>> All right, well good morning everybody. Pardon me. So just a comment or two
on the homework that's being returned to you now. Well first off, if you haven't
picked yours up it's just laying up here at the front. You know once again, most
of you did just fine. You know the first several problems were review problems.
A lot of students solved problem 675 instead of 678 and you know at first I just
didn't understand, but then I realized I probably didn't write it down as neatly
as I should have on the board. Many of you did 78, but many of you did 75; so I
wasn't prepared to grade number 75; so I just put 78 with a little question mark
by it. But nonetheless I think the only other error of significance was that
many of you forgot that I had asked you to do particular problems using variable
specific heat, not constant specific heat as it mentioned in the book. And then
also several of you on those last couple of problems, was it 38 and 52, the book
specifically says accounting for variations in specific heat with temperature,
which his the authors way of saying solve using variable specific heats. Again,
many of you solved using constant specific heat; so you'll see a big red X
through those problems and you know, it could very well have affected your grade
getting a check instead of a plus. Remember plus is three points out of three,
check is two out of three; so if you didn't do those last two problems
correctly, I mean those two problems are really the longest problems in the
entire homework set. Again, it could have cost you a point on the homework, but
nonetheless in general I'm pleased with the homework and again the homework
solutions are posted outside my office. One other thing I wanted to note on
problem 9-90 from your current homework set it may not be entirely obvious how
to solve it. I would note that on that particular problem you're given the
actual temperature at 4A and what are you trying to find? I guess you're trying
to find the temperature of .3. You know you've never really been given a problem
exactly like that before. And all I would really note is that basically the
problem has to be solved iteratively. I'm not going to give you too may hints on
that but you actually have to guess the value of T3 and then look up the
relative pressure and the [Inaudible] at that point [Inaudible] then have to
plug it into the equation for pressure ratio equals the ratio of the relative
pressures and that's going to give you your relative pressure of .4S then you
can find the [Inaudible] at 4S and then you can apply that to the efficiency
equation and the unknown there will give you the [Inaudible] at 4A and then see
if the temperature at that point 4A is the same as the one given. If it's not
then you pick the wrong value of T3, so start over again. Pick a new T3, go
through the exact same procedure, 3-4, 4S to 4A using efficiency and then check
again to see if .4A has the same temperature as the one that's given; so that's
how you would approach that problem. So this is given and it's an iterative
solution, you can call it trial and error if you prefer. Same thing. So any
questions on that? All right, good. So with that let's just go back to where we
were on Wednesday. Remember that we're talking about the Brayton Cycle and we're
now looking at variations to the Brayton Cycle, so it's not the simple Brayton
Cycle anymore. I had mentioned that there's three variations that we're going to
look at, all of which are designed to improve the thermodynamic efficiency and
indeed they all will. First we could utilize some of the waste heat, exhausted
from the turbine in that cycle rather than just dumping it into the environment
as heat output. We're going to send it through a regenerator and we're going to
pick up some of that heat as we move out of the compressor such that when we
finally go into the combustion chamber we have a higher temperature, and
therefore we add less heat as we move from the point of discharge from the
regenerator that is .5 up to .3; s our heat input is lower and our work is going
to be the same, so we end up with a higher efficiency. The second thing was
using intercooling along with compression and stages and that's really what
we're talking about now. This is all in your notes from last time; at least I
think it is. A schematic diagram as well as a TS diagram that represents the
intercooling process, so we're going to compress in stages, so two stages,
right? One to two and three to four, those compression processes take place at
the same pressure ratio. The inlet to each compressor is going to be at the same
temperature. We also note that the intercooling process from two to three is
done at constant pressure. Nonetheless if we do this compression in stages with
cooling in between each stage then we should minimize the compressor work input
and therefore with less work input we've got more network output. And again, the
efficiency should increase. So that's what we'll continue talking about now and
then the third one is where we actually allow the work process, the work output
process to be done in stages through multiple turbine stages and we have some
form of reheating in between. So we'll get to that in just a little bit and then
I'm going to give you an example problem that basically throws all of this
together. And so last time again we talked about the intercooling process, we
understand the basic process now. The equations are not really any different
than they were before so we still have our thermal efficiency, which is still
the network over the heat input. The network is still just the difference
between the turbine and the compressor work, divided by the heat input and again
I'm writing this on a per unit mass basis, all lower case letters. But it could
be capital letters, it could be dots. It doesn't really matter but this is the
basic thermal efficiency equation. What about the work input then? I mean the
work input which is the work associated with the compressors is really the only
thing that we're addressing here, right? We'll talk about the turbine work later
and heat input later as we combine regeneration, intercooling and reheat. But
for now it's really just the work input that's changed by using the intercooler
and that work input is just going to be the sum of the work input required by
each of the two compressors, so it's just going to be H2 minus H1, plus H4 minus
H3 and of course these are magnitudes, right? So this is the magnitude of the
compressed work input. But let's also remember that we're treating the working
fluid here as air as an ideal gas and as such the [Inaudible] are only functions
of temperatures. We know that just by looking into our table A17. But
nonetheless H1 and H3 are the same because T1 and T3 are the same, right? H2 and
H4 are the same because T2 and T4 are the same; so really this just becomes two
times H2 minus H1. And again since H2 equals H4 and H1 equals H3. So this is
really relatively straightforward as far as being able to calculate the
compressor work input. Now I might note further that in reality if we had an
infinite number of compression stages that would truly maximize compressor work.
And we can't do that, right? Well I guess one could try but it will take you
forever to build such a device, right? And plus it will be so expensive it
wouldn't' really be worth it; so having anything more than one intercooler with
two compression stages really is about the limits of our economics if you will.
So we're not going to look at other situations. But certainly you could have two
intercoolers or three intercoolers. The only thing that would be different would
be that the pressure ratio across each stage would be the Nth route, that is N
being the number of compressors rather than square root. So we know that the
total pressure ratio when we take the square root will give us the pressure
ratio across each individual stage. If we had two intercoolers with three
compressors we'll still know the total pressure ratio. Well we take its cube
root and that will give us the pressure ratio across three stages. If we had
four stages, the fourth root; so just keep that in mind, although you really
never have to use that fact, but certainly you want to remember that in going
through the analysis of compression from one to two or from three to four we
have to use a pressure ratio per stage, right? We're still going to solve this
like any other compressor. We're still going to say that the ratio of pressures
for that stage is equal to the ratio of relative pressures or we're going to use
the fact that the temperature ratio from one to two equals the pressure ratio to
the K minus one over K power. Either way you do it that's how you would go from
.1 to .2. You don't even have to do three to four because one and two is the
same as three and four. So just go through one compression process using the
pressure ratio across one stage and then just multiply by the number of stages,
which in your case is always going to be two. Anyway are there any questions on
this particular process or why it's relevant or important to have the extra
compression stage? All right. So this now brings us to our third and final
modification, which deals with reheat. Now here when we talk about reheat we're
really referring to the expansion in turbine. And I do call it turbine
expansion. It is an expansion process, right? The pressure is dropping as the
air moves through it. The gas should expand some, so lower temperatures, it's
got lower density; so it is an expansion process. But nonetheless like the
compressor where we can compress in stages in order to minimize a compressor
work if we do our turbine work, that is work output in stages we can maximize
the work output, that is maximize the turbine work. Now between each stage you
have to add heat, essentially just the opposite of what we talked about in the
compressor, right? You compress and then you remove heat if we have inner
cooling with reheat you expand turbine and you reheat back up to the temperature
that you had going into that first turbine. So how does reheat work? Well before
I write all these words let me just illustrate. So we have our turbine and I'll
put a big T for turbine. And then we're going to have some sort of heat
exchanger where we're going to have some heat input and then we're going to
expand through the next turbine. I'm not going to use one, two, three and four
here. I'm going to use six, seven, eight and nine. Although frankly the numbers
don't really matter. I mean you can use A, B, C, D or pick an alphabet anywhere
in the world. You just have to keep track of what point is what, right? Anyway
this heat exchange in the middle here this is the reheater. And if we were to
look at this on a TS diagram rather than a schematic you know, we still want to
show our minimum and maximum pressure lines. We're going to start at an elevated
temperature, right? I mean we've added heat already. We've got the high
temperature air, combustion gasses in the real world but high temperature air
for us. And now we're going to expand in stages so let's show the intermediate
pressure line too. So we go from six and we expand down to seven. We then reheat
along the constant pressure line up to .8 and then we expand a second time, do
more work output as we finally get to .9. So the turbine work is really going to
be the sum of the work from each of the two compressors, I'm sorry, each of the
two turbines. Okay? So this is what's happening. Now again I should note that
this is really the ideal case because we're assuming in this particular case
that the turbine is ideal. That is isentropic. But it doesn't have to be, right?
We can certainly solve a problem like this or like a compressor problem for the
non-ideal case too, right? We do have to use efficiencies, isentropic
efficiencies associated with the compressor and with the turbine. So let's just
move ahead - so how do we do this. What's the importance? So I'll just say
similar to compression in stages with inter-cooling - We can maximize the
turbine work output if we expand in stages. And to do this properly you expand
in stages and again, like the compressor with equal pressure ratios - Across
each stage - So with equal pressure ratios across each stage and reheat in
between the turbines up to the same temperature that we had going into the first
turbine. And that's exactly what I've illustrated over here on the TS diagram,
right? We have the same temperature at eight and six. We have the same
temperature, therefore at seven and nine. I mean after all the pressure ratios
have to be the same. And therefore, you're temperature ratios have to be the
same; and therefore we simply would find that if we had the same inlet at six
and eight, we have to have the same outlet temperature at seven and nine; so
again very, very similar to what we saw in the compressor. So this is how we
would maximize the turbine work. Again if we expand in stages with equal
pressure ratios across each stage and reheat in between the turbines up to the
temperature that we had entering the first turbine. I should probably note for
completeness that the reheat has to be done at constant pressure. So with all
this in mind then first of all what would be the pressure ratio across each
stage and also what would be the equations we would use? So once again similar
to what we saw for the compressor. I'm not going to actually go through the
derivation, but if we know the total pressure ratio, that is Pmax over Pmin then
we just have to take the Nth route where N is the number of turbine stages.
Again in this class two is going to be the maximum number of turbine stages
we're going to deal with and therefore just a single reheater in between. Again
it doesn't have to be that way, but it certainly would generally be that way in
the real world. So I'll just say we can find that pressure ratio across each
stage, it's just going to equal the square root of the total pressure ratio and
of course we know that the total pressure ratio is just the ratio of the maximum
to the minimum pressure within the cycle. So that's what we would have as far as
the pressure ratio that we would be required to use. Again, just to make sure
it's in your notes, we would note that the temperatures at the inlet are going
to be the same. And the temperatures at the outlet of each turbine stage are
also going to be the same. So what would the work be? Well the turbine work is
just going to be the Nth will be changed as we go from six to seven. Plus the
Nth will be changed as we go from eight to nine. And furthermore since the
temperatures are the same then the Nth will be the same so H6 and 8 are the same
and H7 and H9 are the same. And therefore - well clearly the turbine work is
just going to equal two times the Nth will be changed across one of the turbine
stages. Okay? Now again how you would actually calculate those would certainly
depend on whether we're using constant or variable specific heats. Certainly if
we're using variable specific heats then we'll just use the [Inaudible] data
directly out of our air tables, table A17. On the other hand if we're using
constant specific heat then all of our [Inaudible] changes just become CP times
temperature changes, ultimately when we plug everything into our equation for
thermodynamic efficiency the CP's are going to cancel and everything will just
be in terms of temperatures. Now rather than illustrate all this for just this
one specific case of reheat, what I like to do now is combine everything. This
would really be the most complete non-ideal cycle, well I should say most
complete cycle and we'll talk later about whether it's ideal or not ideal and
how that changes our mathematics. So any questions on this then? Okay, so
remember again our purpose - the purpose is to maximize the work output. We do
that with reheat and of course now we're going to add to it, minimizing the work
input by compression in stages and minimizing the heat input. Now let me know
one other thing here before I move on pass reheat, when we talk about the heat
input to the cycle we do have heat input now from an external source, right?
It's not like well even the inner cooler had heat rejection to an external sink,
but it never appeared in our calculations. At least the way I use thermodynamic
efficiency with network over heat input we don't really care about how much heat
output there is. It's not something we have to pay for. It's not something we're
concerned with. But on the other hand with regards to heat input we definitely
have to add heat from external source; so one should really go back here and
recognize that there is additional heat input. There's so little space in here.
I guess I'll do this, heat input. The arrow should point towards that line from
seven to eight. So we do have to include that in thermal efficiency. Now you may
wonder, well wait a minute. If the whole purpose is to improve the efficiency
why are we adding extra heat? I mean adding extra heat is going to actually
lower the thermodynamic efficiency, right? I mean we're increasing the
denominator of our efficiency equation, efficiency is going to have to drop but
fortunately the added work you get by this expansion in stages will more than
make up for the added heat that we have to pay for during the heat input process
and reheater; so the efficiency still should go up. All right. So let's just
make that final note - Heat input and our thermal efficiency equation will
include the reheater and really as well as the heat that's added in the
combustion chamber. So please don't forget to include that when you're looking
at the various equations. And again I'm going to bring all this together now by
looking at all three of our improvements. So any questions before I move on?
Good. So let's just call this Brayton Cycle with everything. That's not really
the official name of this Brayton Cycle, but its good enough. So for the
schematic diagram, so we have state point one and from state point one we're
going to go into a compressor. And we come out of the compressor at .2 and we
have to provide our inner cooling. So here's our Q out. We cool down to .3 the
same as T1 and we go through our second compressor. And now we come out of this
compressor and instead of running this air directly into the combustion chamber,
we run it through our regenerator. So here's out big heat exchanger, we call
regenerator. And let's see we come out of the compressor at four, we go into our
regenerator and we come out of regenerator at five. Now we go into our
combustion chamber so CC for combustion chamber. This is our heat input. We come
out of the combustion chamber and we go into our turbine. However this is just
our first turbine stage; so let's say we have six coming out of the combustion
chamber. Now we go into turbine number one. We do some work, we then have to
reheat - By the way I really should have labeled this intercooler over here for
completeness. So we go through our reheater where we have more heat input. And
we go through the second turbine; so let's see seven coming out of the first
turbine, eight out of the reheater, nine coming out of our second turbine. And
now instead of letting this exhaust into the environment or instead of rejecting
heat from our cycle we go right back into the regenerator, we remove heat from
this exhaust at - to .10 and then we finally have our giant heat exchanger in
the sky, the environment and that brings us right back down to .1. Okay so this
is our heat output - Now for completeness here as well let's also show the work
input to the two compressors. And let's also show the work output from the two
turbines, okay? So there's your full schematic diagram of a very complete cycle
with inner cooling and reheat and regeneration. Now that might look a little
complicated but it really isn't. I mean separately we've already looked at the
regenerators; separately we've looked at the inner cooler with compressor and
the reheater with the turbines; so we have all this already in our notes. We've
thought about it, it's been discussed. You might think that the TS diagram is
going to look incredibly complicated, but it really isn't. I mean keep in mind
that everything occurs either as isentropic work processes or as constant
pressure heat exchange processes. So we're only going to have vertical lines and
slope lines representing pressure. So let's just show our cycle again with our
minimum and maximum pressure. But let's also show in here the intermediate
pressure that we know is going to be associated with both the inner cooler and
the reheater. So we'll start down here at .1, so this is not going to be simple
but I will note that it is still the ideal case we're looking at. We're not yet
going to be talking about isentropic efficiencies, although we will be doing
that very shortly. So we start at .1, we compressed to .2. We inner cool to .3
at the same temperate as one. We compress up to .4 at the same temperature as
two. We exchange heat in our regenerator up to .5. We add more heat in our
combustion chamber up to .6. We expand first down to .7, we then reheat up to
eight. We expand a second time down to .9. We now move through the regenerator
and we exchange heat until we get to .10. And then lastly we just end up here at
.1. So the cycle isn't too bad. Certainly you can see now why I chose to discuss
the turbine and reheater from .6, .7, .8 and .9 because that's consistent with
this particular cycle. Now what would be the thermodynamic efficiency? Well it's
the same. It's just the equation; it's just the network of heat input so that
doesn't change. It's still turbine minus compressor work over the heat input.
What does change however is the fact that we now have more than one work term
and more than one well I should say more than one work output term for the
turbine? We have more than one work input term to the compressor and we have
more than one heat input term now. So using the same numbering scheme that's
here on these diagrams, what would our efficiency be? Well let's see, the
turbine work output is going to be the sum of the work output from the two
turbines, but it's the same, right? Just like we saw. So it's really two times
eight, six minus eight, seven. And then we subtract from this the required work
input and that's going to be twice the compressors work, so that's two times H2
minus H1. And then in the denominator it's our heat input and again we have heat
input twice, right? We have heat input from five to six, which is going to be
our combustion chamber. But then we also have heat input from seven to eight,
which is our reheater. So H8 minus H7. And this then is our thermodynamic
efficiency if we have - well I mean this is the good general equation but this
would apply specifically to the case where we would have variable specific
heats. Again all our [Inaudible] data is going to simply come from table A17.
We're going to treat the compression process and the expansion process exactly
as we've already done, so it really doesn't require more discussion at this
point. There is one other thing though that we do have to be aware of, which is
that since regenerator we still have a regenerator effectiveness. And
regenerator effectiveness is still defined the same way, right? The actual heat
input to the air coming out of the compressor going through the regenerator,
divided by the maximum possible. The problem though is that our state points are
different. That is they're numbered differently than when we talked about this
previously. Again the numbers shouldn't matter to anybody, it's the processes
that you need to understand. So what is the actual amount of heat exchange? Well
that's as we go from .4 to .5 so this is just H5 minus H4 and then what would be
the maximum possible? Well it's as if we could heat from .4 all the way up to
the temperature of the exhaust coming out of the last turbine stage. That is all
the way up to .9; so H9 minus H4. The definition of the regenerator
effectiveness hasn't changed, the only thing I've done here is used my updated
thermodynamic state points. And that would be that. So any questions on any of
this? Yeah it should be pretty straightforward. Once again I do want to note
that if we had variable specific heats we use these equations presented
directly. And if we have constant specific heat then we just replace the
[Inaudible] change with CP times temperature change and of course the CP's will
appear in each term in both numerator and denominator so they cancel and your
equation will only be in terms of temperature. So quite frankly you can just
take this equation and just replace the H's with T's and that will be the
equation that you would use if we have constant specific heat. And the same
thing would apply to the regenerator effectiveness, right? You've [Inaudible]
will be numerator and denominator, those equal CP times temperature changes. The
CP's cancel and you end up with just the temperature. So again you could just
replace the H's with T's. T for temperature and that's going to then give you
the equation that you would require if you had constant specific heat. Okay, now
the only other thing then that needs a little bit more clarity is what if we
have the non-ideal cycle? Okay, so we're still going to be looking at this
ranking - I'm sorry Brayton Cycle with everything. But how the equations now
being different and how would our problem solving techniques be different? Well
quite frankly the equations don't really change much at all. Again it's just the
analytics that is the math of the problem that's going to be a little bit
different. The equations will change ever so slightly. I mean quite frankly and
maybe I'll redraw the TS diagram. Quite frankly the only real differences that
we're now going to have to use two different state points at two and four, 2S
and a 2A. We'll have a 4S and a 4A and we'll have two different state points at
7 and 9. We'll have a 7S and a 7A and a 9S and a 9A. So the basic diagram is
going to be the same but we'll go from one to 2S in an isentropic way. We'll
then go from 1 to 2A utilizing well the efficiency, right? The compression
isentropic efficiency. We'll then - inner cool back to .3, the same temperature
as 1. So again, these are going to be the same. We'll then compress to point -
well we're really compressing to .2 - I'm sorry 4S and then 4A. So again the
dashed line represents the process we have to analyze, but not the processes
occurring, right? We assume it's isentropic. That gives me my [Inaudible] at 4S
or temperature of 4S if we're using constant specific heat. We apply the
isentropic efficiency to get to 4A. Then we move through our regenerator again
up to 5. To our combustion chamber at 6. And then we do the same thing, we
expand to 7S, but really we're going to 7A. We then reheat up to 8. Once again
we expand to 9S but really we're expanding to 9A. Again use the efficiency of
the turbine and then we have heat rejection - I'm sorry, then we go back into
the regenerator. And we come out at .10 and then finally we have our heat
rejection back down to .1; so it's the same diagram, right? It's only different
between 1, 2, 3 and 4. And then again 6, 7, 8 and 9. Now the basic equations are
not going to be any different except that we have to make sure that we use the A
terms as applicable. So in other word the thermal efficiency is not going to
just be written the same way, right? We're expanding from 6 to 7A. Not 6 to 7
anymore; so we really need to show two times eight six minus eight 7A. We
compress from one to 2A now, not just 2S. So let's do times H2A minus H1. And as
a denominator, well again we have no differences until we get over here to .7
and then at .7 well again we have to use 7A. So we still have the heat input in
the regenerator from 5 to 6, I'm sorry, not in the regenerator, in the reheater.
I'm sorry I take that back. 5 to 6 is combustion chambers; so we still have heat
input in the combustion chamber from 5 to 6; so that doesn't change and then we
have heat input in the regenerator - heat input in the reheater. Okay and that's
taking us from 7 to 8. So we have H8 minus H7A. All right, so ever so slightly
different, but not that much different, right? Anyway and then of course the
regenerator effectiveness is also going to be effected - we're going to have to
use 4A's and 9A's instead of just 4's and 9's, so H5 minus H4A over H9 minus -
I'm sorry H9A minus H4A and this is going to be our regenerator effectiveness
equation. Now of course, when you see the A's that implies that we have to do
the isentropic analysis first. You know we know that we have to use the relative
pressures to go through both the compression and the expansion process in order
to get points 2S and 4S, or 7S and 9S, right? Then we use the isentropic
efficiency of the turbine. And the compressor to go from well 2A, I'm sorry 2S
to 2A or 4S to 4A or 7S to 7A, 9S to 9A. We've talked about this before. We
haven't used the same numbering scheme again, but again that's really
immaterial, right? It's all about understanding the processes and how to go
through them and you know maybe as a suggestion maybe you want to solve some
problems with different numbers, just kind of swap all the points and just make
sure you can go through something understanding the individual processes and not
just memorizing equations with numbers on them. I mean what if I gave you a
problem on an exam that had an inner cooler but no reheater, or a regenerator
and a reheater but no inner cooler. I wouldn't be using the same numbering
scheme. You know for every device we remove we're going to have one less or
maybe two less state points. So you know understand the processes, don't just
memorize equations. So let us now go through an example problem that will pull
all this together. Yeah it's a little bit long because we have a lot of
processes to analyze, but if you can do this example problem or at least
understand the example problem you should be able to do any problem I would
throw at you. So this is from your textbook with a slight revision. It's chapter
nine, problem 121. And turn this on. So this is problem 9-121. I'll put E8 for
the eighth edition of the book. I'm starting to do it this way because some of
the examples I'm going to use moving forward are from the seventh edition or the
sixth edition. I mean I've got some going all the way back to the second
edition, but that's only because I've been here 32 years. But nonetheless we're
going to modify this a little bit. This is going to be with regenerate
effectiveness of 75%, a compressor efficiency of 80%, and turbine efficiency of
85%. Okay and I guess technically we would call this part B of this problem.
Because part B already tells you that you have a regenerator with effectiveness,
but it doesn't give you anything with regards to the isentropic efficiencies. In
fact, if you read it, it says consider it an ideal gas turbine cycle. But this
is not ideal, okay? So that's why I say with these so please note again, this is
not an ideal cycle. Nonetheless we have our gas turbine cycle, not ideal and
there's two compression stages and two expansion stages. In other words, we have
inner cooling and reheat. We're given the pressure ration across each stage of
the compressor, which often is not given, sometimes they'll give you the total
pressure ratio or say what the maximum and minimum pressures are. And then you'd
have to use the square root of the total pressure ratio to get the compression
ratio across each stage. But it does give you the stage pressure ratios now, so
just keep that in mind. Read your problems carefully, it's very easy to make a
mistake at the very first step of the problem and use the wrong pressure ratios
and then everything is wrong. All right, they've also given you the temperature
entering each stage of the compressor and each stage of the turbine as 300 and
1200 kelvin respectively. Now again they don't really have to say each stage,
because we know that the temperature is the same going into each stage, so they
could have just said that the inlet temperature of a compressor is particular
value or turbine particular value and you just know that they're the same for
each stage. Anyway they want us to bind them [Inaudible] ratio as well as the
thermal dynamic efficiency of the cycle. And again in part B, with my
modification we have this other data. So let's just write out what we have. We
know that the pressure ratio across each stage is given as three. We're given
the temperature into the compressor so that's 300 kelvin, so that's T1. And by
the way I'm going to use exactly the same numbering scheme they have over here
on the far right hand side of the board. This is a non-ideal Brayton Cycle with
everything on it. So T1 is what's been given. We're also given the temperature
at .6 which is 1200 kelvin. And of course T1 and T3 are the same; T6 and T8 are
the same so you can note that here as well. We're trying to find the back work
ratio. And remember the back work ratio is just a ratio of the work output from
the compressor, divided by the work input - I said that wrong. The work input to
the compressor divided by the work output from the turbine. So that's what we're
going to have to find as well. And then of course we'll find the thermal
efficiency; so there's your problem. The question really is how do we approach
the problem. I think one thing I want to do is see if I can raise the screen up
just a little bit, just so we can see the rest of the schematic diagram over
there. I mean I can't adjust where the light is so yeah, do the best you can. So
maybe what I'll do is I'll just kind of leave this up here for now. I can't even
do that, I don't have enough room. That's the problem with these new classrooms
with these jumbo size screens. It would be nice if the screens were over on the
edge somewhere so I could use the white board. But it takes up half the space
and then I can't really do anything; so I guess I will have to start erasing
this. It's in your notes anyway. So what's the appropriate approach for this
problem? Well it's the same approach that we've always used, right? We'll start
at the entrance of the compressor; we'll go through isentropic compression
process. That will give me the information at 2S and now use the efficiency to
get to our results 2A and the have to move over to the turbine. I can't deal
with the regenerator yet because I need to know what the discharge is from the
turbine, so then I go over to the turbine stages and I'll do the same thing,
right. I'll find my data at .6 that is at T6, I'll find the relative pressure at
.6. I'll go through the expansion to 7S to get well my data at 7S. I'll then use
the isentropic efficiency to get the data at 7A. And note 7A is the same as 9A
so I know what's coming out of the turbine. And then I go back to the
regenerator. I do the regenerator effectiveness last. The regenerator
effectiveness will allow me to find the [Inaudible] leaving the regenerator,
right? Over here at .5 from this equation. I've already done the compressor so I
have 4A and I've done the turbine, 9A so this gives me H5. And that's all I
need. Once I have that then I just apply it to the isentropic efficiency - the
thermal dynamic efficiency equation. I have all my thermodynamic states and
problem is solved. So there's nothing different here, we've handled each of
these processes individually, now we're going to do it collectively. Now we're
going to do it collectively. So let's begin. I'm only going to do this using
variable specific heats. Certainly you could do is using constant specific
heats. It would not be as accurate and you're not going to get exactly the same
results but still it's worth noting that this is the only method that I'm going
to use. So let's just begin. So first of all at T1, we have to go into table A17
and we need two things. We need [Inaudible] and the relative pressure. So the
[Inaudible] at 1 is 300.19, [Inaudible] per kilogram, that would be T1 at kelvin
and the relative pressure is going to be 1.386. Now we have to go through the
isentropic compression so we note here that the pressure ratio across each stage
is really just going to be P2 over P1, which is PR2 over PR1. Again this is not
entirely correct because this only applies to isentropic process, right? So
don't forget your S. This will then allow me to find PR2S, so P over P1 is equal
to three. We have PR1 from above. So this gives me PR2S and it's going to equal
4.158. Can you guys see these numbers over in the far corner? Anyway I can't
hold it there but it's 4.158. So I'll just move it. So now we have that number,
so now we go back in at PR2S that is back in table A17 and this gives me H2S.
and of course we have to find this by [Inaudible] and we get 411.26, kilojoules
per kilogram. And now we apply the isentropic efficiency of the compressor which
is given as .8. And of course this is just going to be the ideal over the actual
work. And the work terms are just [Inaudible] terms, we have H1, we have H2S. So
we get H2A then and this is going to equal 439.02 kilojoules per kilogram. And
this is the same as H4A. So we know H1 and H3 are the same. I probably should
have also noted that this is equal to H3 up here above. But nonetheless we now
have all of our [Inaudible] from one to four. And we do exactly the same thing
for the turbine. Really I should say for the turbine stages. So again same
thing; so now we're going to start at 8.6, so at T6, into table A17. We need the
[Inaudible] and the relative pressure at six. So eight six is 1277.79 kilojoules
per kilogram. PR6 is 238.0. Now we go through the expansion so the pressure
ratio per stage is still equal to three and you know P6 over P7, but it's also
PR6 over PR7S. So again we can go through our math, just plug in PR6 from above.
And we can get the relative pressure at 7S, which is 79.33 and then by
interpellation go back into table A17 we get H7S and this [Inaudible] is going
to be 946.36 kilojoules per kilogram. And again now we just deal with the
isentropic efficiency of the turbine. This is given as 85%, we know that the
definition is inverse from that of the compressor, so it's the actual level of
the ideal work. So H7A - I'm sorry H6 minus H7A over H6 minus H7S and then what
we want to do is we simply want to plug in the numbers from above and solve and
we get the actual [Inaudible] leaving at turbine that was H7A of 996.07. And
again kilojoules per kilogram. Now once again we know that each turbine has the
same inlet so up above you can just note that H6 and H8 are the same. And down
below H7A and H9A are the same. So the next step it looks like I'll have to
erase the TF diagram now. The next step is to solve for the regenerator
effectiveness in order to get the temperature - well I'm sorry the [Inaudible]
that leaves regenerator. So again there's all the single unknown, right? In fact
I just erased the equation that we're going to use. I guess I didn't really have
to erase it. So H5 minus H4A over H9A minus H4A. Once again everything we
already know from above, so I'm not going to bother to plug in the numbers. You
can do that at home. We end up with H5 of 856.81 kilojoules per kilogram. That's
the hard part; the rest is just finding the data that we need. So we need the
compressor work, we need the turbine work so that we can find the back work
ratio and we need the heat input additional to find the thermal dynamic
efficiency. So the compressor work is just two times H2A minus H1, again I'm not
going to bother to plug and chug, but you can plug all the numbers in yourself
so we get 277.66 kilojoules per kilogram for the compressor work input. We have
two times H6 minus H7A as the total turbine work output. So again if we go
through the math this is 563.44 kilojoules per kilogram. The heat input is from
our combustion chamber as well as the reheater so H6 minus H5 is the combustion
chamber. And then H8 minus H7A represents the reheater. So again we have all the
data already here on the board, plug it all in and we get 702.7 kilojoules per
kilogram. We find out back work ratio, the compressor work input divided by the
turbine work output. So just take the ratio of the two terms above and we get
.493. That might sound like a lot, I mean fully half of the work being produced
by the turbine is being used up by the compressor. In other words only half of
the turbines work output is actually network that could be used to generate
electricity or thrust or whatever your work output is. But that's not unusual
for these types of systems. It's very common for a significant portion of the
turbines work to be required by the compressor; so there's one result. And then
lastly and the thermodynamic efficiency, network over the heat input we'll find
this is .407 or 40.7% so there's the results for hi particular problem. Now some
of you might think that a 40% thermodynamic efficiency is pretty lousy, it's
actually excellent. If we look at a typical steam power plant that exists out
there in the world and fully 80% or so of our electricity is generated from
steam electric power plants the best of them, the very best of them maybe 34%
efficiency. A gas turbine like this using the Brayton Cycle has efficiencies in
the 40% range. It's significantly higher. So these are actually very good
numbers. The nice thing about having achievable efficiencies in the 30-40% range
is that there's always going to be work for mechanical engineers. I mean you're
always going to have to try to improve the efficiency to do whatever you can
whether it's through materials or processes or controls or thermal dynamics or
fluid flow, whatever - you're always trying to make improvements, right? I mean
you know the company that come up with the next latest and greatest product is
going to be the one that wins if you will, in the business world. And that's a
good thing, right? Of course it's also good from an environmental point of view,
the higher our efficiency than the less heat output, the less thermal pollution
if you will and presumably the less fuel we would have to burn to produce the
work that we need. So that's good or pollution; so for all sorts of reasons it's
good to keep working on improving efficiencies. Now just as a matter of note
here if one were to solve this problem without a regenerator - Assuming that we
still have the compression and the turbine in stages, what we would find is that
the thermal efficiency is going to drop to about .368. The back work ratio is
going to go down significantly to about .335. And you know you might think well
that's good, right? If we're using less work for compression than perhaps we
have more network output. But that's not true, okay? Without the generator we
have more heat input, yeah we might have less work input but we also are going
to require quite a bit more heat input and the efficiency is going to drop,
okay? And again it's not just dropping by four 100's or if you will it's not
just dropping by 4%, it's dropping by a full 10%, right? We compare what is
achievable at almost 41% to what's producible now at 37%, I mean that's a 10%
drop; that's a significant number. So think about it in relevant terms and
you'll see this is a pretty good improvement by adding the generator. Anyway oh
- Without a regenerator I should have also said and with assumed compressor and
turbine efficiencies of 100% -- in other words that would be part A of the
problem as it was actually presented in the textbook. Again we have significant
differences. So nonetheless are there any questions then at all because this is
really our end of the discussion of Brayton Cycles. Are there any questions at
all on this particular type of cycle? Good, so now we will move on and I have
just the right amount of time to introduce you to my favorite cycle, certainly
the one you'll see most in the world, the Rankine Cycle, which is our steam
power plant cycle. Again vast majority of the energy, electricity that's
produced in this world, not just the US, but the entire world comes from steam
power plants, so we really need to understand how to analyze such a device. So,
this brings us the to the Rankine Cycle. And this is going to be the last of the
heat engine cycles that we're going to talk about. It's not going to be the last
cycle, after we talk about the Rankine Cycle we'll then get into the kind of
cycles that we use for air conditioning, refrigeration processes. So you know
that's going to be not the heat engine cycle, but the refrigeration cycle and
we'll get to that eventually but right now the Rankine Cycle. So this is used
for steam power plants. Okay. Now there's significant similarities between the
Rankine and the Brayton Cycle not necessarily in the way that we utilize our
data or how we use that in our equations but just like the Brayton Cycle we're
still going to have constant pressure heat exchange, so the heat input and the
heat rejection is still going to be done at constant pressure. So that's
certainly the same as the Brayton Cycle, also at least for the ideal case we're
going to have isentropic work. So the work input is going to be isentropic, the
work output is going to be the isentropic; so again just like the Brayton Cycle
and of course we use the non-ideal cycle, which we'll certainly do then we'll
still have to assume first isentropic, but then we'll have to apply the
isentropic efficiency in order to get the actual work associated with this
cycle. But nonetheless these are some similarities. The names of the devices
that we use are certainly going to be a little bit different than we do in the
Brayton Cycle. So let me just show a simple schematic. So this is going to be
for the simple Rankine Cycle. So first of all we'll start at some state point,
this is called one. And we'll note that this is used for steam power plants. So
we're not talking about air cycles anymore at all, you can just completely
remove that from your mind at this point. We're using water, right? Steam is a
form of water so we're actually going to take liquid water at .1 and rather than
send it through a compressor for our work input, we're going to send it into a
pump. Now again a pump and a compressor perform exactly the same function, we
just call it a pump because it moves a liquid rather than a compressor which
moves a vapor, but it's the same process. We're going to leave the pump state .2
and we're going to go into a heat exchanger of some sort. Now granted this is
not going to be a combustion chamber anymore, here is where we're going to turn
our liquid water into steam; so this is an actual boiler. So this is where we
have our heat input to the cycle, so we come out of the boiler state .3, so this
is our high temperature, high pressure of the cycle. We're now going to go
through a turbine, we still call this a turbine. So this is where our work
output is going to be. So work output is turbine work. By the way over here when
I put work input I probably should have put WP as well just for pump, but
nonetheless so we're back to the turbine. We have our work output and we leave
at .4 and then we have to reject heat into the environment. Now this is water,
right? It's not air. We're not going to just in the real world, we're not going
to just dump the water out into the environment, so we have to take that water
and cool it down. And in fact, as it comes out of the boiler it's steam, as it
leaves the turbine it's either steam or a two phase mix. But nonetheless we have
to send it through a device that's going to condense that steam back into liquid
water again and of course we use the word condenser, right? It's condensing the
steam, we call this condenser. Okay, now in the condenser there's going to be
some heat rejection, so there's the heat output. How this is done is not
something we're going to talk about immediately. It can be done through a
variety of means, but typically this is a big heat exchanger, you've got the
steam on one side of some tubes, inside a shell. On the other side of the tubes
you have cooling water from the local ocean or a river or a lake or it could be
from a device, like a cooling tower which we'll actually talk about near the end
of this quarter. But it's a heat exchanger again and we just condense back into
a liquid. So this is the basic Rankine Cycle and shown on a schematic diagram.
No more complicated than a Brayton Cycle for sure. In fact, in many ways the
analysis is going to be a little bit easier. We're not going to have to deal
with relative pressures and all that. We're just simply going to take our
[Inaudible] data and our steam tables, table A4 through A6. So this is the basic
process. It would certainly be appropriate to show this on a TS diagram. Now
here I would like to show it relative to the saturation lines. So there's my
saturation curve, my critical point up at the top. State point one without
exception is going to be a saturated liquid, okay so always a saturated liquid.
This represents the low pressure of the cycle, so let me also just show a couple
constant pressure lines, again show the maximum and minimum, but of course this
is different than our Brayton Cycle and our Brayton Cycle who was just air as a
gas, it would occur over here in this area and that's why we always had these
angled lines. Now we have the saturation region, we have constant pressure at a
uniform temperature as we go through the saturation region. So that's why I have
the curved lines that become flat for a while and they curve back up again.
Nonetheless we compress isentropically to .2, right? We then add heat and
constant pressure until we get up to 7.3. We then expand isentropically until we
get back to .4 and then we get right back over to .1 again. So we've got our
pump work input, our turbine work output, our heat input and our heat output and
there's your basic, simple and ideal - so I better write that both, simple and
ideal Rankine Cycle. Okay by the way we'll certainly adjust this later. We'll
look at non-ideal cycles, we'll look at cycles that are not simple. We'll look
at a form of regeneration to try and improve the efficiency, but that will be
later. So this is the basic Rankine Cycle and that's really all the time I have.
Question?
>> Are those from one to two and three to four, streamed down?
>> Yes, they're isentropic and I don't know, maybe my curves aren't as vertical
as they should be, but these are clearly isentropic processes. So anyway here is
our basic cycle. We'll get into this in a lot more detail next week; so have a
nice weekend everybody. Once again don't forget to pick your homework up at the
front. If any of you missed my hint on problem 9-90 then just talk to one of
your classmates and I will see you all then.
For more infomation >> Une astuce simple pour vous débarrasser de l'herpès|Wiki Santé - Duration: 2:56. 
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