>> All right, well, good morning, everybody. So, first, just a reminder that
some of you didn't pick up your homework last week. Please don't forget to pick
it up. Also, don't forget that last week's homework is all due on Wednesday, so
please bring that with you. And, of course, the new homework set from chapter
nine, this is the homework for the entire week. You know, now that we're not
doing review anymore, you know, I'm not going to have these really large
homework sets like I did the first couple of weeks. But, nonetheless, these are
important problems. So, please make sure you write them down. And please note,
also, that the comment I have there in the big square brackets only deals with
problem 119. So, I don't want you to solve it exactly as identified in the book.
I want you to do it that way. So, please make sure you add those terms. Now, at
this point, you don't even know what epsilon means. We'll get to that, of
course, before the end of the week. [Inaudible] are just isentropic efficiencies
for compressors and turbines. But, nonetheless, that's the homework assignment
for the week. And let's just get back to what we were talking about last time.
So, please remember that we're still dealing with gas power cycles here. We're
still dealing only with the air standard type of cycle. We still understand
that, as we saw with the air standard cycles, we could either choose to solve
them a little bit less accurately, that is, using constant specific heats, or a
little more accurately, which is to use the variable specific heats. And, of
course, along with that, is data from table a 17 for air. Now, we've discussed
at this point just two cycles, right? Both are reciprocating engine cycles, one
being the Otto cycle, the other being the diesel cycle. We've seen example
problems for both. So, hopefully this material now makes sense to you. Those
cycles are reasonable models for real internal combustion engines, okay, both
gasoline and diesel engines. But now we need to start looking at other models,
and these other models are going to be applicable for other types of engines.
The next one, actually, the next two that we're going to look at are cycles that
really are very, very limited in their practical use. In fact, I'm not even
going to discuss them except for just a little bit of very, very brief theory.
These are called the Stirling engine and the Ericsson engine, okay? So, we'll
just start with the Stirling. And, quite frankly, I probably shouldn't call it
engine at this point. I mean, granted these are all heat engine cycles, so it's
not incorrect to call it an engine, but we're really just talking about the
thermodynamic cycles now. Now, the Stirling cycle, like the Ericsson cycle, are
rather unique in that they're gas power cycles, yet the heat transfer takes
place at constant temperature. So, heat transfer at constant t. Now, as you
think about that, that's practically impossible for us to do in the real world.
It's not that it can't be done, and certainly there are some Stirling engines
that have been built, but if you have a gas and you're transferring heat into
it, then typically your temperature is going to rise, right? Or, if you're
transferring heat away from it, then the temperature is going to drop. To try to
transfer heat and maintain constant temperature is very, very difficult. In
fact, at least in my opinion, the reason so few Stirling engines are actually
out there is really because of that fact. There really are not good ways to make
Stirling engines. Now, again, that doesn't mean there aren't some Stirling
engines that exist out there. There's some solar Stirling engines. They're
interesting devices. But, again, they're not particularly practical, so I'm not
really going to go through them in any detail. I will note that the heat
transfer does occur at constant temperature. I would also note that the work
which is done is actually going to be done at constant volume for a Stirling
cycle -- Which I might note is also rather problematic, okay? When you're doing
work, typically you've got some sort of expansion taking place. You're basically
increasing the volume, that's typically what would happen. If you're doing work,
that is, you're taking energy out of the fluid, then typically that would imply
that the pressure is probably dropping, as is the temperature, and the specific
volume is actually increasing. But, again, in the Stirling cycle we have to set
this engine up so that work is done at constant volume. Again, this is very,
very difficult to do in practice. So, again, for these reasons I'm not really
going to spend a whole lot of time on this. I will plot the cycles on both the
T-s and the p-V diagram. On a T-s diagram we would typically have something that
would look like this. Okay, so, we're basically starting, at least in this
cycle, at some elevated temperature, .1, and now we're going to add heat at this
constant temperature. So, this is our heat in. Then we're going to have some
work output. We're then going to reject heat also at constant temperature. And
then we'll require some sort of work input. On the p-V diagram we would start at
1 and go to 2 in this fashion, okay? So, the volume is actually going to be
increasing. Then we're going to go through the constant volume work process. So,
there's constant volume, just a vertical line. You know, then we're going to
have our next constant temperature process where we're going to reject heat. And
then, lastly, our work input from 4 to 1. Okay, so, work in and work out, heat
in, and heat out, okay? So, you can pretty clearly see what's happening here,
all right? Again, constant temperature, horizontal lines on the T-s diagram for
the heat transfer processes, and constant volume for the work processes that we
can see pretty clearly as vertical lines on the p-V diagram, okay? And,
nonetheless, that's really all I want to discuss with regard to this. In fact, I
might note that even in your textbook the author really doesn't cover these
engines in any significant way. They do develop an equation for thermodynamic
efficiency, but it's a very simple equation. And, again, in this class I'm not
even going to worry about that. All right, so, what about the other, the
Ericsson engine? Or, again, I should call it Ericsson cycle. So, really, the big
difference between the Stirling and the Ericsson is the way the work is done.
The heat transfer is still done at constant temperature -- But the work is done
at constant pressure. So, once again, this particular type of cycle is not
particularly practical in the real world. In fact, quite frankly, I've never
really seen any engines that are designed based on the Ericsson cycle. I assume
that they exist out there. I mean, every thermal book that I've seen talks about
the Ericsson cycle, but you just never hear about an engine that operates on
this cycle. Once again, though, let's just show it on the appropriate T-s and
p-V diagrams. Once again, we'll start with state .1 at our high temperature.
Heat transfer as we go on the horizontal line from 1 to 2. Then we have constant
pressure work from 2 to 3. And then constant temperature heat transfer out of
the system from 3 to 4 and then work input from 4 to 1, okay? So, it actually
looks very similar to the T-s diagram for the Stirling engine. The big
difference is that these are actually constant pressure lines and they're not
constant volume lines. Anyway, let's just show the heat and work terms. And then
let's show the same thing on the p-V diagram. So, here our .1 is going to be
over here. We're going to add heat at constant temperature. This brings us out
to .2. We're then going to have work at constant pressure over to.3, and then
more heat transfer at constant temperature up to 4. And then, lastly, our work
input from 4 to 1. So, work in, heat in, work out, heat out. And these are the
basic processes associated with the Ericsson cycle, okay? So, ultimately, if one
were to calculate the thermodynamic efficiency, and this would apply to both
Stirling and Ericsson -- You could actually see this derivation in the book, but
it's a very simple equation. It's just 1 minus t low over t high. So, this is
how we'd calculate thermodynamic efficiency for this type of cycle. Now, note,
this would be for the ideal case. Certainly, if you have a nonideal case, which
is definitely not even mentioned in your textbook, you know, then the analysis
is going to be quite a bit more complicated. The thermal efficiency is not going
to be just a function of temperatures. But, nonetheless, this at least gives you
something to think about with regards to these two cycles. And what that I'm
actually going to move on. Lots my notes. Thank you. So, are there any questions
on these cycles? Probably not much to ask since we haven't really discussed
much. Now, let's move on to a cycle that is of use to us. In fact, it's a cycle
that was covered, hopefully, in your ME301 class, and this is the Brayton cycle.
So, first of all, just by a show of hands, how many of you have never covered
the Brayton cycle? It just didn't get covered in your 301 class? Okay, so
everybody's been exposed to that. And you're supposed to be exposed to just the
simple Brayton cycle in your 301 class. And, hopefully, you were also exposed to
both the ideal and not ideal Brayton cycle. But let me go through just some
basics here so that we remember what the Brayton cycle is all about. And, in
fact, while I'm talking about this, these are just some, well, vendor literature
that has some nice diagrams on the inside that actually shows cutaways of a gas
turbine engine, which is what the Brayton cycle is used for modeling. So, let me
just pass one around each side. So, as we're talking about this you can see what
a real engine would look like. And, basically, these are jet engines, okay?
Whether they're stationary jet engines, in other words, they're stuck to the
ground and they're just spinning an electric generator, or whether they're jet
engines on the wing of an aircraft, it really doesn't matter. From the
standpoint of analysis, they're both analyzed exactly same thing. Frankly, we
don't really care how the network is used, just that it is somehow used, right?
We, as mechanical engineers, would probably be more interested in a stationary
gas turbine engine. On the other hand, if you're an aerospace engineer, then
perhaps you'd be more interested in a jet airplane and that kind of an engine.
Note that on a jet airplane you're just spinning a turbo fan. That fan creates
the thrust needed to move the airplane through the air. On a stationary
application, you're spinning an electric generator. But, again, that's just the
network, that really doesn't matter. So, Brayton cycle. Now, the Brayton cycle
is not a reciprocating engine cycle, right? It just isn't. This is a series of
steady flow devices that are all stuck together, end to end if you will, the net
result being the ability to produce work having given some heat input. I mean,
that's what heat engines are all about, right? So, this is not a reciprocating
engine cycle. This uses many steady flow devices, okay? And perhaps the best way
to begin something like this is just to show a nice, relatively simple
schematic. So, again, this is just the simple Brayton cycle. And in a typical
simple Brayton cycle we'd start at state .1 and we would enter devices call a
compressor. So, I will often just use a big capital c, but I'll just write the
word compressor for now anyway. So, work input is certainly going to come into
this compressor. In other words, the air that's moving through the compressor is
going to have work done to it and it's going to compress the air, right? It's
going to go to a higher pressure and temperature. So, that's our work input.
Sometimes we would just write this as w's of c, c meaning compressor, rather
than just writing the word in all the time. So, c here represents the
compressor's work input to the cycle. Now, the air that comes out of the
compressor has to have heat added to it. So, here we have our big heat
exchanger. And, certainly, there's going to be heat input. By the way, I'm just
using lowercase letters q and w. It's not necessary. You can use capital
letters. You can use letters with dots on top of them, as long as you understand
that we're talking about heat input or work input or whatever. But, nonetheless,
the heat input in our cycle is assumed to be just heat input in a heat exchanger
of some sort, okay? In the real world, that's certainly not the case, right? In
the real world we have a combustion chamber. We're actually mixing fuel with the
air and we're burning it until it achieves a very high temperature, and that
high temperature gas is now going to move into the turbine section of this
system. But, of course, this is an air standard cycle, right? We're only dealing
with air standard cycles. So, we simply have air that's circulating through the
cycle. We're simply assuming that however much heat that would otherwise be
liberated by the combustion process is simply being added by the heat exchange
process. In fact, isn't this the same thing that we assumed when we were looking
at the Otto and diesel cycles? Yes, of course, it is, right? These are closed
cycles where there's just air moving in a loop. There's no intake. There's no
exhaust. There's no fuel and air mixtures that are burning. They're relatively
simple air standard cycles. Nonetheless, the heat input goes into this device.
I'm still going to call it the combustion chamber. Sometimes we might just call
it a combustor, and often I'll just use c c as an abbreviation for combustion
chamber. But, again, from our analysis, it's not really combustion chamber
anymore, it's just a heat input device, a big heat exchanger. All right, so, now
we leave this device at .3 and now we're going to enter the turbine. So, bit t
for turbine. Of course, the turbine is a work output device, the very high
temperature, high-pressure gas that's leaving the combustion chamber. It's going
to go into the turbine. It's going to cause it to spin. It's going to give up
its energy. Its pressure and temperature are going to drop; therefore, its
specific volume is going to rise, or fuel extensity drops. And we leave this
particular device then at .4. By the way, the work output, as it's out from a
turbine, is often just shown as w t instead of writing the word work out. So,
this is a turbine work output. And then, lastly, what leaves the turbine is
still going to be at an elevated temperature. We have to reject heat from it. So
here is another heat exchanger, and this heat exchanger is, again, not really a
heat exchanger in the real world. This is just the environment. You're taking
those exhaust gases from the turbine and you're simply dumbing it out into the
environment. The environment is so huge, compared to the small amount of exit,
or the small amount of output from that turbine, that we assume that turbine
eventually -- I'm sorry, that exit from the turbine eventually cools down until
it's at the same temperature as the environment, okay? So, this is the
environment. Again, in the real world that's the environment, but in our world
it's just a heat exchanger, and this is where we're going to reject heat from
the system. So, here's our q out term, okay? So, this is the simple Brayton
cycle. Now, another thing that I might note is that in our real world you should
at least understand what these engines look like, even though we're modeling
them in this nice, simple, closed manner. In fact, that's why I've passed those
two handouts around. You can see that a real gas turbine kind of has this kind
of a look to it. On this end of the turbine, these are the various stages of the
compressor. So, these blades are all compressor blades. So, this is our
compressor. As the gases move through, again, in our case it's air, but they're
really gases, you know, air is a mixture of various gases, but we move through
the compressor. We'll actually add fuel. And this is our combustion chamber.
Okay? So, again, in the real world fuel is injected, the fuel and air burn. And
then, as we continue, again, noting this is a single shaft, so the compressor is
spinning, but it's spinning under the influence of the turbine, and now we'd
show all the turbine blades on this end. So, this is our actual turbine. And
then we have our discharge. So, this would represent .1 over here, entering the
compressor, right? This would represent .2 as we leave the compressor and enter
the combustion chamber, right? So, maybe I'll also show here's my work input. We
go to the combustion chamber. Here's the heat input. So, we pass through the
combustion chamber. Combustion is finished and we get to state .3. And then at
state .3 we pass those gases across the blades of the turbine. The turbine
spins. Now, this is, again, a single shaft, right? So, the work produced by the
turbine, much of it is actually being used by the compressor. The difference
between the two is the network output, and then this shaft would be connected to
something else, okay? If this were a stationary gas turbine engine, then it's
connected to an electric generator. We could show a generator attached to the
backside. Sometimes the generator is on the frontside. If it's an aircraft
engine, then the shaft will be hooked to a turbo fan. In fact, the fan is never
going to be on the backside. The turbo fan is on the frontside, and that's
actually what you see when you're looking into a jet engine. You know, usually
they'll have something like a smiley face or a ying-yang or something painted on
the front side for no apparent reason, just to be funny, but that's the turbo
fan you're looking at. That's spinning under the influence of the network from
this Brayton cycle, and it's, well, producing thrust, right? The fan blades are
moving air at an incredible speed. It passes actually around this section. It
accelerates, shoots out the back, provides thrust. Again, it's not really
important from our analytical point of view, but you should understand what
we're thinking of and what we're talking about. Anyway, so we leave at .4 over
here. And then, again, in the real world, we're done. It's not really a cycle,
right? We have new fresh air coming in at the ambient conditions, and we have
new exhaust going out. But, in our world, we have the whole environment -- And
we're simply removing heat from our system into the environment, and that brings
us right back to the thermodynamic state at .1. So, this is our gas turbine
engine. This is our Brayton cycle. Now, one other thing I might want to note,
just as a reminder, is that the work that's done in a Brayton cycle is done at
constant pressure. So, constant pressure work. And this would be both the work
input and the work output, right? The work output from the turbine is done at
constant pressure. The work input to the compressor is done at constant
pressure. So, we have constant pressure work. And then, also, note that in the
ideal case we have isentropic. I'm sorry, I just totally wrote the wrong thing
there. It's the heat input that's done at constant pressure, right, not the
work. The work is done as the turbine spins, or as a compressor spins, and those
are, in the ideal case, isentropic processes. So, I should've said that we have
constant pressure heat transfer. So, sorry about that work thing. Constant
pressure heat transfer, okay? And in the ideal case, anyway, we have isentropic
work, okay? Now, will talk about both the ideal and the nonideal Brayton cycle,
okay? In the ideal case, well, it's isentropic work. In the nonideal case, what
do you suppose we have to use? Well, isentropic efficiencies for these steady
flow devices. This is something that was covered in chapter nine, I guess. No,
chapter seven. Right, chapter seven, of your thermal book. So, we're going to
have to figure out, not just how to deal with the ideal case, but also the
nonideal case, and we'll get to that here shortly. So, with all of this in mind,
let's now show the thermodynamic property diagrams. Okay, I'll make it easy on
myself. Okay, so, we have a T-s and a p-V diagram that should be shown. And
we'll start with a T-s diagram. So, in a typical Brayton cycle, and, again, this
is going to be our simple ideal case, we'll start here .1. So, this is going to
tend to be the lowest pressure and lowest temperature portion of the cycle. This
is just the ambient air in the real world, right, that comes into the engine. In
our case it's down here at .1. And now we go through isentropic compression,
that brings me to .2. So, again, this is isentropic. I'll just put d s equals 0.
And this process is the work input process, right? So, work input. Why don't I
just put w c now, because we know that the work input is through a compressor,
okay? On the p-V diagram this whole process would go from 1 to 2, like this.
Okay, so, as you compress, certainly the pressure is going to rise, but also the
volume is going to shrink. So, that would be the first process. Now, we have our
heat input process. So, the heat input is done at constant pressure, although
being done at constant pressure doesn't mean that it's done at constant
temperature, right? It's a constant pressure process, but as we add heat the
temperature is going to rise. So, we get to .3. This can also be shown as a
horizontal line on the p-V diagram, right? Constant pressure heat input. So,
here's our q in. And, now, in the ideal case, again, we go through the expansion
process through the turbine. And I often call it expansion. It is expansion,
right? The volume is increasing. The gas, in this case, our air, is actually
expanding. So, turbines do cause the air to expand and produce work at the same
time. Anyway, so we go from 3 to 4. This would be our work output from the
turbine. So, again, I'll just put w t for turbine. And, again, this is
isentropic, so I'll just put a d s equal 0. In fact, maybe on the line from 2 to
3, just for clarity, why don't I just put d p equals 0 since there is no
pressure change from 2 to 3. I mean, that's obvious just looking at the p-V
diagram, it's a horizontal line, but, still, I'll put it on the T-s as well. All
right, so, back to the p-V diagram. The isentropic process allows the pressure
to drop back down to the original pressure at .4. And then last, but not least,
we have our heat rejection, our heat output. So, this goes from 4 to 1. This is
also done at constant pressure. So, by the way, when we look at the real world,
there's nothing that our exhaust from this engine is going to do to change the
pressure in the environment. I mean, the environmental pressure is the
environmental pressure. So, clearly, that heat transfer process is done at
constant pressure. Nonetheless, from 4 to 1 we have another horizontal line. And
that's what the process would look like on these two diagrams. Now, again, I
could show the various work terms, compressor work input, turbine work output,
heat input, heat output. But, nonetheless, these are the diagrams that you would
generally use, you know, to help illustrate these types of cycles. Now, what
about the thermodynamic efficiency? Like all problems involving heat engine
cycles, it's a thermodynamic efficiency that's really of interest to us. At
least that's one of the things that's of interest to us. This gives us our best
measure of how good that engine is at turning heat into work. I mean, that's
thermal efficiency, right? It's the network over the heat input. So, the more
efficient the engine, the more heat is being turned into work, and that's best
for us, right? We're able to sell more work to our customers. So, perhaps, we
make more profit. Or, we can think of it as buying less fuel from our suppliers,
so we have lower expenses. But that's the whole idea behind efficiency, right?
The higher the thermal efficiency, the more economical your engine is, and
you're trying for that, right? I mean, you're engineers. You're trying to
minimize what your clients are going to have to pay for the work that they're
producing. You're trying to maximize their benefit. In other words, the work
that they're able to sell to their clients. So, this all makes sense. So, what
would the thermal efficiency be? Well, it's no different than the thermal
efficiency for any other heat engine. It's just the network over the heat input.
And, of course, the network is just the difference between the work out of the
turbine and the work into the compressor. And, again, divided by the heat input.
So, this is something that we would, you know, be using on a regular basis.
Also, let's note that this could, again, be written in a variety of ways. You
know, if we have, you know, capital letters, in other words, if we want the
total amount of work by the turbine less the total amount of work by the
compressor over the heat input, that will be an appropriate equation. Or, it
could be written as a rate equation, right? We can have the rate of turbine
work, so w.t, in other words, the power produced by the turbine, minus the power
required by the compressor, w.c, divided by the rate of heat input. And, quite
frankly, it's this final form of the equation that is very often used. I mean,
after all, these are steady flow devices, right? There's a certain flow. In a
real engine there's a certain flow moving through it, right? We have a certain
rate the fuel is being added and, thus, burns. So, we have a certain rate of
heat input. We have a certain rate that the air is flowing through the engine,
thus we have a certain power. Rate of doing work is power, right? So, this form
of the equation is very common. And, of course, this form up here on a per unit
mass basis is also very common. So, hopefully, you all remember this from your
previous thermal class. One other thing I just want to define as we start
looking at some of these equations is a ratio that's different than the
compression ratio. When we looked at compression ratio, that applied
specifically to these reciprocating engine cycles. But now we're actually
interested in a pressure ratio. We know that there's a certain maximum pressure
corresponding to heat input and a certain minimum corresponding to heat
rejection, so that pressure ratio is important to us. It's something we use all
the time. So, we'll just define r p as the maximum cycle pressure over the
minimum cycle pressure. And you can see that's either going to t 2 over -- I'm
sorry, p 2 over p 1, or it's going to equal p 3 over p 4, okay? But this is a
pressure ratio -- Okay, it's not a volume ratio, and we need to just recognize
this. This is one of the big differences between our reciprocating engine
cycles, right, where we always use the volume ratio, called compression ratio,
and this, the Brayton cycle, which uses pressure ratios, okay? All right, so,
with all this in mind, what are the equations now for the work and the heat
transfer terms? And I'm just going to show everything on a per unit mass basis.
First, I might just note that these are steady flow devices, right? So, for a
steady flow device. Let's also note that there's not going to be any kinetic or
potential energy change to deal with. Like most engine cycles, we purposely keep
the velocity down. You know, if the speed gets above about 10 feet per second,
which is, what, 3 meters per second or so, then you'll find that even a gas like
air can start to erode the inside services of your engine, including your
turbine and compressor blades, and we just don't want that to happen. So, we
purposely leave the speeds on the low side. As such, we're going to have
essentially no change in kinetic energy. And, for that matter, you know, these
engines are generally horizontal in their orientation. There's no significant
height difference between the inlet and the exit, so there's no potential energy
change. And, of course, in the ideal case, the work that is done is done
adiabatically reversibly, in other words, without any heat transfer, right? So,
in an ideal case there's no heat transfer with the work terms. And, for that
matter, when we talk about the heat transfer terms, the heat transfer process
would also be done with no work being done, okay? So, there's no work during
heat transfer. So, no work during the heat transfer processes. So, what does
that give us with regard to our first law of thermodynamics? Well, these are
steady flow. They're all single stream steady flow, I might note. So, our basic
equation is heat transfer minus work equals change in enthalpy, right, plus
change in kinetic energy, plus change in potential energy. But, for the heat
transfer, there's no work nor kinetic or potential energy. And for the work,
there's no heat transfer nor kinetic and potential energy, so the terms just end
up being functions of enthalpy alone. So, again, this was discussed, hopefully,
at length in your previous thermal class. So, we would note that the heat input
then is just going to be the difference in enthalpy between 2 and 3, so h 3
minus h 2. The work input, which is our compressor work input, is just going to
be the enthalpy change from 2 to 1, really from 1 to 2. By the way, another
reminder, that all of these heat transfer and work terms are all magnitudes,
that is, they're all absolute values, just like they were previously. Again,
we're still talking about the same air standard cycles. I mean, it's a different
cycle, but the same air standard kind of assumptions. Nonetheless, so, we have
our heat transfer. We have our work input. We know that work input is really
negative, just because of our sign convention, but, again, this is the
magnitude. We also have our work out, which is our turbine work. So, this is
just going to be the enthalpy change from 3 to 4, so h 3 minus h 4. And then, I
suppose, if we really wanted to we could just show the network. And, of course,
that's just the difference between the compressor and turbine work -- I'm sorry,
turbine and compressor work, so h 3 minus h 4, and then minus h 2 minus h 1.
And, again, please note that the minus sign within the equation takes care of
the actual sign of the compressor work term. H 2 minus h 1 is the magnitude of
the work. The minus sign gives us our direction. All right, so these would be
the general equations we would use. Now, at this point, we actually need to
start thinking about our two different methods of solution, okay? If we had
variable specific heats -- Well, then we just use the equations that are
presented here on the board. So, just use the equations above. And, of course,
the data you're going to get is going to come from table a 17, or a 17 e. So,
you're certainly going to use the data from a 17. Let's also keep in mind that
you're going to have to use relative pressures now. Remember, that when we were
looking previously at the reciprocating engine cycles, again, Otto and diesel
engine cycles, if we used the method of variable specific heats, then we used
relative specific volumes because we were always given compression ratios,
right, which are volume ratios. So, we had to recognize that for an isentropic
process the ratio of volumes equals the ratio of relative specific volumes.
Here, it's the Brayton cycle. We have pressure ratios now, not volume ratios, so
we need to make sure that we use the relative pressures because, after all, for
an isentropic process the pressure ratio is equal to the ratio of relative
pressures. Yes?
>> So, is p r the same thing as r p over here?
>> No, r sub p is a pressure ration. This is the maximum over the minimum. This
is p r. This is our relative pressure. This is the data that exists in table a
17. And, frankly, it's not even a pressure term. They call it relative pressure,
but it's really an entropy function, and it's only a function of temperature.
Nonetheless, it's called relative pressures because the pressure ratio equals
the ratio relative pressure, so we just use the word relative pressures, but
it's not a pressure term. Anyway, we're going to use relative pressures for the
isentropic work processes, okay? In other words, p 2 over p 1 is going to be p r
2 over p r 1. And, similarly, p 3 over p 4 is p r 3 over p r 4. And, by the way,
p 2 over p 1, it is the relative -- I'm sorry, it is the pressure ratio. So, the
pressure ratio is p 2 over p 1, is p r 2 over p r 1, or that pressure ratio is p
3 over p 4, which is equal to p r 3 over p r 4. So, we're definitely going to
have to use this. Now, I will give you an example problem here eventually that
helps illustrate this, but we're not quite ready for that yet. Okay, so, any
questions just in general about this? The other case, of course, would be -- For
the case where we have constant specific heats. Okay, so, if we have constant --
Specific heats, okay, then what we do? Well, we can't really use these equations
above anymore. Here they are. But, certainly, we understand how to adjust them.
We know that an enthalpy change is c p times a temperature change. We just know
that already, so we'll utilize that. So, let's just look at each term in turn.
The heat input, instead of h 3 minus h 2 is just going to be c p times t 3 minus
t 2. The work input in the compressor is just going to be c p times t 2 minus t
1. Again, note, these are enthalpy changes, so we have to use c p, not c v. C v
only applies if we're dealing with internal energies, and we have no internal
energy change terms at all anymore. That's just for your reciprocating engine
cycles. We have our work out, our turbine work. So, this is just going to be c p
times t 3 minus t 4. And, of course, the network is just the difference between
the two. So, we could just factor the c p out entirely and get t 3 minus t 4
minus t 2 minus t 1. Okay. And, of course, the general equation for the thermal
efficiency, that doesn't change. I mean, that's still just a network over the
heat input, so we'll still use that as is applicable. And then something that
should also be noted is that we can modify these equations a little bit in order
to find a simpler version of our equation for the thermodynamic efficiency. We
did something similar when we looked at both the Otto and the diesel cycles,
right? We had an efficiency equation that was in terms of work and heat
transfer, but through manipulation we were able to adjust that equation so that
it was only in terms of a volume ratio term. That's what we did for the
reciprocating engines. Now, we're going to do it and, hopefully, we'll get an
equation that's only and entirely in terms of just the pressure ratio. So, let
me just go through that very briefly. These are the general equations. I don't
want to change those, but I do want to just make a couple of notes here. If we
look at the work out and the work in terms, what I want to do is I just want to
factor a t 1 out of the work in term, and I'm going to factor a t 2 - I'm sorry,
t 4 -- a t 3 -- no, yeah, t 3 out of this work term. So, for work in we get c p
t 1 times t 2 over t 1 minus 1. And here, for t 3, we get 1 minus t 4 over t 3.
Oh, in fact, I think I did not do that right. I actually want to factor out at t
2, not at t 1. So, let's just modify this compressor work term. So, the t 2
comes out, and then we have 1 minus t 1 over t 2. All right, so that's how it's
supposed to be. Now, I'm just going to continue this little derivation below
here. All right, so we have the c p and we have the t term. So, we have c p
times t 2, and now let's just note that isn't the process from 1 to 2 one of
those isentropic processes? Yes, it is. And for an isentropic process, when we
have an ideal gas with constant specific heats, we would note that the
temperature ratio is going to equal the pressure ratio raised to the k minus 1
over k power. So, you know, this should just be below that in your notes, but
let's just keep that in mind, that t 1 over t 1 is going to equal p 2 over p 1
to the k minus 1 over k, right? This applies only for the isentropic process
when we have an ideal gas with constant specific heats. And, of course, p 2 over
p 2 is just our pressure ratio. And then the same thing between points three and
four, right? So, again, this is an isentropic process and, as such, the
temperature ratio has to equal the pressure ratio. So, the k 1 minus 1 over k
and, of course, p 3 over p 4 is the pressure ratio. So, these now can actually
be subbed in to the temperature ratios that are over in those equations. So,
again, isentropic. Plug in above. So, I just plug them right back in here,
right? So, instead of t 1 over t 2, please note t 1 over t 2 is the same as 1
over t 2 over t 1, which, as we can see, is just the pressure ratio to the k
minus 1 over k. And the same thing for the turbine. So, we have c p and the t 3
is factored out. And then, again, 1 minus. And this is the same, right, t 4 over
t 3 is the same as 1 over t 3 over t 2, which is, again, the pressure ratio to
the k minus 1 over k. And now if we would like to, we can simply throw these
into our efficiency equation. So, I guess I need another arrow here. Plug into
the thermal efficiency equation. Okay, so the network over the heat input. What
else has to go here? So, again, what is our net work? Well, let's just plug in
these various terms. You can see that both terms have a c p in them, and both
terms have a 1 minus 1 over the pressure ratio to the k minus 1. So, I'm going
to factor those out of both terms and we'll just end up with t 3 minus t 2 times
1 minus 1 over the pressure ratio to the k minus 1 over k. So, there's our
numerator. And don't want to forget my c p. And then what about our denominator.
Well, it's the same as those first terms in the numerator, isn't it? The heat
input from 2 to 3 is just the enthalpy change, which is the c p times the
temperature change, and can you see that all that cancels and we end up with
exactly what we were looking for, which is an equation that's only a function of
the pressure ratios. So, this is going to be the easiest way for us to calculate
the thermodynamic efficiency for this particular type of situation. Now, again,
keep in mind that this is specifically for the ideal Brayton cycle, okay, not
the nonideal Brayton cycle. Once we move on to nonideal, then everything
changes. And, of course, also, this was -- well, this is only for the case where
we have an ideal gas with constant specific heats. One other thing I might note
-- This pressure ratio to the k minus 1 over k is still equal to the associated
temperature ratio, right? Well, the fact that they could have it up here still.
So, we could just sub in this ratio, t 2 over t 1, so 1 minus 1 over t 2 over t
1 is just 1 minus t 1 over t 2. So, this is another simple equation that we can
use in order to calculate the thermodynamic efficiency. We don't absolutely need
our thermodynamic properties at every state point, we only need either the
temperatures across the compressor, t 1 in, t 2 out, or we need the pressure
ratio across the system, and then we can find the thermodynamic efficiency. Now,
one reason I'm spending so much time going through what I really think of as
mainly review, is because we're going to have to use all of this when we start
looking at modifications to the Brayton cycle, and there are several
modifications that we're going to look at, every one of which is designed to
improve the thermodynamic efficiency. I mean, that's what it's all about again,
right? We talked about that earlier. If we can improve the efficiency then we're
able to get more work out for each unit of heat input, and that's really our
best desire. So, we'll get to that eventually, but not yet. What about the
nonideal case? So, we're still looking at the simple Brayton cycle. In other
words, we still just have a single compressor, a single heat input device,
combustion chamber, a single turbine, a single heat rejection device, you know,
that doesn't change. But the way we analyze it now is going to be a little bit
different, because we're now going to look at the nonideal case. All right, so,
now we're going to look at the simple, but nonideal Brayton cycle. Now, quite
frankly, there's no reason for me to redraw the schematic diagram. I'm only
changing the way that I analyze the cycle. I'm not changing its physical
configuration. I still have our compressor. We still have our combustion
chamber. We still have our turbine. None of that changes. We're just analyzing a
little bit differently. When it says nonideal, basically that means that it is
not isentropic work, okay? And the way we deal with nonideal behavior for a
steady flow device, like a turbine or compressor, is through the use -- through
the use of isentropic efficiencies. So, this is not isentropic work anymore, and
we must use the isentropic efficiency. And, of course, this is going to be the
isentropic efficiency for both those devices. That is, of the compressor, which
is just a to sub c, and the turbine, which we call a to sub t. Okay, so, that's
really the big difference. Now, I do want to redraw the T-s diagram. I'm not
going to bother with the p-V diagram. The T-s diagram really illustrates what
I'm trying to show here. What I would typically do as I draw this is, I would
show the minimum and maximum pressure lines, and on a T-s diagram for an ideal
gas like air, the constant pressure lines angle up in this fashion. So, this is
the low-pressure in the cycle and this is the high pressure. In fact, instead of
using the words high and low, why don't I just say minimum and maximum. So, the
lower line is the minimum and the upper line represents the maximum pressure. We
also know that as we go through the compression process from .1 to 2, we're
going to have to analyze it in two different ways, right? First, we assume that
it is indeed an isentropic process, so we we're going to go from 1 to 2
vertically upwards. And then once we have solved this part of the process, then
we're going to apply the isentropic efficiency of the compressor to determine
the actual discharge conditions from this compressor. So, the actual process
goes from 1 to 2 along the solid line. There's definitely an increase in
entropy, right? We know that if it were ideal it would be like the dashed line,
just a vertical line. After all, ideal would represent isentropic. But this is
not ideal anymore, so the entropy will increase and we'll end up with a state
point 2 above and to the right of state point 1. Now, please note that the
pressure doesn't change. I mean, I know what my maximum pressure is in the
cycle. I know what the minimum pressure is in cycle. I don't need to identify a
different pressure associated with each of these state points, too. It's a
constant pressure at the discharge, right? That's the maximum pressure. It's
just p 2. But what I do need to do, though, is put a subscript on these 2's
because I can't have two different state point 2's on one single problem,
certainly not on one diagram. So, we'll always use 2 s to represent the ideal
discharge for the compressor. S nominally stands for entropy, and we know that
the entropy is constant if we assume that it's ideal, right? Now, again, I've
shown this as a dashed line because that process doesn't exist, right? We're not
actually going through a process from 1 to 2 s. We're analyzing it. We were
assuming that it exists just so we could use the right data in our isentropic
efficiency calculation. But there's no such thing as that process from 1 to 2 s.
That's an ideal process. We're just people, right? We can't make ideal
processes. So, I'll show it as a dashed line. The real line goes from 1 to 2.
And since this is the actual discharge from the compressor, I'm just going to
use a subscript a. Different books write different things. Most books use 2 s to
represent isentropic. Some would just call this state point 2, just 2, or 2 a
like I've done. I prefer a just so we have a distinction between the ideal point
s and the actual point a. All right, so, that's our work input as we go from 1
to 2. So, there's my compressor work. Now, I add heat up to .3. So, here's my
heat input. And now I have the same thing between 3 and 4 as I had between 1 and
2. I will have to analyze it assuming that it's an ideal process as I expand
through the turbine from .3 to .4. This time I'll just write 4 s, right? We know
that would be the ideal discharge, just a vertical line, and hopefully that's
vertical. But, in the real world, entropy increases, so we'll end up with a .4,
the actual .4, which is going to be to the right of 4 s, okay? Still on that
pressure line, right, still on the discharge pressure line from the turbine, the
p minimum line is still at a different point. And then we go from 4 actual back
to 1. Let's not forget to write my turbine work output, and then also heat
rejection, or heat output, from this particular cycle. So, this is what, well,
the T-s diagram would look like -- would look like for this particular case,
okay? A simple, but nonideal Brayton cycle. So, with this in mind, what are the
equations associated with this particular type of process? Well, the equations
aren't a whole lot different. We would still note that the heat input is still
just going to be the heat input from 2 to 3, although I can't just put 2, I have
to put 2 a. So, the actual heat input is from 2 a to 3, so this is just h 3
minus h 2 a, okay? So, these are, again, just the general equations which will
apply to the case of variable specific heat. So, maybe I'll put variable
specific heat. So, if we use variable specific heats then the heat input is just
found by finding the enthalpy change. We know that the work input to the
compressor is also just the enthalpy change. But, again, it's from 1 to 2 a, not
1 to 2 or not 1 to 2 s. So, we'll just write this as h 2 a minus h 1. We also
know that the work out from the turbine is from 3 to 4 a, so that's another
equation that we're going to have to utilize. And, of course, the thermodynamic
efficiency is still just the network divided by the heat input, so it's just a
matter of putting all these various enthalpy terms into the equation below. So,
h 3 minus h 4 a minus h 2 a minus h 1 divided by h 3 minus h 2 a, okay? Now,
certainly, there are ways to adjust this equation. I mean, there's an h 3 minus
an h 2 a in the numerator. I suppose if you wanted to you could just write this
in its alternate form. Just collect the h 3 and the h 2 a terms, that just
becomes 1. And then we're going to subtract from that, well, what do we have
left? H 4 a minus h1. And then in the denominator is still h 3 minus h 2 a. And,
in fact h 4 a minus h 1, isn't that the heat output? And isn't h 3 minus h 2 a
the heat input? And this just tells us that this is the other pretty common form
of the efficiency equation, right? Instead of net work over heat input, another
way to write is just 1 minus heat output over heat input, which is exactly what
this is. So, that makes sense. So, those are the equations we're going to
utilize. Now, again, we're going to have to use isentropic efficiencies. We're
going to have to use, because we're now talking about variable specific heat,
we're going to have to use the relative pressure data. So, you're still going to
have to use that. So, remember, the pressure ratio, which is p 2 over p 1, is
still equal to p r 2 over p r 1, but we had to be really careful here. I can't
write any 2's. I mean, I can put p 2, because I know there's only one discharge
pressure from the compressor. I could call it p 2 a, I could call it p 2 s, I'll
just call it p 2. But, the relative pressure when shown in this context only
applies to the isentropic process, right? I can't use this to go from .1 all the
way to .2 a. This only applies from 1 to 2 s, so this really now has to say that
the pressure ratio equals p r 2 s over p 1, okay? Now, we're going to know the
pressure ratio in most problems. We're going to know the inlet temperature,
because that's just the ambient temperature. So, we can look up p r 1. We could
find p r 2 s just by using this equation, and then once we have p r 2 s then we
could go back into our property tables and the enthalpy at 2 s. So, now we're
going to use p r 2 s in table a 17 to get the enthalpy at 2 s. And then once we
have the enthalpy at 2 s, then we just have to go back to our compressor
efficiency. Our compressor efficiency is defined as the ideal amount of work
associated with that compressor over the actual amount of work associated with
the compressor. And, of course, the ideal work is just h 2 s minus h 1, and the
actual work is just h 2 a minus h 1. And you're going to now use this to solve
for what's now going to be your only unknown, which is your enthalpy at 2 a. So,
we'll use this in order to get the enthalpy at 2 a, okay? So, there's your
enthalpy at 2 a term. It's going to be there. And then further note that the
same thing applies to the turbine, right? We're going to also have to use the
fact that the pressure ratio equals p 3 over p 4, which is p r 3 over p r, but
this time let's make sure we write down the 4 s. I don't need to discuss it so
much. I mean, we talked about it over here for the compressor, now we're looking
at it with regard to the turbine, but this would be the ideal discharge from the
turbine. And then we're going to our p r 4 s, this is going to be the unknown
from this equation, right? We'll know the pressure ratio. We'll typically know
our temperature at .3, that represents the maximum temperature in the cycle.
That's the temperature achieved by the combustion process. It's a designed
temperature limit that the designers will definitely know in designing the
combustion turbine, which is another name for these gas turbine engines. So, t 3
is always known to us, right? You don't want to exceed t 3. If you do, you might
melt the entire engine, or, these days, a lot of engines are made out of
composites. They don't really melt, they just essentially turn to dust and
disintegrate. We don't want that. Anyway, we're going to use p r 4 s in order,
that is in table a 17, in order to get h 4 s. And then we have our turbine
isentropic efficiency. Now, keep in mind that the definition of isentropic
efficiency for a work output device, a turbine, is the inverse of that for a
compressor. It's the actual work over the ideal work. So, the actual work is
going to be h 3 minus h 4 actual, the ideal work h 3 minus h 4 s for ideal. And
we already know from above h 4 s, we'll know h 3, there's only one unknown,
which is the actual enthalpy coming out of that turbine. So, we use this in
order to get h 4 a. So, now we have all our enthalpies and we can now find the
thermodynamic efficiency, or, really, any other data or properties that we need.
So, any questions at this point? Okay. Which really then just leaves us with one
more special case, which is now constant specific heats. Okay, so, let's note,
first of all, that no need to change my diagram. No need to change the schematic
diagram or the realistic diagram. We're only changing the methods of solution,
so nothing really changes. Also, let's keep in mind that these general equations
from variable specific heats came from the first law of thermodynamics. They
still apply. The only difference is, of course, we're going to replace the
enthalpy change with c p times the temperature change, just like we did a few
minutes ago for, well, for the ideal cycle, right? Nothing is changed. So, let's
just rewrite some of these terms. The heat input is now going to be c p times t
3 minus t 2 a. The work input, which is our compressor work, is going to be c p
times t 2 a minus t 1. Sometimes you can put a subscript a next to this, I mean,
like I've done over here, if you just want to remind yourself that this is the
actual work. It's not absolutely necessary, but it doesn't hurt, so w c a. The
work output is the work produced by the turbine. Again, this is the actual
turbine work. So, this is c p times t 3 minus t 4 a. And then, of course, the
equation for thermodynamic efficiency isn't going to be really any different.
So, the thermal efficiency is still just the network over the heat input. In
other words, the compressor -- I'm sorry, the turbine minus the compressor work
over the heat input. So, again, we could just plug in all these different terms
from above. Please note that you can have a c p term in all terms in the
numerator and denominator, so they're just going to cancel out directly, and we
just get t 3 minus t 4 a, and then minus t 2 a minus t 1. And then just divide
this by, well, it would be the heat input term, the denominator, t 3 minus t 2
a. Now, again, if you want to, you can see there's a t 3 minus t 4 -- I'm sorry,
a minus t 2 a term in the numerator as well as the denominator. So, if you
wanted to you could manipulate this a little bit, but, frankly, generally, we're
not going to. So, why don't I just leave it in this particular form, and that's
then how we would find that the thermodynamic efficiency. Now, you might wonder,
well, is there a simplified version of this equation that would allow me to find
the efficiency only as a function of the pressure ratios? And the answer is
simply, no, you can't do it that way, okay? These are nonideal processes now. We
absolutely need to find the temperature, the actual temperature, leaving both
compressor and the turbine, and we have to use those temperatures directly in
these various equations, okay? And then one other thing just to point out, in
the same way that we used relative pressures to deal with the isentropic -- I'm
sorry, to deal with the isentropic efficiencies of both compressor and turbine,
we're not using table a 17 anymore, right? These are ideal gas constant specific
heat problems now, so we're just going to have to use the equations, right?
Again, the same kind of equations that we just looked at a few minutes ago. So,
again, let me just make this note. So, the temperature ratio from 1 to 2 is
going to be equal to the pressure ratio raised to the k minus 1 over k power.
Okay, and, in fact, this isn't quite right yet. We can't just leave .2 hanging.
It's either 2 s or 2 a, and, clearly, this only applies to the isentropic
process, right? I mean, this this equation was specifically derived for an
isotropic process involving an ideal gas with constant specific heats. You can't
use it for the actual process. So, this is actually t 2 s over t 1 equals a
pressure ratio of the k minus 1 over k, okay? So, this equation will allow us to
find -- T 2 s, right? And then we want to use the compressor's isentropic
efficiency equation. Now, again, the isentropic efficiency is the ideal over the
actual work, but those work terms are really just enthalpy change terms, right?
We know that now. And, of course, we know that the enthalpy change is a c p
times a temperature change, and we have c p in numerator as well as a
denominator. So, the c p's are going to cancel, and this is only going to be a
function of the temperature. So, this is just going to be t 2 s minus t 1 over
the actual temperature at .2 minus t 1, and we use this in order to find t 2
actual. And then once we have t 2 actual, well, that's something that we're then
going to use directly in our equations, right, either for work or heat input or
thermodynamic efficiency. And, similarly, for our turbine, we know that t 3 over
t 4 s is going to equal the pressure ratio raised to the k minus 1 over k power,
right. So, we're going to use this in order to find t 4 s, and then we use the
turbine's isotropic efficiency, which is now the actual over the ideal work.
And, of course, the actual work is just the enthalpy change, and enthalpy change
is c p times the temperature change, and the c p's cancel and everything I
mentioned just a moment ago. So, this then just becomes t 3 minus t 4 actual
over t 3 minus t 4 s. And we use this to find t 4 actual. And then having t 4,
well, again, that's the last thing we really need and we can calculate any other
terms that we need, work, heat, efficiency. And, of course, any of these terms,
or, I should really say, all of these terms, we can multiply the mass flow rate
by the q or the w terms, right? The q or the w terms are lower case, right?
They're per unit mass. So, if we simply multiply them by the mass flow rate,
well, that gives me the heat input rate, right, heat per unit time or work
production rate, that is power per unit time. So, every one of these equations
we talked about today, that is the q and the w equations, can certainly be
multiplied by a mass flow rate in order to get power or rates of heat transfer,
which is, again, another thing that we would often see in these various
problems. So, I'm out of time. Next time I'll go through an example problem. And
have a good one. Please don't forget to bring your homework with you next time.
Also, I do have a couple of old homework sets here. Mumby [assumed spelling] and
Sargecian [assumed spelling]. And, also, remember I've assigned new homework for
the week, so don't forget to write it down. Anyway, see you guys on Wednesday.
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