>> Alright, again, good morning everybody. I'm not going to take roll. The
course is so large, I'm not going to deal with it. Plus, I think everybody who
was on the wait list is now in the class except for maybe one person. Is there
anybody who is still trying to add? So I guess everybody got in. There were some
drops. Great, so, welcome to my thermodynamics class. Let me just remind you
that I've put a whole bunch homework problems on the board last time. So please
make sure that you have written those down and recall that they're always going
to be due on Wednesdays. So that will be due this Wednesday. Let me also note
that I've activated the Blackboard system. I will be doing example problems over
the course of the quarter, and so of the examples are from previous editions of
the textbook. So I've actually made a JPEG image of all the example problems, I
just, you know, photographed them out of the old books. And those are also
available on Blackboard. So, you know, that might be something you want to look
at or make a copy of perhaps. I mean, you don't have to do it now. I have
digital copies too and I'll represent them on the board with the old projector
when I give those examples in class. But, you know, if you don't want to write
all those down, then at least you'll have access to that. Also, I've made a copy
of the handout that I will pass out during exams. Basically, it's the property
tables from the appendix of your book. Those are also available on the
Blackboard site. And of course [inaudible] there too. So, there's a variety of
information. And eventually when these videos are published I will put links
there as well. As well as the various videos. And hopefully I'll have those just
within a few days. Last year when I did this in my 301 thermo class they were
always available to me, you know, within like one or two days. This quarter
we'll see, nonetheless, but for the better I suppose for those that they want to
reduce their lectures or just want to have some fun on a Friday night and tired
of Netflix videos, or Amazon Prime has videos and Hulu has videos, but what can
be better than thermodynamics? So, I'm just going to continue the review from
last time. You might recall that at the end last time we were talking about
ideal gases and specifically ideal gases with constant specific heats. I had
presented the equations of what my unit of enthalpy, or internal energy or
entropy changes. Now I want to just continue this. So we're still talking about
ideal gases and we're still talking about constant specific heats. And we're
just going to continue now by simply talking about isentropic processes. Now,
the way that I give [inaudible] specific heats and ideal gas under an isentropic
process, basically you'll just take the entropy change equation from last time.
Isentropic means constant entropy, right? And just note that the final and
initial entropies are going to be the same. So we can derive the following. I'm
not going to derive, they should have been derived already for you in your first
thermo class. But we'll be able to show that the temperature ratio is going to
be equal to the pressure ratio raised to the K minus 1 over K power. K being
just ratio specific heat, which is about the last thing I wrote on the board
last time. Oh by the way, I said initial and final with regards to one and two,
this would apply both to closed systems as well as single and open systems. So
one and two could represent the initial and final state for a closed system. Or
it could represent the inlet state and exit state for an open system, in a
control volume problem. So, just be aware of that as we're looking at all of
these equations. In fact, in this class as you'll be dealing with flow through
thermodynamic cycles for much of the course, one and two generally will
represent the inlet and the exit conditions from a particular device. Okay, a
turbine, compressor, a pump, something like that. Nonetheless, we can derive
this particularly equation. One could also note that this is related to volume
ratio. So this could also been shown as V1 over V2 to the K minus 1 power. So,
this is an equation we can use for an isentropic process. One can look at the
pressure and volume terms and do some hocus pocus and find that the pressure
ratio is going to equal the volume ratio to the K power. Again, I'm not going to
go through the mathematics, but this is what we find right now. This is all
isentropic processes if we have constant specific heats. And of course the
equations that were presented were of constant specific heats in general. Now,
let's go to the next step and we're still dealing with ideal gases, but now
let's look at what are called variable specific heats. So, in this particular
case with variable specific heats, we can't for instance say that the enthalpy
change is just equal to a constant value of CP of the temperature change. We
actually have to recognize that you need to take an integral between point one
and two of CPDT. This is what variable specific heats means. Variable means that
the specific heats vary with temperature. Okay. I mean, specific heats are
functions of temperature only for ideal gases, but they're not constants now.
Okay, we assume they're variable and in fact this is really a more accurate
method if we use the variable specific heats then you're going to be expected to
use this method as we get into our material here. Really, the new stuff we're
going to start next week. So, as we do this, let me also note that there's a
similar equation for internal energy change, initial is going to be CB intake as
interval from one to two is CBDT [assumed spelling]. I guess more specifically,
this should probably write T2 and T1, since after all we are integrating
temperature, right? So, these equations can be utilized. Now, we really have two
choices when we're trying to for the internal energy or enthalpy change for an
ideal gas of variable specific heats. We could go into our appendix table A1,
part C. Which gives you the polynomial equation as a function that is specific
heats as it comes in the temperature. And actually go through the integration.
But quite frankly, nobody really wants to do that. It's easier just to note that
we do have some gas tables. And since the only gas we're going to deal with in
this class is air, specifically I would note that we can find data in the air
tables which is table A-17. So, you could actually go into the air tables at
particular temperature and look up the values of enthalpy or internal energy
based on variations of specific heat with temperature. Now, that's typically
what we're going to do. Okay? We're not going to go through a long integration.
Why bother? It's already been done for you and that data's presented in an
appropriate appendix. So, this is something we're actually going to be talking
about at length here shortly. For those of you who had a different instructor
for 301, you may have already talked about variable specific heats and
understand already how to use table A-17. But if you had me in Thermo 301, then
we never talked about table A-17. In fact I specifically said, don't worry about
it, we'll get to it in the next quarter. Just concentrate on constant specific
heat to understand the basic processes. And now we're going to start looking at
some more detailed information. Certainly more active. In fact, as the book
talks about analysis based upon variable specific heat, it calls it exact
analysis. When it talks about analysis based on constant specific heat it talks
about approximate analysis. So we're going to be dealing, well with both, but
certainly the emphasis is going to be on the exact amounts using greater
specific heats and using data from A-17. Now, I should also show the entropy
change equation here, because we are going to have to find entropy change. And
again, we're not doing constant specific heat. So basically we just have the
equation equal to from one to two CPDT over 10, then minus R natural log
conclude one. And when we have the ability to modify this. So, this equation is
already in your notes from last time. Certainly you probably did that in 301.
But now what we want to do is go one step further. So the first thing we want to
do is just open up this integral, what we can do and what we remember doing this
in, you know, in your integral calculus class. We can just turn this into an
integral from zero to T2 of CPDT over T. And then minus and equal from zero to
T1, CPDT over T. So that's the same as this first one. And then minus R natural
log T2 over T1. Okay. Now, please note that these terms here are only functions
of temperature as well. So, they're actually presented for you in table A-17, or
really any of the other gas tables. Like from table A-17 all the way up to 25,
or 6 or something like that in your textbook for other gases. And again, we're
only going to deal with air. So, what we do is we define a new term here, we'll
just call it S superscript zero. Different authors of different books have
different names for this. Sometimes it's called a compression function,
sometimes an isentropic compression function. I personally just like to call it
an entropy function. But still, we're going to let the interval from zero to any
temperature of CPDT over T just equal this entropy function. And therefore, the
entropy change for any particular process is just going to be S superscript zero
at point two, which is really T2. Minus S zero at T1 and then minus R times the
natural law of the pressure ratio. So, this is an equation that we may need to
use occasionally as we're trying to find an entropy change through a process
that involves the energy of gas. That is, you know, considering variable
specific heats. Next what I would like to do is note that in many, many cases
we're interested in analyzing ideal or if you will, isentropic processes. So
continuing with ideal gases and variable specific heat, what if we have an
isentropic process? So we already looked at that over here, for constant
specific heat. These are the equations that were developed. But now we need to
use variable specific heat, so we're using the equations that are shown here
above. So basically, it's isentropic so there's no entropy change. The left side
of the equation is just zero. And then the entropy function at two, minus that
at one, minus R log of the pressure ratio. And now what I'm going to do is I'm
just simply going to rearrange this and again do just a little bit of
mathematical hocus pocus. Nothing special. But, if we go through the
mathematical rearranging we're going to end up with the following that P2 over
P1 is then equal to the entropy function two, over the S constant R. [inaudible]
Glad I brought my own. I've carried this around with me for about two years and
I've never used it. But, I knew there would be a day eventually. I just have to
remember to bring it back with me. Anyway, so E entropy function over R. Divided
by E to entropy function of 1 over R. Now, you know you should be able to see
pretty easily how we get that. You know, I'm just going to move the entropy
terms, the entropy functions to the left. I'm going to have to raise both sides
of the equation above E. E is going to cancel out the natural log, which is why
I have the pressure ratio. And then of course the E remains with the entropy
function of it. So, this is the equation you want to use for an isentropic
process. Now, let's note that these two terms here have their own name. And we
give it the term, P substitute R. This is called relative pressure. And the
relative pressure is defined as E to the entropy function over R. So, with this
in mind, we therefore can plug everything into the equation above. And we get P2
over P1, is going to equal the relative pressure two, over the relative pressure
one. Now, keep in mind that these relative pressure terms are again only
functions of temperature, right? These entropy functions are only functions of
temperature. That over here, I should really put a third line in this term,
because it's the definition. That's how we define the term entropy function. So
I forgot that third line. But nonetheless, these are only function of
temperature. So we find there are more than PR data is going to exist in the
same table. Table A-17 or in the others. So, PR data is in table A-17. Okay? So,
the fact is we never really do have to go through this integration process.
Right? All we have to do is know what the temperature is at either point one or
point two, and then just go into the book and that allows us to find these
relative pressures. So, this is something again, that I did not cover in my
thermodynamics class. And you know, shortly I'll talk about well, how do we
realize this information? One thing we can do further is, following. So,
furthermore, let's just note that from your gas equation state, our RP equals
RT. In other words, R equals PV over T. This by the way is specific volume. And
note that R's a constant, right? It's going to apply at point one or point two
or any point. So then this is also going to be equal to P1 to B1 over T1. To
equal P2, T2, over T2. I mean, it's a constant. I divides at any point. We can
now rearrange this, so we get B2 over B1. And then you can see that the pressure
temperature terms move over to the other side. So, we have T2 over T1 times P1
over P2. And then we would simply note that P1 over P2 equals, well, PR1 over
PR2. Right? So, this is then T2 time PR1. Divided by T1 times PR2. And then just
continue this rearranging. We would like to put the point two terms and point
one terms together. So let's just show this as T2 over PR2, divided by T1 over
PR1. And then these terms, this ratio of temperature to relative pressure, it
too has it's own name. So, that's going to be called PR, which is the relative
specific volume. Okay. It's not even a specific volume term. It's just another
entropy function, isn't it? Or really, a temperature function based on the
relative pressure which is of course based on the entropy function. So it's only
a function of temperature. So VR is going to be divide [inaudible] and it's just
T over P. And then we'll want to do is note that for S V2 over V1, is going to
equal, this specific volume to the relative specific volume for it. So this is
another term that we're going to use very frequently as we solve different types
of processes. So, again, keep in mind this implies only isentropic processes,
right? These equations imply. If you're doing a problem and again we'll work
these problems eventually, but if you're doing a problem with constant specific
heat with isentropic process
[ Knocking ]
if you're going through a problem with a ideal gas that's [inaudible]. Then,
this equation and this equation apply. So, make sure you read the problems
carefully. You know, you might note that on the homework that I assigned on
Monday I put some little asterisks that I identified certain problems as
constant specific heat and there's a variable specific heat. So make sure you
use the appropriate method. Again, they're only a function of temperature so VR
data is also in table A-17. Okay. Now, what would a typical problem look like
that's isentropic? So, just so you know what you might find uses for. Okay, so
again, isentropic processes, ideal gases, variable specific heats. Typical
problem. Okay, so what would a typical problem look like? So, in a typical
problem we're going to know state one. It's going to be the initial state The
closest of the four of them would be the inlet state, for a single stream steady
flow. Open system problem, but we're typically in state one. And either the
final pressure or the final specific quality. Or I should say, or the exit
pressure over exit specific quality. So, these are typically going to be known
to us. Now we can note that FT1, you simply go into table A-17 and you can look
up PR1 for VR1. We're not going to do both. It really depends on whether we know
the final pressure or the final specific volume. If we know the final pressure,
we're going to be interested in using the relative pressure. If we know the
final specific volume, then we'll be interested in using relative specific
volume. So, we're going to go into the table at T1, we're going to find the
relative pressure at point one. And then we're going to use P2 over P1 equals
PR2 over PR1, in order to find, well, let's see. We know the first state, we
know P2, we know PR1. Well, we obviously use this to find PR2. Or, if we know
specific volume, then what we need is V2 over V1, equals VR2 over VR1. And we
use this again to find the only unknown, which would be VR2. And now that we
have PR2 or VR2, relative pressure and specific volumes, then it's just a matter
of going right back into table A-17 at that value of PR over 2 or VR2 and just
looking up the data we need. That we need to find temperature, we need to find
internal energy or the exit temperature or internal energy or enthalpy or
entropy or whatever it happens to be. Right? So the last step is then to use PR2
or VR2, in order to find data at state 2, again in table A-17. Okay? Now, is
this really any different as far as the procedures we use? Compared to when we
have constant specifics heats? Well no, in constant specific heat we needed
specific equations, right? Like P2, I'm sorry, like T2 over T1 equals P2 over P1
and then K minus 1 over K. That's how we're going to get from point one to point
two if we use constant specific heats. If we use variable specific heats, it's
not quite as simple, but the procedure is not much different. You know state
one, you know either P2 or V2. And now we're just going to use the table with
the A-17 data with relative pressures or relative specific volumes, in order to
find our thermodynamic property data at state point two. So, that's the
procedure. I will note that you're most certainly going to have to use
interpolation to find the data and then very likely going through this process
that you're going to find that have you of PR2 or VR2 that are specifically
listed in the property tables, so you're going to have to interpolate. So make
sure that you remember how to interpolate and shouldn't have any problems. Okay.
So are there any questions on this material? Again, this is new, but it's not
complicated. It's just a different way to approach the problem. In fact I would
guess that many of you have already read about this because this is the method
that is discussed in the textbook, even though I didn't require it in my first
course in thermo. It was still there, I had plenty of students that used this
method to solve problems, even in the 301 thermodynamics class. But now we're
going to have to use it for sure. So again, no questions? I will move on. So,
we're done now, we talked about [inaudible] gases. New topic. Now we're just
going to continue to review and talk about first law. So, I'm sure you know that
essentially how your problem was solved, was a first law problem, right? We're
always interested in processes. We analyze them using the first law, which can
be called conservation of energy. The first law certainly has different forms.
It really just depends on whether we're dealing with a closed system or an open
system, as to which version of the first law we're going to use. So, if we
happen to have a closed system then the version of the first law we would use is
this one. That in this equation, it might be a little bit different topic, we
would put the work turned on the right hand side as a positive. Or we could move
it to the left hand side as a negative. That's what I'm going to do here. So, Q1
and 2 minus work 1 and 2. This is representing a closed system. So 1 and 2 is
initial and final state. So, we added minus the work done by the system, is
simply equal to change in the energy, right? And that energy could be internal,
a kinetic energy or potential energy. So we will put all three on the right hand
side of the equation. So we have U2 minus U. There's the internal energy. We
have kinetic energy, one half m v 2 squared minus V1 squared. And then we have
potential energy, which you may rather call the gravitational potential energy.
But this would represent the first law for a closed system. Please note that
this is the change in kinetic energy. Often I'll just abbreviate it the whole
term as a delta ke and this represents the change in the potential energy. So,
sometimes in these problems I'll just write delta ve but we know what these
terms are, okay? I might also note that if we just divide through by the mass,
we have a mass specific version of this equation. So just divide up the mass and
we have a lowercase q 1 2, minus lowercase w 1 2, equals. And then we would note
that the total internal energy is just a mass times a specific internal energy.
So, the mass is divided out. Then we're just a mass with the change in the
specific internal energy. So again, these are all per unit mass terms, right?
And then we have, kinetic energy change, of course the master has gone away. And
we have potential energy. So, this is what the equation could look like. As to
which one we use, it really just depends on the problem at hand. Right? Some
problems give you data that is total, for instance it gives you energy that is
at kilojoules or BTU's. Well then, you can use the first version of the
equation. Others are mass specific. [inaudible] So, you used to be kilojoules
per kilogram or possibly BTU's with mass, if we have the [inaudible]. So these
are the equations that would apply for a closed system. Now keep in mind that as
we begin talking about cycles next week, we will start with the closed system.
In fact, what we will start with first next week is the various cycles that we
apply to turn combustion engines, which are really just piston and cylinder
devices, right? It's a piston cylinder device. The mass is fixed, they're closed
systems. So, that's the equation. Or actually those are the equations we might
use as revitalized engines or process of involving piston cylinder devices that
are engines. Alright, now that we have an open system. So, an open system can
also be called a control volume. Now, specifically I'm only going to deal with
the single streams, they do both processes here. We're not going to really look
very much at processes that involve more than one stream. We will eventually
look at some heat exchangers, but that's not going to be until near the end of
the class. But, usually we have a single stream process that involves a pump or
a compressor or a turbine. Nozzle or diffuser, a throttle. Those are all single
stream thing to a processor, right? So, for these processes, the equations a
little bit different. We have Qdot. And usually put CD just as a representation
of the control volume, just to remind us we're talking about an open system,
that is going to be flow into a fixed volume, to our control volume. So the
volume within the pump will be a control volume. The volume within the turbine
is a control volume. The volume of the compressor, that's the control volume
that we're dealing with. And this represents the rate of heat input. Now please
note that they're open systems, they're processes that involve flow. So we
should have everything on a rate basis. The rate of heat transfer. For a closed
system it's just the total heat transfer and the total work, as you go from
initial to a final point, right? Or total heat transfer of the work on a
[inaudible] basis. Here were interested in flow. So, it's going to be the rate
of heat transfer and the rate of work production that has to be included in the
equation. So here we have the equation that the rate of heat transfer is going
to equal mdot times. And then we have these various terms. It's the same terms
we have over here, well slightly different. We have an enthalpy change. We have
a kinetic energy change. And we have a potential energy change. And this is
going to be an appropriate form of the first law. And I don't want to forget my
work term. Really, I should call it power, the rate at doing work is the power.
Okay. So, here we have an appropriate first law equation for an open system. We
could also divide now by a mass flow rate. And the rate of heat transfer over
the rate of mass flow is just the [inaudible] per unit mass again. So we use a
lowercase q. So this would be the mass and the mass is going to equal H2 minus
H1. So that's v2 squared minus v2 squared over 2, plus G T2 minus E1. And then
plus the work we didn't pass. So the units are going to be kilojoules per
second, which by the way equals a kilowatt or it could be in BTU's per second,
that would be pretty typical for the British system. Okay. On unit mass basis,
it's the same as we have over there, kilojoules per kilogram or BTU's per pound
mass. Now, a few things to note. These two equations for dealing with closed
systems, they sure look very similar and many students are tempted to just kind
of arbitrarily use whichever one they're more comfortable with. But they're
different, right? For problems that involve flow, we use the enthalpy, right?
For problems that are closed systems, we use the internal energy. So you can't
just arbitrarily, you know, swap out internal energy for enthalpy, it just
doesn't work that way. You have to use the right equation for the right problem.
If it's a closed system problem, you use internal energy. There's one exception
if you have a constant pressure process, but we're not going to talk about that.
If you have an open system problem then you definitely have to use the enthalpy
within the equation. A few others notes. First of all, I have used a kind of
cursive letter V to represent velocity, so please don't confuse my velocity
notation with that of specific volume. Specific volume as we've seen already
today, is more like a little pointy curved letter V. It always looks more like
cubes that are new, doesn't it? But, here I'm going to use a cursive letter V so
make sure you understand the difference in notation. In the textbook they just
use bold face print, so it's obvious what velocity is. But I can't do that, I
don't have a bold faced pen. So this is the way I do that. Additionally, I would
note that on mass basis, if I have a work term, I use kind of a lowercase w with
a rounded bottom. It really almost more like the Greek letter omega. But again,
on the board, I can't show any real distinctions between a capital W and a
lowercase w. So my capital W's always have pointy bottoms and the lowercase w's
have rounded bottoms. And that's just the notation that I'm going to use. So,
again then be aware of it. You know you guys who had me in 301, you know this
already. For the rest of you, now you know. Also, we should talk about the
signed conventions here. In some textbooks and some instructors, rather than
putting Q and Q, you have Q-in and Q-out and Work in and Work out. So your
equation would say something like P-in minus, I'm sorry, P-in minus V-out minus
work out, minus work in, is going to then equal the change in energy terms. I
prefer not to do that. I'm just kind of one work term, or one [inaudible] term
and then we have to use the appropriate sign convention. So, if we have a
positive heat transfer, in other words heat input, means we have positive value
of Q. And if we have heat output or rejection if you will, then we have negative
value of q . Or lowercase q . It depends on the topic, right. With work, the
sign convention is actually the opposite. Okay. Work output, in other words,
work done by.
[ Coughing ]
This is positive. Work input, or if you will, work done to the system, this is
going to be a negative value. Okay. So again, positive work. And this would also
apply to work per unit mass. It would also apply to the rate of doing work, that
is power. So, just without you're aware of that. So, that's the sign convention
we're going to use in this particular class. Which is consistent with everything
I'm doing. Let me also just note lastly, that power is certainly related to the
work per unit mass. It's just the mass flow rate times the work per unit mass.
And so, that's something you would remember. Now, where do we use this set of
first line equations? Well, really everywhere. In all sorts of problems. But,
what I would like you to do is just kind of emphasize those problems, those
types of processes that you use in this class. And they generally deal with, new
flow devices. Things that use steady flow devices. The same ones that I keep
mentioning. Pumps, turbines and compressors. We're dealing with a cycle within
that cycle, we have pumps, turbines or compressors. Pumps provide the work
input, compressors provide work input, turbines provide the work output. So,
these various terms we're going to have to apply specifically to those devices.
So, let me just note a few things. So, for the following devices, turbine and
pump and compressor, we have the following versions of the first law. Now, keep
in mind this is all first law. Right? We're still talking about the first law.
All of these different devices that I mentioned here, what would the equation
be? Well, first let's note certain assumptions. Technically when we have a
device like this, you know, we're spending anywhere from hundreds to millions of
dollars, right? A large steam turbine in a big central station power plant can
easily cost 100 million dollars. You want to insulate that thing well, you don't
want to lose heat because of the environment. So, generally we would consider
these to be adiabatic devices. Not adiabatic in heat transfer. And that's
something we can achieve by prepare insulation or these get pretty close using
proper insulation. So if we deal with an adiabatic process, there's no heat
transfer. I'm sorry, there's no -- yeah, that's right. There's no heat transfer
term. So that terms going to drop out of the first law basis. Also, you would
generally neglect any changes in kinetic energy and potential energy. Please
keep in mind that in the real world you don't want high velocities moving
through these very expensive pieces of equipment, because that's going to erode
away the inside of your turbine or your turbine blades or your pump propellers.
So, the speeds are purposely kept low so there's really not going to be a
significant kinetic energy change and there's not going to be any specific
height change. I mean, if you have a term and this heat goes in and out at
pretty much the same elevation, so these are not bad assumptions to neglect both
kinetic and potential energy. So, what are you left with? Well, for the turbine,
the power that's produced by the turbine is just going to be the mass flow rate
multiplied by the enthalpy change, H1 minus H2. We're dealing with a pump, then
the pump work is going to be the mass flow rate and then times H2 minus H1. I
will make a note that this is the magnitude. Or if you prefer, call it the
absolute value. I mean really, the equation should give you the same thing as
above, right? Just m dot H2 minus H1. But keep in mind that pump work is work
input through the system and as such, it's going to have a negative value,
right? Based on our sign convention. So we're just going to swap the order of H1
and H2. And this is going to be the magnitude of the pump here. Further I might
note that if we're indeed dealing with an adiabatic process and the pump, just
like when adiabatic is also reversible, if it's an isentropic process then we
would note that the enthalpy change can be replaced by the specific volume of
this pressure change. So, [inaudible] times specific volume inlet, times P2
minus P1. So, often we're going to deal with these ideal pumps. And then lastly
for the compressor, same thing applies as it does in the pump. The power is just
going to be the mass flow rate times H2 minus H1. Again, this is just the
magnitude. And we cannot apply this isentropic equation without a pump. This
only applies to a pump. And the reason is because pumps move liquids and this
equation only applies to liquids, okay? So, let's just put a little liquid right
underneath the pump. But note that compressors don't move liquids. I mean, even
though the devices perform exactly the same function and actually look very
similar in the real world, compressors move gases or two types of mixtures. We
just call it a compressor. But still, the [inaudible] here only applies to
liquids, in other words, only pumps. So, these are the equations that we're
typically going to use when we deal with these three specific devices, pumps,
turbines and compressors. And will see it in pumps, turbines and compressors a
lot. So, moving ahead then. When we analyze a typical pump, turbine or
compressor, you know, these particular tools use steady flow devices. What we
often do is want to know how the device performs compared to an ideal pump,
turbine or compressor. Okay? And for those situations, we define what's called
the isentropic efficiency. So, isentropic efficiencies sometimes can be called
adiabatic efficiencies. It just depends on where you're looking. But let's talk
about isentropic efficiencies. Now please keep in mind that isentropic
efficiency is not the same thing [inaudible] efficiency. The isentropic
efficiency is nothing more than the ratio of the absolute amount of work to the
ideal of work associated with one of those three devices. So, the isentropic
efficiency for a turbine is just AST [assumed spelling]. By the way, this is
data that will always be given to you. When you guy one of these big devices the
company who has manufactured it is going to have tested it out. They will tell
you exactly what the efficiency is over a wide range of operating conditions. A
wide range of flow rates, a wide range of pressure changes or temperature
changes. So you don't have a performance characteristic curve, which you might
just call a map, because that's what we call them. But you'll have a map for the
turbine, pump or compressor. You'll know what these are. In this class, I'm
never going to give you a map, I'm just going to tell you, or they often tell
you in the problem statement, that at this operating point this is the
isentropic efficiency. But in the real world, the efficiencies going to vary
just depending upon the operating conditions. But nonetheless, the isentropic
efficiency of the turbine is the actual work over the ideal work. Okay. What
we're going to do is just use a couple of subscripts. A represents actual, for
actual. And S represents ideal. Now, why [inaudible] have like I or something
like that. But keep in mind we're talking ideal work, we're talking about an
isentropic process, right? An adiabatic reversible. That's isentropic. And an
ideal turbine would be one that is indeed isentropic, right? That can't be
achieved, right? In the real world there's always losses, but it does represent
the theoretical limit. So as we analyze our turbine and indeed the pumps and
compressors, we're going to have to determine the ideal amount of work that that
turbine, pump or compressor would have under the situation of the states
associated with that process. But then we're going to have to use the isentropic
efficiency to calculate the actual work. Again, hopefully you remember this from
your thermo class. So that's why we use the S, right? Isentropic means constant
entropy, S for entropy. So I'm going to always put the subscripts for these
processes. Now again, we could divide by mass and we get the actual over the
ideal work we use that. We can divide by time or if you will take the time, and
we get the actual power divided by the ideal power. They all apply. Which
version you use, again, just depends on what [inaudible]. If we look at the
pump, we have A and P. Now for a pump and a compressor, these are work input
devices and as such we're going to invert our definition. This is going to be
the ideal work over the ideal actual work. The reason we do this is because we
always want to make sure that our efficiency terms are less than one, or less
than 100%. Realize that in a pump, you know, you're putting work into this
system. The actual amount of work is not only the amount of work it takes to
let's say, energize the fluid that's moving through the pump. But you also have
to provide a little bit of extra work to overcome the losses associated with the
bearings and the fluid flow through that pump. So, in the real world, the actual
amount of work that is required by that pump, is going to be greater than the
ideal amount of work that will be required by that pump, because we have
friction and other losses. With a turbine, it's essentially the same, but
reverse. We get work output. The energy lost by the substance moving through
that turbine, that's the actual amount of work. But the reality is, that work
output also has to overcome the frictional losses. So the real amount of work
that is the actual work that we're getting out of the turbine is not the same as
what you would think of as, you know, the enthalpy loss. It is enthalpy loss,
right? Associated with that flow. You also have the friction. So the actual work
is less than the ideal work. Some of that ideal work, I'm sorry, that ideal work
also has to have additional work to overcome the friction losses. And as such,
the actual work that we get out of that turbine is going to be less than the
ideal amount of work that we get that turbine. So, nonetheless as we continue
with the pump, we would just write this as ws over wa . Or work per unit mass.
So work per unit mass. Or this could also be the ideal power over the actual
power, the pump. And it turns out that in this general form, the same thing
applies to the compressor. Okay? The compressors will work input devices, we
define the isentropic efficiency in the same way. So, for a compressor, I mean
we use a to sub c, to see for compressor. [inaudible] But it's exactly the same
as above. So, I'll just put, a bunch of dittos. So we know that everything's
exactly the same. If you prefer, just go over here and write, equals a sub c.
This is a [inaudible]. Alright, again this is all, obviously this is all
covered, obviously all of this is all covered in the other class. So I'm not
going to go into this in too much detail. But we do need a little bit more
detail. Now, let's again note that for a typical stay put device, like a
compressor. They're well insulted or neglect any heat losses or neglect any
negative potential energy change. So, what would this look like if we went
further into these equations? So, let's start with the terms. Let's just show a
TS diagram. And why don't we just show a couple pressure lines if you will.
These are just represented lines of pressure. This would be the inlet pressure
and this would be the exit pressure. And let's just say that that's going state
point blank. So, the wheel turbine is definitely going to have an increase in
entropy. Right? The actual turbine has to have an increase in entropy. We
learned about entropy change and the increase enthalpy principle. The entropy
has to increase from a wheel device. So, in the real world as we go through the
turbine, we're going to end up at state point two, that's to the right of state
point one. But then of course below it because of lower pressure. This would be
2A. The ideal discharge point from the turbine is going to be directly below
point one. And again, ideal means isentropic. No entropy change. And the
horizontal axis is entropy, so clearly this has to be a vertical line and this
will be point 2S. Okay. I like to show the point 2S at the end of a dotted or a
dashed line, okay? Because quite frankly, that doesn't exist, right? We use it
as analytical tool. First we're going to assume that the process is isentropic
and then we're going to apply the isentropic efficiency, and we'll then be able
to the actual change associated with that substance within the turbine. So, we
do have to utilize point 2S, but we can't actually achieve point 2S. We're just
humans, right? We don't have the ability to create ideal processes. We just
can't do it. We need to understand that we can't do it. It represents a
theoretical movement. Now, if we look at a pump. Oops, I guess I better give you
the equation before I talk about the pump, right? So, what does this mean from
the standpoint of the equation? Well, let's go back to our isentropic efficiency
equation for our turbine. What is the actual work going to be? I'm just going to
write this on a per unit mass basis. Well, it's going to be, the enthalpy
change, right? So, our per unit mass basis, the actual work is just going to be
H2 actual. I'm sorry, I have that reversed. H1 minus H2 actual. So, remember the
numerator is the actual work. And the denominator is the ideal work. Here we're
going from point 1 to 2S. So this is H1 and this is H2 S. So this is the
equation we would use for a turbine. Let's look at a compressor now. Again,
showing the TS diagram. And again, let's show a couple of lines. Now note that a
compressor where it's fission pressure, so the inlet pressure is going to be the
lower pressure. The [inaudible] pressure is going to be the higher pressure. In
this case, here's point one. The actual exit from this compressor is 2A. The
ideal exit is 2S, directly above point one. The isentropic efficiency of the
compressor is the ideal over the actual. So ideal, well, basis over here, it's
just going to be H2 minus H1. Although, the numerator will be H2 S minus H1. And
the denominator is going to be H2 actual minus H1. Okay. And again, I could have
written this as a total quantity. You know, total enthalpies. I could have
written this as rate equations. I could have like m dot times H1 minus H2, right
here again. Dot time H1 minus H2, S is in denominator. I'm just not going to do
that. If you use those other forms, the mass of the mass flow rate terms,
they're just going to cancel out of the numerator and denominator anyway. And
you'll just end up with this equation nonetheless. So you're always going to end
up with the equation on a given mass basis. In other words, you're always going
to use specific enthalpy data in solving these equations. And then lastly, we
have our pump. So again, a little diagram. Moving from the lower to higher
pressure, from P1 to P2. There's my [inaudible] here's my actual 2. Here's my
ideal point 2. So the total equation is going to be H2 S minus H1, over H2,
actual minus H1. However, I am going to modify it, right? Just like I showed
over here, that we have an isentropic process associated with a pump. Then the
enthalpy change can be converted into specific volume and pressure change. So,
this is going to be, this volume at one, P2 minus P1. And then in the
denominator we'll just have to leave the other blank terms. H1 actual minus H.
So this is the equation we would tend to use for a pump. Now by the way, we
wouldn't have to use this version if we had good compressed liquid data in our
textbook, right? In our table A-7 for water for instance, but again, table A-7
starts at such a high pressure, it becomes practical useless to us. So if we
want to solve a problem involving a liquid, we just don't have the liquid data.
What other choice to we have but to use this version? Because this is data we
can just look up in the situation tables. So, anyway. Let me now move on. Now,
when we start looking at gas power cycles next week, they only deal with gases.
So let's look at these equations and modify them based on ideal gas assumption
and let's look at other specific heats. So, let's have a look at ideal gases.
And specifically we would just be looking at turbines and heat pressures. So,
what would be the turbines isentropic efficiency? Well, we know the basic
equation, right? So this is H1 minus H2 A or H1 minus H2 S. This was derived
just a moment ago or presented just a moment ago. I didn't really derive it. But
what we want to look at now is the case of constant specific heat. So if we have
an ideal gas with constant specific heats, well we know what the enthalpy is.
It's just CP times delta T. In other words, I should have said the enthalpy
change is just TP times delta T. So, this is just CP times T1 times T2 actual,
over CP times T1 minus T2 ideal. And it is constant specific heat, right? So
clearly the CP's are going to cancel and we just end up with nice simple
equations, only a function of temperatures, isentropic efficiencies. It's D1
minus D2A, over D1 minus D2 S. That's really all there is to it. And if we look
similarly at a compressor, then well, again basic equation is right over here.
H2 S minus H1, over H2 actual minus H1. Again, this could be just converted into
CP delta T term and then the CP's are just going to cancel. So this is just
going to be T2 S minus T1 over T2 actual minus T1. So these are the equations
that we use for the turbine or for the compressor. Now, why didn't include a
pump here? Well, pumps always move liquids and this is specifically dealing with
an ideal gas. So it's simply not actual. Okay. For gases, we deal with turbines
and compressors, no pumps. Alright, so you know, what would be a typical problem
where we would utilize this information? So, typical problem. So, you know,
typically we're going to know state one. And we're going too typically know the
exit pressure from that device. Again, this would imply to determine the
compressor. So, we know state one and P2. The next thing we need to do is we
need to assume that it's isentropic. And when we assume the process is
isentropic, this will allow us to get state point 2S. And I mean, it depends on
whether we have constant or variable specific heats as to what the specifics are
here. You know, with constant specific heats, remember we have those equations,
right? T2 over T1 equals P2 over P1 over K. Although now we have to be pretty
careful because we don't just have a point 2 anymore. We have a point 2 A and a
point 2 S. So we're typically going to use the fact that the temperature ratio
equals the pressure ratio to the K minus 1 over K. But keep in mind, this
specifically applies to an isentropic process. If we have an isentropic process,
we don't just call it 2, we not call it 2X. So it's exactly the same equation
that I dealt with at the very beginning of this lecture. But I'm just replacing
the 2S, I'm sorry, replacing the 2 with the 2S to make sure that we have the
right equation. Okay. So, this is going to allow us to get state 2S, in fact
specifically T2 S, right? And once we have T2 S, then you simply apply the
isentropic efficiency equation. So A to T or A to C, depending on whether we're
dealing with a turbine or a pump. And we use this in order to get our one and
only remaining unknown. Well, let that be the temperature, the actual
temperature at the exit. So, this is going to allow you to get T2 A. And once we
have these various temperatures, then of course we can go back to our first law
equation. We can find our enthalpy change, we can find our work, we find
whatever it is that is being required of us in our problem. Now, I will note
that if you happen to have a non-ideal gas, I'm sorry, I didn't say that
properly. If we have a non-ideal gas with variable specific heats, that's a
little bit different, right? This is constant specific heats. If we have
variable specific heats, well everything here applies but the equations are a
little bit different. Right. We're still going to know state one and the
pressure of the discharge. In other words, state two. But we're still going to
have to assume that's it's isentropic to get from state point one to state point
2S. But now we're not going to use this anymore. What are we going to use? We're
going to use relative pressures, right? And that's what we talked about in the
middle of this lecture. So you can use the equation P2 over P1, equals PR 2 over
PR 2. You know state one, you know T2. You can find PR2. You can then find T2
and then find the enthalpies. So, if it had to be variable specific heat, so
variable Specific heats. In this particular case, instead of this step, again
we're going to know state one. And P1, I'm sorry, we're still going to assume
isentropic. In order to get state 2 S. But we're not dealing with the equation
now using the compressions. So T2 over P1, it's PR 2 over PR 1. Again, we know
what P2, we know state one. And therefore, look up PR 1 and use this as PR 2.
And then let's have PR 2 and we just continue with table A-17. And then by H2.
Now again, you have a little bit of an error here, right? This only applies for
an isentropic process. You have to be very careful to make a distinction between
the actual point two and the ideal point two. This is isentropic so this isn't
really giving me PR2 is it? You need PR2 S. So let's add an S to the PR2. This
gives me PR2 S. You can find H2 S from Pr2 S. Again, just by looking it up in
table A-17, right? And then lastly, you simply apply the ideal output efficiency
equation in order to get H2 actual. Remember, if you have a process that is not
an ideal gas with the constant of the variable specific heats or maybe you just
need some basic equations for isentropic efficiencies. But use the actual
enthalpies. The enthalpies were not in the use, CP double T's. In other words,
were not going to use the temperatures. Okay, so it's the same process. It's
just a different way of utilizing the data. So, alright. I've got just about one
more minute so let me just give you a very brief intro to the next topic. By the
way, Friday is a holiday, right? It's Cesar Chavez day. You might think that
it's Good Friday when in fact it is. But that's now why we're having the
holidays. In the state of California we get Cesar Chavez day. So no class
Friday, until Monday. In fact, I'm not going to bother to start this. We will
continue next time talking about heat engine or duration cycles. So, that will
probably take me 20 minutes and then we start brand new material that we'll
cover for the rest of the class. Anyway, any questions about today? Alright,
don't forget you're going to show proof that thank you met the prerequisites, in
other words 301. If you were in my 301 class last quarter or last year. Don't
bother showing me, I have your grades, I know you passed. I'll just mark it. But
the rest of you, bring something to [inaudible].
For more infomation >> Review tướng hayate(ninija Long Ảnh) sắp ra mắt liên quân - Duration: 5:00. 
For more infomation >> Mr. Majnu Movie Review | Akkineni Akhil | Nidhi Agerwal | Venky Atluri | SouthCineUpdates - Duration: 1:30. 
For more infomation >> Apple Watch Series 4 - Review - Duration: 3:00. 
For more infomation >> Bermain VR Epic Roller Coaster Ride - Oculus Samsung Gear Review dan Ekspresi #7 - Duration: 6:11. 
For more infomation >> RAVNICA ALLEGIANCE | Tribal Cards Review | MtG - Duration: 8:42. 
Không có nhận xét nào:
Đăng nhận xét