b>Look for a key adjective that might give you one of your unknowns.</b> <b>In this case, insulated.</b> <b>This means Q equals zero.</b> <b>No heat escapes through insulation.</b>
21:10
<b>Gives us the correct answer is A.</b> <b>The gas lost most of its internal energy because it</b> <b>had to use its own molecular vault to push the piston up.</b>
21:23
<b>Huge tip.</b> <b>Keep a keywords list of all the words your textbook</b> <b>and professor use to indicate something important.</b> <b>Like insulated means Q equals zero as the first one on your list.</b> <b>Don't mind me, just adding another reminder to your list.</b> <b>Let's do one more example problem to drive home the sign convention.</b> <b>This is the ultimate sign convention trap on the FE exam.</b> <b>Let's list our terms based on our power plant rules.</b>
21:57
<b>There's no mass going in or out of the</b> <b>tank, so this is a closed system energy balance.</b>
22:04
<b>30 kilojoules is negative work.</b> <b>Since the paddle wheel is doing work to the system, just like a power plant pump,</b> <b>positive work would require work by the system to the surroundings.</b> <b>So heat adds...</b>
22:23
<b>What was that?</b>
22:26
<b>So heat adds 15 kilojoules, so this is positive heat.</b> <b>Just like the boiler in the power plant.</b>
22:35
<b>And what about all those adjectives?</b> <b>Rigid means there's no other work terms, no</b> <b>boundary work, which we'll get to in later chapters.</b>
22:45
<b>Well insulated means no other heat terms.</b>
22:50
<b>So the double negative turns into addition.</b> <b>The correct answer is D.</b> <b>Both the heater and the paddle wheel were dumping energy into the closed space,</b> <b>so the internal energy had to go up.</b>
23:06
<b>Every time energy changes forms or crosses a boundary,</b> <b>nature extracts a tax.</b> <b>You never get a perfect exchange.</b> <b>We quantify this performance using thermal efficiency,</b> <b>denoted by the Greek letter η.</b>
23:26
<b>And η is desired output over required input.</b> <b>Efficiency is always a dimensionless fraction between zero and one, or 0% to 100%.</b>
23:38
<b>For our Rankine power plant, remember the sign</b> <b>convention since the pump adds energy to the system,</b> <b>its work term will be negative, so it gets subtracted.</b> <b>A perfect cycle would convert all boiler</b> <b>heat to turbine work, an efficiency of 100%.</b>
23:58
<b>But real cycles never get there.</b> <b>A coal power plant will have efficiency around 40%,</b>
24:06
<b>meaning 60% of the energy is rejected by the condenser as heat out.</b>
24:14
<b>Nuclear is even worse, like 33% efficiency.</b>
24:19
<b>On the submarine, when we weren't spying on people,</b> <b>we joked that our mission was to heat up the ocean.</b> <b>Wasn't really joking though.</b> <b>Two thirds of the reactor output was lost as heat from the condenser to the ocean.</b> <b>In another form, we can substitute and rearrange to</b> <b>get efficiency in terms of heat output to heat input.</b> <b>To get 100% efficiency, the turbine would have to use 100% of the energy,</b> <b>so that 0% would be lost in the condenser.</b>
24:54
<b>So if heat output would be 0.</b> <b>But, sneak preview of chapter 5,</b> <b>according to the second law of thermodynamics, this is impossible.</b>
25:06
<b>Now, before the next example problem, there's another useful term of the energy</b> <b>balance when working with power instead of energy.</b> <b>To shift to rate form, take a time derivative.</b>
25:20
<b>If the system is in steady state, then internal</b> <b>energy isn't changing, so box this equation now.</b>
25:31
<b>In a steady state closed system, heat and</b> <b>work have to perfectly cancel each other out.</b> <b>So, next example problem will have efficiency and our new rate form.</b>
25:46
<b>First step, net power output.</b> <b>We use the rate form of the closed system energy balance.</b>
25:55
<b>W.net equals Q.in minus Q.out.</b> <b>So 800 minus 480.</b>
26:04
<b>Network equals 320 megawatts.</b> <b>Now for efficiency, there's two ways to get there.</b> <b>Use whichever one your exam gives you the data for.</b> <b>So method one, efficiency is W.net over Q.in.</b> <b>The desired output over the thing you have to pay for input.</b> <b>320 over 800, so efficiency is 40%.</b>
26:31
<b>Method two, efficiency is 1 minus Q.out over Q.in.</b> <b>So 1 minus 480 over 800, and that still gets us back to point four.</b> <b>40% efficient.</b> <b>The 480 megawatts, that represents 60% of input energy.</b> <b>It's all rejected as heat from the condenser.</b> <b>These are the big white fluffy clouds from the cooling tower.</b> <b>It's not pollution, it's just wasted energy.</b> <b>And notice, you can solve this problem</b> <b>without ever knowing what fuel the plant burns,</b>
27:11
<b>how the boiler works, or anything about turbine blade curvature.</b> <b>The energy balance doesn't care.</b> <b>By defining my system around the overall power plant, I only needed to know about</b> <b>the energy balance crossing the boundary.</b> <b>Not about anything happening inside any of the individual components.</b> <b>And that is the first law of thermodynamics, conservation of energy.</b>
27:40
<b>So kinetic and potential energy can often be ignored.</b> <b>Internal energy is new.</b> <b>It's sort of like temperature energy, but not exactly.</b>
27:51
<b>The sign convention for heat and work is the</b> <b>easiest mistake to make in this entire course.</b> <b>So it's always worth a few seconds to double check your signs.</b> <b>And lastly, professors are evil.</b> <b>And we like to hide important information in the problem statement.</b> <b>Insulated, rigid, adiabatic, and a bunch</b> <b>more terms you'll learn in future chapters.</b> <b>Keep a list of these somewhere you can find easily, you will need it.</b>
28:26
<b>And now, if you have ever thought to yourself, "I hate equations.</b> <b>Why can't I just look everything up in a table?"</b> <b>I think that all the time.</b> <b>Then I have got great news for you.</b>
28:48
<b>In the last chapter, your TA serenity helped us balance energy.</b> <b>Now we need to know what that energy is inside our working fluid.</b> <b>That means evaluating properties.</b> <b>And that starts with understanding what phase the fluid is in.</b>
29:07
<b>We work almost exclusively with pure substances,</b> <b>materials with a fixed uniform chemical composition.</b> <b>Water nitrogen R134A.</b> <b>In a power plant, the working fluid is almost</b> <b>always water, and it changes phase throughout the cycle.</b>
29:28
<b>Three phases.</b> <b>You know them as solid-liquid gas.</b> <b>For water, we usually call the gas form vapor.</b> <b>But for our power plant, there's actually five keywords you need to know that</b> <b>describe this liquid-to-gas transition that happens</b> <b>in the boiler, which I can demonstrate in the lab.</b>
29:52
<b>Let's make this visual.</b> <b>Start with regular normal water.</b> <b>This is compressed liquid, also called subcooled liquid.</b> <b>Same thing, two names.</b>
30:05
<b>Turn on the heat.</b> <b>When it gets to 100 degrees Celsius, we are at a saturated liquid.</b>
30:12
<b>Now, any more heat added, and there it is.</b> <b>The water crosses the saturation curve and starts converting to steam.</b> <b>So inside the steam generator is a mixture of both liquid and gas.</b> <b>We are in between the saturation curves.</b> <b>Now, the steam leaving this tube is entirely gas.</b> <b>It is a saturated vapor.</b> <b>This is the right side of the saturation curve.</b> <b>This steam generator can't superheat, but imagine I</b>
30:49
<b>captured this steam and applied even more heat to it.</b> <b>That would keep pushing us further to the right of the curve.</b> <b>That superheated steam is certified, bonafide, top choice steam that is exactly</b> <b>what power plant turbines are looking for.</b> <b>And one more term for the two phase region that exists inside the steam generator,</b> <b>quality, which will appear in equations as x.</b> <b>Quality ranges from zero to one, essentially the percent of water that is vapor.</b>
31:26
<b>Zero means it's entirely liquid, a saturated liquid.</b> <b>X equals one means 100% vapor, a saturated vapor.</b> <b>Anything in between is a wet mixture.</b> <b>Quality only has meaning inside the two phase region.</b> <b>There's no such thing as the quality of a</b> <b>compressed liquid or quality of a superheated vapor.</b> <b>This definition makes no sense.</b> <b>The internet says cats are liquid, but clearly my quality is 100%.</b>
32:03
<b>So there has to be a mistake.</b> <b>Liquid is 0%, gas is 100%.</b>
32:09
<b>Cats are solid, serenity needs to stay off the internet and go touch grass.</b> <b>You know I'm not allowed outside.</b> <b>Let's visualize this on the most important</b> <b>diagram in property evaluation, the PV diagram.</b> <b>Pressure on the vertical axis, specific volume on the horizontal.</b> <b>For a pure substance, there's a curved dome shape.</b> <b>This is the saturation dome.</b> <b>Not exactly an upside down parabola.</b>
32:42
<b>Usually you should draw it steeper on the left and shallower on the right.</b> <b>So the peak is left of center.</b>
32:52
<b>It's kind of easy to remember since volume of</b> <b>water doesn't change very much with pressure.</b> <b>But for a gas, it's very easy to change volume of a gas.</b> <b>So everything to the left of the dome is a compressed liquid, regular normal water.</b> <b>Everything under the dome is a two phase mixture, liquid and vapor coexisting.</b> <b>Here pressure and temperature are locked together.</b> <b>If you know one, you automatically know the other.</b>
33:25
<b>That's why they call them saturation pressure and saturation temperature.</b> <b>They're a pair.</b> <b>So everything to the right of the dome is superheated vapor.</b> <b>High specific volume, high energy.</b> <b>And here is how your power plant maps onto this diagram.</b> <b>State one, saturated liquid leaving the condenser,</b> <b>the bottom left edge of the dome with a quality of zero.</b> <b>State two is a compressed liquid.</b>
33:59
<b>When it leaves the pump, it's left of the dome at a higher pressure.</b> <b>State three is all the way in the other</b> <b>corner, a superheated steam leaving the boiler.</b> <b>It's right of the dome and really high up.</b> <b>State four is going to be wet steam leaving the turbine, partially condensed inside</b> <b>the dome with a quality between zero and one.</b> <b>The peak of the dome is the critical point.</b> <b>Above it, there's no boiling bubbles during phase change.</b>
34:37
<b>Liquid and vapor basically become indistinguishable.</b> <b>For water, this happens at 374 Celsius and 22.1 megapascal pressure.</b> <b>Now, here's an important difference between this class and real world power plants.</b>
34:58
<b>Many modern power plants do operate in this supercritical region above the dome</b> <b>because the higher the temperature, the higher your overall efficiency.</b> <b>But we are not really going to cover that in this class.</b> <b>All of the systems and cycles on your homeworks and your tests, they will all be</b> <b>left under or right of the curves, not above the critical point.</b> <b>I don't like that we're skipping real world use</b>
35:32
<b>cases, but I do like that we're making it easier.</b> <b>TA Indiana approved.</b> <b>Before we open any tables, we need one more property,</b> <b>enthalpy, which is capital H or enthalpy per unit mass, lowercase h.</b> <b>So remember, you from like 10 minutes ago, internal energy, that molecular energy</b> <b>or temperature energy, you will only use that internal energy for closed systems.</b> <b>For open systems, you'll use enthalpy instead.</b>
36:08
<b>When fluid flows across a boundary through a turbine, through a pump or a nozzle.</b> <b>Yes, it still carries the same internal energy, but</b> <b>it also has to do work to push itself through the pipe.</b> <b>So that pressure volume term is called flow work.</b> <b>Because we see that combination of pressure and volume in almost every single open</b> <b>system problem, engineers got tired of writing them separately.</b>
36:43
<b>So we combined them into a single property called enthalpy as the same units as</b> <b>energy kilojoules or kilojoules per kilogram for specific enthalpy.</b> <b>And the best part, it's listed in property tables.</b> <b>You don't calculate enthalpy.</b> <b>You look it up.</b> <b>For closed systems, you look up internal energy, U.</b> <b>For flowing open systems, you look up enthalpy, H.</b>
37:14
<b>Right here, right now, this is the big one.</b> <b>There are 34 property tables and charts in the back of my thermo book, 68 actually,</b> <b>since there's both SI units and English units.</b>
37:29
<b>If you can use these reliably and find the information</b> <b>you need quickly, you are on track towards an A in thermo.</b> <b>If you can't, you will be on the struggle bus to "C" town, population you.</b> <b>I've been on that bus.</b> <b>The Wi-Fi is terrible and the seats are sticky.</b> <b>You do not want to be there.</b> <b>Now, if you call the whole collection of 34 tables,</b> <b>the steam tables, everybody will know what you mean.</b> <b>But there's really only a big three tables</b>
38:01
<b>that you'll use more than all the others.</b> <b>These are the three steam tables.</b> <b>Number one, saturated water temperature table.</b> <b>You use when temperature is given and you</b> <b>know or suspect that the fluid is saturated.</b> <b>Each temperature row lets you look up pressure,</b> <b>specific volume, internal energy, enthalpy and entropy.</b> <b>We'll talk about entropy more in chapter seven.</b>
38:30
<b>Now, why are there two volumes and three internal energies, you ask?</b> <b>Well, that's a great question.</b> <b>The left columns with subscript F apply to saturated liquid with zero quality.</b> <b>The right column with subscript G are for saturated vapor with a quality of one.</b> <b>The middle number when there are three is just the difference between the two.</b> <b>It just saves you a calculator work so you don't have to subtract them manually.</b>
39:03
<b>The second table, saturated water pressure table.</b> <b>It's the exact same thing except rows are</b> <b>organized by pressure instead of temperature.</b>
39:14
<b>The third table, superheated water vapor.</b> <b>Each mini table is a different pressure and then rows are different temperatures.</b>
39:25
<b>Now, there's also a compressed water table,</b> <b>but it's small and it's often skipped on exams.</b> <b>The standard accepted approach is approximate compressed liquid properties using</b> <b>the saturated liquid value at the given temperature.</b> <b>Not pressure.</b>
39:50
<b>Because water is incompressible, the pressure doesn't really help.</b> <b>Changing the pressure doesn't change any of the other properties.</b> <b>Use saturated liquid at the same temperature because when you change the</b> <b>temperature of water, the properties do change.</b>
40:08
<b>Let's learn the tables with an example problem.</b> <b>So step one, what's the fluid?</b> <b>Steam, so that is water.</b> <b>Step two, what's the phase?</b> <b>Is it liquid, saturated or gas?</b> <b>In this case, saturated with quality of 0.6.</b> <b>So we are in the two phase region between the saturation curves.</b> <b>Step three, which table do we go to?</b> <b>We know temperature, so we go to the saturated temperature table.</b>
40:38
<b>And step four, interpolate because our system isn't liquid or gas.</b> <b>It's in between, so we need a number in</b> <b>between the two numbers given to us in the table.</b>
40:53
<b>Interpolation is the process of finding values in the</b> <b>table that fall in between rows or in between columns.</b> <b>Like if your temperature were 101 degrees, that's not listed in the table.</b> <b>So you'd have to interpolate between the 100 degree row and the 105 degree rows.</b>
41:14
<b>You'll have to interpolate literally hundreds of times in this course.</b> <b>You have to get good at this.</b> <b>Hundreds?</b>
41:24
<b>They should make an app for that.</b> <b>Oh, there is.</b> <b>But you can't use it on the test because your professor is evil.</b>
41:33
<b>Real talk time.</b> <b>Your textbook has an equation for quality.</b>
41:39
<b>Don't use it.</b>
41:43
<b>Start all your interpolations with fractions.</b> <b>It reinforces physical understanding of what you're actually</b> <b>doing and prevents you from making conceptual or sign mistakes.</b>
41:59
<b>Quality is the percentage of water that is steam.</b> <b>If 60% of the water is steam, we need an</b> <b>enthalpy value that is 60% of the way from HF to HG.</b> <b>So set up your enthalpy values as a fraction that makes 60%.</b>
42:23
<b>Quality is H minus HF over HG minus HF.</b> <b>Total liquid to gas difference in the denominator, only your</b> <b>smaller difference to your system's value in the numerator.</b>
42:40
<b>So plug in numbers 1773, answer C.</b>
42:45
<b>You know, let's go ahead and do that 101 degree example, too.</b> <b>So what is the enthalpy of saturated steam at 101 degrees Celsius?</b> <b>I'll go to the temperature table since I know temperature.</b> <b>I'll grab HG values since I know it's saturated steam.</b> <b>But unfortunately, 101 falls in between 100 and 105.</b> <b>So interpolation time.</b> <b>Since 101 is in between two rows, we interpolate to</b> <b>find the HG value in between the two HG values we can read.</b>
43:23
<b>So I set up two fractions.</b> <b>The left fraction is what you know, temperature.</b> <b>The right fraction is what you want to find, enthalpy.</b> <b>Left fraction, the 101 falls in between 100 and 105.</b> <b>So we set this up to form a percentage, in this</b> <b>case 0.2, since 101 is 20% of the way from 100 to 105.</b>
43:52
<b>For the right fraction, we have H minus the 2675.</b> <b>And in the denominator, the 2683 minus 2675.</b>
44:04
<b>Notice how all four terms in both fractions</b> <b>correspond to the matching term in the other fraction.</b> <b>The 2683.4 is the H value at 105 degrees.</b> <b>And they are both in the same position of their respective fractions.</b>
44:27
<b>So the H I'm trying to find matches the position</b> <b>of the 101 degree temperature I'm trying to find H4.</b>
44:37
<b>So do the algebra and solve.</b> <b>H is 2677.2.</b> <b>This is the enthalpy 20% of the way from 2675 to 2683.</b> <b>I think I get it.</b> <b>But how do I know which table to use?</b> <b>One more set of rapid fire examples.</b> <b>So for each case, identify the phase, select the</b> <b>correct table, and then find the specific volume.</b>
45:08
<b>So for A, we're going to check the saturation table first.</b> <b>At a temperature of 300 degrees Celsius, pressure should be 8.58 megapascal.</b> <b>One megapascal is too low.</b> <b>Low pressure is gas.</b> <b>So check the superheated vapor table.</b> <b>One megapascal mini table, 300 degrees Celsius row, specific volume 0.25799.</b>
45:38
<b>For B, if quality is between 0 and 1, you're guaranteed saturated.</b> <b>With temperature 100, go to this saturated</b> <b>temperature table, and we are going to need to interpolate.</b>
45:54
<b>That gets us to V of 1.42.</b>
45:58
<b>And for C, first check the saturation table.</b> <b>At pressure equals 400 kilopascals, temperature should be 143.6.</b> <b>Our temperature of 50 degrees is too cold</b> <b>to be saturated, so it's a compressed liquid.</b> <b>It's not hot enough to be boiling.</b> <b>I could go to the compressed liquid table, but I don't have to.</b> <b>Just use specific volume at 50 degrees saturated liquid, 0.001012.</b> <b>Remember, you have to use temperature, not pressure, for compressed liquids.</b>
46:40
<b>Recap.</b> <b>First, learn which table to use.</b> <b>Always check the saturation table first if you're not sure.</b> <b>If your system is too hot or too low</b> <b>pressure, then switch to the superheated vapor.</b> <b>If your system is too cold or too high pressure, it's compressed liquid.</b> <b>Second, a warning.</b> <b>Different textbooks will have slightly different numbers in their tables.</b> <b>They're each calculated based on different sets</b>
47:12
<b>of experiments with different best-fit equations.</b> <b>Good news, different tables are pretty much always within 0.1% of each other.</b> <b>So discrepancies don't usually matter.</b> <b>Just use whichever set of numbers come in your textbook, and you'll be fine.</b>
47:31
<b>Never mix and match numbers from different textbooks.</b> <b>In the same problem, though, that will cause problems.</b> <b>But now that you know this, this might</b> <b>explain why sometimes you'll watch a YouTube video,</b> <b>and they're getting slightly different numbers than you.</b> <b>It might just be because they're using a different textbook.</b>
47:51
<b>All right, one more homework problem, and then we'll move on to PIVNERT.</b>
47:56
<b>We're going to have to look up values we need in the steam tables.</b> <b>So state one, I'll check the saturation table.</b> <b>One megapascal is 1,000 kilopascals.</b> <b>So temperature, saturation temperature should be 179.88.</b> <b>But we're hotter than this, so we must be superheated.</b> <b>So for change in internal energy, we'll need final and initial.</b> <b>So the initial internal energy is 2710.4.</b>
48:27
<b>So state two, pressure is down to 100 kilopascals,</b> <b>but we need two pieces of information to fix the state.</b> <b>I only have one.</b>
48:38
<b>Remember, professors are evil.</b> <b>So if you don't have enough information,</b> <b>reread the problem statement looking for those important keywords.</b> <b>In this case, isothermal means constant temperature.</b> <b>So T2 is still equal to 250 degrees.</b> <b>I check my saturation tables.</b> <b>At 100 kilopascal, saturation temperature should be 99.6.</b>
49:06
<b>250 degrees is too hot.</b> <b>So I go back to the superheated steam tables.</b>
49:13
<b>100 kilopascals is 0.1 MPa.</b> <b>So I go to the 1.1 megapascal mini table.</b> <b>Go to the 250 degrees Celsius row.</b> <b>Get a specific volume, 2.4.</b>
49:30
<b>And internal energy, 2733.</b>
49:34
<b>So step three, change in internal energy.</b> <b>Final minus initial, the 2733 minus 2710,</b> <b>23.5 kilojoules per kilogram.</b> <b>Final answer.</b>
49:47
<b>This is why I said before that internal energy equals temperature energy.</b> <b>Isn't entirely correct.</b> <b>This was an example where temperature didn't change, but internal energy did.</b> <b>But it only changed a little bit.</b> <b>So I still like it as a useful approximation or a memory aid.</b> <b>It's Pivnert time.</b> <b>Sort of.</b>
50:15
<b>Steam tables are your go-to for water.</b> <b>But for gases, air, nitrogen, oxygen, carbon dioxide,</b> <b>we have a much simpler tool, the ideal gas equation of state.</b> <b>In chemistry, you call this Pivnert.</b> <b>In mechanical engineering, we don't care about moles.</b> <b>We use Piv-mert with mass.</b> <b>Or even better, just divide by mass and use specific volume.</b> <b>P is absolute pressure, V specific volume, T absolute temperature,</b>
50:55
<b>always Kelvin or Rankine, and M mass.</b> <b>There are two different R's and you have to be super careful which one to use.</b> <b>Make sure your units all cancel.</b> <b>That's the easiest way to verify that you're using the right one.</b>
51:15
<b>The one you need is R, the specific gas constant for that particular gas.</b> <b>To find R, you can look it up directly, or</b> <b>sometimes you have to look up molar mass M</b> <b>and calculate it using the other R, R bar.</b> <b>The universal gas constant R bar, 8.314 kilojoules per kilomole Kelvin.</b>
51:43
<b>So you know R bar isn't the one to use in the PV equals RT equation</b> <b>because it has moles in it.</b> <b>And moles are for chemists, not engineers.</b>
51:57
<b>Don't memorize molar masses or R values, just look them up when you need them.</b>
52:05
<b>So when is the ideal gas assumption valid?</b>
52:11
<b>Low pressure or high temperature, far from the saturation dome.</b> <b>Air at atmospheric conditions, ideal gas, no problem.</b> <b>Steam at high pressure, near saturation, use the tables, not ideal gas.</b> <b>Also, this is important for ratio problems.</b> <b>If mass and R are constant, the ideal gas law simplifies even further.</b>
52:42
<b>So change two properties and you can solve for the third.</b> <b>That's the recipe for most ideal gas problems on the FE exam.</b> <b>Please don't mention recipes.</b> <b>It's been minutes since my last snack.</b> <b>This is an advanced gas law's demonstrator.</b> <b>Inside is just air.</b> <b>We've got two gauges, one for volume and one for pressure.</b> <b>And temperature, three gauges, like I said.</b>
53:11
<b>Watch what happens when I compress the air.</b> <b>Pressure goes up, volume goes down.</b> <b>If temperature holds steady, then quantity PV stays nearly constant.</b> <b>That's Boyle's law, a special case of the ideal gas law.</b> <b>We've got two tools in our chest so far for finding properties,</b> <b>steam tables and the ideal gas law.</b> <b>But there's actually a third and don't worry, it's just going to be three.</b> <b>This is the last one.</b>
53:46
<b>Ideal gas is a great approximation until it isn't.</b>
53:53
<b>Real gases deviate from ideal behavior,</b> <b>especially at high pressure or near the saturation dome.</b> <b>We quantify this with the compressibility factor, which gives us "piv-zert".</b>
54:10
<b>For an ideal gas, Z would just be 1, always just by definition.</b> <b>But for real gases, Z deviates from 1 depending on the conditions.</b> <b>Z is a correction factor that accounts for how non-ideal the gas is.</b> <b>To find Z, we'll use the generalized compressibility chart.</b> <b>It is not as scary as it looks.</b> <b>First, you'll usually actually see two or three compressibility charts together.</b> <b>These are all the same chart, they're just</b>
54:48
<b>like zoomed in or zoomed out from each other</b> <b>with a different range on the x-axis.</b> <b>So you want to use the most zoomed in version that matches your values.</b>
55:02
<b>Now, you'll have to do a bit of math before you can use it.</b> <b>The axes aren't just temperature and pressure,</b> <b>they are reduced temperature and reduced pressure.</b>
55:15
<b>You calculate these by first looking up the</b> <b>critical temperature and pressure in a table</b> <b>and then dividing your actual temperature and</b> <b>pressure by the critical temperature and pressure.</b> <b>Pr is the horizontal axis, Tr tells you which</b> <b>of all the swirly-do curves you need to follow,</b> <b>and then you'll read Z off the left vertical axis.</b> <b>If Z is close to 1, then just assume it's an ideal gas.</b>
55:47
<b>If Z deviates significantly, then apply the correction.</b>
55:52
<b>So three methods, tables, compressibility chart, ideal gas equation.</b> <b>Which should you use?</b> <b>Tables are essentially truth.</b>
56:05
<b>These are the most accurate.</b> <b>Compressibility cart,</b>
56:10
<b>compressi-</b>
56:13
<b>The compressibility cart,</b>
56:16
<b>the compressibility chart could be nearly as good,</b> <b>but it's hard to read a chart to high levels</b> <b>of precision, so it's probably second best.</b> <b>Ideal gas law is the easiest to do, but the least accurate.</b> <b>What's that, sleeping TA serenity?</b> <b>You want more than three methods?</b> <b>All right.</b> <b>Number four, the Benedict Webb Rubin</b> <b>equation has eight empirical constants per gas.</b>
56:48
<b>If a professor assigns you a full BWR problem in an intro thermo course,</b> <b>that professor is wrong and should feel bad.</b>
57:02
<b>You can tell them, from me, they should be nicer to you.</b> <b>Hey, you assign Reynolds transport theorem to your fluid students.</b> <b>Don't act like you're above all this.</b> <b>Touche.</b>
57:18
<b>We've got nitrogen at low pressure, so ideal gas law assumption is valid.</b> <b>Normal atmospheric pressure is about 100 kilopascals,</b> <b>so 200 kilopascals, it's only two atmospheres, that's not that much.</b>
57:38
<b>I cancel out mass and R, since they don't change, and sub in what we know.</b> <b>We're missing V1 and V2, so reread the problem.</b> <b>Rigid tank.</b> <b>That means V1 equals V2, so we can cancel those two out also.</b> <b>Calculator work, T2, 600 Kelvin choice D.</b> <b>If nothing else changes, double the pressure causes double the temperature.</b>
58:09
<b>This problem type appears on nearly every version of the FE exam.</b> <b>Rigid equals constant volume.</b> <b>Start with the general equation, cancel what's constant, and solve.</b> <b>If you see rigid on an exam, your next thought should be,</b> <b>volume is constant, so volume cancels.</b>
58:32
<b>Next example, let's look at the difference between approximation and reality.</b> <b>So, method one, the ideal gas assumption.</b> <b>I'll just look up R in my appendix, and I get specific volume, 0.01039.</b>
58:50
<b>Method two, using the compressibility factor correction,</b> <b>I first have to find the reduced</b> <b>parameters, reduced temperature and reduced pressure,</b> <b>in order to use the compressibility chart.</b> <b>So, I also look up in my appendix, PC and</b> <b>TC, the critical pressure and temperature.</b> <b>Do my division, get 0.812 for pressure, 1.085 for reduced temperature.</b> <b>Then, if I locate PR 0.812 on the bottom axis of the compressibility chart,</b>
59:25
<b>and move up vertically, and look for the reduced</b> <b>temperature line of 1.085, which there's not one,</b> <b>so it's in between the 1.05 and the 1.1 line, then read across to the right side,</b> <b>I get a Z factor of approximately 0.74.</b>
59:49
<b>This means carbon dioxide at this state only</b> <b>takes up 74% of the space an ideal gas would.</b> <b>That's a big difference.</b>
1:00:00
<b>So, PIVZERT, plug in the numbers, get 0.007688 meters cubed per kilogram.</b>
1:00:12
<b>If you use the ideal gas law to design this storage container,</b> <b>your volume calculation would be 35% too high,</b> <b>resulting in an unsafe or wildly inefficient layout.</b> <b>That's why I always design containers the old-fashioned way, by sitting in them.</b> <b>Alright, T.A. Serenity needs you to wake up</b> <b>and help me out with the chapter 3 summary.</b> <b>3 critical skills from chapter 3.</b>
1:00:43
<b>Number 1, identify the phase. Use the saturation temperature or pressure as your</b> <b>reference. Is your state above, below, or right inside the saturation dome?</b>
1:00:58
<b>Number 2, use the correct table. Saturated tables are for states inside the dome.</b> <b>The superheated table is to the right, and you</b> <b>don't really need the compressed liquid table,</b> <b>just use saturated liquid at your temperature.</b> <b>And number 3, know when the ideal gas assumption applies.</b> <b>Low pressure, high temperature, far from the saturation dome.</b> <b>If you're not sure, use Z, the compressibility</b> <b>chart, to correct the ideal gas law if you need to.</b>
1:01:37
<b>The tables feel overwhelming at first, but</b> <b>every problem follows the same three-step logic.</b> <b>Identify the state, pick the table, and then</b> <b>read the number or interpolate to get your answer.</b> <b>So, repetition is the teacher here. You'll</b> <b>need to interpolate more than a hundred times this</b> <b>semester. Learn it properly from the start, and</b> <b>it will save you many hours of your life later on.</b>
1:02:07
<b>So, enough theory, let's talk real application. Next chapter is closed systems.</b> <b>[Music]</b>
1:02:21
<b>Alright, TA Serenity. In chapter 2, we learned the closed system energy balance.</b> <b>Delta U equals Q minus W. Now, we need to</b> <b>actually calculate W. Specifically, moving boundary work.</b> <b>The kind you get from a piston pushing on a gas.</b> <b>This is the most common work term in any closed</b> <b>system. It shows up in internal combustion engines,</b> <b>refrigerators, compressors, the cylinder of a steam</b> <b>engine. Anywhere a piston is squishing or expanding</b>
1:02:58
<b>a fluid. If this box is a closed system, when I push</b> <b>up on the lid, I'm doing boundary work. So,</b> <b>remember the old work is force times distance.</b> <b>Well, on a piston, force is pressure times area.</b>
1:03:19
<b>So, multiply that area by the distance you push</b> <b>the cylinder, and what you have looks like a volume.</b> <b>And here is the boundary work equation.</b> <b>Your old friend calculus is back. Don't panic.</b> <b>First, a units check. Pressure times volume.</b> <b>Pressure in Pascals is Newtons per meter squared</b> <b>times volume in cubic meters. Gives Newton meters.</b> <b>Which is joules. So, yes, it works out to energy.</b> <b>Second, and this is the big one, boundary work</b>
1:03:57
<b>equals the area under the curve on a PV diagram.</b> <b>But watch your signs. If the path is going to</b> <b>the right, that's expansion, that's positive work.</b> <b>If your path on the PV diagram is going to the</b> <b>left in compression, that's negative work. This</b> <b>matches the power plant sign convention from</b> <b>chapter two. Work done by the system is positive.</b> <b>Boundary work is path dependent, not just for</b> <b>signs. Two processes can connect the same start</b>
1:04:35
<b>and end states and produce completely different</b> <b>amounts of work because they trace a different</b> <b>path. So, unlike pressure or temperature, work</b> <b>is not a property of a state, it's a property of</b> <b>the process connecting two states. But we do still have to evaluate that integral.</b> <b>You'll see four common shapes for these</b> <b>processes. The first isochoric, constant volume,</b> <b>like a rigid container. It's drawn as a</b>
1:05:13
<b>vertical line, so there's no area underneath a vertical</b> <b>so work is zero. Done. Zero work on the system, zero work for engineering students.</b>
1:05:27
<b>The second, isobaric, constant pressure, a</b> <b>horizontal line. This case, the area is just</b> <b>a rectangle. Work is equal to base times</b> <b>height. Third is polytropic, the general workhorse.</b> <b>Any process that follows the form PV to the n</b> <b>equals c, where n is the polytropic exponent</b> <b>and c is a constant. I'm not going to derive the</b> <b>whole thing. Hey, calculus is your friend, not mine.</b>
1:06:04
<b>But amazingly, it does eventually simplify down</b> <b>to this easy formula that doesn't even have the</b> <b>constant in it. Though, your problem</b> <b>statement probably won't give you both P2 and V2,</b> <b>so you'll have to use the constant to find</b> <b>either P2 or V2, whichever one you're missing.</b>
1:06:28
<b>The fourth one is isothermal, a constant</b> <b>temperature. PV equals MRT, where MRT is constant.</b> <b>This is what happens when your polytropic</b> <b>process has n equals 1. And because you can't divide by</b> <b>zero, for an ideal gas only, we can substitute PV</b> <b>equals RT into the integral and you get a natural</b> <b>log. This isothermal case shows up</b> <b>constantly on the FE exam. One other keyword reminder,</b> <b>adiabatic for no heat. That is a polytropic</b>
1:07:06
<b>process with n equal k, where k is the specific heat</b> <b>ratio, which we'll meet later. Another new</b> <b>variable? I'll wear my fanciest tie. That's time</b> <b>I forgot, and entropy never respected me again.</b> <b>Here's the most relatable closed system in your</b> <b>life, a car engine cylinder. This is the model of</b> <b>a four-stroke gasoline engine. Intake, the piston</b> <b>moves down, fresh air-fuel mixture flows in.</b> <b>Compression, piston up, volume drops, pressure</b>
1:07:47
<b>spikes. The work during compression is</b> <b>negative. Work done on the gas. Power costs money. Then,</b> <b>ignition. A big addition of heat followed by the</b> <b>power stroke. The piston gets pushed back down by</b> <b>the expanding hot gas. That work is positive.</b> <b>Work done by the gas. This is what's actually</b> <b>turning the tires. Then last is the exhaust</b> <b>stroke. The piston goes back up, pushing out the burned</b> <b>gases. Remember the Rankin cycle for our power</b>
1:08:28
<b>plant? This engine is following a different cycle,</b> <b>the Otto cycle. If I plot all four strokes</b> <b>on a PV diagram, they form a closed loop.</b> <b>Area under the top expansion process is</b> <b>positive work. Area under the compression process is</b> <b>negative work. The area enclosed equals net work</b> <b>of the cycle. Uh, what about that random horizontal</b> <b>line down at the bottom? Those are intake and</b> <b>exhaust processes. It's two lines right on top</b>
1:09:08
<b>of each other. They cancel each other out.</b> <b>So that's it. The entire engine, four strokes,</b> <b>thousands of RPM, all the noise and fire, it</b> <b>boils down to an enclosed area on a PV plot.</b>
1:09:26
<b>Step one, identify the process. Polytropic, PV to the N equals constant.</b>
1:09:35
<b>Step two, I need V2 and I don't have it directly,</b> <b>but I have everything else. So I use PV to the N</b> <b>equals constant, meaning PV to the N initial</b> <b>equals PV to the N final. So V2 is 0.1008 meters</b> <b>cubed. And this is compressed. Volume dropped to</b> <b>one fourth. So yes, that makes sense. Sanity check</b> <b>passed. Step three, plug into the polytropic</b> <b>work equation. And I get work negative 68.3</b> <b>and negative, which makes sense. We compressed the</b>
1:10:21
<b>gas, so work was done on the gas, not by the gas.</b> <b>If you got plus 68, you flipped a sign</b> <b>somewhere. Let's stitch boundary work back into the closed</b> <b>system energy balance from chapter two. For</b> <b>most closed system problems in this course,</b> <b>the only work term is boundary work. So delta U</b> <b>equals Q minus boundary work. So a quick checklist</b> <b>problem. One, calculate boundary work as area under</b>
1:11:01
<b>the PV diagram curve. Two, find delta U from tables</b> <b>or coming up next, you can find it using</b> <b>specific heat. Step three, solve for whatever's left.</b> <b>And then step four, feed Serenity more</b> <b>snacks. Serenity, have you been using my tablet?</b> <b>How dare you accuse me without any evidence?</b>
1:11:32
<b>All right. Well, don't ever say I don't follow procedures.</b>
1:11:43
<b>Oh, and if you have multiple processes that form</b> <b>a closed loop, a cycle, then all of your changes</b> <b>in internal energies from each of the process</b> <b>have to all add up to zero since you end up back</b> <b>where you started. So all of your delta U's</b> <b>for each individual process have to add to zero.</b> <b>Sometimes that last little trick is needed to</b> <b>find your last unknown. So that's it. That's the</b> <b>template for every problem in this chapter.</b>
1:12:14
<b>Even step four, feeding Serenity extra snacks.</b> <b>She assures me that step is not optional. For</b> <b>water and steam, we look up internal energy in</b> <b>the tables, but for ideal gases, air, nitrogen,</b> <b>oxygen, most textbooks have tables for these also,</b> <b>but there's another way. You can use specific</b> <b>heat. How much energy does it take to change the</b> <b>temperature of something by one degree? That is</b> <b>the definition of specific heat. There are two of</b>
1:12:48
<b>them. Cv, specific heat at constant volume and Cp,</b> <b>specific heat at constant pressure. Both have units</b> <b>of kilojoules per kilogram Kelvin. Now for</b> <b>an ideal gas, and this is the magical part,</b> <b>internal energy and enthalpy depend on</b> <b>temperature only, which means we get two clean equations.</b> <b>And warning, always use Cv when finding</b> <b>internal energy. Always use Cp when finding enthalpy.</b> <b>It doesn't matter whether your problem is</b>
1:13:33
<b>constant volume or constant pressure. Think of these</b> <b>subscripts, V and P, as historical leftovers</b> <b>that don't really apply anymore. Are you finding U</b> <b>or H? That's all that matters. Finding U, Cv,</b> <b>finding H, enthalpy, Cp. So you're usually interested</b> <b>in Cv closed system, Cp open system. And good</b> <b>news, you only have to decide for gases, liquids</b> <b>and solids. Since they are incompressible, they</b>
1:14:11
<b>don't have Cv, only Cp. Just one. So you'll never</b> <b>get them mixed up for liquids or solids. Hey,</b> <b>Indy, guess why I've been having trouble getting a job?</b> <b>Because you're lazy and irresponsible? No, it's</b> <b>because cats are liquid, so I don't have a Cv.</b> <b>So two more relationships to relate terms</b> <b>you've already heard in this video. Cp is equal to Cv</b> <b>plus R. That's kind of interesting. Gas</b>
1:14:43
<b>constant appears here, but Cp is always bigger than Cv.</b> <b>Why? It's because when you heat a gas at</b> <b>constant pressure, some of your energy has to go into</b> <b>pushing the piston out. That's boundary work.</b> <b>At constant volume, all of the heat goes straight</b> <b>into raising temperature. So you need more</b> <b>energy per degree at constant pressure to account for</b> <b>this work that you're also doing. That ratio K,</b>
1:15:22
<b>the specific heat ratio, for air, K is about 1.4.</b>
1:15:28
<b>For monatomic gases like helium, 1.67. K shows</b> <b>up in adiabatic processes and compressible flow.</b>
1:15:41
<b>This is why you'll see lots of homework</b> <b>problems in your textbooks using air in a polytropic</b> <b>process with N equal to 1.4. With N equal to K,</b> <b>this is an adiabatic process. No change in heat.</b> <b>It's one of the easiest cases to model, but</b> <b>it's not always this way. A few minutes ago,</b> <b>we did an example problem with N equal to 1.3.</b> <b>That was not adiabatic, so there would have been</b> <b>a heat gain or loss in that system that we just</b>
1:16:15
<b>didn't calculate. So there wasn't just boundary work.</b>
1:16:21
<b>Oh, and one warning, Cv and Cp actually drift slightly with temperature.</b> <b>Across a small delta T, you can treat them as</b> <b>constant, but if your temperature is changing by</b> <b>hundreds of degrees, then you can get a</b> <b>better estimate by averaging your initial and final</b> <b>specific heats. It's not perfect, but it's usually good enough.</b>
1:16:49
<b>So how do you actually measure specific heat for a real substance? Calorimetry.</b>
1:16:56
<b>Let's do a material identification experiment. I</b> <b>have a chunk of hot gray metal at a known mass.</b> <b>I'm going to drop it into a known mass of room</b> <b>temperature water inside this insulated foam cup.</b> <b>The water heats up, the metal cools down. They</b> <b>meet at some equilibrium temperature in the middle.</b>
1:17:23
<b>No work is being done, and the cup is insulated, so Q across the boundary is zero.</b>
1:17:30
<b>Energy is conserved inside the cup, so all of</b> <b>the heat lost by the metal equals all of the heat</b> <b>gained by the water. So MCP delta T for the metal</b> <b>equals MCP delta T for the water. Some call this</b> <b>equation MCP delta T, the MCAT equation, since the</b> <b>delta looks like an A. I approve of all cat-related</b> <b>memory aids. So since I know CP for water, every</b> <b>term in that equation is known except for CP of the</b>
1:18:05
<b>mystery metal. Solve for it, and then check a</b> <b>material's property table to find the metal with</b> <b>the closest specific heat. And it looks like</b> <b>this is probably aluminum. And hey, there's an</b> <b>AL stamped right on the top of this cylinder. What</b> <b>a coincidence. And me, I thought his name was Al.</b>
1:18:27
<b>Nah, cut that joke. It's not funny. Anyway, this</b> <b>is how every value in your textbook specific heat</b> <b>tables originally got there. Tens of thousands of careful calorimeter measurements.</b>
1:18:41
<b>At the exam style question, closed system, piston cylinder, air, low pressure,</b> <b>ideal gas assumption is good. So delta U is</b> <b>going to equal Q minus W, find delta U, find W,</b> <b>and then we'll get Q. So step one, boundary work.</b> <b>It's isobaric, so a horizontal constant pressure,</b> <b>so work is area of a rectangle, 10 kilojoules.</b>
1:19:12
<b>So step two, internal energy change. For an ideal</b> <b>gas, delta U will be MCV delta T. We'll</b> <b>find the mass using PIVMERT. So PV equals MRT,</b> <b>rearrange M is PV over RT, 0.1161 kilograms.</b>
1:19:33
<b>Step four, find T2. We're at a constant pressure,</b> <b>volume doubles. So from the ideal gas law, V over</b> <b>T is equal to V over T at time two. So if we double</b> <b>the volume, we double the temperature. T2</b> <b>equals 600 degrees. So delta T, 300 degrees.</b>
1:19:55
<b>So we plug in 0.1161 times 1.005 times 300. Wrong, it is a trap.</b>
1:20:07
<b>The problem gave me CP, not CV. Remember, your</b> <b>professor is evil, unless you're in my class.</b> <b>I trick you because I care. I want you to learn.</b> <b>Other professors trick you because they are mean.</b> <b>Don't trust them. You have to use the relationship CV equals CP minus R. So 0.718.</b> <b>Will be CV. So now plug that in 25</b> <b>kilojoules and then solve Q equals delta U plus work.</b>
1:20:46
<b>Q is 35 kilojoules B. Now there is a humongous</b> <b>shortcut for this problem also, which we just</b> <b>haven't quite got to yet. But previewing what</b> <b>we'll do in the next chapter, which is solve for Q</b> <b>directly for constant pressure. Specifically,</b> <b>there's a one line move at constant pressure.</b> <b>Q is MCP delta T. The boundary work is already baked in to CP.</b>
1:21:19
<b>So 35 kilojoules, one line. That's it. Cat</b> <b>equations are known for their efficiency.</b> <b>And that is the secret of CP. It's defined exactly so that at constant pressure,</b> <b>Q equals MCP delta T. Boundary work is</b> <b>included automatically. So constant pressure heating</b> <b>problems become one liners. The same trick</b> <b>works for constant volume, a vertical line on a PV</b> <b>diagram with no area underneath. So work is zero</b>
1:21:57
<b>and delta U would then be Q equals MCV delta T.</b> <b>Also just one line. Recognize the process. Use</b> <b>the matching specific heat. Skip half the algebra.</b> <b>That is the real speed weapon on the FE exam. So</b> <b>chapter four, closed systems for things to remember.</b>
1:22:23
<b>Boundary work is area under the PV curve.</b> <b>It's path dependent, not a property. This one</b> <b>polytropic formula covers most processes.</b>
1:22:36
<b>Specific heat lets you skip the tables for</b> <b>ideal gases. And last, the constant pressure shortcut Q equals MCP delta T.</b> <b>It works for both closed systems and as</b> <b>you'll see in the next chapter, also open systems.</b>
1:22:56
<b>All right, enough engines. I am a nuclear engineer and this is a power plant class.</b> <b>Next chapter, we open the system back up.</b> <b>Mass crosses the boundary, turbines, pumps,</b> <b>nozzles, heat exchangers, every component in</b> <b>the Rankine cycle we promised you in chapter one.</b> <b>Welcome to Jurassic Park. I mean open systems.</b>
1:23:22
<b>All right, if you want to keep your dinosaurs</b> <b>in a well-functioning amusement park, closed</b> <b>systems are great. Mass stays put and only energy</b> <b>crosses the boundary. But open up the floodgates</b> <b>literally and you get open systems. We</b> <b>usually call these control volumes. Same thing, open</b> <b>system and control volume, same thing. Mass</b> <b>flows in, mass flows out. We're finally analyzing</b>
1:23:54
<b>the real components of the Rankine cycle. Turbines, pumps, nozzles, heat exchangers,</b> <b>the actual machines that make a power plant work.</b> <b>The energy balance still applies, but we need to</b> <b>one new conservation law first. I'm not fat, I'm</b> <b>conserving mass. If mass is crossing the boundary,</b> <b>we need to track it. Mass is conserved. It doesn't</b> <b>just appear or disappear. But just like energy in</b> <b>chapter two, we want this in rate form. So we use</b>
1:24:30
<b>mass flow rate, m dot, units of kilograms per second.</b>
1:24:36
<b>Three ways to write it. Density times area of</b> <b>the pipe times velocity. Or since specific volume</b> <b>is just the inverse of density, you do area</b> <b>times velocity divided by specific volume.</b>
1:24:54
<b>Use whichever of these forms matches the givens</b> <b>in your problem. And a quick warning of capital V.</b>
1:25:04
<b>Capital V usually means volume. Lowercase V</b> <b>usually means specific volume. What the heck was that?</b>
1:25:15
<b>But then there's velocity, which most textbooks</b> <b>also call capital V. Could be capital or lowercase.</b> <b>Understand the meaning of your equation. Don't</b> <b>just read the letter, double check your units.</b> <b>Because V, there's a lot of different Vs. So</b> <b>this is the mass rate balance for a control volume.</b> <b>In English, the rate of mass accumulating inside</b> <b>a box equals what flows in minus what flows out.</b>
1:25:52
<b>If TA Serenity eats one churu and doesn't poop it out, her mass increases.</b> <b>You do not want to be near the litter box when it's m dot out time.</b>
1:26:06
<b>For steady state, which is the vast majority</b> <b>of your problems, mass inside the control volume</b> <b>isn't changing. The left side is zero. So</b> <b>box this equation. Now for a single inlet,</b> <b>single outlet like a pipe. It's even</b> <b>simpler. Just mass in equals mass out. That's mass</b> <b>conservation. So now back to energy. Goodbye Serenity. Energy is my domain.</b>
1:26:41
<b>When mass crosses a boundary, it carries energy</b> <b>with it, internal, kinetic, and potential. But</b> <b>it also has to push itself through the pipe,</b> <b>doing flow work. Remember chapter three. So that's</b> <b>exactly why we invented enthalpy. H equals U</b> <b>plus PV. Closed systems get U, open systems get H.</b> <b>Do not mix those up. The full energy balance</b> <b>for a control volume is a beast. Keep work,</b> <b>enthalpy, kinetic energy, potential</b>
1:27:19
<b>energy. Let's tame it. If you're in steady state,</b> <b>then DEDT, the rate of energy change is zero.</b> <b>For a common single inlet, single outlet case,</b> <b>mass flow rates are also equal. So you can</b> <b>factor out the m dot. So box this equation.</b> <b>This is the steady flow energy equation,</b> <b>SFEE. It's the workhorse of chapter five. For most</b> <b>components, kinetic and potential energy changes</b> <b>are negligible. So your SFEE collapses even further.</b>
1:27:58
<b>You'll use this version 90% of the time. 60% of</b> <b>the time, it works every time. I learned that from</b> <b>cat based cologne. So you know it's accurate.</b> <b>Now here's where the SFEE gets friendly. Each</b> <b>Rankine cycle component has its own set of</b> <b>simplifications. Learn the pattern and the math</b> <b>becomes fast. Turbines and compressors. These move</b> <b>fluid so fast through the blades that heat transfer</b>
1:28:33
<b>doesn't have time to happen. So they are adiabatic. Q dot equals zero. Kinetic and</b> <b>potential energy changes are small compared to</b> <b>work. So you can also drop them. What's left?</b> <b>Is work is just m dot change in enthalpy. For</b> <b>a turbine, H1 is bigger than H2. So W dot is</b> <b>positive work out of the system. The power plant sign convention from chapter two.</b>
1:29:05
<b>For a pump or compressor, H2 is bigger than H1.</b> <b>W dot comes out negative. That's work into the</b> <b>system. Same equation, the signs sort themselves</b> <b>out. Now don't memorize these individual component</b> <b>equations. Just use the main energy balance</b> <b>equation every time. If you understand how these</b> <b>components work, you'll be able to just cross</b> <b>off the terms you don't need. And the one equation</b> <b>will work for every piece of machinery. So it's</b>
1:29:40
<b>way less to remember. A nozzle accelerates a fluid</b> <b>by squeezing it into a smaller area. A diffuser</b> <b>does the opposite. There's no shaft, no moving</b> <b>pieces at all. So no work. W dot equals zero.</b> <b>It's typically adiabatic. So Q dot is zero too.</b> <b>Little or no change in height. So what's left</b> <b>is essentially converting enthalpy to velocity</b> <b>or the other way around. Now warning, every</b> <b>student messes this up the first time. Except for you.</b>
1:30:17
<b>You're awesome. You're going to nail it. If</b> <b>your velocity is meters per second, then V squared</b> <b>gives you joules. But enthalpy is usually in</b> <b>kilojoules. You'll need to divide your V squared</b> <b>over two term by a thousand to convert it also</b> <b>to kilojoules. This is the one time kinetic energy</b> <b>actually matters in this course. The whole</b> <b>point of a nozzle is to trade enthalpy for speed.</b> <b>At snack time, I trade dignity for speed.</b>
1:30:53
<b>Heat exchangers. Two fluids passing each other,</b> <b>exchanging heat through a wall. There's no shaft.</b> <b>So W dot equals zero. Kinetic energy and potential</b> <b>energy terms are also negligible. Now you have a</b> <b>choice and pick wisely. Professors love this trap.</b> <b>They are evil. Not me. Just all of the rest of us.</b> <b>Draw a control volume around just one fluid stream.</b> <b>Then Q dot equals M dot delta H for that stream</b>
1:31:30
<b>only. Or draw your control volume around the whole</b> <b>heat exchanger. And now Q dot equals zero.</b> <b>Because all the heat stays inside the box.</b> <b>Then in that case, the heat lost by one</b> <b>fluid equals the heat gained by the other fluid.</b> <b>So same idea as the calorimetry demo from</b> <b>the last chapter. Same energy bookkeeping,</b> <b>just different equipment. If I draw the control</b> <b>volume around the entire kitchen, I can't get in</b>
1:32:11
<b>trouble for stealing snacks since they're</b> <b>all still in the kitchen after I eat them.</b> <b>I promise you turbines in chapter one. So here</b> <b>is a model of a turbine. This is a multi-stage</b> <b>steam turbine. Real power plant turbines look</b> <b>more or less like this, only 50 feet long and cost</b> <b>millions of dollars and are made of metal</b> <b>instead of plastic. So exactly the same. It's got three</b> <b>stages with two stators in between. So why</b>
1:32:45
<b>multiple stages? Because no single set of blades</b> <b>can efficiently extract all of the energy from</b> <b>high pressure steam. Each stage drops the pressure</b> <b>a little, it extracts some work, and then</b> <b>hands the fluid to the next stage where you have</b> <b>stationary blades to redirect the steam at the</b> <b>perfect angle to hit the next set of moving blades.</b>
1:33:15
<b>I'm putting energy in as airflow. Each stage</b> <b>takes a bite and by the time the air exits,</b> <b>most of the pressure and kinetic energy is gone, converted to shaft rotation.</b>
1:33:31
<b>In your energy balance equation, the turbine is</b> <b>a single black box. H1 in, H2 out, and work comes</b> <b>out the side. But this is what's happening</b> <b>inside. There's multiple stages of coordinated energy</b> <b>extraction. I'm still mad that Duke Energy</b> <b>rejected my cat-powered turbine power plant's design.</b>
1:33:59
<b>All right, FE exam style problem. We've got</b> <b>turbine, constant volume. We'll assume steady state.</b> <b>Start with the SFEE. We're adiabatic, so Q dot is</b> <b>zero. It's a turbine, so we'll ignore kinetic and</b> <b>potential energy. So what's left? W dot is M dot,</b> <b>change in enthalpy. So two enthalpy lookups in the</b> <b>steam tables. State one, we're given is five</b> <b>megapascals and 500 degrees Celsius. Normally,</b> <b>I would check the saturated table first, but if</b>
1:34:34
<b>state two is saturated vapor, then state one is</b> <b>going to have to have even more energy. So state</b> <b>one is probably a superheated vapor. So I check</b> <b>there first, find the five megapascal mini table,</b> <b>look for 500 degrees Celsius, and boom. There it is,</b> <b>3434.7. State two is a saturated vapor at 50 kilopascals. So the quality is one,</b> <b>since it's a saturated vapor. I go to the</b> <b>saturated pressure table, 50 kilopascals row,</b>
1:35:06
<b>and get Hg column for the saturated gas, 2645.2.</b> <b>Remember, to look up values in the steam tables,</b> <b>you always need two pieces of information.</b> <b>At state one, I had pressure and temperature.</b> <b>At state two, I had pressure and saturated vapor,</b> <b>meaning quality equals one was my second piece of</b> <b>information. So now plug everything in and</b> <b>answer C, 7,900 kilowatts. So notice the algorithm,</b>
1:35:46
<b>identify the component, apply the right simplification two table lookups, plug in.</b> <b>That's the recipe for every steady state open</b> <b>system problem in this chapter. Get fast in this</b> <b>exact workflow. You'll see it probably five times</b> <b>in the thermo section on the FE exam, super common.</b>
1:36:08
<b>Now, almost everything in this chapter is steady</b> <b>state, but there are a few classic problems where</b> <b>it's filling a propane tank or discharging a</b> <b>pressure vessel, charging an HVAC accumulator.</b> <b>Mass inside the control volume is changing with</b> <b>time for these cases, you're filling something.</b>
1:36:29
<b>The DMDT term that we've been zeroing out,</b> <b>the change in mass in the control volume,</b> <b>it's alive now. For these, we need to</b> <b>integrate the rate equations from time one to time two.</b> <b>Let's show how this can simplify as an example</b> <b>problem. So unsteady flow, the tank starts empty,</b> <b>M1 equals zero. Rigid means no work, insulated</b> <b>means no heat. There's no exit, so M out is zero.</b>
1:37:07
<b>So our mass balance, M2 equals mass in. So</b> <b>the energy balance, Q is zero, W is zero,</b> <b>there's no exit flow, and initial energy is zero because the tank started empty.</b>
1:37:24
<b>So M2U2 is mass in times H in.</b>
1:37:31
<b>So since M2 is equal to mass in, the mass</b> <b>cancels and we're just left with final internal energy</b> <b>equals enthalpy that entered. That's the</b> <b>magic move for these problems. The final internal</b> <b>energy in the closed system tank equals the</b> <b>enthalpy that passed through the control volume</b> <b>border in the supply line to enter the system. So</b> <b>we look up enthalpy in, if the supply was 1 MPa,</b> <b>300 degrees Celsius. Saturation at 1 MPa is 179,</b>
1:38:10
<b>so this is superheated. So we go to the superheated</b> <b>table, 1 MPa, 300 degrees Celsius, H in,</b> <b>3051.6. So U2, final internal energy, 3051.6.</b> <b>To find temperature 2, our final state, pressure 2</b> <b>is 1 MPa, U2 is 3051.6, so we go back to the 1 MPa</b> <b>We scan the U column internal energy for</b> <b>3051.6. It falls in between 400 C and 500 C.</b>
1:38:52
<b>So we interpolate left fraction is U, right</b> <b>fraction is T. And notice the placement of</b> <b>all four terms on both sides. Each internal</b> <b>energy corresponds to the corresponding temperature</b> <b>in the same place on the other fraction. So</b> <b>pop quiz, if you fill a tank with 300 degrees</b> <b>Celsius steam, what temperature will it be? 456</b> <b>degrees, duh. Well, what the heck is even that?</b> <b>The final temperature is higher than the 300</b>
1:39:30
<b>degree supply line because as the steam pushes its way</b> <b>into the tank, it does flow work on the gas</b> <b>already inside. That flow work becomes internal</b> <b>energy, which increases temperature. This is</b> <b>a classic counterintuitive result. It's also a</b> <b>favorite of all those evil professors who aren't</b> <b>me. You never let me catch the red dot. If that's</b> <b>evil, I don't know what is. Chapter 5, open</b> <b>systems or control volumes, two names, same thing.</b>
1:40:13
<b>Four things to take with you. Mass is</b> <b>conserved, just like energy. Mass in equals mass out,</b> <b>unless there is accumulation, filling or emptying a tank.</b>
1:40:28
<b>Two, use enthalpy, not internal energy. In open</b> <b>systems, flow work is already baked in to enthalpy.</b>
1:40:38
<b>Three, each component, turbine, heat exchanger,</b> <b>nozzle, they all have a simplification pattern,</b> <b>but don't memorize 50 different equations for every different possible machine.</b> <b>If you understand what the machine does and how</b> <b>it works, you only need to know the energy balance,</b> <b>and then you just cross off all the terms</b> <b>you don't need. But be extra careful with heat</b> <b>exchangers. Choose your control volume wisely.</b>
1:41:11
<b>Sometimes you want to draw it only around one</b> <b>of the fluid flows. Sometimes you want to draw it</b> <b>around both. And finally, for unsteady tank filling</b> <b>problems, U final equals enthalpy supplied. Easy</b> <b>game. Next chapter, the second law of thermodynamics,</b> <b>which is also you don't talk about</b> <b>thermodynamics. Stop making Fight Club references. That movie is</b> <b>27 years old. Fine. The second law of thermodynamics governs thermal efficiency,</b>
1:41:48
<b>and it introduces entropy. Entropy is the</b> <b>thing that everybody pretends to understand,</b> <b>but they actually don't. So good luck with that.</b>
1:42:02
<b>Seek. Evolve.</b>
1:42:06
<b>All right, your TA Indiana is back because he's</b> <b>all about energy. The first law is about quantity.</b> <b>Energy is conserved. But the second law is</b> <b>about quality and direction, which is way more</b> <b>complicated. Yay. According to the first law of</b> <b>thermodynamics, a cup of hot coffee sitting on</b> <b>your desk could spontaneously absorb heat from</b> <b>the cooler room and boil itself. No energy would be</b> <b>created or destroyed in that process, but it</b>
1:42:42
<b>never happens. Processes have a natural direction</b> <b>to understand why we need the heavyweight</b> <b>anchor of mechanical engineering, the second law of</b> <b>thermodynamics. But before we hit the law itself,</b> <b>we need two definitions. First, a thermal energy</b> <b>reservoir. This is a hypothetical body with a</b> <b>massive thermal energy capacity that can absorb</b> <b>or reject an infinite amount of heat without</b> <b>changing temperature. Think of the atmosphere</b>
1:43:18
<b>or the ocean. If a reservoir supplies heat, we call</b> <b>it a source. If it absorbs heat, we call it a sink.</b> <b>Second, a heat engine. These are cyclic devices</b> <b>designed to convert heat into work. Every single</b> <b>heat engine on earth does four things. It</b> <b>receives heat from a high temperature source. It converts</b> <b>part of that heat into work, usually by a</b> <b>spinning shaft. It rejects the remaining waste heat</b>
1:43:59
<b>to a low temperature sink. And then the cycle</b> <b>repeats. A power plant Rankin cycle, that is a</b> <b>heat engine. An Otto cycle for the engine in</b> <b>your internal combustion car, another heat engine.</b> <b>You pay for QH worth of fuel and your power</b> <b>plant sells W net worth of electricity. Or your car</b> <b>converts W net into miles. QL is exhaust. It's</b> <b>necessary by physics. Am I a heat engine? Step one,</b> <b>cry for food. Step two, convert food into naps.</b>
1:44:45
<b>Step three, reject hair balls onto the carpet.</b> <b>Step four, repeat. That will never get through</b> <b>peer review. To measure how well a heat engine works,</b> <b>we use thermal efficiency. Like we said in</b> <b>chapter two, efficiency is always what you want,</b> <b>divided what you have to pay for it. For a</b> <b>heat engine, you want net work output. You pay</b> <b>for the heat going in. So what's the maximum possible efficiency? It's not 100%.</b>
1:45:25
<b>But we need some theory first before we can</b> <b>calculate it. We need the second law of thermodynamics.</b> <b>Now, there are two classic statements of the</b> <b>second law. Hold up. Two classic statements?</b> <b>How many statements of the second law of thermodynamics are there? Well,</b> <b>Wikipedia lists at least a dozen. I want one. Don't move. I'll be right back.</b> <b>Well, hurry up. I'm not waiting for you.</b>
1:46:06
<b>Well, so one of the reasons the second law is</b> <b>so hard for students is because there's no single</b> <b>concise description of it. There's dozens of</b> <b>different versions that really all say the same</b> <b>but just in different ways. But there's one</b> <b>that's the most popular, the Kelvin Planck statement.</b> <b>In English, no engine, no matter how cleverly</b> <b>designed, can take heat from a single hot source</b> <b>and convert 100% of it to work. There must</b>
1:46:41
<b>always be some heat rejected to a cold sink.</b> <b>Hot things don't give up their energy for no reason. Energy flows from hot to cold,</b> <b>and your heat engine can be a tax collector</b> <b>pulling out energy along the way. But a 100%</b> <b>efficient heat engine is physically impossible.</b> <b>Oh, since I'm 100% efficient in everything I do,</b> <b>that means I'm not a heat engine. Glad we</b> <b>cleared that up. Let's make this abstract law</b>
1:47:17
<b>completely undeniable. This is a classic steam</b> <b>engine. It's a living, breathing heat engine.</b> <b>Right here, these burning fuel pellets act as</b> <b>our high temperature source, dumping heat into the</b> <b>boiler. The water absorbs that energy, boils</b> <b>into high pressure steam, and expands inside the</b> <b>cylinder, pushing the piston to do work. W net,</b> <b>which spins the heavy flywheel. Ah, but look here</b> <b>at the exhaust valve. See that steam escaping?</b>
1:47:55
<b>That is QL. That is the mandatory tax required</b> <b>by the second law. The engine cannot extract work</b> <b>unless that steam expands out into the cooler room,</b> <b>which acts as our low temperature sink. If I plug</b> <b>that exhaust hole completely, the piston locks up</b> <b>instantly and the wheel stops turning, which</b> <b>I'm not going to do because I don't want to break</b> <b>anything. But point stands. No heat rejection, no work. That's Kelvin-Planck.</b>
1:48:33
<b>Don't like Kelvin-Planck? Don't worry, I got another one. The Clausius statement.</b>
1:48:39
<b>Drop an ice cube in your coffee. The ice cube</b> <b>melts and the coffee cools. Heat flowed from hot</b> <b>to cold naturally. No help is needed. Now,</b> <b>imagine the reverse. The ice cube gets colder</b> <b>while the coffee gets hotter, spontaneously.</b> <b>That would not violate the conservation of energy.</b> <b>But that never happens. What about a</b> <b>refrigerator, you ask? They pull heat from the cold inside</b>
1:49:13
<b>and dump it into the hot kitchen outside. How? Well, you plug it in.</b>
1:49:22
<b>The Clausius statement says, "Heat won't flow</b> <b>uphill on its own, but with external work input,</b> <b>it can." My refrigerator won't work unless I plug</b> <b>it in. And I won't work before my morning churu.</b>
1:49:41
<b>A refrigerator and a heat pump are the same device. Same components, same cycle.</b> <b>The only difference is which reservoir you</b> <b>care about. A refrigerator, energy flows from cold</b> <b>to hot in order to make the cold space colder.</b> <b>The useful output is QL. A heat pump, energy flows</b> <b>from cold to hot in order to make the hot space</b> <b>hotter. The useful output is QH. The first law for</b> <b>a cycle is the same as before. QH is equal to QL</b>
1:50:24
<b>plus network in. But we don't measure efficiency</b> <b>for refrigerators and heat pumps. Why? Because</b> <b>the ratio comes out greater than one. And engineers</b> <b>got tired of explaining to non-engineers that</b> <b>400% efficiency is actually okay in this case.</b> <b>So we renamed it coefficient of performance.</b>
1:50:50
<b>And there's a clean relationship between the two.</b> <b>Coefficient of performance of a heat pump is</b> <b>the coefficient of performance of a refrigerator</b> <b>plus one. The extra plus one is because every</b> <b>joule of work you put in eventually gets dumped</b> <b>as heat into the hot reservoir. It's like a buy</b> <b>one get one deal. You pay to pump the cold air</b> <b>and you get the compressor's waste heat</b> <b>absolutely free. I use that exact same plus one formula for</b>
1:51:22
<b>treats. The coefficient of a perfect meal is</b> <b>always the number of snacks I just ate plus one more.</b> <b>Heat pumps are low-key amazing at heating</b> <b>buildings. Way better than an electric resistance</b> <b>heater because heat pumps steal energy from</b> <b>outside air for free. And your only cost is the</b> <b>compressor. All right. FE exam style problem.</b> <b>Step one find the work using first law on a cycle.</b> <b>QH equals QL plus work. So 200 kilojoules.</b>
1:52:05
<b>Step two coefficient of performance. This is</b> <b>what you want divided by what you have to pay</b> <b>in order to get it. You want the cold space to be</b> <b>colder so the numerator is QL. You pay for the</b> <b>work input. Gives a coefficient performance of</b> <b>three. Answer B. For every one joule of</b> <b>electricity you pay for three joules of heat are pulled from</b> <b>the refrigerator. Not too shabby. Indy. Indy</b> <b>you're not going to come back for this part?</b>
1:52:42
<b>He talks about perpetual motion machines all the</b> <b>time. Engineers inventors and your TA Indiana over</b> <b>there they have spent centuries trying to break</b> <b>the first and second laws. They have all failed.</b> <b>They will continue to fail. The US Patent Office</b> <b>has a standing policy. They will not even review</b> <b>a perpetual motion machine application without a</b> <b>working prototype. And they will never get one.</b> <b>Their failures come in two different flavors.</b>
1:53:16
<b>A PMM of the first kind violates the first law.</b> <b>It creates energy from nothing. The output</b> <b>is greater than the input. These are sometimes</b> <b>branded as free energy devices. A PMM of the</b> <b>second kind violates the second law. It doesn't</b> <b>create energy but it converts heat to work with</b> <b>100% efficiency. It's an engine with no exhaust.</b>
1:53:45
<b>Or it moves heat uphill from cold to hot</b> <b>with no work input. Both kinds are impossible.</b> <b>Do not fund your TA Indiana's Kickstarter. We</b> <b>already determined the second law sets legal limits</b> <b>for efficiency. Every engine you'll ever</b> <b>analyze falls into one of three buckets. Reversible,</b> <b>the theoretical best. Frictionless. Perfect. It</b> <b>doesn't actually exist anywhere in the universe</b> <b>but it's our ceiling. You could undo it</b>
1:54:23
<b>completely without leaving any trace on the system or</b> <b>your surroundings. Irreversible, this is the</b> <b>real world. Every actual engine, it has losses,</b> <b>friction, mixing. Things that can't be undone.</b> <b>Like saying "I love you." You can't un-say it.</b> <b>Once it's out there, it's out there. And last,</b> <b>impossible. Efficiency exceeding the reversible</b> <b>maximum. Or 100% conversion of heat to work.</b> <b>It violates the second law. It does not exist.</b>
1:55:05
<b>It can't exist. Reversible, irreversible,</b> <b>impossible. Three kinds. If you solve a problem</b> <b>and you calculate an efficiency above the</b> <b>reversible maximum, you made a math error.</b> <b>That is not your professor's fault. Probably.</b> <b>Maybe. Some professors are lazy and don't actually</b> <b>check the problems they assign so also that could</b> <b>have happened. So what makes a process irreversible?</b>
1:55:37
<b>Friction is the big one. Unrestrained expansion of</b> <b>a gas. Heat transfer across a non-zero temperature</b> <b>difference. Mixing of different fluids. Current</b> <b>flowing through an electrical resistor, which we</b> <b>call I-squared R-losses. Or chemical reactions.</b> <b>If any of these are present, and they always are,</b> <b>your process is irreversible. Real processes are</b> <b>always irreversible. The reversible process is the</b>
1:56:14
<b>asymptote that can never quite be reached. My</b> <b>nap is reversible. I lie down. I stand up. No</b> <b>trace left behind. Except for the mountain of cat</b> <b>hair that she leaves behind. So, irreversibility</b> <b>confirmed. In 1824, a French engineer named</b> <b>Sadi Carnot asked the question, "What does the most</b> <b>possible heat engine look like?" His answer is</b> <b>the Carnot cycle. Four reversible processes between</b> <b>two temperature reservoirs. Isothermal expansion</b>
1:56:59
<b>at the hot temperature. That's heat added. Then,</b> <b>adiabatic expansion. No heat, just a</b> <b>temperature drop down to the cold temperature. Then,</b> <b>isothermal compression at the cold</b> <b>temperature, where QL heat is rejected. Lastly, adiabatic</b> <b>compression. No heat, but temperature climbs back</b> <b>up to the hot temperature. If work is area under</b> <b>on a PV diagram, then I fully admit this</b> <b>diagram looks like a nightmare to integrate. But don't</b>
1:57:39
<b>panic. In the next chapter, we will introduce</b> <b>the TS diagram, temperature entropy. And there,</b> <b>the Carnot cycle is a beautiful, clean rectangle.</b>
1:57:55
<b>So, stick with me until then. It's coming. But for</b> <b>the main takeaway is that all four steps are</b> <b>reversible. Therefore, it's impossible to actually</b> <b>build this engine. But the math is beautiful.</b> <b>And it sets the efficiency ceiling for everything.</b> <b>Box this one. That's the maximum thermal efficiency</b> <b>of any heat engine operating between two temperature</b> <b>reservoirs. Two non-negotiable details. The</b>
1:58:32
<b>temperature must be absolute. Kelvin or Rankin,</b> <b>not Celsius or Fahrenheit. Get this wrong, and</b> <b>your answer turns to garbage. And your professor</b> <b>will look at you with deep sadness. An important</b> <b>historical trivia. Lord Kelvin actually used the</b> <b>heat engine to create the Kelvin temperature</b> <b>scale. Kelvin temperatures were literally invented</b> <b>for this. Don't forget to use them. And</b> <b>second, the design lesson. A bigger temperature gap</b>
1:59:12
<b>means a bigger maximum efficiency. So,</b> <b>engineers either crank T-hot up with better materials or</b> <b>supercritical steam, higher furnace</b> <b>temperatures, or drop T-low down like a deep ocean condenser,</b> <b>or colder cooling water, or drawing a vacuum.</b> <b>Right? Both T-hot increases or T-low decreases.</b> <b>Both increase your efficiency ceiling higher.</b>
1:59:48
<b>Now, this same form of efficiency can also be applied</b> <b>to refrigerators and heat pumps for the</b> <b>maximum possible coefficient of performance with two</b> <b>sources and sinks. So, these equations should</b> <b>look familiar because we've seen them before.</b> <b>But now we're just using them with temperature.</b> <b>Last time you saw them, those T's were Q's for heat.</b> <b>So, you use temperature to find the</b> <b>theoretical maximum. You use heat when you're calculating</b>
2:00:24
<b>the actual performance of your real system. The</b> <b>Carnot cycle is theoretical. You can't actually</b> <b>reach it, but the Stirling cycle is its real world</b> <b>cousin. Two isothermal processes plus two constant</b> <b>volume processes. It's the closest practical</b> <b>thing we have to a Carnot engine. And look at that.</b> <b>A burner heats the bottom. That's the hot reservoir.</b> <b>The room air on the top, that's the cold reservoir.</b>
2:01:02
<b>The temperature difference drives the cycle.</b> <b>Pistons move air back and forth between the</b> <b>hot and cold ends. Work comes out of the</b> <b>flywheel and the flywheel spins a small generator that</b> <b>lights up the LEDs. There's no combustion inside</b> <b>the engine. There's no fuel injection, no explosions.</b> <b>Just work from a temperature difference.</b>
2:01:39
<b>Stirling engines powered some early refrigerators.</b> <b>They fly on deep space NASA probes for thermal</b> <b>management. And they're studied today for low-grade</b> <b>waste heat recovery. They are efficient, quiet,</b> <b>and elegantly simple. But they still never quite</b> <b>reach Carnot. Where's Indiana? I'm doing all the</b> <b>physics and all the comedy. He better be picking</b> <b>up snacks. So next example problem, step one,</b>
2:02:15
<b>find the actual efficiency. So for this we use Q</b> <b>and we get 50 percent. Step two, find the Carnot</b> <b>reversible maximum possible efficiency for these</b> <b>temperatures, which is 62.5 percent. And compare</b> <b>them. Since 50 percent is less than 62.5 percent,</b> <b>this engine is operating below the reversible</b> <b>ceiling. That's allowed, that's good. That's the</b> <b>entire real normal world. So answer is B,</b> <b>irreversible. If the actual had been exactly 62.5,</b>
2:02:57
<b>then it would have been reversible. If it</b> <b>had been 70 percent, that would be impossible.</b> <b>So congratulations, 70 percent, you've broken</b> <b>the laws of physics. And possibly your homework</b> <b>rate. So this problem type appears in some form</b> <b>on nearly every FE exam. Three steps, calculate</b> <b>the actual, calculate the Carnot, and compare. All</b> <b>right, next a little bit bigger practice problem.</b> <b>We're building up to it. So a heat pump problem,</b>
2:03:30
<b>we want to keep the hot side hot. So the useful</b> <b>output is QH. Step one, convert temperatures to</b> <b>absolute. We are Kelvin only. Step two, use the</b> <b>reversible coefficient of performance to find out</b> <b>the best possible case. 14.75 is our best coefficient</b> <b>of performance maximum possible. That is huge.</b> <b>For every one kilowatt of electricity you use,</b> <b>you deliver 14.75 kilowatts of heat into the</b> <b>house. But this is the reversible ceiling. Nobody</b>
2:04:15
<b>actually gets quite to that level. Real</b> <b>residential heat pumps come in around coefficient performance</b> <b>three or four. That's still way better than an</b> <b>electric resistance heater, which is permanently</b> <b>stuck at a coefficient of performance of one. All</b> <b>right, step three, find the required work input.</b> <b>So the house is losing heat at 75,000 kilojoules</b> <b>per hour. So the heat pump must deliver heat at</b>
2:04:46
<b>same rate to hold temperature steady. So Q hot</b> <b>75,000. Work in, right, just rearrange about 5,000</b> <b>kilojoules per hour. Convert that to kilowatts by</b> <b>converting the hours to seconds. 1.41 kilowatts.</b>
2:05:07
<b>So the minimum power to heat this house</b> <b>reversibly is about 1.4 kilowatts. A real heat pump with</b> <b>a performance of four would draw about 5.2</b> <b>kilowatts. An electric resistance heater would draw</b> <b>20.8 kilowatts for the same heat delivery.</b>
2:05:31
<b>And now you know why heat pumps are taking over</b> <b>residential heating. I am a heat pump. I</b> <b>extract heat from sun beams and deliver it to laps.</b> <b>Coefficient of performance? Infinite, since I require</b> <b>zero electricity. But you require constant snacks.</b> <b>That's biofuel, not electricity. Totally</b> <b>different. Chapter six, the second law of thermodynamics,</b> <b>four things to take with you. One, heat engines need a temperature gap.</b>
2:06:10
<b>No cold reservoir, no work. That's Kelvin</b> <b>Planck. Two, heat flows downhill on its own from hot to</b> <b>cold. But to go uphill from cold to hot, that</b> <b>costs work. That's Clausius. That's the entire reason</b> <b>refrigerators and heat pumps exist. Number three,</b> <b>engines use efficiency. Refrigerators and heat pumps</b> <b>coefficient of performance. But they're all still</b> <b>useful output divided by paid input. And fourth,</b>
2:06:49
<b>Carnot efficiency is the absolute ceiling for</b> <b>heat engine between two temperatures. But when</b> <b>calculating it, use absolute temperatures only.</b> <b>Honor Lord Kelvin's legacy. I'm back. And I've got</b> <b>the best second law statement. While it's impossible to achieve Carnot efficiency,</b> <b>it's even more impossible to exceed it. It's</b> <b>grammatically nonsense. I hate it. I love it.</b> <b>It helps me remember the difference between</b>
2:07:27
<b>impossible because of friction and impossible</b> <b>because of math and physics. The Carnot ceiling</b> <b>is less than 100% because of math and physics.</b> <b>Real engines operate less than the reversible</b> <b>Carnot ceiling because of friction and vibration.</b> <b>Irreversible. That's what TA Indiana's statement says. So to me, it's perfect.</b> <b>Good job, buddy. Your statement's perfect.</b> <b>Students love it. But speaking of irreversible,</b>
2:07:59
<b>buddies, these are also why we need the</b> <b>next chapter. Because to actually quantify the</b> <b>irreversibility to put a single number on how</b> <b>far we are from perfection, we need a new property.</b> <b>The most feared and misunderstood property in all of thermodynamics. Entropy.</b>
2:08:25
<b>Whoa.</b>
2:08:28
<b>All right, TA Indiana. Real quick. Before</b> <b>entropy, there's a third law of thermodynamics.</b> <b>It says that entropy of a perfect crystal at</b> <b>absolute zero is exactly zero. That's the whole</b> <b>law. We need a reference point for entropy</b> <b>and the third law hands it to us for free.</b> <b>So mention it on a test, get the point, move</b> <b>on. That was a lot easier than the second law.</b> <b>And we're not even done with the second law yet.</b>
2:09:04
<b>Here's the real reason everybody struggles with</b> <b>entropy. The second law was complicated</b> <b>because there's a million different definitions,</b> <b>but at least they all agreed with each other.</b> <b>There are different ways to say the same thing.</b> <b>But for entropy, there are actually two accepted</b> <b>definitions and they do not fully agree with each</b> <b>other. Whoa, careful. So the classical definition,</b>
2:09:34
<b>the classical definition based on the experimental</b> <b>definition of Clausius. For a reversible process,</b> <b>change in entropy is a ratio of heat transferred</b> <b>divided by the temperature of the system. If</b> <b>you extend the implications of this work, you get</b> <b>other formulations of the second law of</b> <b>thermodynamics, including entropy of an isolated system</b> <b>never decreases. That's a hard law. No</b>
2:10:05
<b>exceptions, ever. And guess what? If we draw our box around</b> <b>the entire universe, that is an isolated system.</b> <b>So the entropy of the universe never decreases.</b> <b>That's the second law of thermodynamics. But the</b> <b>second definition, the statistical definition from Boltzmann</b> <b>entropy is a count of all the microscopic</b> <b>arrangements that could correspond to a macroscopic</b> <b>observation. More possible rearrangements, higher</b>
2:10:44
<b>entropy. So your TA serenity, freshly groomed with</b> <b>her fur still on her body. That's a relatively</b> <b>low entropy state. There are far fewer ways for her</b> <b>on her body than like scattered all over the</b> <b>house. Spread the same hair across the couch, my shirt,</b> <b>in my food. There's way more possible</b> <b>arrangements, therefore higher entropy. But here's the key. In</b> <b>this framework, entropy decreasing spontaneously</b>
2:11:22
<b>isn't actually illegal. All the cat hair all over</b> <b>clothes and furniture, it could spontaneously</b> <b>still end up back on serenity all on its own.</b> <b>Boltzmann just says it's so astronomically unlikely we treat it as impossible.</b>
2:11:45
<b>Those two definitions don't say the same thing.</b>
2:11:50
<b>Clausius says entropy decreasing is</b> <b>impossible. Boltzmann says astronomically improbable.</b>
2:12:01
<b>For all the air in the room to spontaneously</b> <b>rush to one corner for no reason, Clausius says</b> <b>this can't happen. Boltzmann says it could happen. It just won't.</b>
2:12:17
<b>Physicists and philosophers have been</b> <b>arguing about how to reconcile this discrepancy for</b> <b>over a hundred years. There is no consensus yet.</b> <b>So if you've always felt like you almost understood</b> <b>entropy but not quite, that's not you. The</b> <b>definition itself is scientifically unsettled</b> <b>at its very foundation. Now for me, I love poker.</b> <b>I love probability. But I do not like the Boltzmann</b> <b>interpretation of entropy. I'm a Clausius man.</b>
2:12:55
<b>Give me rules and order. Because we're engineers.</b> <b>We build real things in the real world. You don't</b> <b>need a full microscopic theoretical framework to</b> <b>use entropy on the macroscopic level. Our</b> <b>great grandparents already did all the math for us.</b> <b>We just look up entropy in a table. The steam</b> <b>tables we've been using for the past few chapters,</b> <b>entropy is in those. It's S. It's those last</b> <b>three columns we've been ignoring until now.</b>
2:13:31
<b>Entropy is complicated, but it's just another</b> <b>property. So look it up in the table just like</b> <b>every other property. And definitionally, that's</b> <b>everything you need to know to pass your thermo</b> <b>course and pass the FE exam and call out any</b> <b>know-it-all classmates or co-workers. So we'll leave the</b> <b>fundamental interpretation to the physicists and</b> <b>philosophers. They can have it. In chapter three,</b> <b>we drew the PV diagram, pressure, volume. Now</b>
2:14:05
<b>meet its cousin, the TS diagram, temperature on the</b> <b>vertical, entropy on the horizontal. Same</b> <b>saturation dome, same left side compressed liquid, right side</b> <b>superheated vapor underneath the dome</b> <b>two-phase mixture. Why do engineers love TS diagrams?</b> <b>Two reasons. One, the area under the curve on a</b> <b>TS plot equals heat transfer Q. Just like area</b> <b>PV diagram was work. So T times DS, that's heat.</b>
2:14:46
<b>And two, remember how in the last chapter I said</b> <b>the Carnot cycle was going to clean up perfectly?</b> <b>Well here it is. Now we're talking, even I could</b> <b>integrate this. So two horizontal lines at T hot</b> <b>and T cold, the isothermal heat transfer steps,</b> <b>and then two vertical lines are the adiabatic</b> <b>reversible steps. A perfect rectangle, area inside</b> <b>equals net work. Area under the top is QH.</b>
2:15:20
<b>Beautiful, easy. A vertical line on the TS diagram</b> <b>means constant entropy. We call that isentropic.</b> <b>An isentropic process is one where entropy doesn't</b> <b>change delta S equals zero. So for that to</b> <b>happen, two conditions must be true. Adiabatic,</b> <b>so no heat transfer, Q equals zero, and</b> <b>reversible. No friction, no mixing, no irreversibilities.</b>
2:15:56
<b>Adiabatic plus reversible equals isentropic.</b> <b>So let's see an isentropic process in an example</b> <b>problem. All right, it's isentropic, so S2</b> <b>will equal S1. Now at state one, we're given four</b> <b>megapascals and 400 degrees Celsius. Now</b> <b>saturation at four megapascals is 250.35 degrees Celsius.</b> <b>We are hotter than that, so it must be</b> <b>superheated. So I go to the four megapascal mini table, 400</b> <b>degrees Celsius row, get entropy one, 6.77. Now</b>
2:16:34
<b>state two were 50 kilopascals and entropy of 6.77.</b> <b>So I check the saturated pressure table at 50</b> <b>kilopascals, see that SF 1.09, SG 7.59. Since our</b> <b>entropy 6.77 falls in between them, state two is</b> <b>a wet two phase mixture. So we've identified the</b> <b>phase. So I set up the quality fraction, same</b> <b>technique as chapter three, just with entropy</b> <b>instead of enthalpy. I say that the quality is</b> <b>0.87, 87%. So the closest answer, B. That's actually</b>
2:17:21
<b>the workhorse pattern for the Rankine cycle.</b> <b>The turbine exit from your power plant is almost</b> <b>always a wet mixture. And isentropic analysis</b> <b>with S2 equal to S1 is how you find quality.</b> <b>So lock this pattern in. But real world machines</b> <b>aren't isentropic. Turbines, pumps, compressors,</b> <b>nozzles. If we insulate them or water passes</b> <b>through them really quickly, we are approximately</b> <b>adiabatic. But they aren't reversible. Real</b>
2:18:00
<b>components fall short. And we measure how short</b> <b>with isentropic efficiency. My isentropic</b> <b>efficiency at locating snacks is exactly 1.0.</b> <b>I am the Carnot cat. So let's look at our</b> <b>Rankine cycle power plants on a TS diagram.</b>
2:18:24
<b>For a turbine, you want big work output. So</b> <b>the isentropic case, a vertical line gives you</b> <b>the maximum. Reality gives you this line that</b> <b>goes slightly to the right. H4 actual will be</b> <b>higher than H4 isentropic. Meaning we did not get</b> <b>as much energy out of the system as theoretically</b> <b>possible. So actual work over ideal work. Real</b> <b>turbines have isentropic efficiency around 85 to</b> <b>90%. For a pump or compressor, you want minimum</b>
2:19:07
<b>work input. The isentropic case is still a vertical</b> <b>line. The real world case also ends up to the</b> <b>right. So H2 actual is larger than H2 isentropic.</b> <b>But the efficiency fraction is upside down</b> <b>compared to the turbine version is to ensure</b> <b>that the larger term is in the denominator. This</b> <b>time, ideal work over actual work. So the trick</b> <b>on these problems is solve the isentropic case</b> <b>first. And then if you know entropy, say at points</b>
2:19:45
<b>1 and 3, then you know S2 entropic and S4 entropic. They're the same as S1 and S3.</b>
2:19:55
<b>Let's actually measure the gap between theory</b> <b>and reality. So this is a heat engine cycle.</b> <b>Air is sealed inside the small cylinder. It's</b> <b>connected to a piston that lifts a 200 gram mass.</b>
2:20:13
<b>From 1 to 2, we add mass.</b>
2:20:23
<b>So we are now at a lower volume and a slightly higher pressure.</b> <b>Next from 2 to 3, we're going to switch from the cold over to the hot.</b>
2:20:41
<b>You'll see volume increases, but pressure stays basically the same.</b>
2:20:50
<b>Now from state 3 to state 4, we're going to remove the mass.</b>
2:21:04
<b>That's just going to be a slight increase in volume and a decrease in pressure.</b>
2:21:13
<b>And then to complete the cycle back to stage 1, go back to the cold bath.</b>
2:21:24
<b>Now you can see pressure basically stayed the same, but volume decreased.</b>
2:21:31
<b>And that is a heat engine. The area enclosed is equal to net work done,</b> <b>exactly like the Otto cycle from chapter 4. We</b> <b>can measure it, so we can also measure QH going in</b> <b>to hot water. If you divide them, you get</b> <b>the actual efficiency. Then we can compute the</b> <b>Carnot efficiency just using the two bath</b> <b>temperatures. Carnot efficiency is the maximum</b> <b>possible. It will always be higher than the</b>
2:22:04
<b>real world value, always. The gap between those two</b> <b>numbers, that's irreversibility. That's entropy</b> <b>generation. Friction in the piston, heat transfer</b> <b>across the finite temperature difference, the gas</b> <b>not having time to fully equilibrate, equilibrate,</b>
2:22:26
<b>equilibrate. I don't think any of those was</b> <b>correct. But every real cycle has irreversibilities.</b> <b>Aww, this engine is sad because it isn't</b> <b>Carnot. Every engine on earth is sad because it's not</b> <b>Carnot. Welcome to engineering. For steam</b> <b>and refrigerants, just look up S in the table.</b> <b>Same workflow as for U and H. Now for an ideal gas,</b> <b>we have two formulas. Same logic as specific heat,</b> <b>CV and CP in chapter four. Just pick</b>
2:23:01
<b>whichever one matches the information you have.</b> <b>So temperature in Kelvin always and natural logs,</b> <b>not log base 10. So use the first equation if you</b> <b>have temperatures and specific volumes. Use the</b> <b>second if you have temperatures and pressures.</b> <b>Both give the same answer if you have all three,</b> <b>use whichever one you want. They'll both give you</b> <b>exactly the same answer. For an isentropic ideal</b>
2:23:31
<b>gas, where delta S equals zero, the second equation</b> <b>simplifies into what's called isentropic</b> <b>relations, where K is the specific heat ratio from chapter</b> <b>four. So this is your fast tool for isentropic</b> <b>compression and expansion of air. Ideal means</b> <b>perfect, gas means fart. Whenever your engineering professor says ideal gas,</b> <b>pretend he's describing the perfect fart. Class will never be the same again.</b>
2:24:08
<b>So just like we had energy balances, chapters two</b> <b>and five, there's an entropy balance for any system.</b> <b>Same accounting structure with one critical</b> <b>addition. So for a control volume in rate form,</b> <b>entropy accumulates in the CV due to heat</b> <b>crossing the boundary, due to mass carrying it in,</b> <b>mass carrying it out, and irreversibilities, which</b> <b>generate it internally. The term S dot gen is total</b>
2:24:43
<b>entropy generation rate caused by the irreversibilities</b> <b>like friction. And here is the golden rule of the second law.</b>
2:24:55
<b>S dot gen can never be negative. If S dot gen is</b> <b>greater than zero, the process is irreversible and real.</b> <b>If S dot gen is equal to zero, the process</b> <b>is completely reversible and ideal. If your</b> <b>calculator spits out an S dot gen that is less</b> <b>than zero, you made a math error or invented a</b> <b>perpetual motion machine. Probably the first one.</b> <b>My patience around the food bowl has zero entropy</b> <b>generation. Therefore, my hunger is reversible.</b>
2:25:35
<b>Feed me. Homework problem, entropy generation.</b> <b>Step one, define the system. It's an open steady</b> <b>flow compressor. So set up the rate form of the</b> <b>entropy balance. Step two, apply the key adjectives</b> <b>from the problem statement. Steady state means that</b> <b>the accumulation term will be zero. The problem</b> <b>also says insulated, which means Q dot will be zero.</b> <b>I can divide the entire equation by mass flow rate</b>
2:26:07
<b>and switch to S2 minus S1, since we subtract it to</b> <b>the other side, and we get our per unit mass</b> <b>generation. So step three, use the ideal gas</b> <b>entropy equation, since we're dealing with air.</b>
2:26:28
<b>Step four, substitute the given numbers into the</b> <b>equation. Step five, evaluate the natural</b> <b>logarithms, being careful because there's fractions and</b> <b>parentheses. Boom. And the final value is positive,</b> <b>meaning the compressor is entirely possible in the</b> <b>equation, and includes some irreversibilities. If</b> <b>the exit temperature had been lower, our entropy</b> <b>generation would have dropped below zero, which</b>
2:26:59
<b>would alert us immediately to an impossible design.</b> <b>I keep checking my entropy. It's still going up. That's good, right?</b> <b>Indie, this is a thermometer. Do you have to go to the vet?</b>
2:27:14
<b>Chapter seven, entropy, four things to take with</b> <b>you. One, the third law of thermodynamics. It just</b> <b>sets a baseline of entropy equals zero for a</b> <b>crystal at absolute zero temperature. As far as</b> <b>problem solving goes, it's not really relevant.</b> <b>At least not like the first and second laws. Those</b> <b>get used all the time. Second, look up S in the</b> <b>same steam tables you've been using all semester</b> <b>but use equations to find S for ideal gases.</b>
2:27:47
<b>Third, isentropic means adiabatic plus reversible.</b> <b>Set S2 equals S1. Solve for the ideal case, then</b> <b>apply isentropic efficiency for real components.</b> <b>And number four, oh wait. And number four, entropy generation is never negative.</b> <b>S gen is greater than or equal to zero. It's the second law in math form.</b> <b>Negative S gen means you made a mistake. So I</b> <b>have a comment challenge for you. Leave a comment</b> <b>down below with what thermodynamics</b>
2:28:29
<b>terminology would best describe your professor. Example,</b> <b>Carnot. They are theoretically perfect but impossible to find in the real world.</b> <b>Or isochoric does no work. I know some professors</b> <b>like that. So maybe superheated. They're full of</b> <b>positive energy. So be creative. I can't wait</b> <b>to read your comments. This is going to be fun.</b>
2:28:54
<b>Try not to be too mean but it's okay to be a</b> <b>little mean as long as it's in the spirit of good fun.</b>
2:29:01
<b>All right, T.A. Serenity. It's final boss time.</b> <b>The Rankine cycle going back to the power plant.</b> <b>I too am ready for the final boss. The final boss at the treat factory.</b> <b>We started this video with the Rankine cycle.</b> <b>You looked at it, you nodded politely, you had no</b> <b>idea what any of it meant. Pump, boiler, turbine,</b> <b>condenser, four arrows in a loop. Cute. Now that</b> <b>T.A. Indiana has joined us, now we've got</b>
2:29:30
<b>the full toolbox to actually understand it all.</b> <b>So it's time to actually analyze the machine</b> <b>that powers most of human civilization. I got this.</b> <b>So four components, four control volumes, four</b> <b>applications of the steady flow energy equation.</b> <b>So each one you have actually done before, we're just stringing them together now.</b> <b>So state one, a saturated liquid at the bottom</b> <b>left of the dome. Pump from one to two. It compresses</b>
2:30:07
<b>cold liquid water to higher pressure. This is</b> <b>adiabatic. Kinetic energy and potential energy</b> <b>are negligible. The process line is straight</b> <b>up if the pump is isentropic, 100% efficient.</b> <b>Otherwise, it slants slightly to the right, which</b> <b>increases entropy. The line is really short because</b> <b>liquid water is merely incompressible. So temperature hardly changes as the pump</b> <b>increases pressure. But also, since water is</b>
2:30:46
<b>essentially incompressible, there's a shortcut</b> <b>equation. It's just specific volume times pressure</b> <b>rise, usually only a few kilojoules per kilogram.</b> <b>Compared to everything else in a power plant,</b> <b>pumps are cheap. Now the boiler from state two</b> <b>to state three. Again, coal, nuclear,</b> <b>anything. It adds heat to convert high pressure liquid</b> <b>into high pressure superheated vapor. Temperature and entropy both go way up.</b>
2:31:17
<b>Temperature curves upwards in the liquid and gas</b> <b>region, but under the saturation dome is always</b> <b>horizontal. Temperatures don't change during</b> <b>boiling. There's no work in the boiler. So this</b> <b>equation is just Q dot in. This is where you pay</b> <b>for your fuel. This is the biggest number in the</b> <b>cycle. Next, the turbine from three to four. Big</b> <b>spinning blades, multiple stages, the steam expands,</b>
2:31:51
<b>drops pressure, and does work. It's adiabatic and</b> <b>we can usually ignore kinetic and potential energy.</b> <b>Now the line would be straight down if it were</b> <b>isentropic, 100 efficient turbine. Of course,</b> <b>that's impossible in the real world. So this</b> <b>line slants slightly to the right. Entropy has to</b> <b>increase. State four should normally be a liquid</b> <b>gas two phase mixture underneath the saturation</b> <b>curve. So some condensation happens inside the</b>
2:32:26
<b>turbine. And this is the moneymaker. It's the</b> <b>whole point of building the power plant. The</b> <b>turbine is connected to the generator which produces</b> <b>electricity. So last, condenser from state four</b> <b>back to state one. A heat exchanger. It rejects</b> <b>out to the environment. There's no work.</b> <b>The horizontal line back to state one since</b> <b>temperature doesn't change during</b> <b>condensation. Now Q dot out is required by the second law.</b>
2:33:00
<b>You cannot avoid it. This can never be zero.</b> <b>Hey, hey, ho ho. The second law has got to go.</b> <b>Four equations, four enthalpy differences.</b> <b>That's how the electricity is produced that</b> <b>charges your cell phone. FE exam problem,</b> <b>information overload. I'm out. I surrender.</b> <b>Bury me in the litter box. That's too many</b> <b>numbers. Amateur. When they put the whole buffet in front</b> <b>of you, you don't eat the salad. You locate</b>
2:33:37
<b>the premium salmon paté and ignore the rest.</b>
2:33:42
<b>Read carefully. The question asks for heat in</b> <b>the boiler. The boiler is just states two to three.</b> <b>So we only need two numbers, H2 and H3, plus mass</b> <b>flow rate. Ignore everything else. It's just there</b> <b>to distract you and slow you down. The FE exam</b> <b>is full of traps. The writers think to themselves,</b> <b>if I were a student, what mistake might I make on</b> <b>this problem? Then they make all the wrong answers</b>
2:34:19
<b>the answer a student would get if they made that</b> <b>mistake. Rankine cycle problems on the FE exam are</b> <b>loaded with unnecessary distractors. Ignoring</b> <b>distractions is not my strong suit. So skill of</b> <b>the chapter and honestly skill of the whole course,</b> <b>identify the component, pick the two enthalpies that</b> <b>bound it, and then plug into the SFEE. Done. Big</b> <b>picture time TA Serenity. Thermal efficiency of the</b>
2:34:55
<b>power plant. It's the same definition we've been</b> <b>using since chapter two. What you want divided by</b> <b>what you have to pay for. You want net work out.</b> <b>You pay for heat into the boiler. The second law</b> <b>guarantees this number is less than one. And</b> <b>remember from chapter six, the ceiling is set by</b> <b>temperature gap. So to push efficiency up, you</b> <b>raise T hot or drop T cold. So real world scenario.</b> <b>Two US submarines, one up north spying on</b>
2:35:34
<b>Canada, one down south spying on the Bahamas.</b> <b>Which nuclear reactor would operate more</b> <b>efficiently? The one in Canada or the one in the Bahamas?</b> <b>The one in Canada. The colder the ocean, the</b> <b>lower TC. So the more efficient your power plant.</b> <b>Bring me back some poutine. All right,</b> <b>example problem. So part a, the isentropic cycle.</b> <b>Start with pump work. We have liquid water.</b> <b>So we'll use the shortcut that work of the pump</b>
2:36:09
<b>is 8.07 kilojoules per kilogram. Now for turbine work, we'll have to look up H3,</b> <b>3399.5, entropy 3, 6.7266. If it's isentropic, then</b> <b>S4 is equal to S3. So state four at 10 kilopascals</b> <b>is a wet mixture. So we're going to have to interpolate. The left fraction is the</b> <b>entropy we know. The right fraction is the</b> <b>enthalpy that we need. We get H4S, the isentropic</b> <b>enthalpy, 2130.3, which gives work of the turbine 1269.2 kilojoules per kilogram.</b>
2:37:01
<b>So to calculate heat in, to find H2, we'll use</b> <b>the SFEE, which reduces down to H2 is equal to H1</b> <b>plus the work of the pump. So that gets us to</b> <b>199.88. So heat in from the boiler, H3 minus H2,</b> <b>3199.6, which gets us to then efficiency, which is going to be work of the turbine</b> <b>minus work of the pump, that's net work in</b> <b>the numerator, all divided by heat input in the</b> <b>boiler. So it's what you want divided by what</b>
2:37:42
<b>you have to pay for it. Gets us to an efficiency</b> <b>39.4%. And notice how tiny the pump work is compared to the turbine work,</b> <b>8 versus 1269. It's less than 1%. This is why a</b> <b>lot of textbooks will just throw pump work away</b> <b>in early problems. Don't do that on the FE exam,</b> <b>but understand why people think they can get away</b>
2:38:11
<b>Now part B, a real turbine with 85% isentropic</b> <b>efficiency. So the pump stays isentropic, only</b> <b>the turbine changes. So from chapter seven, we can</b> <b>find that the actual work of the turbine would be</b> <b>1078. So the heat in doesn't change, the pump</b> <b>work doesn't change, only the turbine output drops.</b> <b>So the efficiency is now 33.5%. Does this mean</b> <b>the sad engine from chapter seven is now a sad</b> <b>power plant? Sorry, Indy, but yes, it is a sad,</b>
2:38:51
<b>irreversible power plant. And this sadness costs</b> <b>real money. If a real 33.5% efficient 1000</b> <b>megawatt plant were operating isentropically at 39.4%</b> <b>efficiency, it would be producing 1176</b> <b>megawatts. Those missing 176 megawatts of power in North</b> <b>Carolina, that costs the power plant $8800 every</b> <b>hour, $211,000 every day, $77.1 million per year.</b>
2:39:34
<b>If engineers can optimize the turbine efficiency</b> <b>from 85% to 86%, then a 1000 megawatt power plant</b> <b>will make an extra $5.1 million every year. So</b> <b>real engineering is just chasing tiny percentages?</b> <b>Pretty much. Attention to detail is the name of</b> <b>the game. Four takeaways from the Rankine cycle.</b> <b>You have four components with four equations,</b> <b>pump, boiler, turbine, condenser. Each one is</b> <b>just the SFEE with the right terms canceled</b>
2:40:15
<b>out. Two, pump work is very small. Liquid is</b> <b>nearly incompressible. So most of the action is in</b> <b>the boiler and the turbine. Three, the TS diagram</b> <b>is your friend. Vertical pump, horizontal</b> <b>boiler, vertical turbine, horizontal condenser,</b> <b>the enclosed areas equal net work. And four,</b> <b>isentropic efficiency is where real world losses live.</b> <b>A turbine at 85% drops overall plant</b> <b>efficiency by several points. Small component losses</b>
2:40:57
<b>add up to large dollar consequences. All right,</b> <b>so y'all have seen TA Serenity sleep through like</b> <b>this whole video. Whereas Indiana has always got</b> <b>all sorts of kitty business to do, he is my best</b> <b>thermo study buddy. So I have a question for you.</b> <b>Who is your best thermo study buddy? Whoever you</b> <b>are thinking of, send them this video. If you</b> <b>don't, they're going to be bugging you all semester long</b>
2:41:25
<b>asking you a million questions. But if you send</b> <b>them a link to this video, I'll explain it all</b> <b>for you. And with all the time you save, you can</b> <b>get extra sleep, play more video games, whatever</b> <b>you want. Have a great day. Good luck on your next test.</b>
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