[ Music ]
>> Okay, so good afternoon to all of you.
Just a couple things before we start today.
Today's obviously the last day of class.
Just a reminder, not that you need the reminder,
but your final exam is Monday at 11:30.
So please remember that's a different time
than the normal class time
and we'll be working right here in this classroom.
I will talk more about the final exam near the end today.
There will also be a question and answer period.
I'll give you some review.
We'll do teacher evaluations.
But the first thing I want to do is just finish
up the example problem that I started last time.
And that was 996 from the 6th edition.
Now we had actually already seen the problem.
It's still over here but I'm going
to get rid of that right away.
We understand in this problem that is non ideal
yet still simple air standard Brayton cycle.
There's a compressor acting from 1 to 2.
There's a heat input device,
a combustion chamber acting from 2 to 3.
We're got our turbine from 3 to 4,
that's our worked output device.
And then we reject into the environment from 4 to 1
but we really do not need to quantify that.
So this is the basic cycle, right.
This is a non-ideal cycle
so I have shown non ideal compression from 1 to 2a.
I've also shown the non-ideal expansion from 3 to 4a.
But remember we still have to use this tool.
We still have to assume that it's a non-ideal, I'm sorry,
we still have to assume that's it's an ideal
that is isentropic process from 1 to 2s.
And then use the isentropic efficiency to go from 2s to 2a.
And we have to do this same thing for the turbine.
We have to assume that it's ideal from 3 to 4s,
isentropic if you will, and use the appropriate equations
to get T4s and then apply the isentropic efficiency
to get the temperature at 4a.
So that's the basic process.
Once we have those actual temperatures we'll just plug
them into the appropriate equations.
All these equations were presented last time
and we can get the results that we're looking for.
So perhaps we could get rid
of the problem statement over here now.
And then we'll have better access to the board.
All right, so first things first.
We're almost always going to start a problem like this
at state .1, the input to the compressor.
We know that for an isentropic process the ratio
of temperatures, T2s over T1 is equal to the pressure ratio,
to the K minus 1 over K power.
This applies to our isentropic compressor.
T2s is unknown but T1 is known.
The pressure ratio is known, right,
we're given 2,000 kilopascals at a maximum
and 100 kilopascals as our minimum.
So this is our 2,200, in other words 20 pressure ratio and has
to be raise to the K minus 1 over K power.
By the way .4 over 1.4 is something like .2857
or something like that.
So that might be something you want to remember.
You can just go through the math.
Anyway this allows me to find the ideal temperature
that would leave this compressor
if indeed it was an ideal device of 706.1 kelvin.
Actually I think we did this last time.
Now we need to continue by using the compressor's
isentropic efficiency.
This is our relationship between the ideal work
and the actual work.
We know these are entropy changes, right.
We know that the entropy change for an ideal gas
with constant specific heat is Cp times temperature change.
And the Cp's and the numerator and denominator would cancel
so this becomes nothing but a equation based
on temperature change.
So this would just be T2s minus T1 over T2 actual minus T1.
So this is given to me as .8.
T2s is known, T1 is known.
We do need to make sure
that we're using the appropriate units.
Certainly we have to use kelvin in the equation above.
Here we can find the temperature difference either in kelvin
or Celsius, it will be the same but frankly
since all the numbers are
in kelvin let's just leave them that way.
And we end up with the temperature at 2a,
[inaudible] known now of 807.6 kelvin, right.
So now we have that and we're going
to do exactly the same thing for the turbine.
Although we're going to use the turbine's efficiency instead
of the compressor's but it's still the same basic process.
So again we assume that's it's isentropic first.
So T2 over T4s is equal to the pressure ratio to the K minus 1
over K. T3 is given so converting it
into kelvin it's 1,000 over T4s.
And again this equals the pressure ratio
of 20 to 04 over 1.4.
I believe I mentioned once upon a time
that you should be using your calculator,
save memory button a lot.
I mean you've already done the Rp to the K minus 1 over K
when we did the compressor analysis above.
So I mean you might as well just store that in memory
and save yourself a minute when you're solving the problem.
Nonetheless we can go through our mathematics.
And the T4s becomes 424.9 kelvin.
Now this would be the temperature
if it were an ideal cycle but it's not, right.
Again you're given isentropic efficiencies.
So we use the turbine's isentropic efficiency,
that's .9.
And we note that this is the actual
over the ideal work for the turbine.
Again this is the nfop change, the nfop change
or Cp times temperature changes.
The Cp's cancel so this is going to be T3 minus T4 actual
over T3 minus T4s ideal.
So again this is .9.
T3 is known, T4 actual is the unknown.
So oops, just plug in all the numbers that we have
and we solve for the actual temperature leaving
that turbine.
And we get 42.4 kelvin.
So now we have all of our temperatures.
And once we have all
of our temperatures there's really not a whole lot more
to do but just apply those to the appropriate equations.
If we're trying to find the thermodynamic efficiency,
that's going to be a function of the network.
The network of course is the turbine work minus the
compressor work.
And the efficiency's going to be a function of the heat input
which is, well just Cp times the temperature difference
across that heat input device.
So we know all that now.
All right, so the thermal efficiency then is going
to be network over heat input or if you will the difference
between the turbine
and compressor work over the heat input.
Now I could either separately calculate each
of these different work terms or I could just plug
in the equation, you know, Cp delta T for each of these work
or heat transfer terms.
And then just let everything cancel out.
You know the Cp's have all cancelled out,
I would only have a function of temperature
and that might be the easy way to do it.
I'm not going to do it that way simply because I think
at this point you're still learning the material
and it's probably a little bit more useful
to show you the equations, right.
So the turbine work is going
to be Cp times the temperature change across the turbine.
And in fact here I need to be a little more careful
to use the proper subscripts, right.
The network for the real cycle would be the actual work
produced by the turbine minus the actual work required
by the compressor.
So it would make sense just to put the subscripts A just
to remind us that these are the real numbers,
that actual numbers that we're using.
So Cp times T3 minus T4a is the actual work produced
by the turbine.
Cp times T2a minus T1 is the actual work required
by the compressor.
And Cp times T3 minus T2a, you know, the actual discharge
for the compressor is 2a.
So that's the actual input to the combustion chamber, right.
So we have to use a 2a here.
Now again we have the value of Cp, we have the value of all
of these temperatures.
Just plug in all the numbers.
We get 520.2 kilojoules per kilogram for the work out.
We get 510.1 kilojoules per kilogram
for the work in to the compressor.
And then the heat input is 193.4 kilojoules per kilogram.
Once we have this information now we can find the
thermodynamic efficiency.
[ Pause ]
So again just plug in the numbers that are right here
on the page and go through a little mathematics and we end
up with .0522 or 5.22%.
Just a reminder we cannot use the equation
for the non-ideal cycle that applied
to the ideal cycle, right.
For the ideal cycle you could use the equation 1 minus 1
over the pressure ratio to the [inaudible] over K, that's fine.
But that is not an ideal cycle, right, this is the non-ideal,
non-ideal Brayton cycle.
So we need to use this version of the equation.
Now again we could have just plugged in the terms above
and cancelled everything out just leaving us temperatures.
And that would be an alternate way to do it.
Lastly the network is just the difference between the turbine
and the compressor work.
So we can just take the difference between the two above
and we end up with 10.1 kilojoules per kilogram.
So this is the network per unit mass for this particular cycle.
Now one reason that I also solved it this way is
because this problem may not be the end of a problem
that would be similar to this one.
For instance what if I'd asked you for the rate of heat input?
Or what if I asked you for the mass flow rate?
Or what if I asked you for the actual power, you know,
the rate the work is being done?
Then we need some other relationships, right.
So for instance one thing I could have done
in this particular problem is given you the mass flow rate.
So if Mdot is given then we could find such things, oops,
we could for instance find the net power, right.
The net power is just the mass flow rate times the network per
unit mass.
We could have also found again the heat input rate.
That's just the mass flow rate times the heat input per
unit mass.
Some problems will be a little bit different than this.
What if I gave you the rate of heat input or the power?
Then you could use these same two equations above
but work backwards to find the mass flow rate, right.
So if Q.h or W.net or for that matter W.t or W.c,
if any of those are given then you could find,
then you could find the mass flow rate using well the
equation similar to above.
[ Pause ]
All right, so there's lots of different problems
that would all be related.
But you have to be really careful
when you're solving them.
The equations are always going to be the same.
You have the full set of equations.
None of this has changed really since we were talking
about this material back in Chapter 7.
But the problems are all a little bit different.
So again be careful, read them carefully,
make sure you're using the equations properly.
Are there any questions on this particular problem?
Yes.
>> Why is it that the minimum temperature corresponds
to .1 [inaudible] .3?
>> Well there's many ways to answer that.
I guess the easiest way is to just look at the Ts diagram.
We're always starting with atmospheric air.
That's what's used in the compressor and that's going
to be the lowest point in the cycle, you know,
what we have right here in the environment.
Once you compress it you get
to a higher temperature and pressure.
And then once you add heat to that
at constant pressure the temperature goes even higher.
It's the output from that compression chamber that's
always going to represent the highest temperature
in the process.
Once you do work you're going to end up at .4
but that's still going to be
at a higher temperature then the surroundings.
You're still going to have to reject heat into the environment
to drop that temperature back down to .1.
So, you know, .1 is always the low point,
always the low temperature in the cycle,
.3 is always the high temperature in the cycle.
It's just based upon the way that these cycles work.
So is there a question here too?
Okay. The author does do that a lot in the book, right.
The author will give you a minimum to maximum temperature
and just assume that you know what that means.
Well again just look at the Ts diagram.
The Ts diagrams are all going to look like this.
They're not going to be significantly different.
You're always going to have 3 as the high point temperature and 1
as the low point temperature, I mean always.
That doesn't mean that one can conceive of other cycles
that might work a little bit differently.
But this is a Brayton cycle, okay.
The Brayton cycle is the gas turbine, the jet engine cycle.
It uses atmospheric air at .1, that has to be the low point.
After combustion that's what goes into the turbine.
That .3 has to be the high point
and that's the way the cycle works.
Any other questions before we move on?
Yeah.
>> So will we always be given
like the minimum maximum temperatures?
We won't be given like Ts,
T2s and like T4s [inaudible] to have to find.
>> I can't promise anything.
All I can say for sure is that all the problems are going
to be a little bit different.
I think the one thing you might want to do as you're studying
for your final is look at the problems that were given
to you whether it's examples or homework problems and think
about how you might want to solve them backwards.
What if you started with the end result and worked backwards just
to give you a little bit of practice?
Again it's all the same equations
but I can certainly conceive a problems that would be sideways
or backwards compared to what a typical problem would be.
So yeah, I won't promise that I'm not going to do that.
I mean I think that there's plenty of problems
that I could give you that, you know, you'd be able
to show me you understand this material
or what they actually are, well you'll see on Monday, right.
And remember too that I did promise you a problem
like this on the final exam.
So this really brings together material
from Chapters 5, 6 and 7.
If you can do a problem like this
that means you can solve a problem
for a study flow process.
You can use thermodynamic efficiencies.
You can use isentropic efficiencies.
You can use ideal gas relationships.
I mean it kind of brings them all together.
So that's why I rather
like these Brayton cycle kind of problems.
It just kind of shows a little bit of everything
that we've been covering for the last month.
Any other questions before we move on?
All right, great.
Well this is the last example problem then
that I'm going to do.
And the other things that are left to do today are review
and questions and answers, talk a little bit about the final.
So I think what I want to do is let me start with the review.
My intention is not to fill up the board here with equations.
In fact I really don't want to write
down anything unless somebody has some specific questions.
I'm just going to kind of talk through what we've been doing
for the last 2-1/2 months
and then hopefully that all makes sense.
If anything I say brings
up a question you can either interrupt me or just wait
until the end and ask it later in the Q&A area or Q&A section.
It's really up to you.
So a little bit of review then.
So first of all I tried to impress upon you the importance
of thermodynamics for solving what will be some important
types of problems in the future, right.
The very first day we actually talked about heat engines.
We talked about refrigeration cycles.
Even though you had no idea what those were or what kind
of equipment one would use, I at least wanted to point you
in the right direction.
Then we really got into the material starting
on the second day.
So basic concepts, terminology, definitions,
I mean there's a whole new language
of thermodynamics, right.
This is where we talked about such things as state
and property and equilibrium.
We look at various properties.
We looked at extensive and intensive properties.
We just tried to get a feel for the language if you will.
We learned about the word process.
We learned about the prefix iso that would be used
when something is constant.
So there's a lot of terms that we looked at.
The properties we looked at included such things
as pressure, temperature, density or specific volume,
the inverse of density.
And those are all measurable properties.
But we also know that there's other properties
that we cannot measure that are equally important
and equally useful.
Such things as energy and [inaudible] and entropy.
So there's all sorts of properties.
Now we look at the concept of energy.
We I think all generally understand
that energy is really just a measure of our ability
to do work or our ability to transfer heat.
But we also realized that energy's in different forms.
You could have an energy stored internal to your system.
We call that internal energy.
You could have energy based on the motion
of the system, kinetic energy.
You could energy based upon its height
and gravitational field, the potential energy.
We also note that energy can be transferred
in a variety of ways, right.
We can transfer energy as work.
We can transfer energy as heat.
But ultimately the energy transfer has to equal the change
in energy of our system, right.
So that's where we eventually started to talk
about the first law but we're not really quite there yet.
Once we understood the idea
of these different properties we started looking
at pure substances.
Hopefully it was clear that there's a lot of property data
out there but when you start dealing
with mixtures then it gets confused and there isn't
that much data out there for the wide variety of mixtures
that exist in this real world.
So we're just looking at pure substances.
Well at for now we are, right.
In Me302 we'll start looking at mixtures
but for now we were just looking at pure substances.
It was hopefully made clear
that if you know any two intensive properties then
that sets the thermodynamic state
and all other properties are known to us.
So we can get those other properties either
through well equations like the ideal gas equation of state
or through tables like the property tables
for refrigerant 134a or water that are in your book.
Or even using graphs we can find relationships
between properties.
For instance the compressibility chart.
But nonetheless we were able to understand
that property data exists in various forms.
This is when we started talking
about the different phases of existed.
How we know that phases are just the way we observe these
substances out there in the world.
And that we can also have mixtures of phases.
So we could have solids, liquids or vapors,
those are the three phases.
We could also have a mixture for instance we dealt a whole lot
with liquid vapor mixtures, right.
It was hopefully made clear that it's mixtures of liquids
and vapors that really represent the majority of the processes
that we're going to deal with in this world.
And in fact as we started to go
through problems hopefully you saw that that was indeed true.
You're often taking let's say liquid water
and you're boiling it into a vapor.
Sometimes you have a two phase mixture during that process.
You'll often have a refrigerant that's being well,
being ends up having heat added to it or remove from it
so that you have again liquids or vapors.
So the majority of the work that we're actually dealing
with once we finish this discussion
of pure substances was property data specifically deals
with liquids and vapors.
Now for water or 134a we've got really good tables, right.
We have, well mostly really good tables.
We have superheat tables, so that's everything to right
of the saturated vapor.
We've got mixture tables or saturation tables.
So that's where we would use quality in order
to find the thermodynamic properties
for these two phase mixtures.
We also have compressed liquid tables although we only have a
water compressed liquid table.
There's not one for R134a.
And the property table for compressed liquid water starts
at such a high pressure that it's practically useless
at least to us it's practically useless.
But that's okay because one of the things I also mentioned
at this point was that saturated liquids, let me reverse that.
Compressed liquids have approximately the same
properties as the saturated liquid
at the same temperature, right.
And that would apply to entropy or internal energy
or entropy or specific volume.
Any of the properties we'd be dealing with on a daily basis
for compressed liquids we just take saturated liquid data
at that temperature and that's just fine.
We also looked at ideal gases at this point, right.
We looked at the ideal gas equation of state.
It must have had at least three
or four different versions, right.
PV equals Rt, Pv equals Mrt, Pv equals Nrt,
sometimes V is total volume, sometimes V is specific volume.
Sometimes V is [inaudible] specific volume.
Just make sure you follow the units through if you're confused
at all about which form of that equation is correct or proper
and the units will always tell you whether you're doing
it right.
Nonetheless we looked at ideal gases at that point.
We also looked at real gases.
We saw that there's a test one could make
to determine whether a gas is an ideal gas.
If it is an ideal gas, great.
You could use the various ideal gas relationships
that were yet to be covered.
If it's not an ideal gas, then if it's a real gas,
well then we would have to use a compressibility chart.
There are other equations of state for real gases
that related pressure temperature and volume.
But we're not going to deal with any of those.
So the only one I ever had you look
at was a compressibility chart.
So all that was just in the first three weeks of class,
just all the basics necessary
to really begin looking at real problems.
Finally then we began looking at real problems.
So this brings us to Chapter 4.
At this point the first law was developed
for you for a closed system.
In fact Chapter 4 is closed systems.
Chapter 5 is open systems, right.
So we looked at the closed systems.
And for closed systems the first law is pretty straightforward,
in fact it's always straightforward.
Heat transfer minus the work equals the change
in internal energy plus the kinetic energy change plus
potential energy change.
I mean that's the first law.
And every problem we dealt with really
for the next two weeks was just a variation
on solutions to the first law.
Lots of different types of problems.
Some constant volume problems, some constant pressure problems.
Like if we have a frictionless vertical piston cylinder device
where the piston is free to move
and maintain the constant pressure.
So a lot of constant pressure problems.
We had problems that were called polytrophic where P times V
to the nth power is equal to some constant.
So we saw how to calculate work specifically
for that type of problem.
And of course there's no work for a constant volume problem
because the work is integral Pdv
and if there's no volume change there's no work.
For a constant pressure problem the work is just pressure times
the change in volume.
For a polytrophic process it depends
on whether the exponent N is equal to 1 or not equal to 1.
But, you know, some very specific equations for work.
With regards to heat transfer, the heat transfer is something
that would always either be given to you
or be the unknown we're solving for.
In this class we never dealt with heat transfer.
We never dealt with the different modes of heat transfer
like conduction, convection, radiation or how
to calculate heat transfer or heat transfer rates.
That will be left to your 415 class
which you'll get to eventually.
Anyway so with all sorts of closed system problems,
not only do we look at problems associated
with our pure substances like water or the refrigerant
but we also looked at ideal gas problems.
And that's where we saw that the internal energy change is equal
to Cd times the temperature change.
We also looked at solids and liquids at this point.
And there it was shown that Cp and Cv were equal
and we just called that heat capacity.
And as such the internal energy change will actually equal the
[inaudible] change which is equal
to the heat capacity times the temperature change.
So we looked at various problems associated with solids
and liquids again as well as the ideal gasses.
And that took us right through about the middle of this course.
Then we started to look at the open system problems.
So what I did at least in this class was I derived the first
law for an open system from the first law for a closed system.
And then we started to look at some of the special cases.
We realize that the most common is the steady flow type
of process.
So the first law was then further developed specifically
for these different types of processes.
First [inaudible] flow processes, right,
so we have heat exchanges, mixing chambers, throttles,
pumps, turbines, compressors, nozzles, diffusers, throttles,
I mean I think that's the list of these steady flow devices.
Once the equation was developed for the general first law
for the steady flow process then we looked
at the even more specific special case
of the single stream study flow process like a turbine, pump,
compressor, nozzle, diffuser, throttle.
And most of the problems we looked at actually fit
into that category, right,
single stream, study flow processes.
So we looked at that.
We also looked at multiple stream study flow processes
again like the heat exchanger, like the mixing chamber.
For all these the basic equation is pretty much the same, right.
Basically the heat transfer minus the work equals the sum
overall inlet streams of the mass flow rate times the Nfop
at the inlet, well plus kinetic and potential energy.
And then minus the sum overall exit streams
of the mass flow rate times the Nfop and kinetic
and potential energies for each exit stream.
I mean that was the general equation.
Fortunately we never looked at anything that had more
than just two inputs or two outputs
like a heat exchanger, right.
So it became a much simpler equation.
And then after we looked at that I might note that we looked
at some problems that involved water or the refrigerant.
We looked at other problems that dealt just with ideal gases
where we would have to use for the Nfob chain Cp delta T. Just
like for the internal energy change
in the previous closed system problems we used Cv times delta
T to give us our change in internal energy.
Let me not again that for an ideal gas the internal energy
change is Cv times delta T.
And for an ideal gas the Nfob change is Cp times delta T.
Students ask me all the time how do we know whether
to use Cp or Cv?
That shouldn't even be asked really.
If you're finding internal energy it's always Cv.
If you're finding Nfop is always Cp.
Just go back to the definition of internal energy
or I'm sorry go back to the definition of specific heat
of constant pressure or constant volume
when we first started looking at ideal gases and look
at that derivation [inaudible] are related.
Cv and U are part of the same definition, right.
H and Cp are related.
So if you need internal energy you find Cv.
If you need Nfop you find Cp data
and you work on it from there.
You have a question?
>> Yeah. So does some data basically like in the Chapter 4
where our closed system was, you know,
Q minus W equals [inaudible].
It's an ideal gas [inaudible] and just Cp.
[Inaudible] open system
where it's delta A [inaudible] Cp delta [inaudible].
>> But make sure that it's an ideal gas.
And I might note that if you have something
like water vapor use table A6, you always want
to your super E tables.
If you have refrigerant vapor then use table A13.
You may have an ideal gas for superheated water vapor
or superheated refrigerant.
But why use that assumption?
I mean when you're using ideal gas I mean the word ideal means
that you're essentially assuming something
that can't really be, right.
And is it really an ideal gas?
No, nothing is really ideal in this world of ours.
So it's always better to use the actual thermodynamic property
data when it's available to you.
And we have good water and refrigerant data.
So always use it.
Anyway. Any other questions before I continue?
All right.
So when we're looking at open systems again we looked
at these different types of processes both for ideal gases
and for the different pure substances.
We also looked at the uniform flow problems.
These were sometimes called charging
or discharging problems.
So the first law was a little bit different.
But mainly because the internal energy is changing
within our system, right.
If we're talking about a steady flow process there's no change
within the system.
The thermodynamic state is always going to be the same.
You can something different going in versus going out.
But within the system there's no changes.
However for the uniform flow problems charging,
discharging problems, there definitely is a change
within our system, right.
We're emptying or filling a vessel,
as such the properties are going to change within that vessel.
So we definitely are going to have a term that deals
with the initial internal energy [inaudible]
and there will be a term dealing
with the final internal energy, the M2U2 term.
And keep in mind as well the mass is going to be different
in these charging and discharging problems
between the beginning and the end of the process, right.
For a charging problem the mass is going to rise,
for the discharge problem the mass is going to fall.
So we definitely have to keep track of the initial
and final mass M1 and M2 as well as the inlet or the exit mass,
Me or Mi and those then get factored
into our first law equation.
So that took us through Chapter 5.
Now we get into Chapter 6 and this is
where we were first introduced to the concept
of thermodynamic efficiency.
And we specifically looked at the refrigeration cycle
and the heat engine cycle.
The heat engine is a device where we turn heat into work.
Of course the second law tells us that the only way to do
that is to reject heat at the same time.
We also looked at the refrigeration cycles.
And the refrigeration cycle is where we're transferring heat
from a low temperature source
to a higher temperature heat [inaudible].
And of course that can't happen all by itself.
The only way that can happen is
if we provide some sort of work input.
We looked at both the heat pump as well
as the refrigerator freezer, air conditioner type
of refrigeration cycle.
They're the same cycle.
I mean it's just a refrigeration cycle.
Heat added at a low temperature, work in, heat rejected
at a high temperature.
The cycles are the same.
But what is of use to us is different, right.
For a heat pump we're interested in the heat that is rejected
from the cycle into the space we're trying to keep warm.
For your refrigerator, freezer
or air conditioner we're concerned
with heat that's being taken out of the space that we're trying
to keep cold like the food in our refrigerator
or the cold air inside of our house during a hot day
when we're air conditioning that house.
So as such you would have different definitions
of coefficient of performance for those two different types
of refrigeration cycles.
There's only one type of heat engine.
So we only have one thermodynamic
efficiency equation.
But again the coefficient
of performance is our performance characteristic
for the refrigeration cycle.
Ql is used for refrigerator, freezers
or air conditioners, right.
And Qh is used for heat pumps within those equations.
We also at that point looked at the [inaudible] cycle just
to give ourselves a feel
for what would be the best possible cycle
that could operate between any two temperature limits.
We saw that the thermodynamic efficiency or the coefficient
of performance for the [inaudible] cycle is only a
function of temperature.
So that makes it nice and easy.
And then we can actually compare a real cycle to our ideal,
our [inaudible] cycle to determine whether a process
that we're analyzing and this would be some sort
of a cyclical process, right, a refrigerator, freezer,
air conditioner, heat engine of some sort.
You know, we could compare the actual thermodynamic efficiency
or coefficient performance to the [inaudible] cycle value
and see whether that process is real
or ideal or simply impossible.
That's when we first started looking
at the second law of thermodynamics.
Essentially that statement is a statement
of the second law of thermodynamics.
Nothing can have a higher thermal efficiency
than the [inaudible] cycle operating
between two temperature limits.
Nothing can have a higher coefficient of performance
that a refrigeration cycle operating
under the [inaudible] cycle
between those same two temperature limits.
So that was our introduction
to the second law of thermodynamics.
However, that really wasn't sufficient, right.
I mean, we're not only interested in cycles,
we're interested in individual processes.
So this is where the whole concept
of entropy was introduced to you.
It actually [inaudible] for you the existence
of the property entropy.
And the way we interpret entropy is as a measure
of randomness or disorder.
But that interpretation is really meaningless.
What we want is to use entropy to see
if certain processes can or cannot occur.
So we looked at entropy changes.
And what we found is that the total entropy change
for a system, that we can also call the entropy generation,
has to be positive.
If that total entropy change is positive then the process
is possible.
It satisfies the second law.
If the total entropy change is negative then that's impossible.
It does not, it violates the second law.
And the way to do that was to look at the entropy change
of the system plus the entropy change of the surroundings.
Now the entropy change of the system was simple, right.
Entropy being a property has tabulated values
that are inside the property tables in your book.
So if you want to find the entropy change
of the system just look up the data.
The entropy change in surroundings is even easier
because we recognize that the surroundings are
so huge compared to the system that's being analyzed.
That it's essentially a ideal isothermal process and as
such the entropy change is just Q over T. So we just had
to quantify the heat transfer so that we could then find the Q
over T for the surroundings.
And that will give you when added to the entropy change
of the system the total entropy change
for that particular process.
So that's what we dealt with with regards to entropy.
Of course we also looked at pure substances as well
as ideal gases to find ways to calculate these,
go through these problems.
I think I made it clear
that these second law problems were actually just first law
problems in disguise, right.
We still have to find the heat transfer
so we can find the entropy changes in the surroundings.
And how do you find the heat transfer
but by solving the first law of thermodynamics.
So all these second law problems,
the entropy problems really are just first law problems
with a little bit added onto the end
to determine whether the entropy change,
the total entropy change, the entropy generation
if you will is indeed positive.
Now we also looked at isentropic processes at this point.
You know, we realized that there might be a process
with no entropy change.
And that isentropic process would represent the ideal
process, the reversible process that's taking place.
So we looked at ideal study flow devices.
We look at ideal turbines, we looked at ideal pumps,
we looked at ideal compressors.
And then the concept
of isentropic efficiency was introduced to you.
We realize that the way we would actually determine the
thermodynamic state of the substance leaving one
of these steady flow devices, pump, turbine or compressor,
is to first assume that it is isentropic.
That isentropic process will then give you the value,
either the entropy or the temperature leaving
that particular device.
But I put the subscript S on there
because that would be the ideal properties leaving that device.
And then you'd be given an isentropic efficiency
for that pump, turbine or compressor.
And then you would apply that
and that would convert your ideal data
into your actual data.
I mean really what you're finding here is a relationship
between the ideal work and the actual work, right.
The ideal work is going to be a function
of the [inaudible] change for a compressor.
It'd be like say H2s minus H1.
The actual work is also a function of the entropy change
but it's going to be H2 actual minus H1 and then the ratio
of the two is your isentropic efficiency.
So once we got into isentropic efficiencies we realized
that the way to solve those problems is to assume
that it's an ideal process first.
Find the ideal discharge state, .2s, .4s, whatever it happens
to be and then apply the isentropic efficiency
to get the actual discharge state.
And then at that actual discharge state well then you
can find your work, you can find your heat transfer,
you can find maybe well other information
as is necessary or important.
So that was pretty much the end of Chapter 7.
One thing I will note is that I also covered that section
on reversible steady flow work where it turns
out for a liquid undergoing isentropic process the entropy
change which is your ideal work is absolutely equal
to the specific volume times the pressure change.
That not only applies to a liquid and only
for an isentropic process.
And even though I gave you a homework problem
or two that's not really something that's going
to be emphasized.
That will be part of ME302 when we start analyzing pumps as part
of a cycle that's actually very similar to the Brayton cycle
that we've been dealing with over the last week.
You can have a cycle like the Brayton cycle that uses water
or steam and we call it a ranking cycle.
And that's going to be something that will be covered in ME302.
But a ranking cycle requires a pump and that's
where you're going to have to use
that specific volume times delta P relationship to get the work.
Lastly then we dealt with gas power cycles.
The only gas power cycle that we dealt
with was the Brayton cycle.
We look at just the simple Brayton cycle.
Please not once again that I did not assign many sections
from Chapter 9, only the first few that deal
with the general gas power cycles and talks all
about the whole idea of being an air standard cycle.
Then section 98 which is your simple Brayton cycle.
There's a lot of other information about other types
of cycles that you're totally not responsible for.
So just deal with the Brayton cycle.
And keep in mind that the Brayton cycle deals
with the same things that we've already been learning
about all quarter long, right.
Well not all quarter long but certainly since Chapter 5.
Then that was something that I just explained
at some length earlier in today's lecture
so I don't think I'm going to go through that again.
That then is just a summary of the course.
And by the way I'm just going off of my syllabus here.
I don't have pages, you know, spread out in front of me
that tell me what I need to cover today.
This is just basically my recollection
of what we were covering in this class.
So hopefully you recall the same thing.
And you show me so in your final exam
and hopefully I'm pass everybody and that'll make me very happy.
So we'll see if that happens.
Has it happened before?
I don't know.
I've taught this course for a long time and no, never ever.
But you guys are lucky.
You had at least one student that stopped showing
up a few weeks ago, so maybe that will the one
and only F in the class.
Okay, so we'll see.
So at this point are there any questions then on any
of this material, homework problems, basic problems?
I haven't yet talked about the final
but I'll get to that shortly.
No, I think I just said one notecard for this, yeah, yeah.
Should I talk about the final now?
Did you have a question on material?
Okay, good question.
I like to think a [inaudible] process is kind
of a catch all, okay.
Keep in mind that when you're dealing
with a polytrophic process it really just tells you
that the relationship between pressure
and volume is a continuous polynomial function.
We go from point 1
to 2 following any of a number of lines.
And as long as it's continuous
in this fashion then there will be some exponent N
and some constant C. That would give you the exact equation
or at least pretty close to the exact equation
for the actual process that takes place.
With that in mind you would have
to be told it's a polytrophic process.
You couldn't just guess that.
There's only one situation
where you would know it's polytrophic without being told.
And that's if you have an ideal gas undergoing an isothermal
that is constant temperature process.
In that case mass R and T are all constant so Pv equals Mrt,
Pv equals a constant and that satisfies this equation,
Pv to the 1st power equals a constant.
So for an ideal gas undergoing isothermal process,
yes it's polytrophic with N equal to 1.
But otherwise you would always have to be given the value of N.
In fact we don't even need C when we do the integral of Pdv
and calculate the work.
It turns out that constant's going to cancel that anyway.
So the only thing we really need are the values of P and V
at the initial and final points, 1 and 2 as well
as the numerical value
of N. Sometimes N is given, sometimes it's not.
Sometimes you'll be given pressure and volume data
and then you just have to use P1V1 to the N equals P2V2
to the N and calculate N that way.
But you still need N so you can calculate the work.
And then that work will be part of your first law.
And the presumably you'll use the first law
to find some final state, some heat transfer, whatever.
Any other questions?
Yeah.
>> It it's polytrophic are you always allowed
to use P [inaudible]?
>> Yep, yep, P times V to the N is a constant and that means
that that equation applies at every single point.
So you can use it at point 1, you can use it at point 2,
you can use it at any of the intermediate points
if you knew what they were.
But yeah, that would apply at any state point.
Yep? Well we talk about fusion we're talking
about the phase changing to a solid and a liquid.
And I haven't really gone into any problems like that.
All the phase change problems we looked at involve vaporization.
So we can say heat of vaporization, in fact Hfg,
you know, the difference between Hg
and Hf is often called the heat of vaporization.
And if we had a fusing process, you know, for melting ice
or for solidifying water into ice, you know,
that heat of fusion is just well the difference
between the solid and the liquid phase.
But we haven't looked at any problems like that.
And I can't imagine I'm going
to give you something like that, no.
Yeah?
>> Can you go over the [inaudible]?
>> Okay. So the first law is a general equation and we have
to respect the various signs that are within that equation.
But there's also a [inaudible] convention
that has to be covered.
Now if you look at this material in your textbook,
the author likes to talk about heat in,
heat out, work in, work out.
That's fine but I prefer just to think about it as heat or work
with the appropriate sign.
If heat is added heat into the system is positive.
And if heat is lost heat rejected
from the system is negative.
Work is actually the opposite of that.
If work is being done by the system, you know,
if the system is actually pushing onto its surroundings
and doing work, in other words work output, then that's going
to be considered positive.
And work input, you know, if we're pushing on the system
or maybe up to something like a compressor, right,
where work is being done to the fluid moving
through that compressor then
that would be considered negative.
So heat in is positive, work out is positive.
Heat out is a negative, work in is negative, yeah, okay.
And then keep in mind too that when we looked
at the thermodynamic efficiency equation or coefficient
of performance equation, in that case only the heat transfer
and work terms are always taken as absolute values.
And then the sign is actually placed into the equation itself.
So for instance if we have a thermal efficiency
as the difference between the turbine work and compressor work
over the heat input, well we know the compressor work is
negative because it's work required by the system.
I don't show it as W turbine plus W compressor
with W compressor as a negative value.
I just show it as W turbine minus work compressor
with work compressor as the absolute value
of the work that is required.
So yeah just make sure
that within your thermodynamic efficiency
or COP equations you use magnitudes or absolute values.
>> And for [inaudible] the work [inaudible].
>> Like a mixing device or something like that?
Well yeah, because they would involve work input.
Yeah, so any work input from a mixer.
Any work input from resistance heating or an electric resistor,
yeah those are always definitely negatives
because they're inputs.
Any other questions?
Yeah.
>> Do you assume that the gas is ideal gas?
In what circumstance can we make that assumption?
>> If you have, well let me just put it this way.
The only time we even looked at the difference between ideal gas
and a real gas is if we were dealing
with a specific volume and, you know,
[inaudible] compressibility chart.
But the only method that we ever use to solve first law problems
for gases was to assume that it was an ideal gas.
When you assume it's an ideal gas then you can use the fact
that the internal energy is the integral of CvDt
and that entropy, well change
in entropy really is the integral of CpDt.
So if you're solving a first law problem which then by extension
if you're solving a second law problem then you are always just
going assume it's an ideal gas because that's the only method
of solution that we have.
So you're never even have to take a B on that.
If you're going to assume it's an ideal gas just assume it
and use ideal gas relationships.
You don't have to check whether it's an ideal gas or not
because it's the only method we have.
And that should be sufficient, okay.
Okay, yeah.
>> But if it's not ideal then [inaudible]?
>> Well if it's not an ideal gas we just haven't looked
at the methodologies to find the entropy data
or the internal energy data.
I mean if we had property tables similar to say our,
well our water tables or refrigerant [inaudible] tables
where we can just look up property data like Hu, etc.,
then we would use those.
But the fact is for gases we just don't.
Now there is a set of tables I think it's A17 to about A27
or so in your book and those are gas tables.
But those are still ideal gas tables, okay.
Those are ideal gas tables with variable specific heat.
And that's something that we just have not talked
about in this class.
So I know that the author describes using that method,
using the data from table A17 for various problems
in Chapter 4 and, even in 7
but that's not something we've ever done.
So yeah, we're always just going to assume ideal gas
and use the ideal gas relationships.
Okay.
>> Wouldn't you get an ideal gas in the final [inaudible] to try
to find it [inaudible].
>> Well right but I mean on that first test I just wanted
to make sure people were paying attention
and understood there was a difference
between a real gas and an ideal gas.
At this point we're only dealing with ideal gases.
So I can't imagine that I'm going
to give you a problem just says is this gas
or real or an ideal gas?
I mean those are real simple problems.
And the problems you'll see
on the final are really more similar to the kind of problems
that we've seen, well over these last seven or so weeks.
So let me talk a little bit about the final.
Then I need to leave at least five minutes or ten
at the end to do evaluations.
So all right, so first of all there's going
to be just four problems on the final but some
of them might be multiple part.
The final will cover material from the entire class
but there will be emphasis on the material
that you have not yet covered.
And I did already tell you
that you'll definitely get a Brayton cycle problem.
And it might be a non-ideal Brayton cycle problem.
So you definitely want to make sure
that you understand all this material.
You will be allowed to use a notecard, just one notecard,
3 by 5 written on both sides with equations only.
You're really not allowed to write sample problems or step
by step methods of solution or anything like that.
I will note that it might be rather difficult
to put everything you've learned on one little notecard.
So don't fill it up with things that you already know about,
just fill it up with things that you don't want to memorize.
You know, there are certain equations that are too long
to memorize and I think should be kind of careful
with your 3 by 5 card.
You'll have the full two hours.
I will give you a property table handout like your midterms.
If you have your own property table handout you can certainly
bring that with you.
You know, mine is on both sides, kind of two pages per sheet.
You may want to print one off in color, just one page per sheet,
single-sided if it'll make it easier
for to read or go through it.
But that's up to you.
And all that is available online,
you know, through Blackboard.
All lectures through Wednesday lecture have already been put
onto the Blackboard system.
So if you need to review anything and you want to go
through those lectures they're available through Blackboard
and they're all just YouTube videos.
So those are available to you.
Are there any questions, anything I didn't mention?
>> Today's will be by sometime this weekend [inaudible]?
>> Today's, well lately I've been getting these things
within like an hour or so of class.
So, you know, hopefully, I mean I can't guarantee it but,
you know, hopefully they'll be up before the end of the day.
So I would plan on it.
>> The Brayton cycle.
>> You know, I don't mind if you have a sketch
or two for yourself, yeah.
So if you want to show a Brayton cycle, yeah you can do that kind
of a sketch if you want to show some cycles.
You know, that's not a big deal.
I'd let you do that.
You know, Pv diagram, Ts diagram.
Why don't we just say yeah, if you want to have diagrams
as well as equations then that's okay with me.
Okay. Again the final is at 11:30, not 1:00 o'clock, right.
So make sure you come here at the right time.
And it's on Monday.
My finals week office hours were different.
I put those on the board here last time.
If anybody didn't get those come see me afterwards or talk
to one of your classmates.
But I'll be in my office today until 3:30,
I'll be there again at 8:30 until 9:00.
But I do have a final right before yours Monday morning.
So I'll only be there from 8:30 to 9:00 Monday.
Okay, so now it's time for teacher evaluations.
[ Music ]
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