# Thermodynamics: 1st Law for Open Systems, Uniform Flow; 2nd Law, Heat Engine & Refrig (15 of 25)

[ Music ]>>Well, good afternoon. [inaudible response]

So as we start today. First let me just note that I have your new

homework set for the week. So these problems are — this week’s material

I’ll do next Wednesday. And remember that

you have — what? About seven or so? Eight problems from last week

that are due on Wednesday. So don’t forget to

bring that with you. And I have some more homework

to be picked up here as well. Just a note or two on the

homework that you’re doing now; one student had asked the

question just a minute ago, and I thought I’d

answer for all of you, if you have a mixing chamber, and in a typical mixing

chamber the pressure’s going to be the same everywhere

in the mixing chamber. Think of a mixing chamber just

like a big, open container with two pipes coming in

and one pipe going out. There’s really no possible

way for the pressures to be different in any of

those particular fluid streams. If let’s say the

discharge pressure were less than the inlet pressure then

the fluid would be pushing back up stream and up those

inlet pipes, right? You have to have

the same pressure at each section of

a mixing chamber. So when you’re looking at — there’s one more problem that

dealt with mixing chambers. It doesn’t specifically tell you that the outlet pressure is the

same as the two inlet pressures, but that’s really

what they’re implying. So please make sure

that you recognize that, you know the two inlet and

exit pressures are always going to be the same for a

mixing chamber unless you’re told otherwise. But for most problems they will

always be exactly the same. Let me also remind you that I

had reassigned problem one — I’m sorry, problem 32 from

the previous homework set to the homework set

that’s due on Wednesday. And remember that to solve that one you need

two Sevelking’s [phonetic] Equations. You have your mass fluid

equation, which you’ll combine with your ideal gas

equation of state and that’ll give you

one equation in terms of the unknown temperature

and velocity. And then you’ll have

a second equation, which is just your first

law of thermodynamics, which will also have an unknown

temperature and velocity. So two equations, solve for two

unknowns, so that’s the hint that I gave you a week ago, but

I’ll just give it to you again because it was, you

know arguably a somewhat confusing problem. Anyway, let’s just move

on to new material now. Oh, one other thing

I wanted to mention. I’ve been updating

Blackboard so I think we’re up to lecture number

13 now that’s available as a recording on Blackboard. Also, I’ve included all the

example problems that I’m going to be working on — the

ones we’ve already done, but also the ones that you’ll

see over the next month or so. I really never thought to

make these available to you until some students were

saying that it’s kind of hard to read a problem that’s on

one board and my lecture that’s on the other board up here, and this way we should

all have access. So I put those up just

right before class today. They’re all in image

format so you should be able to access the image on

your Smartphone, your IPad, or anything you bring

with you to class. As long as you can access

Blackboard then just go to “course information” and there’s a new topic

there that’s called “example problems.” So they’re all shown there. Again, they’re all images. So anyway, moving on. [laughter] So we’re actually

done, believe it or not, with the steady flow problems

associated with the first law of thermodynamics

for an open system. And we’re going to move on now

and talk about another type of problem, which you call

uniform flow problems. By the way, another thing that

I learned last time, afterwards, was that some students are

having trouble just reading the writing that I have

because the thickness of the pen was not

really that thick. Again, I didn’t realize

that was a problem. I guess it’s more of an issue

once we have one board going to the computer and then

the other board that’s just a lecture. Those of you sitting over here

probably have some difficulty reading over here. So I’m using a thicker

pen starting today and hopefully that’ll

make it easier for you to read the material that’s

going up on the board. All right, so uniform flow. First of all, we’re

still talking about process involving first

law or a control volume, or an open system if you will,

but we’re no longer talking about steady flow problems. And there’s a couple

of main differences between the uniform flow and

the steady flow problems. So first one difference is that the state will actually

change inside the control volume — and again, I’m just going

to use CV as an abbreviation for control volume, okay? — although it remains uniform

throughout the control volume. So the state will change

inside the control volume but is uniform throughout. And then the other difference is

that the state of the entering or exiting substance is

going to remain constant, but the flow rate will vary. So we’re saying that the inlet

and exit of them is constant. The flow rate — I’ll just

rate this as [inaudible] — but the mass flowrate varies. Okay, so this is a

little bit different than our steady flow problem

in that [inaudible] problems that we use this method for

are also going to be different. These types of problems are

typically used for filling or for emptying a vessel, like if we’re discharging a

fire extinguisher or for filling up our propane tank down

at the U-Haul Store. Those would be kind of typical of these particular

types of problems. So again, these are

primarily used for filling or emptying vessels. We often call these charging

or discharging problems. This is about charging

and discharging problems. And what I’m going to

do is we looked both at the continuity equation

as well as our first law and modify them from the general

form to be more accommodating of these types of problems. So let’s just move on with that. All right, so first

thing’s first. Let’s just remember the

continuity equation. And the continuity equation

basically tells us that a change in control volume — mass

with respect to time — is going to equal the difference between the rate

that mass flows in. So that’s the sum of all inlet

streams and that’s the rate that the mass flows out. So that’s the sum

of all exit streams. I left out DET. Now this could be

written in different ways. Sometimes we put the two

terms on the right hand side over onto the left-hand

side of the equation, but this should make

sense to you, right? The difference between

what enters and exits is what

remains, right? So if the entering flow

streams have a greater mass than the exiting, like

let’s say you’re filling up the propane tank, then

the change of the mass of the control volume

will be positive. If we’re emptying the

propane tank then the mass that leaves is greater

than the mass that comes in and we have a reduction in

the mass control volume. So this is one term

we’re going to note, or one equation we’re going to

note, but let’s also again keep in mind what it says above;

the mass flow rates will vary. You are definitely going

to have a different rate that mass is going to flow

in verses what it flows out and in fact in many

problems we’re going to have nothing flowing

in or nothing flowing out. Again, if we’re talking about

like filling up a propane tank, you don’t empty the propane

tank while you’re filling it. If we’re talking about a fire

extinguisher being discharged, you’re not filling it while

you’re discharging it. So it’s very easy for one

or more of these terms to go to zero and we’ll talk

about that here shortly. Anyway, if we take this

continuity equation and now just integrate it with time then basically

the DMDT term for the control volume

— well that — you differentiate that

and you just get the mass. So the mass evaluated

between the initial and final state is

going to be m2 minus m1. So that’s what you get on the

left hand side of the equation. And on the right hand

side of the equation — now remember, the dot

represents the time derivative. So if we’re integrating over

time then we’re getting rid of the time in the denominator

and this ends up then with a sum of our exit streams

of me minus sum of overall inlet streams of mi. And note that i and e and 1

and 2 are not the same things. Okay? Unlike the single

stream steady flow problems where I’ve replaced i

with a 1 and e with a 2, here they mean different

things, right? Here 1 and 2 represent, you

know the final and initial state in the control volume, yet the i and the e represents what’s

entering or exiting, right? So e represents what’s

flowing out of control volume and i represents what’s

flowing into the control volume. So no more equating 1 and i, 2

and e. They’re different now. Anyway, here is a sensory — a version of the

continuity equation — that we’re going to use for

these types of problems. Let’s now look at

the first problem. If we look at the first law —

and I’ll just write this in, well, a very general form — you would get the rate of e transfer plus the sum

overall inlet flow path. So m dot i, then hi plus ei

squared over 2 plus 2i — and these are all i’s here. And this is an equal rate

in change in the energy of the control volume. And remember, this term

would have gone away if we were talking about

a steady flow process because with a steady flow

process there is no change within the control volume,

but now there is a change in the control volume. I mean we’re always uniform

throughout the control volume, but it’s still changing

over time. So it is a change in energy

overtime with respect to — well, for the control

volume itself. And then we also

have the exit terms. So sum overall exit flow

paths, dot e and he. That’s ve squared plus 2 plus

g2e and then that’s [inaudible]. So this is the general

form of the first problem. Again, it’s not the

single stream steady flow, not the steady flow, it’s

just the general flow. Okay? So now what I’d like

to do is just make a note. So if we’re looking at

the control volume — The energy associated with

the control volume is — well, we see the

definition of energy. Energy is going to be the

internal energy plus the kinetic energy. So we’ll take me squared 2

plus the potential energy. So I’ll just write that as tz. That’s right, mtz. So this can be rewritten

as the mass — m’s the internal energy — I’m sorry, specific

internal energy — then plus e squared over 2

plus t. And then what I would like to do is I would like

to differentiate this. I mean, after all this is

going to have to go back in to the first [inaudible]

equation above. So if we take the change

in energy plus the time for that control volume, now

this is [inaudible] of mass. [inaudible] energy as

kinetic [inaudible] Well, at this point

what I’m going to do is [inaudible]

couple of steps all at once. I’m going to go back to

my first law equation, but now I’m going

to integrate — The first law — me over time. So we just go back up to the

equation above here and looking at this [inaudible]

equation here — And let’s take that and

[inaudible] over time. Note that as I do this,

at the same time I’m going to be plugging in this

equation that’s directly above. So this basically

goes in up here, but then it too gets integrated. So I just wanted to make sure

that you’re aware of that. So let’s integrate the first law

over time and let’s just be sure to include the equation

above in these equations. So if we do so, what

are we going to get? So let’s just take these

one term at a time. First of all the integration

of q dot over time, well that’s just q right? I mean q dot is qt. If we integrate the qt

we just get q again. So that’s just transfer

the control volume. If we integrate the

next term, keep in mind that these uniform flow problems

there is no change in the state at the inlet or at the exit. Whatever it is at the beginning

of the problem, it’s the same as it is at the end

of the problem. So there’s no change in

any of these terms, right? There’s no change in the

empathy, velocity, the height. The state is the same. The only thing that that

integrates is just the mass flow rate term. And whatever the mass flow

rate is really is dmpt, so if we integrate that then

we just get the mass again. So this summation

then [inaudible] mi, note that the dot

is gone here again. And then time is hi plus

[inaudible] q [inaudible]. And then on the right hand side,

well here we’re going to have to be a little bit careful. First of all [inaudible]

but here we’re looking at the [inaudible] term right? So we’re integrating that —

well, if we integrate that, again this is an e. The integral

just becomes the mass time internal energy plus the

kinetic input times the energy and we have to evaluate that

between points one and two. So what we get here then is

m2 and then e2 plus e2 squared over 2 plus 2d2 and

then minus m1. Oops, that’s not

supposed to be i and 1 plus e1 squared

over 2 [inaudible] 1. So that’s the change

in the internal energy or the control volume

integrated over time. Okay? And then we have to add to it the integral

of the exit term. So again this is

one of my equations, but I have to keep

spreading it down here. So next we have some overall

exit speeds and the mass of the exit and the [inaudible] of the exit plus t

squared over g plus gt. Okay? And then last but not

lease we have [inaudible] term or w dot term. [inaudible] w plus

w [inaudible]. So this long, long, long equation right here is

the first law for uniform flow. Now, as you’re looking at this

equation you realize there’s only like 18 separate terms. Now how can that be to

solve the problem, right? Obviously I’m being facetious. We don’t have problems that

use this entire equation. What we typically do is

simplify this into problems where we have some — well, let’s say useful

problems to solve. So I mentioned already

some of them right? I talked about fire

extinguishers and I talked about filling and

emptying a propane tank. What’s unique about those is that they have only a single

inlet or a single exit and what we can do

is we can take a look at the equation above,

then just modify it based on this special case

of a single inlet and exit plus some

other simplification. And you’ll see that

by the time I’m done with all this we’re going to

get rid of this lengthy equation and it will be replaced by

something that’s a lot more — let’s call it user friendly. So what do we want to do then? Let’s look at this special case. And the special case is going

to have either a single inlet — Or exit, not both

at the same time. Either one or the other. And let’s note that some further

simplifications we’re going to make is that we’re going to

elect all kinetic energy change and potential energy changes. Okay? For these types of

problems you’re filling or emptying a vessel, its

speeds are going to be low. Not only is their going

to be low speed associated with what we’re filling or

emptying, but then there’s going to be essentially no velocity

associated with what’s inside that particular container. And there’s not going to be

any height change associated with either of these processes. I mean the exit is

where the exit is. The entrance is where

the entrance is. The tank sits where

the tank does. So there’s no potential

energy changes either and if you look only

at a single in or out at a time then we have

a great simplification. So let’s look at the case

of first the single input. So let’s say we’re filling

up a fire extinguisher. Well, let’s relate this to

the general equation above. Right, so we’ve got the

general equation above. I mean, you can’t really put

everything there all at once, but note that; wow, eight of

the terms cancel out right away. Then you’ve got four kinetic

energy terms, all the d squared and the 2 terms, right? We also have four

[inaudible] e terms. So those all go away. And if we’re dealing with only

a single input then the entire exit stream goes away. There’s hardly anything

left for our equation right? What we get is the problem. We get the heat transfer

minus the work — I’m just using work term for the

positive on the right hand side over the negative on

the left hand side. And then we have

our inlet flow path, but only the [inaudible] right? There’s no [inaudible] So — and

there’s only one of them right? We can even drop the

summation and end up with mihi and then plus m2 minus m1e1. Absolutely everything in place. Absolutely everything

else goes away, right? So this is the first

[inaudible] or uniform problem where there’s only

a single input. I will note that when we solve

problems involving uniform flow, we still have to

solve a continuity equation simultaneously. And if we go back to the

continuity equation and get rid of the term that

deals with the exit — this obviously is

[inaudible], but if we look at the continuity equation

then the continuity equation is really also pretty

simple, right? It basically tells me

that the mass that flows in is just the difference

between what we end with and what we start with. I mean really you have

to know that’s called the continuity equation. It just has to do with

common sense, right? What flows in, this

is [inaudible] to the final and the initial. I mean, that should make sense. You can go back to the

continuity equations and get rid of the summation signs and

note that whatever mass is of t [inaudible] only

[inaudible] based on what flows in and out. So that’s going to have

to be used simultaneously. Yeah? [ Inaudible Question ] Good question. Yeah, it should be negative

on that side, but [inaudible]. You’re right because

I had it as a positive on the left hand

side, so as I move to the right hand side it’s

going to have to be negative. So that’s a good observation. That’s a typo. So, yeah put a minus

sign there to make sure that we’ve done it correctly. So thank you for that. Anyway, so this is an equation

that we’d use for single input and then we have

equally simple equations if we have a single exit. Only a single exit and no

inlet at the same time. So we go back up

to our equation. [inaudible] first law, and we still have our heat

transfer minus the work. Again, work moves over to

the left hand side and then on the right hand side,

well we have our mtvt and then minus m1 u1. All of [inaudible]

terms go away. All the inlet terms go away. It’s using only a single

exit and then the last term on the right hand side will

be exit term [inaudible] plus [inaudible] and the sign

I think is correct here. This should be fine. So this is the first

line equation. [inaudible] And that is — again, we’ll

have to look at the continuity. But again, we can go to

the continuity equation or we can just, you

know use our logic. If flow means leaving

[inaudible] have some mass flowing out, the mass that flows

out had better be the difference between what we start

with and what we end with. I mean, where else is it going? Right? So clearly

it’s just m1 minus m2. Now, again you go back to

the continuity equation and write this or you can

just use logic and say; well, obviously what’s flowing out has

to have started on the inside. The difference between what

you start with and end with — m1 minus m2 — had better

equal what’s flowing out. So these are the equations,

the two that are underlined, that you can use for either the

charging or discharging problem. You can see that the one that’s

[inaudible] isn’t that charging? So we’re charging

the fire extinguisher or we’re charging our propane

tank when we’re filling it up. So single input; these

are the charging problems. And single exit, this is

when we’re discharging. [inaudible] Thinking of just filling or

emptying a vessel, charging or discharging, single in,

single out, I mean think of them however you want,

but these are the types of problems that [inaudible]. Now, there’s no more

theory to talk about. Really, I think all

we need to do is go through an example problem and help illustrate how one

solves these kinds of problems. So, are there any questions? [inaudible] the theory? All right, so the example

problem that we’re going to do is problem 125

in chapter eight. So it’s on the computer and

let’s get that [inaudible]. Okay, so this one is from

the [inaudible] again, all this is on Blackboard now. So [inaudible] provide

students with [inaudible]. Let me read it to you anyway. [inaudible] So, it says

that we have a concrete tube and a rigid tank that contains

r134a at 14 degrees Celsius. At this state 55 percent

is in the vapor phase. In other words, [inaudible]

quality. And the rest in the

liquid phase. Obviously, what else

could it be, right? This says the tank is connected

by a valve to a supply line where refrigerant

at 1.4 mega pascals and 100 degrees Celsius

flows steadily. It says the valve is now open

slightly and the refrigerant is about to enter the tank. When the pressure reaches 1

mega pascal all refrigerant in the tank is in the

liquid phase only. In other words, you finally have

turned that two phase mixture into just all vapor, which

is saturated vapor and that’s when the valve is closed

and the problem ends. So it wants us to find the

temperature of the tank, the mass of the refrigerant

that has entered the tank and q to be transferred between the

tank and its surroundings. So let us start this one

with the usual slide. All right, so we have a tank

that looks bigger than this tank and it is connected

to a supply line and they give you all

sorts of information. So they do tell me the

volume and it’s rigid. So this is common. Most of these problems

will be rigid tanks, but you know I’ve seen problems

where they’re not rigid. You know perhaps you have a

[inaudible] cylinder device and you’d have to include

some sort of boundary work. But most of the time

they’re rigid. So it tells me that the

bottom is 0.3 cubic meters. Its refrigerant will

be a. It gives me the initial temperature. [inaudible] plus 14 Celsius. And the [inaudible]

is 55 percent. It says that the supply line

[inaudible] the supply line has an inlet pressure

of 1.4 mega pascal and [inaudible] temperature

of 100. And it says that the

refrigerant flows steadily in the supply line and that’s

really just the author’s way of saying that there’s no

change in inlet [inaudible]. But frankly, all these

uniform problems will show you that the inlet conditions

are uniform or the exit conditions

are uniform. There’s no change in them. The author didn’t really

have to tell us that, but it doesn’t hurt

I guess to do it. It says the valve is open, the

refrigerant’s entering the tank and it finally gets to a final

pressure of 1 mega pascal. All the refrigerant’s

in the vapor phase only. So 0.2 [inaudible]. And they would like us

to find the temperature and mass [inaudible]. And then transfer [inaudible]. Now [inaudible] call it c.

But this is the heat transfer to the control volume

during this particular task. All right, so I suppose

[inaudible] now how do we even solve this problem? Well, this is really

a charging problem and there’s just

the single inlet. So it’s pretty clear

what the version is of the first [inaudible] about basically [inaudible]

equal [inaudible]. So this is the version of the

first problem that [inaudible] and of course you’re also going

to have use some continuity. So [inaudible] Now do we

have the ability to solve for anything up here yet? Yes. First of all you might

notice that we know the initial and final mass as long as we

know the volume and the initial and final state — and

we do know the initial and final state, right? We have the temperature

and the quality in SOU [inaudible]

graph would allow you to find the initial

specific volume. And we know that the pressure

[inaudible] at the final state. So we should be able to

look at our graphing tables and find [inaudible]

volume and know that the initial mass

is just the volume of h as the specific volume. And we know that the

final mass [inaudible] is that same volume [inaudible]. So that’s something very common

to these types of problems, just to find this

relationship between the volume and the specific volume

in order to get the mass, but we do this not to

get the [inaudible] mass. Right? We do it to get the

initial mass or the final mass. The difference between the

two in the continuity equation to solve for the inlet mass. And nonetheless, at least we

see that we can find all this. Let me start putting

little checkmarks here. I know the volume and I know

the initial and final state of the [inaudible] volume. So yes [inaudible] m1 and m2. And finally these are not dots that are going [inaudible]

Anyway, these are my checkmarks. So we know the inlet mass. Okay, so let’s just

check off inlet mass. By the way it’s a rigid vessel. There’s no work associated

with it. There’s no mixing devices. There’s nothing. So we can always put zero

there on the work page. And again, getting back

to what we were talking about a moment ago, we

know our initial state. So we observe it by [inaudible]. We know our final state. We’ll observe it by [inaudible]. We know our inlet state, both

pressure and temperature. So we observe it by hi. And, as we just mentioned

a moment ago, we know m1, we know m2, we know mi. There’s nothing left

to solve for but the heat transfer, right? So if we’re trying to solve

for the heat transfer, really it’s a matter of

finding the specific volume and inputting [inaudible] data

— I’m sorry, [inaudible]. We can now plug it in to

all three of the equations and show them near the bottom

and we can get [inaudible]. By the way, as far as p2, I don’t know that’s

really chapter two or three probably isn’t it? If it was saturated

vapor it could pressure, the temperature is obviously

just the saturation temperature, right? So I don’t know why they

ask us that but [inaudible]. So, any questions guys? Let’s just keep in mind

that as we’re talking about refrigerant all of our

data comes from day 11, 12, or 13, just depending

on what the state is. Let’s just solve the problem. We’ll start [inaudible]

property data. So — So at p1 and [inaudible] 1

it’s a two phase mixture. We know that we have

these [inaudible] 11. It’s a two phase mixture

[inaudible] temperature. So I’m just going to go

down to the 14 degree line. I don’t know that I need

to take the property tables and put them up here

[inaudible]. Oh, we know how to do this. [inaudible] So, you’re

looking for two things, right? You need the internal

energy of one and you also need the specific

volume so we can find the mass. So p1, there’s vf. [inaudible] at 14

degrees Celsius it’s [ Inaudible ] Find mu one, you

get the same thing. [ Inaudible ] So, again it’s just a matter

of filling in this other table. 70.56 is [inaudible] 167.3 [ Inaudible ] As long as we have

it we might as well as [inaudible] So

m1 is the volume. 0.3 cubic meters is

the specific volume. [ Inaudible ] By the way, again

this is kind of sloppy with a big pen, but [inaudible]. We get mass of 12.36. Now let’s move on to state 0.2. We’re doing much the same thing, although state 0.2

is a saturated vapor. So here [inaudible] vapor. Now it can go to [inaudible]

the saturated pressure table [inaudible] at this pressure, 1 mega pascal, which

is [inaudible]. [ Inaudible ] So eg is 250 and 71. [inaudible] and as long as we have the specific

volume data stated we may as well find target mass. So 0.3 divided by

[inaudible] 0.76 And, last but not least

let’s find our [inaudible]. So [inaudible] you’ll see that

this is [inaudible] vapor. So we take away 13 we go to

the 1.4 mega pascal site table. We read down to the 100

degree Celsius line. I mean again we’ve done

this many times before, and we just pull out the value. We get [inaudible] 232. Next we may as well find m2 — I’m sorry, [inaudible] mi which is the difference

between m3 and m1. So we do that we get 2.40. Okay? [inaudible]

We’ve also been asked to find the temperature, 0.2. I mean honestly I should

have done that above right? As long as I drug in

the saturation table to get 0.2 I should

have found it there. So I’m just going to

go back up and read through [inaudible]

1 mega pascal. So we can read that

right off the table. It’s 30.94 degrees Celsius. And then the last thing we need

to do is solve the first law and find the unknown

heat transfer. So fill in the first law

[inaudible] And now — you know, this is written

into your notes above. We don’t have a big

blackboard here. I drew a big arrow

across half the board, but [inaudible] the first law. [inaudible] I’m not

going to rewrite it. I’ll just go with

the [inaudible] So we have the negative

[inaudible] it’s really pretty straight forward. Maybe to save time I’ll just

let you plug away yourselves. Etcetera, etcetera. If you go do the mathematics

you’ll end up with 899 joules. And you will find that

this is a positive number. If you go through the

mathematics it’s positive. So heat is actually being

added to this system. It is [inaudible]. So are there any questions

as to [inaudible] problem? Hopefully you don’t mind that

I didn’t actually plug away at all the number here. [inaudible] notes. [inaudible]. I will be doing this a lot more

as we progress with the class. You know I like to show

you, you know about units and conversion factors

and all that, but you’ve seen most

of that already. So I’m generally just trying to save time now

and not [inaudible]. But I went through all

the important steps today. Is there a question?>>Yeah [inaudible]

between what you’re finding at the subscript i for the

inlet and — verses 1 or 2?>>Well, let me just

pull this down. You know we have a

tank that’s connected by a valve to a supply line. So the supply line is your inlet

conditions and those are going to always be the same. And that’s your i. But what’s

in the tank is different, right? I mean we’re starting

with something at a markedly lower temperature and [inaudible] vapor

to the mixture. Those are your initial

[inaudible] and then as you open that valve and fill up

whatever’s in that tank, but then see you

get the [inaudible] So those are [inaudible]

condition. So there shouldn’t

be anything confusing between what represents

the inlet and what represents the

initial filent [phonetic] state within the flow volume itself. The inlet is not

[inaudible] flow volume. The inlet is just a pipe

that’s connected to it to provide [inaudible]

substance [inaudible] but it’s not the inlet,

it’s not the exit. It’s the same substance. But [inaudible] So yeah, make sure you read whatever

problems [inaudible] clearly and then so you know

the difference between the [inaudible]

conditions and the filent conditions. One thing I also wanted to

comment on; if we have a problem that involves an exit sometimes

those problems are a little bit confusing. For instance, let’s say you

started with a two phase mixture and we had an exit pipe in

the bottom of that tank. Well, if you are a

two phase mixture, because of the higher

density of the liquid portion of the two phase mixture the

liquid is going to always stay at the bottom of that tank

and if you have a valve at the bottom of the tank then

you’re actually only allowing the saturated liquid portion of

the two phase mixture to exit. On the other hand, if you

have exactly the same problem, the two phase mixture, but

your valve is on the top, well again the density of the

two phase saturated liquid — the saturated portion

of the mixture is such that the vapor’s going

to always rise to the top. So if your valve’s on the top

then it’s only the saturated vapor that’s going

to be leaving. So when you look at all of the

involved discharging for many of those problems you’ll

have to pay close attention to where it’s discharging. You’re actually pulling the

saturated liquid portion off the bottom or are you pulling the

saturated vapor portion off the top? So again that’s just

a [inaudible]. Anyway, any other problems? Any other questions

about this problem? Or [inaudible] after five?>>So is it always going to be

the same substance either way?>>Oh, absolutely. Right, these are still only

pure substance problems. It’s going to be water or –>>[inaudible]>>Yes, it’s never going

to be anything else. [inaudible] Any other questions? Right. Well, then let us

move on to chapter six. This is everybody’s favorite, which is the second

law of thermodynamics. I guess the first thing

that I want to say is that this material

can be quite confusing and it gets even more

confusing as we start talking about entropy over

the next few weeks. If you’re not showing

up to class and you’re not hearing

the lectures and you’re not hearing

my take on the whole idea of the second law and entropy

then you’re really doing yourself a disservice. Most students who just rely on reading the textbook

are generally just not able to get it entirely. The second law doesn’t

have to be complicated and hopefully you’ll see as

I present it that we’re going to use the second law

in a very specific way for a very specific

type of problem. But still, you do need to show

up to class on a regular basis. So if you’re not here today —

I’m probably kind of talking to the wrong people — but if

you’re not going to be showing up to certain classes in

the next couple of weeks, make sure you get the

notes from somebody. Make sure you understand

the notes. If you need to come into

my office and ask me to interpret those notes for you

so that nobody’s left behind. All right. So, the second law

of thermodynamics. Well, first of all

the second law’s only around because of the first law. The first law tells us that

for any process energy must be conserved. But it really doesn’t

give us any restrictions. It just tells us that as

long as we can keep track of all the energy [inaudible]

paths then entropy process is possible, but that’s

not really true. I mean, for instance let me

just give you an example or two. Let’s say we have a

— I don’t know — some substance, a gas maybe,

and let’s say we put some sort of a mixing device and

we add a lot of work — mechanical work — we

add energy into it. As we’re adding energy

input it’s pretty likely that that gas is

going to heat up. Now, that can happen, but we’re

going to reverse that process and we take a gas that has a

mixer in it can we cool that gas and will that start the

mixer into rotation? I don’t think so. But the first law doesn’t tell

us that that can’t happen. The first law just

tells us that as long as the energies all work out, as

long as you have the same amount of work energy coming in

and that equals the amount of heat that’s being

transferred, then the first law

works out just fine. But we know that it can happen. Right? We base it

on our observations. We know that just because the

first law says something could happen doesn’t mean

it can actually happen in the real world. That’s really what the

second law’s all about. The second law provides us

restrictions, if you will, as to the nature of

certain processes. It basically tells us which direction processes

can occur in. Right? So we need

the second law. You know just because

you can solve something with the first law doesn’t

mean it’s going to work. So that’s what the nature

of the second law is. That doesn’t tell you how

to use the second law. It’s just telling you what

the second law is all about. Okay? So again, the second

law basically tells us — What direction certain processes

are capable of occurring in. So, you know this law gives us

— this is called restrictions. As to what processes

that we [inaudible]. Some people like

to think of these as what direction

they can occur, but we know that not

everything can happen. [inaudible] Another good

example is heat transfer from something cold

to something hot; we know that heat is only

transferred from the hot thing to the cold thing, but there’s

nothing in the first law that says that that

has to happen. Right? The first law just

tells us that as long as the heat transfers and we’re

[inaudible] energy change, all that, why can’t

you transfer the heat from something that’s

cold to something hot? But we know that that

can’t happen, right? We know that if we take a

hot thing and we touch it up to a cold thing the

heat is always going to go into the cold thing, right? The cold thing slowly

gets warmer and the hot thing

slowly gets cooler. If you touch a hot

thing to a cold thing and watch the hot

thing getting hotter and the cold thing

getting colder? Well, why not? The first law doesn’t

say you can’t do that, but we can’t do that. We know that that’s not

the way it really works. So that’s what the

second law’s all about. It’s nothing more

complicated than that. Well, it is because

you have to use it and mathematics is

more complicated, but in essence that’s

the second law. Now, what’s typically done — what I’m going to do here is

give you two examples of systems which pretty clearly illustrate

the need for second law. So two processes [inaudible]. So one of those processes is

what they call heat engine and the other of those

processes they call a refrigeration [inaudible]. Now, quite frankly the

heat engine is a cycle — sometimes it’s also just

called heat engine cycle. But these are two different

types of processes that I want to illustrate the

new [inaudible]. And we’ll just go through

them one at a time. And the first one that

we’ll talk about is going to be the heat engine. So let me make a sketch

of the heat engine. So there’s my heat engine. And the way heat

engine works is you have to have some heat source. And if you provide a heat

source then we should be able to do a certain amount

of network output. Now, there’s many

devices that do this. I mean, like your

automobile engine is one. We do burn fuel to create

heat and that does work. Steam powered train

is another one. You burn fuel to create heat. You turn water into steam,

you let that steam run through a steam turbine, and

that rotates and [inaudible]. So we know that this

is not exactly wrong, but it’s also not complete. Anybody who’s ever worked with

any of these heat engines knows that the only way for

you to turn heat engine into work energy is to simultaneously reject

some heat out of the system. Okay? Why are power plants

always situated on big bodies of water like an ocean or a like

or why do they have that big, tall tower called

a cooling tower? They need to reject

heat somehow. What about your engine? There’s a radiator, right? Even if you don’t have

a water cooled engine, like you have your motorcycle

or maybe, you know VW Bug or something like that. You still have heat rejected, you just have water

are called fins, basically metal protrusions

on the side of the engine that facilitate heat transfer. But you have to have

a way to reject heat. If you don’t believe me

then just go to your car, drain the radiator, and then

[laughter] drive your car for a while and you’ll

see what happens. It’ll work for a while

until the engine seizes up and you pretty much have

to throw away your car. You don’t want that,

but you’re engineers, you know that you need

to reject heat, right? This is just common sense that

this is based on observation that you can’t just

turn heat into work, you also have to reject heat. That is what our

experience shows us, right? To produce work we must be able

to extract heat from one source and reject heat to a

heated [inaudible]. Right? That’s what our

experience is telling us. Nobody’s ever going to be able to observe anything

different, ever, ever, ever. Okay? So that’s an

illustration of the heat engine and what our experience

tells us. Our experience says — when I

say our experience I’m talking about collective

experience from years based on observation and

experimentation. But our experience says

that to produce work — We must extract heat from

a source and reject heat in what we call a sink. We can call it a heat

source and a heat sink, that would also be appropriate. So this is just an

example of a heat engine. And again, this illustrates

the absolute need for a second law

of thermodynamics. The first one says

that we can turn heat into work all day long. It doesn’t say anything

about rejecting heat, but we know in the real

world we have to reject heat. So we got to use the second law. The second law will allow us to determine whether

the heat engine — and we’re analyzing [inaudible]. Let’s look at the other

process [inaudible]. And this is the refrigeration

cycler. And in the refrigeration cycler, again in the illustration what

a refrigeration cycle does is we extract heat from a low

temperature heat source. A heat source at low

temperature, and we reject heat into a high temperature

[inaudible]. So here’s our heat source and

this is that high temperature. Now, if you think about

the refrigeration cycle, I’m talking about things like

an air-conditioning system, a refrigerator, a freezer, and this is exactly

what they do, right? Let’s look at the refrigerator. The refrigerator, you want

to run it about, what? 35 degrees Fahrenheit. The kitchen’s going

to be at maybe 65 or 70 degrees Fahrenheit. You are actually taking heat

out of the food in order to keep it cold, but

you have to return that heat somewhere, right? You reject that heat out into

the kitchen and that’s okay. Right? I mean you’ve all stood

there with your bare toes in front of the refrigerator

and you feel that warm air kind of blowing on your toes or maybe

you feel the top or the back or the sides of the refrigerator

and it’s warm, right? It’s warm because the

refrigerator is taking heat out of the cold food

and rejecting it into the warmer outside air. But we know, again,

based on observation that the new rule

doesn’t work that way. If you could simply pull

heat out of something cold and send it into

something hot well, you know you just invented

a perpetual motion machine of the type and we won’t’

get into that right now, but you know it just

doesn’t work, right? Again, if you take

the cold thing next to the hot thing

you’re never going to make the hot thing hotter

and the cold thing colder, which is exactly what would

happen if you’re taking heat out of the cold thing and

putting it into the hot thing. It’s not how a refrigerator

works. Well, because we have to

provide more input, right? Within that refrigerator

is the refrigeration cycle. We’re actually taking the

heat out of something cold and then doing work to it

so there’s always going to be some work input. And it’s that work input then

allows us to transfer heat from the cold source

to the cold sink. So again, our experience

tells us that the only way you’re

going to transfer heat from the cold thing to a hot

thing is to apply more heat. Now, again if you don’t

believe me just go unplug your refrigerator at home and see

how good your cold food is going to stay cold without

the work input from the electric motor

spinning a compressor. This will not work. So that’s what our

experience says, right? So our experience — Basically says that to

extract heat from a cold source and reject it to a warm sink

we need to provide work. Reject it into a warm space. Anyway, you could use

the word [inaudible]. You must provide work. Again, we need the

second law, right? The first law doesn’t

help us out here. The first law doesn’t care if we

add work as long as the amount of heat coming in from the

source is somehow equivalent to the heat that’s

rejected to the sink, plus any energy change

[inaudible]. Then the first law tells us

that that should work just fine. But we really know

that it doesn’t. So it basically is these

observations and our inability to show otherwise that

lead to the second law of thermodynamics

whereas the first law of thermodynamics denies

the possibility of creating and destroying energy, the second law denies

the possibility of using energy in

a particular way. So that’s the relationship between the first law

and the second law. Let me repeat that; it’s these

observations and our inability to show otherwise that

lead to the second law. Whereas the first law denies

the possibility of creating and destroying energy, right? Conservation of energy. The second law denies

the possibility of using energy in

a particular way. Now, this is all

observations, right? Hopefully you all accept

these observations. If you don’t accept

the observations, well you know you’re young,

you’re learning to be engineers, you’ll have to accept

them eventually. [laughter] It’s also

good being an engineer because you don’t always have to completely understand

the theory. That’s what physicists

are for, right? We really just have to know

how to use the theory in order to create the machines

to advance technology and better human

life and all of that. I mean that’s really what

engineering is all about. Some of you may not always

have the mathematical skills or conceptual skills to

fully understand the theory or the mathematical

relationships, but don’t let that get you down. I mean, you’re an engineer. You have to accept

things as being true. If you don’t understand any of

the theory that we’re talking about from here on out,

well join the club. Most people don’t understand the

concept of entropy, but you have to know how to use the equations

that’ll come with that. So that’s what I’m

emphasizing in this class. So two statements

of the second law. There’s actually two common

statements of the second law. The first one is called the

Kelvin-Planck statement. And what the Kelvin-Planck

statement says is this; it says no process

is possible — Whose sole result — Is the absorption of heat — From one reservoir — And [inaudible] So the sole result

is the [inaudible]. Okay? That’s the statement. What does that mean? Well, to us that means that in

order for us to convert work to heat we have to have

a heat sink, right? So this basically tells

us that we need — this is a statement that the

role is use a heat source. What’s the other statement? The other statement of the second law is called

the Clausius statement. And what the Clausius

statement says is that no process is possible — That results solely — In the transfer — Of heat. From the cold — To a hot reservoir. In other words, we

need work input. Okay? We need work input. So the “we need work input,”

the “we need heat sink,” as I stated above, those are

not part of the statement. That’s just what we

observe from this, right? We know from the

Clausius statement that you can’t just

transfer heat from the cold to a hot thing. So isn’t this Kelvin-Planck

statement basically just giving us some restrictions

on the heat engine? And isn’t the Clausius statement

really just a statement that gives us restrictions

on refrigeration cycle? So these two statements of the second law are exactly

what we’ve been talking about before, okay? This upper one refers to heat

engine and this lower one refers to refrigeration cycle. And I messed it up. My writing is messy. I can’t hardly read it. This says heat and this

says refrigeration. All right? So what we really have to do

now then is look inside the box. If we go back here to the basic

box that is refrigeration cycle and the basic box that is heater

cycle and say, “Well, okay. How do we do this then?” We know that we need

to reject heat somehow. What device do we need? We know that we need

to find work input for the refrigeration

cycle somehow. What devices do we

have inside the box? I’ll give you a hint. These are cycles, right? These are both called

cycles for a reason. We’re going to start

with a substance at a particular thermodynamic

state and then we’re going to allow it to go through

different processes until we get right back with the

same state that we started with. So that’s the next step

and I’m out of time. So we’ll work on this

more on Wednesday. So if this seems like we’re kind

of in the middle of something, then we are right in

the middle of something and you probably don’t want to start your homework

yet on this material. We just haven’t gotten

to that point. So anyway — I mean you can

start working on the homework that deals with uniform

flow problems, but not the chapter

six material for sure. [ Music ]

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