[ Music ]>>What I want to do

today then is just begin with another example problem. In fact I’ll end with an

example problem as well. Basically that’s the

same thing as this. This is just my copy

in my notebook. So I’ll move this now. So we’ll do two example

problems today. And hopefully after you see

these two examples you have a much better feel for the

nature of the kind of problems that you’re going to be

dealing with in this class. By the way the midterm

exam is going to cover through section 42 which is,

you know, problem solving for a first law for

closed systems. And again it’s through

today’s material, through the homework

that’s due next week. All right so we’ve looked a

couple example problems already. Both of those were sealed

rigid vessel types of problems. And now we’re going to look at something a little

bit different. I actually put this on the board

last time but today we’re going to actually solve

through this problem. This does say problem

132 but it’s actually from the 6th edition, we’re

up to the 8th edition now. So you’re not going to

find this in your textbook. So I put it up here and

I’ll just leave it up there for a little while longer as I’m

starting to set up this problem. So as usual let’s

read through it. So we have a mass .2

kilograms of saturated R-134a. And it’s contained in a

piston cylinder device which is at kilopascal. It says initially 75% of the

mass is in a liquid phase which of course means 25% vapor. It says now heat is

transferred to the refrigerant at constant pressure until the

cylinder contains vapor only, all right. Now I will note that the author

could have just simply said that this was a frictionless

vertical piston cylinder device which automatically means that it’s a constant

pressure process. Instead the author gave the

initial pressure and then said that heat transfer occurs

at constant pressure. So either way it’s a

constant pressure process. Anyway show this process

on the P-v diagram, determine the volume

occupied by the refrigerant at the initial state,

the work done and the total heat transfer. All right, so that

is this problem. So 4-132 edition 6. Okay. All right, so the

first thing I always like to do is draw a

sketch and make sure that I show all the information

that’s known and unknown. So this one is a

piston cylinder device. So basically I would

show a piston cylinder. We’re given the mass,

so 0.2 kilograms. It’s R-134a, so I

note the substance. It says it’s saturated and then

it tells me the percentage mass in the liquid phase,

oh I’m sorry. First it tells me

the initial pressure so I’ll just put P1 is

equal to 200 kilopascal. All right now it tells

me that initially 75% of the mass is in

a liquid phase. And we all know that the

author did this to us on the last problem, right. We need the quality which is the

percentage vapor phase not the percentage liquid phase. So 75% liquid obviously

means 25% vapor. So the quality is given us .25. It says heat is transferred. It doesn’t tell us which

way heat is transferred just that heat is transferred. So I’ll show Q with kind

of ambiguous arrows. And then it says transferred

at constant pressure until the cylinder

contains vapor only. All right, so constant pressure. So I’ll just say P1

is also equal to P2. And now it starts as a two

phase mixture at the point where it contains, where

it contains vapor only is where we hit the

saturated vapor line, right. So we definitely know that

we’re at saturated vapor at .2. So 2 is saturated vapor. And we want to show

this on a PV diagram. And we want to find the

volume occupied initially. And we want to find

the work and we want to find the heat transfer. All right, so this pretty

much sets up the problem. Now let’s just start

one by showing it on the PV diagram even though

I don’t necessarily have to do this immediately. It does make a certain

amount of sense to do so. We note that we start as a reasonably high

quality two phase mixture. And since we’re moving

at constant pressure and this is a PV diagram then

it means that we have to moving to the right, right, moving along the

constant pressure line. And at the point where it

finally becomes all vapor that’s right there on the

saturated vapor line. So there’s my state point two. If this problem said

that heat was added until it becomes super-heated

well maybe we could perhaps show an entire constant pressure

line as we look like that and that way we show .2

being somewhere up here above saturated vapor line. But not for this problem, right. I’m sorry, I didn’t

do this right. This is a PV diagram,

I’m thinking TV diagram. Constant temperature lines

kind of work like that. And nonetheless it

doesn’t matter. If it’s super-heated vapor then

clearly we’ve added heat beyond the point it’s a saturated vapor and that’s not what

we want, right. This problem ends when the

cylinder contains only vapor and there you have it. Yes? [ Inaudible Student Question ] You know what? I didn’t really say

that properly. Yeah, 75% would be high quality but that’s the percentage

of liquid. Twenty five percent is a

relatively low quality. Honestly it’s still

probably going to be near the right hand

side because after all even through the vapor’s only 25%

by mass, it’s specific volume at this pretty low

pressure is going to be about 1,000 times greater than the specific

volume of the liquid. So even though it’s only 25% by

mass it might be more like 80 or 90% by volume and this

is a volume axis here. So it is very possible at .1

is far to the right like this. I don’t really know

for sure until I go through the numbers but, you

know, at least this is accurate. You know, horizontal

[inaudible]. All right, so what

next do we need to do? Well let’s make sure

we know what to assume, what not to assume. First of all there’s no kinetic

energy change as is the case with essentially all

these problems, right. There’s no velocity associated with the substances

in that cylinder. What about the potential energy? Clearly there’s going to

be a height change, right. I mean the volume is increasing so the piston does

have to move upwards. However, what you find is that

potential energy changes have to be what, I’m sorry

let me rephrase that. Height changes have to be very

significant before there’s any meaningful and measureable

change in potential energy. Usually you’d have to have a

height change of, you know, something on the order

of several meters before that potential energy

is not negligible. So here we’re talking about a

really small mass of material. I don’t know what

the total volume is but it’s not that great. And, you know, even if the cylinder rose

significantly there’s still not going to be several

meters of height change. So it makes sense to assume

that there’s no change in the potential energy, okay. So that’s typical of

these kind of problems. I might also note I

have no way to quantify that even if I wanted to. I don’t know if this is a short

squat cylinder which is going to increase a fraction or if it’s a tall skinny

cylinder that’s going to increase significantly. They don’t tell me anything about the geometry

of that cylinder. So even if I wanted to find

the height change I have no way to do so. So I just have to assume

that there’s no change in potential energy, okay. Now there definitely will

be some heat transfer. There’ll definitely

be some work. We know that work in this

problem is all boundary work, right, they’re not

electric resistance work, there’s nothing like that. And for a constant pressure

process this is mass times pressure times change

of specific volume. Remember this is boundary work but for constant

pressure only, all right. So that’s how we’re

going to find the work. And then we’re always going

to need the first law. The heat transfer is the change

in internal energy plus changes in kinetic and potential

energy plus the work. And by the way if I was

able to quantify the kinetic and potential energy, I

would definitely write out the equations but

here we know they’re zero so I mean why bother. We’ll just show delta

KE and delta PE and just show them

equal to zero. So this is the basic equation

that we need to solve for. So now the question is

can we find everything that we’re looking for? Not quite yet. What about the volume

at point 1. The only way to find the

volume is to know the mass and the specific volume. And fortunately I

can find that, right. So I’ll also note that to find

the intial volume is just mass times specific volume that

is initial specific volume. I’m given the mass

and I need V1. So with these equations we’ll

ask ourselves do we have enough information now to

solve the problem? So before I even solve anything

we just say well yeah, right. We know that, as long as we

know my two state points, state points 1 and 2, I

can into my property tables and I can find any

property I want. Now quality counts

as a property, right. So we know the pressure and

that the quality is 25%. So yes we know state

.1, we can find V1 as just Vf plus quality

times Vfg. We can find U1, just the

quality times Uf, I’m sorry, just Uf plus the

quality times Ufg. So we know state point 1. State .2 is given, I’m sorry, pressure 2 is given,

it’s the same as 1. What about state 2? Well again we need two

properties and we have them. We know that the pressure is

still 200 but we also know that it’s a saturated vapor

meaning that the quality’s 100%. We know that at state .2 V

is just Vg, U is just Ug, right, saturated vapors. So we have both [inaudible]

states. We can find everything

shown down here on the page. So now it’s just

a matter of going through a little

bit of mathematics. I will note that because

we’re talking R-134a and we’re in the SI system then

we’re going to have to use tables A11 and A12. A13 is a super-heat table but we don’t have super-heated

vapor anywhere in the problem. So we’ll just use A11 and A12. And in fact we really don’t

have the temperature anywhere in the problem either. We’re given pressure

at points 1 and 2. So really it’s just A12

that we’re going to use or our saturation

pressure table. So all the data’s going

to come out of there. Now we can work on this problem. So let’s just find our

thermodynamic property data first. And we’ll start at state .1. So at P1 and at the quality

at 1 we go into table A12. And now let me get rid

of the problem statement and put the property tables up. [ Pause ] Okay, so A12, all right. So here’s table A12, the

saturated R-134a pressure table. And all the data we need is

going to be at 200, right. So we go to the right table. We know the pressure. We read across. We need both Vf and Vfg. Of course we don’t have Vfg,

it’s not printed in the book. I don’t know why the author

didn’t just add one more column and give you the Vfg

data but he didn’t. So V1 is Vf plus the

quality times Vfg but we know Vfg is

just Vg minus Vf. [ Pause ] All right, so this is the

equation we have to solve for. So let’s just pull the

numbers out of our table. You can see them

right over here. So we go down to

the 200 psi line, I’m sorry, kilopascal line. We read across and Vf is

.0007532, the quality is 25%. By the way I should

put X1 over here just to make sure we know this

is the initial quality. Anyway, quality’s .25. And then Vg is .099951 minus Vf. So point triple zero 7532. And if we just go through the

mathematics we end up with V1 of .02555 cubic meters

per kilogram. So there’s V1. And we do U1 exactly

the same way. This is Uf plus the

quality times Ufg and there is Ufg

data in this book. So again just go to the 300,

I’m sorry, 200 kilopascal line and read across to the

5th and 6th columns. Uf is shown as 38.26 and then

plus the quality times Ufg which is 186.25, oh yeah. And then again the units are

all kilojoules per kilogram. We go through a little math. And we get 84.82

kilojoules per kilogram. By the way one thing I want to note sometimes I may have

some incorrect numbers in some of my problems that I’m

solving in class or even some of the problems I’ve

posted on the Blackboard or on the Board S in my office. You know, conventions are held

like every five years or so by the international community

and new research is presented. And the tables change, you know. New data becomes available for refrigerants or

water or whatever. So periodically the various

governing bodies usually say Sme. We’ll then print

new property data. So if you look at the property

tables from the 6th edition of this textbook which

is I think about 7 or 8 years old new compared to

the property tables that are in the 8th edition, the numbers

will be slightly different. Not all of them. The refrigerant tables didn’t

change, the water tables changed or maybe it was the

other way around. It doesn’t matter. I fixed everything on this

problem and I’ve tried real hard to fix everything on all

the problems that I’ve used in the past but I might

miss something a little bit. I have noticed that some of you are indeed using

[inaudible] property tables presumably from some

old book that you have. I know that because some of your

numbers are subtly different. You know, maybe just off down

in the 3rd or 4th decimal place. But some of you are using

numbers that aren’t correct. So that’s okay, I’ll

still accept that. But I just want you to

be aware the numbers do change periodically. All right, so back

to the problem. Let’s also find the

volume at this point. So that’s the mass times

the specific volume. So .2 kilograms times .02555

cubic meters per kilogram. And that’s going to give me

.00511, oops, cubic meters. So kilograms cancel and there’s

the first thing we’re looking for, the volume. All right, next let’s find the

thermodynamic property data at state .2. So at P2 saturated vapor. We’re still in table A12. So V1 is equal to Vg and

again we just go to the table. In fact Vg’s already here. It’s the same .09995

that was up at the top. So .0999.>>It should be M2

right, not P2.>>Oh yeah, yeah,

yeah, I’m so sorry. It is .2, I mean, I

put P2 saturated vapor but then I wrote and said V1. This is .2 saturated vapor. So V2 is Vg so it’s .099951

cubic meters per kilogram. And similarly U2 is

Ug which is then going to be 224.51 kilojoules

per kilogram. So we have all the data that

we need for the problem. Just two things left, right. We can find the work, we

plug it into the first law. We find the heat, we’re done. So work first. So the work and I’ll just

rewrite the equation. So we have .2 kilograms, our

pressure is 200 kilopascals, then .099951 minus plus .02555

meters cubed per kilogram. You know, as usual you

look at those units and the units clearly

don’t look correct. I mean, where’s the

kilojoules and all this. But as usual let’s just remember that a kilopascal is the

same thing a kilonewton per square meter. And a kilojoule is the same

thing as a kilonewton meter. So we can just cancel

out all the units here and everything cancels. Mass cancels, cubic meters

cancel, kilopascals cancels, kilonewtons cancel,

so we’re just left with kilojoules, right. So as long as you’re in

the SI system solving for work there’s no unit issues. Everything’s an identify

with regard to these conversion factors

so it works out easily. Caution, in the British

system it’s not going to work out this easily. It will definitely not

work out this easily. You know, you’ll have Psia

for your pressure units and then cubic feet per pound. A Psia cubic foot is not

the same thing as a BTU. You’re going to have to go into

the conversion factors which is in the inside back

cover of your book or in my case the last

page of the handout. And you’ll find the

conversion factor you need. In fact it’s right here where

the cursor is, why cursor turned to an x, I don’t know. But it’s right here. It says 1 BTU equals

5.40395 Psia foot cubed. So all the conversions you

need are in the book somewhere. But follow through

with the units. If you don’t follow through with

the units there will be errors. All right, no unit issues. We go through the mathematics and we find the work

is 2.98 kilojoules. And it’s a positive number

which is supposed to be, right. The volume is increasing. The piston is moving upwards. Work is being done by

the system as it pushes into the piston and

moves it upward. So work done by the

system is positive. And that’s a positive number. So that gives us

some confidence. Lastly heat transfer. All right, so heat transfers

found with the 1st law. Again I’ll rewrite the equation. So it’s really just

a matter of plugging in all the numbers

we already have. So .2 kilograms, U2 minus

U1, so 224.51 minus 84.82, those are both kilojoules per

kilogram plus 2.98 kilojoules. The units work out just fine and

we go through the mathematics. And we end up with

30.9 kilojoules. So that is this problem. Any questions? There is an alternate way to find the heat

transfer isn’t there? Remember that is we have a

constant pressure problem we can use [inaudible] instead

of internal energy as long we do it properly. I would note that for a

constant pressure process. [ Pause ] We still have the

same equation, right. I mean this is our

basic equation. However, the boundary work

is mass times pressures times change in specific volume. We remember that U2 and P2V2

is H2 and U1 plus P1V1 is H1. So it turns out that if

we wanted to we could just as easily have found

the total heat transfer for a constant pressure

process only and this is a constant

pressure process, right. But we could have found it just

as easily by finding H2 and H1. I could’ve gone back to state .1 and I say H1 equals Hf

plus quality times Hfg, H2 is just Hg, right. And I could have found

the heat transfer in a slightly more

direct fashion. Now you may wonder

well why didn’t I do it if it’s a little easier

to do it that way? Well the reason I didn’t do it

is because the author asked me to calculate the work. In fact to calculate the

work anyway what does it really matter? I mean I’m going to

need to find most of this thermodynamic

property data. It doesn’t really

save me any time. As long as I have to

find the work I might as well just use

the internal energy. But you can do it both ways. In fact I would suggest

do it both ways. For constant pressure problems if you don’t get the same answer

you’ve done something wrong. So it’s a good learning tool. Anyway, questions? Good. So this brings us

to a difficult problem. As if this last one

wasn’t hard enough, right. These are not really that complicated once

you’ve tried them. There’s a lot of them and they’re all a

little bit different. But this next one is a little

more complicated than most. So it’s this problem here. [ Pause ] Well, there we go. So this is out of the current

edition of this textbook. And this is problem 4-139. So let me read it. You have a piston

cylinder device which initially contains

.35 kilograms of steam at 3.5 megapascals, superheated

by 7.4 degrees Celsius. By the way that’s

not the temperature. That’s the number of degrees above the saturation

temperature. When it tells you

superheated by a certain number of degrees that’s

the temperature above the saturation. So we know we can go into the

steam tables at any pressure and find the corresponding

saturation temperature. The temperature initially is

7.4 degrees higher than that. So make sure you’re

aware of that. It says now the steam loses

heat to the surroundings and the piston moves down

hitting a set of stops at which point the cylinder

contains saturated liquid water. So the first part of the problem and what makes this hard

is it’s a two part problem. But the first part of the problem honestly is

not a whole lot different than the last problem. You know, you start

with a piston. Unlike the last problem

where you add heat and the piston moves up,

so far this problem starts with something superheated

and you remove heat and the piston moves down. So it’s the same problem as

the last problem but only up to that intermediate point. And this is where

things get difficult. It says the cooling

now continues until the cylinder contains

water at 200 degrees Celsius. By the way when the author uses

the word water he doesn’t mean saturated water, he

doesn’t mean liquid water, it doesn’t mean vapor

water, it could be anything, it could be any phase. So all we know is that it’s at

200 degrees at that final point. So the [inaudible]

will continue. We finally get to 200

degrees Celsius and that’s when the problem ends. Now isn’t the second part of the

problem really the same problem as the one we did at

the end last time? It’s just a constant

volume problem isn’t it? Once that piston comes down and hits the stops it

can’t move anymore. The volume is constant

inside that cylinder. The mass is clearly constant. So this is a constant

volume process. So this particular problem

is a two-part problem. You do the first part, let’s

call it from one to two which is a constant

pressure process. And then you do the second part

which is from two to three, let’s just call it last .3 which

is constant volume problem. All right, so 4-139 and

we know that from one to two is constant

pressure and from two to three is constant

volume, okay. So in other words P1

equals P2, V2 equals V3. I’ve seen problems like

this that are reversed where it starts sitting

on the stops as a constant volume problem. And then when enough heat is

added then the liquid starts to vaporize and the piston

starts moving upwards. So, you know, I can

certainly look at this problems

in both directions. All right, so we now have some

information about the problem. First of all any

questions just conceptually? Everybody should understand how

this is just a constant pressure problem followed immediately

by constant volume problem. At the end of the constant

pressure problem is the beginning of the constant

volume problem, right. All right. So now let us create

the right sketches. [ Pause ] All right, so here’s my

piston cylinder device. And then at some point

down here are the stops. This represents state .1. As the piston moves down and

sits on top of the stops here, I’ll just show this as a

dash line, then that’s going to represent state .2

and it’s also going to represent state .3, right. All right, so this is

just the illustration. So now let’s write down what

we know and what we don’t know. Oh by the way there’s

one thing I should point out about this problem

that I forgot to mention. I’m not really happy with

the way the author worded the problem. He really should have used

the word frictionless vertical piston cylinder device. Now granted there’s

a diagram in the book so it should be very obvious. But it may not be obvious to you as you’re first learning

this material. The fact that it’s a

frictionless vertical piston cylinder device means that it’s

a constant pressure process, all right. We know that. We’ve seen problems

like this before. But they don’t tell me that. The previous problem said

that the piston moves under constant pressure. So it was obvious. In this problem the

author has not done well. So if I give you a

problem like this or any piston cylinder

problem on the exam, I’ll make sure you know whether

it’s a vertical frictionless piston cylinder device or just

some random piston cylinder device that is a

moving horizontally at some odd angle set to

the pressure’s not constant. So I’ll make sure

it’s clear to you. In this problem anyway it’s

a constant pressure process from one to two. All right, back to the problem. So we know the mass

as .35 kilograms. We know that it’s steam,

in other words H2O. By the way when I write my

own problems I just use H2O, I don’t even use the word

water because I don’t want to confuse anybody, so I’ll

just say H2O and let you figure out what the phase is. All right, so it’s

H2O, it’s superheated by 7.4 degrees therefore T1 is

equal to T set plus 7.4 degrees. And we’ll figure out what

that value is shortly. But at least we know that T1 is

not the saturation temperature. It’s a certain number of degrees above the saturation

temperature, okay. Steam loses heat to

the surroundings. So there’s heat transfer

out of the system. And the piston moves down so

there is some heat transfer from 1 to 2. Eventually hits the stops at which point the cylinder

contains saturated liquid water. So 2 is sat liquid, all right. Again we know now that as

the cooling continues the temperature’s going

to drop some more. The volume’s going

to remain constant. But finally we get

to .3 and it tells me that T3 is 200 degrees

Celsius, all right. And let’s see, oh yeah I forgot

to write down the steam pressure at .1 so up here at 1 I should

have said that P1 is equal to 3.5 megapascals that was

given in the problem statement. All right, okay. So what are we trying to find? We’re trying to find

the final pressure and the final quality

if it’s a mixture. And let’s keep in mind that

the final is .3 not .2, right. And then we also would note that

we’re trying to find the quality if indeed it’s a

2 phase mixture. It also asks us to

find the boundary work. It also asks us to find

the amount of heat transfer when the piston first

hits the stops. So that’s really Q12. But it also asks me to find

the total heat transfer. So that’s the sum of the heat

transfer from 1 to 2 and 2 to 3. So Q total is Q1-3 which by

the way is equal to the sum of the heat transfer for

the first part from 1 to 2 and then plus the heat

transfer from 2 to 3. Okay so this is something

that we’re trying to find for this particular problem. So before we actually go

through let me show this also on a appropriate, how

to use a TV diagram. I’ll just put this

over here on the side. There’s my saturation line and there’s a constant

pressure line. All right, so we know that we

start as a superheated vapor. It’s not that superheated. I mean only 7 degrees above

saturation so it’s going to be pretty close to

the saturated vapor line. It says that .2 is

when it hits the stops and it’s all saturated liquid. So it’s a constant

pressure process, we’re going to follow

pressure line down to the saturated

vapor line all the way over until we finally get to

the saturated liquid line. So there’s state .2. And then we have a constant

volume process, right. Now notice a closed system. The volume’s constant. The mass is constant. The specific volume

is there for constant. And because we’re cooling the

temperature’s going to drop and we’ll end up with state

.3 that’s right there directly below state .2. So here’s our initial setup. Not really quite

done with the setup. Mostly done with the setup. What about some assumptions

to make and some equations and all that? Well like the previous

problem we’ll note that there’s no velocity

associated with the water that’s

inside this container. So there’s no kinetic

energy change. We’ll also note the height

change is probably minimal but we don’t know for sure, just because we don’t

have any geometry data. So we’ll just have to assume that this is pretty

small system. I mean it’s 1/3rd of a

kilogram, that’s not much, so. We can note that we’ll

assume there’s no change in potential energy, right. What about the work? That’s something we

should talk about. Well the work from 1 to 2 is

constant pressure work, right. So that’s just going to be

the mass, well I’m sorry. Let me first stay it’s

just boundary work, right, there’s no electric resistance,

there’s no other forms of work. So the work from 1 to

2 is all boundary work. And this for constant pressure

process is mass times pressure times change in specific

volume, okay. Again constant P only. So that’s the work

we have from 1 to 2. What about the work from 2 to 3. Yeah it’s zero, right. It’s a constant volume process. Again I should first write

that this is the boundary work and no other forms of work. And then that’s just the

integral of PdV from 2 to 3, but there’s no change

in volume so it’s zero. I mean we talked

about this before. You don’t have to write all

this out if you recognize that constant volume problems

have no work then just say work is equal to zero. So this gives us our work terms. And then what about

the heat transfer? Well the heat transfer,

first we’ll look at 1 to 2 and then we’ll move

onto the next part. So the heat transfer

is the change in internal energy plus

the change in kinetic and potential energies

and then plus the work. We know there’s no kinetic

or potential energy changes. So the heat transfer

from 1 to 2 is just this. And of course the work we

found earlier in the equation. So this is going to be how we’re

going to find the heat transfer. By the way since the

author specifically asked me to find the amount

of work I don’t have to use the [inaudible] version

of the 1st law that applies to these closed systems for constant pressure

processes, right. I could write Q equals

simply mass times H2 minus H1 but again there’s no need. I already have the

work so I’m just going to leave it in this form. So there’s the heat

transfer from 1 to 2. Now throughout the

heat transfer from 2 to 3 maybe I’ll just

do it this way. We’ll note that the

heat transfer from 1 to 3 is just the sum of

the heat transfer from 1 to 2 plus the heat transfer

from 2 to 3, all right. We know the heat transfer

from 1 to 2 is above. So M U2 minus U1 and

plus mP V2 minus V1. So there’s the heat

transfer from 1 to 2. What about the heat

transfer from 2 to 3? Well again there’s no

kinetic energy change. There’s no potential

energy change. There’s no work. So the only heat

transfer we have from 2 to 3 is just a change in

the internal energy, right. So this is just M

U3 minus U2, okay. Maybe what I should do to

make it more clear is note that this is equal

to the following. And all three of these

terms are now zero. So we got this. And then we’ll note

that the M2 U2’s cancel and we simply get M U3 minus U1. And then plus the

boundary work from 1 to 2. And this is how we’re going

to find the heat transfer. Thank you. So the equations are

the same equations. If this isn’t clear then

you just need to study more. There’s not a whole lot of

difference between this problem, the last problem, the

two we did last time. They’re 1st law problems

for closed systems. They use the same general

version of all the equations. They’re slightly different. The work sometimes

slightly different, the heat transfer might be

found slightly different. But they’re all the same. All these problems here in Chapter 4 are the same

problem but with variations. Yes?>>So when it says

that it hits the stops when it’s saturated

vapor [inaudible] that V2 is going Vm, right?>>Yes. That liquid, yeah.>>Okay, all right, so

what we want to do next? Not much more to do, right,

except get our property data. Do we have two independent

intensive properties at each of the three state points? And if the answer is

yes then we’re done. So let’s think about it. We know the pressure and we

know the temperature at .1, although I haven’t

calculated the temperature yet but I know them both. So yep state .1 is

known, superheated, pressure temperature

independent intensive. We have state .1. We can find V1, we

can find U1, okay. What about state .2? Yes, right, we know

that the pressure at 1 and 2 are the same. We have the pressure and we

know it’s a saturated liquid. So state .2 is known. We have Vf, we have

Uf is we need it, although we don’t need it. And then what about a .3? Well we know that it’s a

constant volume process. So we know the specific

volume is state .2, that has to equal the

specific volume at state .3. So we know that information. And we’re also told

that the temperature of .3 is 200 degrees Celsius. So again we have two properties. We have volume and

temperature, specific volume and temperature I should say. Again independent

intensive properties, so state .3 is known. So as long as we recognize

that we’re using the 1st law in its variations and we’re

using the data that we all have, the rest of this problem

is really just a matter of pulling data out of

tables and chugging away. So we’ll do that next. Let me pull the steam tables up. By the say SI units

we’ve got superheat, we’ve got two phase mixture. You know, we’re going to use A4, 5 and 6 as we solve

this problem. So that’s worth noting

here as well. All right, so hopefully you

have all this written down. [ Pause ] I’ve never owned

a PC in my life. I’m a Mac person. Although I do one of those

Surface Pro computers which is kind of a,

I won’t say anything. They work nice but

my iPad’s better. Okay, so I had to say something. Yeah, I don’t know, I guess PC’s

are just cheaper aren’t they? Okay, so [inaudible]

there we go. All right, so 4, 5, and 6. So let’s just start

at A5, so here’s A5. We really would like to

know what the temperature is at state .1. So let’s just start there. So the temperature at 1 is

the saturation temperature at the pressure of 1 plus 7.4. So the saturation

temperature at, was it 200? No, 3.5 megapascals,

quite a bit higher. So onto the next page,

there we go, 3,500. So that’s 242.56 plus 7.4 and why don’t we just call

it 250 degrees Celsius for simplicity, all right. So now I have the temperature,

I have the pressure. So now I need to go into

table A6 at P1 and T1. I’m going to table A6. So onto A6. Remember that we have these

sub tables for superheat. So here is the 3.5

megapascal sub table. We read down to the 250 line

which is the third line down. And it’s this data right

here where the cursor is. So we need two things, right. We need specific volume and we

need specific internal energy. So let’s just write

them both down. V1 is .05875 cubic

meters per kilogram. And U1 is 2623.9

kilojoules per kilogram. So there’s my data at state .1. Again I want to remind you that if I’ve made a

math error somewhere or pulled some data incorrectly

out of the tables I would like somebody to let me know

that so that I can fix it after all it’s all

getting recorded. So yes?>>On the tables so if we

needed like a 225 degrees for example, how

would we do that?>>That is interesting question. First of all you wouldn’t

because it turns out that at, well I mean maybe I

should take that back. What you could do is

you could interpolate between the saturated

vapor conditions and your superheat conditions. So you would have to go

in here at saturation which now corresponds

to 242, right. So these are saturation

conditions. These are conditions at 250. So you’re interpolating

between let’s say 242 and 250. Now that’s why I said

225 wouldn’t happen because 225 already put you

into the two phase region. But let’s say the

temperature was 245, right. So 250 obviously isn’t right. The data at 242.56

is not right either, you’re in between the two. So you would just

do the interpolation between those two entries, okay. All right, so next. We have our data at .1. Let’s just find all of our

thermodynamic property data. It’s going to be

easiest that way. So at .2 we have

a saturated liquid at 3.5 megapascals, right. It’s a constant pressure problem

because it only hits the stops when we finally get

to saturated liquid. So at P2 which equals

P1 and saturated liquid. All right, so this

is the pressure data. So we have to use table A5. We don’t need the

internal energy at .2 so, I think we do don’t we

because they have asked me to find the heat

transfer from 1 to 2. So I guess I better

write it down. All right, so we need V2 which

equals Vf and that is .001235. And U2 is Uf which is 1045.4

kilojoules per kilogram. All right, so that’s all the

data we need at state .2. Let’s do the same at .3. So at V3 which equals

V2 correct and at T3 and T3 was 200 degrees Celsius. Well now we’re given

temperature data. So that means we’re going

to have to go to table A4. I don’t really need to the test

because I simply know based on the problem statement

we’re a saturated liquid and then we’re cooling, you

know, you have to move downwards and that’s going to

put you definitely into the two phase region. I could do the test if I

wanted to just to prove that but not really necessary. You may want to do

it just to make sure. Nonetheless what

we would find is that at 200 degrees Celsius

my specific volume is greater than Vf and less than Vg. Okay, I mean we can see that

if we go back up to the table. I mean I have to go up to the

table anyway to get my numbers. But, all right so here’s A4. My temperature is, oops,

I need the other page. Yeah, here we go. So, you know, Vf is .0011

and my specific volume at .2 which equals my specific

volume at .3 is .01235. So .012 is greater than

.011 so clearly I’m above Vf and it’s a two phase mixture. So it’s a two phase mixture. How do we find the data we need? Well first keep in mind

that we really only need, well actually we need

two things don’t we? Really three things. They’ve asked me

for the pressure. They asked me for the quality. But then they also want to know

what the internal energy is. Actually I want to know

what the internal energy is so that I can solve my 1st law. So three things I need. The first one’s easy. It’s two phase. So P3 is just P sat at

200 degrees Celsius. So that gives me 150, I’m sorry,

1,554.9 kilopascals, okay. To find the quality

I’m going to have to use the specific volume

data for the two phase mixture, V3 is Vf plus the

quality at 3 times Vfg. So this is how we

find the quality. V3 is V2, it’s known to me. So this is just .001235

at 200 degrees Celsius. Vf can be looked up. Vf is .001157. The quality is what

we’re trying to find. Vfg is just the difference

between Vg and Vf. So Vg is .12721, so that’s Vg. And Vf is the same

number we have, .001157. Anyway, go through some math. We find that the

quality at .3 is .00062. So it’s extremely low quality. But still it’s a

two phase mixture. So now we have the quality and now let’s find

the internal energy. So U3 is Uf plus the

quality times Ufg. So again just look

up the numbers. Everything should be in

the table over there. Uf is 850.46, actually we had

that from before didn’t we? No we didn’t, we’re

at a lower temperature so we didn’t have

that from before. The quality is .00062. Ufg is 1743.7. And if we just go through

our math here we get 851.55 kilojoules per kilogram. All right, so not much

left to do but to go through our various equations. Next let’s find the work. So the work for this

particular problem is let’s see, we wrote it down before. It’s just our boundary

work from 1 to 2. I mean technically it’s the work

from 1 to 2 plus the work 2 to 3 but theirs is no

work from 2 to 3 because it’s constant

volume, right. So maybe I’ll just write that. Boundary work is what,

work 1, 2 plus work 2, 3 but that’s equal to zero. So that’s work 1, 2 which is

mass times pressure times V2 minus V1. So let’s just plug in the

numbers that we have here. Mass is .35 kilograms, pressure,

I’m going to put the pressure into kilopascals

instead of megapascals so that all my conversion

factors are just identities, just 1’s. In other words I don’t have

to worry about conversions. And then the specific

volume change. So we have .001235

and then minus .05875 and those are meters

cubed per kilogram. Again the units here are

the same units we had in the previous problem. Kilopascals basically convert into kilojoules per cubic

meter is what it converts into. And we end up with

minus 70.45 kilojoules. Now before we move on does

the minus sign make sense? Well it should, right. I mean negative work means work

is being done to the system. In other words the piston

is pushing into the system. The work is clearly

being done to the system. The volume is shrinking. So yes, the negative

sign is proper. And next we want to find

our two heat transfer terms. So again these equations

are already in your notes. So we have change in internal

energy plus the work from 1, 2. So again it’s just a matter

of plugging in the numbers, .35 times what are our numbers? I’ve lost my numbers. Okay, here we go, yeah,

thanks, 145.4 and U1 was 2623.9. I forgot my units. I don’t want to forget my units. And then the work

is minus 70.45. So if we go through

our mathematics we end up with minus 622.9 kilojoules. [ Pause ] And then lastly we find the

heat transfer from 2 to 3. And the heat transfer

from 2 to 3, well again this is just mass

times change in internal energy. Oh I’m sorry, I’m doing 2 to

3, I’m doing 1 to 3, right. I’m trying to find the

total heat transfer. I can find Q23 if I wanted

to but that’s not asked of me so I’m not going

to waste the time. So Q13 is mass times

U3 minus U1. And then plus the work from 1 to

3 which equals the work from 1 to 2 which still equals

the boundary work. So it’s everything

we had before. So .35 kilograms. Yes, and then U3 is 851.55. U1 is 2629, 2623.9 and

then plus the 70.45 again. And this gives me

minus 690.8 kilojoules. So there’s a nice

long fun problem. Yes?>>You could also find the

total [inaudible] 1 to 3 within Q12 plus [inaudible].>>You know, I’m not going to,

I’m not going to say for sure. I might have to go back and look

at the equations and see but, you know, we went through a

derivation of these equations and it would seem that there

are certainly ways to simplify if we needed to or to just

give us a different version. So yeah, I’m not

going to answer that. I would have to write

the equation down and see if it works out properly. But yes, there are certainly

ways you can get your answers that are a little bit different

then the way I’ve done here. For instance sometimes

the students were trying to find the work. They calculate the heat transfer

based on the [inaudible] for a constant pressure process. And then they calculate the

heat transfer using the internal energy plus the work. And they equate them and then

they get the work that way. So they never actually have to

calculate the boundary work. They just solve for

heat two different ways and the boundary work

is one of the unknowns. So yeah, there’s different ways

to solve some of these problems. Questions? All right. Well one of other thing

then that I want to do today and this is just to be

consistent with your syllabus. By the way we’re done

now through section 4-2. And we’re not going to talk

anymore about this material. Everything that we covered up until right now is fair

game on the midterm exam. And again the midterm

is a week from Monday. So you got plenty of

time to study for it. Again it’s a closed book, closed notes exam except

you’ll have property tables that I’ll hand out to you. Anyway so starting now we move on to the next section

of Chapter 4. And this deals with something

that’s called specific heat. So we’re not completely

leaving behind these closed system problems. But we have to ask ourself

what if I have something other than water or the

refrigerant, 134-a? For instance what if

I have, I don’t know, how about a gas like

an ideal gas. How do I solve 1st law

problems for ideal gases? They’re just as applicable

to those types of problems as they are to any

other type of problem. Well the answer is

specific heat. It’s not going to be clear yet what specific heats

are, why we use them. But I will tell you that

we are going to have to solve 1st law problems. Some of these 1st law

problems are going to be involving ideal gases. And because there’s so many

gases there isn’t nearly as much property data available

for this wide variety of gases as we have for say the

refrigerants or for water. So often times you just don’t

have property tables like these, like the ones we’ve been

looking at here on the board. We’re going to have

to use another method. And that involves what’s

called specific heat. [ Pause ] Okay, so specific heat. Now specific heats

are properties. [ Pause ] And there’s actually

two of them, okay. One is called specific

heat at constant pressure. And one is called specific

heat at constant volume. [ Pause ] The one we call specific heat

as constant pressure we’re going to use CP and the other one

that we call specific heat at constant volume

we’re going to call CV. Now specific heats

are properties. But these are different

than most of the other properties

we’ve seen. Remember when we

talked about [inaudible] and I defined [inaudible]

as U plus PV. And I simply said

that since U and P and V are all properties then

any mathematical combination of those properties

is also going to be a thermodynamic property. That’s really what

specific heats are. They’re different combinations

of other properties. We define the specific

heat at constant pressure as the partial derivative

of the specific [inaudible] with respect to temperature

holding pressure constant. And we define the specific

heat at constant volume as a partial derivative of

the specific internal energy with respect to temperature

holding volume constant. The reason we call these

specific heat at constant volume or constant pressure is not

because it has anything to do with constant volume or

constant pressure processes, that’s not at all important. They’re called constant

pressure and constant volume because of the way we take the

partial derivatives, right. When you write a partial

derivative like this that means we’re taking the

derivative of one variable with respect to another holding

this third variable constant, right. Specific heat at constant

volume, we’re taking the partial of U with respect to T but we’re

holding the volume constant. So that’s why we call

these specific heat at constant volume

or constant pressure. So please make sure

you recognize that. That we’re not talking

about constant pressure or volume processes,

we’re talking about new thermodynamic

properties that are combinations

of other properties. And we’re going to have to

figure out how to use them to solve some realistic

problems. Now this will have to

wait until next week. So we’ll talk about it later. Anyway at least you’ve

been introduced to this and we’ll deal with

it in more detail. So I’ve got homework

up here from last time. Please don’t forget

to pick up your work. If anybody has problems with anything please come

see me in my office hours. So far I mostly just have

to sit there by myself in the office hours because

nobody has questions. So that’s great. I guess you know the

material really well. But if anybody has

questions please come see me. Your exam’s in just over a week. And well it’s just one midterm. It’s still worth a quarter of

your grade so it’s significant. So if you have questions

or problems with. [ Music ]