>> All right well, good morning everybody. So, as we begin today, first just a
reminder that the homework put on the board last week is due on Wednesday, so
please don't forget to bring that with you next time. By the way, it doesn't
have to be on engineering paper, you know, whatever paper works good for you.
These days, to be environmentally conscious, you could even use recycled paper.
Just don't write on the side that has other writing on it. But, otherwise, you
know, any kind of paper is fine for turning in the homework. Here's the new
homework assignment for the week, so this is for the entire week. These will be
due next Wednesday. And you know the ones with an asterisk, I just want to note
that I do want you to solve them using variable-specific heat, regardless of
what the problem statement says. You know, the ones with the asterisks, please,
variable-specific heat. And we did talk about that a little bit last time.
Certainly, we'll talk about that more as we get into the new material this week.
Nonetheless, and then lastly, just a reminder that you will have to show me
proof of prerequisites. Everybody who took me last quarter, I already marked you
off. The rest of you, many of you have already shown me. It's only about ten
people or so left who need to show me that proof. So just don't forget. Please
bring up proof that you've made at last a C minus grade or better in ME 301.
Also, let me just note that there's actually one spot left in the class. The
waiting list has been cleared now, and we're down to 44. So if anybody is still
trying to add, first come, first serve on [inaudible] Direct I suppose. Today's
the last day to add. So please make sure you take care of that right away. Some
of you may know somebody who's trying to get into a thermo class this quarter.
Send them a text. Let them know, again, whoever signs up first will get that
spot. So, with that, let us just move on and continue talking about some review.
Please note that most of what I went over last week was just review, right. And
today is going to be mostly new material, but just a couple last things to
finish with regarding this review, and this deals with cycles. We are going to
be looking at cycles, gosh, for at least the first month or so of this class.
So, some sort of a review of thermodynamic cycles is important. Let's note
please that there's two cycles that we've talked about previously. One being the
heat engine cycle, and the other being the refrigeration cycle. We'll talk about
the refrigeration cycle shortly. We'll start with the heat engine. Just keep in
mind that if we have a heat engine, we're going to be adding some heat at high
temperature. And the purpose is then to produce a certain amount of net work
output. Now, we know that from the second law of thermodynamics, well we can't
do this all by ourselves, right. To turn heat into work requires rejection of
heat simultaneously, so we also have to show our load temperature, heat
rejection at this point. Keep in mind that H stands for high or high
temperature. This QH represents the input of heat. And L represents low or low
temperature, and QL represents the heat that's rejected or the heat output if
you will from the cycle. Note that the net work output is going to be the
difference between whatever work is produced by the cycle. Last, whatever work
is required by the cycle. So, with this all in mind, it's the thermodynamic
efficiency that we use as our performance parameter of interest, certainly the
higher the efficiency, the better the heat engine, right. The higher the
efficiency means that we get more work per unit of heat input or perhaps we
could think of it as requiring less heat input for each unit of work. And that's
really what we want, right. We're trying to maximize this. So, this is just the
net work output divided by the heat input. Now please note that this can be
written in any of our three different forms, right. We can write this on a per
unit mass basis. These are all capital letters over here. On a per unit mass
basis we would have lowercase work net output divided by the heat input,
lowercase qh, so you know, whereas these terms are all, you know, kilojoules or
BTUs. These are kilojoules per kilogram or BTUs per pound. We can also take the
time derivative, that should be an equal sign. We could also take the time
derivative and note that the efficiency is also just the rate of net work
output. We call that power. Divided by the rate of heat input. And we use the
dot notation. Of course, the dot represents the first time derivative or, if you
will, per unit time data. Now, let's also understand that over a cycle, the
total amount of heat transfer is also going to equal the total amount of work. I
mean this applies to any cycle. If this wasn't true, then it wouldn't be a
cycle. I mean keep in mind that as you go through the cycle, you have to begin
and end at the same thermodynamic state. And the only way for that to happen is
if the net heat output equals a net work production. So, this certainly applies
to all cases. We'll note in this particular case, if we're interested in the
thermodynamic efficiency, then the net work output, which really represents
right-hand side, would simply have to be equal to the difference between the
heat input and the heat output, which would represent the left-hand side. So,
using the dot notation here, in other words, the final version of this
efficiency equation up at top, the net work is just the difference between Q dot
H and Q dot L, and divide it by Q dot H. And now this term could certainly be
modified. We can just write it as one minus Q dot L over Q dot H. We could
actually go through exactly the same thing on a per unit mass basis or a total
basis. We could also show that this is just one minus lowercase ql over
lowercase qh. Okay? Lastly, let's also note that the net power output, again, is
the difference between two terms, right. There's going to be some work output
that is associated with this cycle. So this will be W dot out. And then, minus
whatever the work input, again, on a per unit time basis, we'd say the net power
output. I'm sorry, the power output minus the power input. And this is therefore
just the net power output. By the way, I could put out here, and let's even that
W up. Okay. So, this is just something we should keep in mind. And when we start
looking at actual cycles, you know, we'll see that there's specific portions in
the cycle or specific devices within the cycle that are producing the work
output versus the work input. So of course, we're going to get into this in more
detail as we get into real cycles. I mean, for instance, if this were a cycle
based on single stream steady flow analysis, for instance, if this were a gas
turbine cycle like you should have covered in your 301 class, you know, there
would be a certain amount of work output from the gas turbine. But there would
be a certain amount of work input from the compressor. So it's the difference
between the two work out minus work in. This can be the net output. And that's
what has to be used in the efficiency equation. All right so, what about a
refrigeration cycle? What would that look like, and what would be the equations
of interest? So, refrigeration cycle now. So, it's still a cycle. I mean we
could show this just as a box with a circle in it. But a refrigeration cycle is
different, right. The purpose of a refrigeration cycle is to take heat out of
something that is cold and transfer it into something that's hot. That doesn't
happen in the real world all by itself, right. In nature, heat is transferred
from the hot things and the cold things, not the other way around. So, when we
operate a refrigeration cycle, we're actually taking heat input from a low
temperature source and then we're able to reject heat output into a high
temperature sink, okay. So in other words, we're taking heat out of the food
that we're trying to keep cold in the refrigerator, right, and we're dumping
that heat into the kitchen. We generally don't care how much heat is transferred
in the kitchen. We feel it, right. We step up in front of the refrigerator. We
feel the warm air blowing across our toes. That's the warm heat output from a
refrigeration cycle. We could have also used an air conditioner to make the same
statements. Nonetheless, this is a typical refrigeration cycle. But again, the
second law of thermodynamics doesn't allow us to simply transfer heat from
something cool to something hot, right. In order for this to work, we have to
have some sort of work input. So, we need to show work input here as well. And
that's going to be part of, obviously, the refrigeration cycle. Now, for a
refrigeration cycle, we don't use the thermal efficiency anymore, right. The
thermal efficiency or what we also just call thermodynamic efficiency only
applies to heat engines. For a refrigeration cycle, our performance parameter is
called the coefficient of performance, just COP. And the coefficient of
performance is simply the desired heat transferred divided by the required work.
So desired heat transfer divided by the required work. And of course, this is
work input. Okay? So this is our coefficient of performance. Now, I will note
that there are different desired heat transfers depending upon the nature of the
refrigeration cycle. In fact, there's two broad categories of refrigeration
cycles, the first would apply to things like a refrigerator or a freezer or an
air conditioner. In this situation, the desired heat transfer, the numerator, is
the amount of heat that we're able to pull out of the space that we're trying to
keep cold, right. I mean that's all we really care about in our fridge, in our
air conditioner, is just keeping cold, right. We don't really care how much heat
gets rejected out into the environment. We're concerned with how much heat we
can pull out of what we're trying to keep cold. So in this particular case, we
use a subscript R, nominally, that's for refrigerator. But again, it still
applies to freezers, air conditioners, ice makers, that kind of stuff, chillers
used on big buildings. So, [inaudible] performance is going to be the desired
heat transfer, which of course, is going to be the heat input from the low
temperature source, right. So this is just going to be Q dot L. And then divided
by the net work. Okay, now I'm writing this on a per unit time basis. It doesn't
have to be that way, of course. We could write it on a per unit mass basis. We
could write QL over lowercase work net. It really just depends on the problem at
hand and what kind of data we have. But certainly, this is the equation we would
utilize for a refrigerator, freezer or air conditioner. Now this can certainly
be manipulated. I'm not going to go through the proof. This was done in your
earlier thermodynamics' classes. But this can also be shown to equal one over,
and then here we would have Q dot H over Q dot L minus one. So this is just an
alternate form for, again, on a per unit mass basis, we can show it as one over
lowercase QH over QL minus one. So, this is the performance parameter of
interest. Now, again, we're trying to have this number as large as possible,
right. What we're trying to do is maximize our benefit, which is the amount of
heat transfer from the cold space, and we're trying to minimize what we have to
pay for. I mean what we pay for is typically electricity to spin a motor which
is spinning the compressor that's your work input term here. So we certainly
want to minimize the denominator and maximize the numerator, and that's the
nature of coefficient of performance. I mean just like it was for thermal
efficiency, right. This is a performance parameter. We want it to be as big as
possible. So, here's a performance parameter for refrigerators, freezers, air
conditioners. However, there's another type of device that's called a heat pump.
Honestly, we don't use heat pumps that much here on this part of the United
States. We have lots of natural gas. It's cheap, it's already in pipelines. It
goes right into your house. And it's actually cheaper typically to heat up a
home using natural gas as burning than to use a refrigeration cycle where the
heat that's rejected is rejected into the house. Nonetheless, a heat pump is a
refrigeration cycle, but what you're effectively doing is taking heat out of
cold outside air. You're essentially refrigerating it, keeping it even colder,
and then the heat that gets dumped out of the cycle at higher temperature is the
heat that gets dumped into your house. So some of you may have, you know, a
single air conditioner/refrigeration unit that's in your apartment or your
house. It also is going to operate as a heat pump in the winter time, right. Air
condition in the summer, heat pump in the winter. All you really have to do is
change the location of a few valves, and you just basically get the thing to
dump the heat out into your house instead of dumping the heat out like you do in
the summertime out of the house. Nonetheless, for a heat pump, it's clearly QH,
which is our desired heat transfer. So the coefficient of performance, and we
use the subscript HP just to make the distinction between that of a
refrigerator, freezer, air conditioner. This is just going to be the rate of
heat rejection at high temperature divided by the net power input. Or again, on
a per unit mass basis we could just write it as lowercase qh over work net. And
again, this term or this equation can also be ever so slightly modified, and we
would get one over one minus, and this will be QL over QH. Or again, I'll write
it as a rate, Q dot L over Q dot H, or [inaudible] one over one minus lowercase
ql over qh. Okay? So this is the equation you would use then for a heat pump.
It's still a refrigeration cycle. There's no difference really in the way that
we analyze this cycle, whether it's a heat pump or a refrigerator, freezer or
air conditioner. We just simply use a different equation because we have a
different desired heat transfer. So any questions on any of this? All right
great. Let me also just make a note here, and I believe I mentioned this last
time, in all these various equations associated with cycles, whether it's
efficiency or coefficient of performance, all the terms that we're using are
absolute values or magnitudes, if you will, okay. You could see that the minus
sign in here includes the rate, not the rate, includes the direction of the heat
transfer. This is heat input. We subtract from it the heat output, okay. So Q
dot L is a positive term. It's a magnitude. Q dot H, of course, a positive term.
So the minus sign appears within the equation. Same thing here with the
coefficient of performance for these various refrigeration cycles. All the terms
are magnitudes, you know, it's the magnitude of the heat input. It's the
magnitude of the heat rejection. Anyway, so just as a reminder. And of course,
this was hopefully made clear to you previously in your first course in
thermodynamics. So this actually then ends the review, and we're about to begin
new material. So any questions on anything we've talked about these last couple
of days? Anything confusing to you? All right. So, now we begin gas power
cycles. Now, in ME 301, you should have covered the Brayton cycle, at least the
simple ideal Brayton cycle, maybe even simple non-ideal Brayton cycle, which is
a gas power cycle. But that's not the one we're going to start with. We're here
in chapter, what, nine now I guess. And in chapter nine I just want to go
through the material in the same order that the author has gone through it in
the textbook. So, we're actually not going to start with the Brayton cycle.
We're actually going to begin with the auto and the diesel cycles. Before I even
get into that, let's just in general talk about gas powered cycles so that we
understand what is the nature of a gas power cycle? So the first thing that
should be noted is that we're talking about heat engines. These are heat engine
cycles, heat input, work output. It's a heat engine cycle. I also want to note
that for this particular category here in chapter nine, we're specifically
talking about a gas being the fluid that's moving through the cycle. The gas is
doing the work. The gas is having required work put to it. So, this gas power
cycle basically has a vapor as the working fluid. Okay. But there's no phase
change. We're not talking about steam cycles where water would ultimately boil
into steam, which could be used in a steam turbine. We'll get to that, but not
yet. That's going to be a little bit later. But for now, it's a vapor that's the
working fluid. And also in this class, we're going to make a variety of
assumptions to make our life easier. Certainly, these assumptions will, you
know, cost us a little bit with regards to accuracy. But as this is really your
first introduction to these types of cycles, it's hardly worth it to use
complicated cycles. So let's keep things kind of simple. And here are the
various assumptions that we're going to make here. First of all, there's no
friction in the connecting pipes, okay. So for instance, let's say, well even
the word connecting pipes is a little bit misleading. But I'll put it this way
anyway. So let's assume we were talking about a Brayton cycle with this
illustration. We know that we're going to have compression in a compressor, and
then the gas is going to travel some distance. And then we'll have heat input in
the combustion chamber. And then the gas will travel some distance. And then we
have work output in the turbine. [Inaudible] is no friction in these connecting
pipes that is in between the individual devices, okay. So maybe I should really
write it that way. Eh, I'll just leave it like this. So there's no friction. And
what this really tells me is that the thermodynamic state leaving one device is
exactly the same as the thermodynamic state entering the next device, okay.
We're not losing anything. So, that's what this really means. So the state
leaving one device equals the state entering the next device. Okay. And if you
want, you could even use the word process as opposed to device. Perhaps that
would be another way to think about it. You know, in the heat transfer process,
we leave that process. We go into a work process. We leave that process, go into
heat transfer process. So it stays the same as we leave one and then enter the
other. Okay. Let's also note that everything is well insulated. In other words,
no heat losses. Okay, so we know that in the real world, things are not always
well insulated. And we may have heat losses into the surrounding. The turbine
might have heat rejected from it, just loss of the surroundings, the compressor,
etcetera, etcetera. We'll assume that everything is insulated very well, and
therefore, there is going to be no heat losses. Okay. Another thing we might
note here is that we're just going to assume that air is the working fluid.
Okay. So, not only are we ignoring any kind of phase changes, which of course,
there won't be for these cycles anyway, as we talked about above. That is, it's
always a vapor. Here we'll note that that vapor is always going to be air, okay.
So, this is typically just called an air standard cycle. So, we're only going to
be dealing with air standard cycles. I might note that, you know, even though
this does, again, well I guess I said this before, this does cost us some
accuracy, but this is really the best way to learn the material. Let me also
note that as we're dealing with air, you know, sometimes we're going to assume
that air acts as an ideal gas with constant-specific heat. Sometimes we'll
assume that the air of an ideal gas is variable-specific heat. But anyway we
look at it, not only is air going to be the working fluid, but also, air is
going to be treated as an ideal gas. Okay. So this also simplifies things quite
a bit. But it does help us learn the material, okay. Another thing we would note
is that the air moves in a closed loop. Sometimes we just call this a closed
cycle. Okay. Now, is that really happening in the real world? No. I mean in a
real gas turbine engine, we don't have anything moving really in the loop,
right. We take fresh air and we compress it. We burn it. We let those high
temperature combustion gases, of course we're treating as air, do work, and then
we exhaust right out into the environment. Then we start with some fresh air all
over again. That's not what we're going to assume in this class, just like you
didn't assume this in your ME 301 class. And this is not only going to apply to
the Brayton cycle, which is the cycle, obviously, for the gas turbine. But it's
also going to apply to the cycles we're going to analyze for internal combustion
engines. Like that's really the first thing we're going to cover. We're going to
start talking about that today. Now even an internal combustion engine, it's not
really a closed cycle, is it? We take fresh air into the intake valves of the
engine. We compress it by the action of the piston. We burn it as the spark plug
ignites the mixture. Work is done. And then we open the exhaust valve and we
expel the exhaust gases. That's what's really happening. But not in this class,
okay, not in your first class in thermo, at least your first class dealing with
cycles. Air moves in a closed loop. There's no exhaust process. There's no
intake process. Okay. It's just air moving in this closed loop, okay. And
furthermore, we would like to note that there's no real combustion. Okay. What
we do is we basically just assume that there is heat exchange equal to the
amount of heat that would otherwise be generated by the combustion process. So
there's no combustion, just heat exchange. Okay. So, these are some of the
things we need to understand with regards to the relatively simple gas cycles
that we're going to deal with. These are all air standard cycles, and again,
this is going to greatly simplify things. Once again, we could either assume
constant specific heats. If we assume constant specific heats with the specific
heat taken at room temperature, we might often call that the cold air standard
assumption. Or we could use variable specific heat. And of course, with variable
specific heats, we use the data that's in table A17. And again, we just started
looking at this last Wednesday. Okay. Now what is the performance parameter of
interest? Well obviously, it's a heat engine cycle, so it's a thermodynamic
efficiency, or again thermal efficiency that we're going to be calculating as we
go through these different gas power cycles. And there are many different cycles
that we're going to get into. So, let me just pause then. Any questions? Let me
just remind you of one other thing. There is a cycle that's called the Carnot
cycle, right. This is an ideal cycle that we cannot achieve, right. It involves
processes that are ideal processes, isentropic processes that we as mere humans
are not capable of producing. Sometimes you look at the Carnot cycle anyway
because it does give us a limit, right. Nothing can have an efficiency higher
than that of the Carnot cycle operating between the same temperature limits.
Nothing can have a higher coefficient of performance than a Carnot cycle
operating between the same temperature limits. We're typically not going to deal
with the Carnot cycle in this class, but you will see some words about it as you
go through this section of the book. For the most part, we're just going to
ignore that. We're interested in the actual performance of a real heat engine
operating on the gas power cycle, not some idealized, you know, cycle operating
between the same temperature limits. So, sure, read about Carnot cycles, but
don't worry about it. You know, you're not going to be required to solve any
problems dealing with that. Plus, you already learned about the Carnot cycle
last quarter, so there's no need for us to really get into it this quarter. All
right so, now we begin the, let's say, initial discussions of the first type of
cycles, and this deals with reciprocating engines. Okay. So when we talk about
reciprocating engines, we're talking about vehicle engines, automobile engines,
truck engines. And in all of these situations, what we have is nothing more than
your good old friend, the piston cylinder device. You studied these quite a bit
in your thermodynamics class, ME 301, and now what we might realize is that we
can actually operate a heat engine cycle based solely on a piston cylinder
device. So what we would have is the cylinder with the piston in it. The piston
is going to move back and forth. The piston is going to move a certain distance
as it goes from the top to the bottom. So, I'm just going to show the same
piston with some dash lines to represent the highest position and the lowest
position. Right here at the very top, we call this top dead center or just TDC
for short. Some books just simply call it top center. And then at the bottom of
its travels, the piston is at bottom dead center or bottom center. And we just
abbreviate that BDC. Okay. So the piston is moving back and forth between top
center and bottom center. Keep in mind that there are certain volumes within the
cylinder associated with these two positions, right. When the piston is at top
center, the volume is actually the minimum volume. And that would be the volume
within the cylinder. By the way, you're never going to have the piston go all
the way up and hit the very top of the cylinder wall. In a real engine, I mean,
you can't do that. There's spark plugs up there. The valves are opening into
that space. There's instruments in the engine that would certainly be damaged if
the piston went all the way to the top. So it doesn't. There's always going to
be a certain minimum amount of volume. In fact, it's really called the clearance
volume. So if you ever read the words clearance volume, that represents the
minimum volume associated with the cylinder in the engine. Okay. At bottom
center, this is the maximum volume. And there's no specific name for the volume
at bottom center. Although there is one other term we should be aware of. It's
called displacement volume. Sometimes we just use the DD for displacement. This
is simply the difference in volume between the maximum position and the minimum
position, okay. So, this is just really the max, which is the volume at bottom
center minus the V min, the volume at top center. And I should note too that
this is the volume that you see printed on the side of all the engines you would
buy. Well, the cars you look at has an engine like a 1.5-liter engine. That's
the displacement of volume that they're presenting, right. It's not the
clearance volume. It's not the total volume at the maximum position that is at
bottom dead center. It's actually the difference between the two, okay. So this
is just another term that you should be aware of. Now, keep in mind again, we're
specifically dealing with closed cycles here. We assume that there's just air in
that cylinder. The air is being compressed and decompressed continually over and
over and over again as a piston moves up and down and we go through a particular
cycle. Again, with the understanding that it's just air. There's no intake of
exhaust processes. There's no real combustion. We're just transferring heat to
the air. We're letting air do some work. So this is clearly a, you know, an air
standard cycle that we're analyzing. Let me also note that the distance traveled
as the piston moves from top center to bottom center is called the stroke. So
again, the stroke is a distance that the piston moves as it moves from top dead
center to bottom dead center, okay. And then also note another term of interest
is what we call the bore. The bore is simply the cylinder diameter. So often,
you would be given data for the stroke and for the bore, and you know, since the
cylinder is just a right circular cylinder, it's pretty easy to use that data,
the bore, in order to find the volume, right. Just pi over four times the
diameter squared. That gives you the area. And times the length will be a
volume. So the bore is nothing more than the cylinder diameter. So again, these
are some of the terms that you should be aware of. Another thing you should be
aware of as we begin talking about this material, is what's called the
compression ratio. So, the compression ratio is R. Please note that the
compression ratio is a volume ratio, not a pressure ratio. As you've covered the
Brayton cycle in your previous thermo class, you often had to deal with the
pressure ratio. This is not a pressure ratio. This is simply defined as the
maximum volume divided by the minimum volume. In other words, it's going to be
the volume at bottom center divided by the volume at top center. So it's really
no more complicated than that. But that is something that we're going to have to
deal with. It is one of those terms that will be important as we try an analyze
our cycle. Okay, so get back to compression ratio. Another term that we need to
talk about is what we call the mean effect of pressure. Now, the mean effect of
pressure is not a real pressure. It's not a pressure that is actually seen in an
engine. It's really just a pressure that's used for comparative analysis as we
compare different engines. What we would do is we would say well, if the net
work that's done by this engine is done at constant pressure, that pressure is
called the mean effect of pressure. So, the mean effect of pressure or we would
just abbreviate it MEP, is the, let's call it the assume constant pressure that
would exist if the net work were done at constant pressure as the piston moves
through the displacement volume. I guess I wrote constant pressure twice. Sorry,
I'm a little bit redundant today. Okay, so at constant pressure as the piston
moves from, well, bottom center to top center or vice versa. Okay. Now, we
should know how to figure that out. So we have a certain net work output. And if
this was done at constant pressure, that pressure being what we now call the
mean effective pressure. Well the work term for constant pressure process for a
closed system is just the pressure times a change in volume, right. So, that's
just going to be the volume at its maximum position, which is bottom dead
center, minus the volume at its minimum position, which is top dead center. So V
max minus V min, okay. In other words, this assumed average pressure, this
performance pressure called the MEP is really just the net work output divided
by V max minus V min, or we could even just write this as the net work output
divided by well, the displacement volume, right. Displacement volume VD is
exactly that difference between the maximum and the minimum volume. So this
again, this is a performance parameter that we will often use to compare
different engines. But it's not like we always use it. You know, engines can
operate at so many different speeds. They can operate at so many different fuel
tare ratios with so much energy being produced. They can operate at so many
average pressures, high pressure, low pressure. Often we use the MEP just as a
way of comparing different engines. Lastly, I might note that if we just divide
by mass both numerator and denominator, then this is going to be the net work
out put on a per unit mass basis. And then the displacement volume on our per
unit mass basis, we can just write this as a lowercase v sub d, or if you will,
instead of the displacement volume, this is displacement volume on a per unit
mass basis. The difference between the specific volume at the maximum minus that
at the minimum position. So divided by the specific volume at the max minus the
specific volume at the minimum position. And again, recognize that V max is the
bottom position, right. That's a bottom center, and V max, I'm sorry, V minimum
is at top center. Okay. So, this is just some of the basics that's going to
apply to all of our reciprocating engines. And now we should begin looking at
different types of engines. Now, there's actually two different reciprocating
engines that we're going to deal with depending upon the type of fuel that we
use. Okay, one type of engine is called a spark ignition engine, which we just
know of as a gasoline engine. And the other type of engine is what we call a
compressive ignition engine, which is what we just know of as a diesel engine,
okay. So those are the two basic engine types that we're going to deal with. And
they actually operate just a little bit differently. The gasoline engine can be
called the spark ignition engine. And it actually requires a spark plug in order
to ignite the air fuel mixture. A diesel engine doesn't require a spark plug.
Just simply going through the compression process of the air as the piston moves
from bottom to top center achieves a high enough air temperature. And when the
diesel fuel is injected into that air, it immediately begins to combust. Well,
maybe I shouldn't say immediately. It takes a fraction of a millisecond. But it
starts to combust essentially right away. And we call this a compression
ignition engine. Okay. And in fact, as we analyze the actual combustion process
and analyze the way in which the heat transfer takes place during that heat
input process, we realize that the two are just a little bit different here as
well. If you have an air fuel mixture, which is actually in a real engine going
to be compressed as a piston moves from bottom to top, as soon as the spark
ignites, the combustion process begins immediately. And the whole thing takes
place more or less when the piston is at top center. So, this is approximately
constant volume combustion. Now, maybe I won't even write the word combustion.
We know there's no real combustion in this class. I mean there's combustion in
the real world. In this class, it's just heat input. So here we're going to have
constant volume heat input representing our combustion process. And for the
diesel engine, what we find is that the fuel injection process takes a finite
amount of time. Yes, you'll begin fuel injection as the piston hits top center,
but it's a fuel injection system. It takes a finite amount of time. The piston
actually starts moving immediately upon the combustion process taking place. And
what we find is that in this type of engine, we have approximately constant
pressure heat input. Now, because of that, we're going to have to analyze these
two different types of engines using two different types of cycles. And of
course, we'll get to those two cycles here in just a couple of minutes. Now,
there are certainly similarities between the two. So before I just break these
two apart and start looking at each one individually, let's just look at the
things that are similar. So, certainly in both of those cases, there's going to
be four processes of interest. Okay. So what are these four processes? Well,
first of all, there's going to be a work input process, okay. In other words,
we're going to have air. If it's a diesel engine, air fuel mixture. If it's a
gasoline engine, again, don't worry about that. We're assuming it's just air for
our analysis. But still, the first process is work input. Basically, this is
where air is compressed as we move from bottom to top center. Okay. You know,
I'm teaching an engine's class this quarter, and there we just use the words
bottom center and top center. Here, I got to remember the D in between. So,
bottom dead center to top dead center, okay. So this is the first process that
we're going to deal with. It's just going to be a work input process. After the
work input takes place, then that's when either the combustion process begins
under the influence of the spark plug, or the combustion process begins as fuel
injection occurs in the diesel engine. So that's our second one. So this is heat
input. Okay. Now, the heat input is always going to begin at top dead center.
And of course, if it's a gasoline engine, it's going to end at top center too,
right. The volume is going to be constant. On the other hand, if it's a diesel
engine, the combustion process is going to take a finite amount of time that the
piston is going to move. So I'm going to say heat input, and I'll just note, it
begins at top dead center with the understanding, of course, that it's not going
to stay at top dead center for the diesel engine, but certainly will for the
gasoline engine. Again, we'll break these out separately, and I'll summarize
this again for you. Now what about the third process? Well, the third process
would be work output, right. The air has been compressed. Heat's been added to
it. We have very high temperature, high pressure gas now inside that piston
cylinder device. The cylinder has a high-pressure temperature gas. It pushes the
piston downwards. So the third process is going to be our work output process.
Okay. So, this is going to occur as we go from, well now, top dead center to
bottom dead center, okay. And then the fourth process is going to be our heat
output, or we can think of it as heat rejection, whichever term you want to use.
Please keep in mind that this is going to occur at bottom dead center. Once the
piston has moved to the bottom, you open the exhaust valve typically and just
let everything just flow out. So, heat output is going to occur at bottom dead
center. It's a constant volume process. Okay. So, these are the basic processes
that we're going to have to analyze within our heat engine cycle. One other
thing I just want to mention briefly, because as I've been describing the engine
and the intake and the exhaust process, it turns out there's actually two
different ways in which we can undergo the intake and the exhaust process. These
have absolutely nothing to do with the thermodynamic analysis. Thermodynamic
analysis, everything we do if you will is only going to be a function of the
amount of work in and out and the amount of heat in and out, okay. But in a real
engine, there's definitely an intake of exhaust process. So I do want to cover
this very briefly, and certainly, it's discussed in your book. But just with the
understanding that it really doesn't apply to any of our analyses, okay. It's
just to give you a better understanding for the nature of different engine
types. And what I'm about to describe will apply just equally to the spark
ignition and the compression ignition engine. In other words, the gas or the
diesel engine. And this just has to do with the way the intake and exhaust
process takes place. So, the intake and exhaust processes could be done two
ways, okay. Now, one way is if we have what's called a two-stroke engine. Okay.
And in fact, as I've gone through the basic discussion, it's really more
representative of the two-stroke engine. You go through the process, right. Your
air is in the cylinder. You compress it. Heat added to it. It does work as it
expands. And then you have the heat rejection process, which really we know is
done through the exhaust process. I mean in the real world we're exhausting
whatever combustion gases are left within the cylinder. And we start all over
again. But that's what I want to talk about here. In a two-stroke engine, you
just simply open the exhaust valve, and the exhaust just shoots out. This all
takes place when the piston is essentially at bottom center. So, the exhaust
comes it. It forces the intake out. There's no separate strokes as the piston
moves up and down. It doesn't. In a two-stroke engine, exhaust and intake occur
at bottom dead center after the work output is done. Okay. Now, these types of
engines are really only used for small applications where weight is important.
For instance, a weed whacker, a moped, you know, those would be pretty typical
of two-stroke engines. A personal watercraft. Although these engines are more
polluting, so some lakes don't allow two-strokes, like Lake Tahoe doesn't allow
two-stroke anymore on the lake because of the pollution. And we're not going to
get into why they're more polluting. But nonetheless, they're very common when
we have weight as our main concern. On the other hand, we could also have a
four-stroke engine. And the vast majority of engines are four-stroke engines.
Your car engines, your truck engines, your boat engines. I mean these are all
four-stroke engines. So what happens here is you actually go through two
separate strokes of the piston. But we still have the thermodynamic cycle,
right. As the engine rotates one rotation, we're still going to compress the
air. We're still going to add heat to it. We're still going to do work as it
moves down the piston or pushes down the piston. But at the end of this process,
we're actually going to go through two more physical strokes. The engine is
going to rotate one more time, and on that rotation, one stroke will push the
piston upwards and force out the exhaust gas while the exhaust valve is open,
and then the four-stroke is where the piston is drawn downwards, and it sucks in
the fresh charge of air through the intake valve. And then you start the
thermodynamic cycle all over again. Now, that's a much more efficient process,
much more efficient at exhausting the combustion gases. Much more efficient at
drawing in the fresh charge of air. But again, from our standpoint, it doesn't
matter. From a thermodynamic standpoint, none of this matters. But again, I want
to make sure you understand it. So here, exhaust and intake occur with two added
strokes. Basically, bottom dead center to top dead center is the exhaust. And
then from top dead center back to bottom dead center, this is going to be our
intake. Okay. So just be aware that these processes do occur, and that's how a
real engine operates. Okay. One last thing to note on this before I move ahead.
The main reason your two-strokes are smaller and lighter is because work is
being done with every rotation of the engine. On a four-stroke engine, work is
only being done with every other rotation, right. In a four-stroke, you have one
rotation which goes to the thermal cycle. And then the next rotation, nothing
really happens from a thermodynamic point of view. We just have the intake and
exhaust process. In a two-stroke engine, not true. You go through one cycle, and
you immediately intake an exhaust. Or really, the other way around, right. You
do the exhaust and intake. And then the next engine rotation, you do another
thermodynamic cycle. Each engine rotation goes through a complete thermodynamic
cycle. So what you can get by with is a two-stroke engine that's essentially
half the physical size of the four-stroke, and even if it operates at the same
speed, even if it operates with the same displacement volume, it's essentially
going to produce twice as much power because [inaudible] power with every engine
rotation rather than power with every other engine rotation. So, it's basically
an engine half the size with the same amount of power, or if you will, twice the
power if it's the same size between a two and four-stroke. You should be aware
of these things. If you're more interested, well take ME 412. It should be
offered every spring quarter. Well, nowadays, spring semester. Right. So any
questions just in general about the nature of these reciprocating engines, how
they're really just piston cylinder devices, and we're just using air as a
closed cycle? Yes.
>> You said the two-stroke and four-stroke engines could be either or of the two
connected engines?
>> Right, right. You could have a two or a four-stroke spark ignition, or you
could have a two or a four-stroke compression ignition engine, right. Okay. So,
that now, as introduction, brings us to our first thermodynamics cycle which is
the Otto cycle. Not auto as in automobile but Otto as in the German scientist
that invented all of this about 150 years ago. So Otto cycle. Now, the Otto
cycle is an ideal air standard cycle. Please note that this is an ideal cycle.
We're not going to be dealing with anything like isentropic efficiencies. We're
going to assume that every process is an ideal process, and that's going to be
the way we analyze this. Again, it's an air standard cycle like all the cycles
we're going to be analyzing here. The ideal air standard Otto cycle is used to
model a gasoline engine or if you will, the sparkedness in engine. Okay. So this
particular cycle, well, it has the strokes we talked about, work input, heat
input, work output, heat output. But now what I want to do is just go into a
little more specifics. So, let's look at these processes on some thermodynamic
property diagrams. And there's really two property diagrams we would typically
use. One would be a PV diagram. And the other would be our good old TS diagram.
Okay. So let's just look at these processes first on the PV diagram. So on the
PV diagram, we would note that the cycle begins at what I'll just call state
point one. This is when the piston is at bottom center, and we're just about to
begin the compression process, the work input process. So at this point,
whatever is within the cylinder is going to be probably close to atmospheric
pressure. The volume is going to be at its maximum because it represents bottom
dead center. So we're going to be way over here somewhere. So that's state point
one. And in fact, as I go through each of these processes, the first process
will be one to two, and this is called compression. So, there's certainly going
to be some work input. So the compression process takes place. The air is being
compressed into a much smaller volume until we get to top center, top dead
center. And of course, during the compression process, the pressure is going to
rise. So we'll end up with a state point two over here somewhere. And let's just
note the volume at both top and bottom dead center are here and here. Now, we
undergo the heat transfer process. Oh by the way, I should not just put
compression. I should say work input and clearly, there's work input taking
place, okay. By the way, I'm just using lowercases W's and Q's and all that. You
can use dots. You can use capital letters. It doesn't matter. Okay, now we have
the heat input process, right. So, this will be state point two to state point
three. I mean in the real world, this is combustion, right. This is where you
burn your air fuel mixture. But for us, this is just heat input. So, this is our
Q in. And this is done at constant volume, okay. This takes place when the
piston is at top dead center. So we have a vertical line as we go from state
point two to state point three. So heat input at constant volume. Okay. Now, we
undergo the next thermodynamic process. This would be from three to four. And
this is typically called the power stroke, or this is where power is actually
developed. So here the piston under the influence of high temperature air, again
combusting gases in the real world. We now push the piston down. Of course, as
we do so, we're creating work. And the thermodynamic state is going to drop. And
we end up with state point four here. The pressure is going to drop. The
temperature is going to drop. The internal engine is going to drop. So this is
our work output term. It looks like I forgot my Q in, so don't forget to put
that Q in over here from two to three. Okay. Anyway, then at state point for,
this is where we have the heat output or heat rejection. So, from four to one,
heat rejection. And again, this is done at constant volume. So we just go
vertically down from four to one. Right. So heat rejection at constant volume,
okay. By the way, I guess up here I could have put work out next to the power
term. Heat rejection at constant volume. We'll just call this Q out. So these
are the four processes that are taking place. Now again, don't worry about the
intake and the exhaust processes. Again, that was just for your understanding.
And it really has nothing to do with our analysis. All right, so a couple of
other things to note here. This would be more evident as I go through the TS
diagram. Remember I had mentioned that this is an ideal cycle. So what happens
to the entropy as we go through the initial compression process from one to two?
Well, it's ideal, right. Work is done, but we're assuming there's no heat
transfer. We're assuming it's a reversible process. It's a very ideal process.
So, the entropy is going to be constant. So state point one, well this is going
to represent the lowest possible temperature in the cycle. So state point one is
over here. When we go through the work input process, it's actually isentropic.
So, the temperature is going to rise as we are compressing the air. But the
entropy is going to remain constant. Okay. So this is our work input, and
perhaps over here, we would note that this is an isentropic process. Isentropic
meaning constant entropy. Now, we have the heat input process. With heat input,
the temperature is going to rise. The entropy is going to rise. And we're going
to end up with a state point three up here somewhere. Now, we're going to go
through the work output process. And again, in an ideal cycle, we're assuming
the work is done adiabatically and reversibly, right. That's what ideal means in
the context of these cycles. So, the work output is another isentropic process,
just like work input. And we're going to end up at state point four, again, at
the end of a vertical line from three to four. So over here you might also want
to note that this process is isentropic as we go from three to four. And then
last but not least, we go through the heat rejection process. Let me show my
other Q's and works. So Q in from two to three. Work out from three to four,
isentropic. And then from four to one, we have our heat rejection or the heat
output. And again, this is done at constant volume. So we know based on our
cycle analysis that we have two constant volume processes, right. We call these
isochoric if you prefer. And then we have two constant entry processes,
isentropic. And we're going to have to use those in order to go through the
analysis of any one particular engine. Okay. So, what I'd like to do next then
is just go back to the third dynamic efficiency equation and make sure that we
understand how to analyze it for this particular type of engine. Again, these
are closed cycles. We have a constant mass of air that's simply continuing to
move through different processes within the cycle. But these are closed systems,
right. So the first law as we utilize it is certainly going to be for a closed
system. All right, so what would my equation be now for the thermodynamic
efficiency. So, let's find the thermodynamic efficiency. Again, this is a heat
engine, so it is a thermal efficiency, not coefficient of performance that we're
dealing with here. We would again note that as we are analyzing these processes
that they're ideal. We're using constant specific heat. Again, I want to
emphasize that we're going to neglect any kinetic and potential energy
associated with any of these processes. So, what would the equation be then for
the thermodynamic efficiency? Now, I gave you various forms of it earlier today,
right. I mean, the net work divided by the heat input is the basic form of this
equation. Again, we might write it as one minus the heat output minus the heat
input. I think I'm going to start using out and in rather than H and L. I mean,
high would be high-temperature heat input. H for high. So I could have that in
the denominator. I could have QL, you know, for low temperature heat rejection.
But I think it might be a little more clearer if I just used in and out at this
point. So this is going to be another version of the equation. I can write it as
total Q out over total Q in. I mean there's just different forms of the same
equation. Now, in this particular case, I have four individual processes, two of
which, of course, are going to have to be included in the thermal efficiency
equation. We have a closed system. There's no appreciable kinetic or potential
energy change. We know that for heat transfer processes, they're going to
[inaudible] constant volume, right. So there's no work, right, for a constant
volume process, no work, no boundary work. So, there's no work. So with all of
this in mind, we can now just substitute in the appropriate version of the first
law. Now, let's just look at this term, you know, using the lowercases. How much
heat is going to be rejected? Well, from the first law, that's just the
difference in the internal energies, right. We know that the equation says heat
transfer minus work equals change in internal injury plus change in potential
and kinetic energy. But there is no kinetic or potential energy change. And
there's no work during these processes of heat transfer because it's constant
volume. So we simply get the difference of internal energy between point four
and point one. So U4 minus U1 is in the numerator. And then in the denominator
is the heat input as we go from two to three, again, with no kinetic or
potential energy and no work we just have an internal energy change from our
first law. So that's just going to be U3 minus U2, okay. So, again, this is just
from the first law for a closed system. And again, assuming that there's no
kinetic or potential energy changes and noting that there's no work. So,
hopefully, everybody can see that. All right, so now what do we want to do with
these particular equations? Well, at this point, we have to, well, decide
whether we're looking at a problem with ideal gas with constant specific heat or
variable specific heat. Certainly, if we use variable specific heat, we're just
going to use the internal energy data that appears in our textbook in appendix
A17, so I don't need to change that equation at all, right. But on the other
hand, if I assume constant specific heat, then we know that an internal energy
change is just equal to CV times the temperature change. So I definitely need to
make some modifications to that particular equation. Before I do it, one last
thing to note, we are at times going to be interested in finding the net work.
Maybe I'm not going to need it immediately, but we certainly know that the net
work is work out minus work in. And again, from the first law of thermodynamics,
we're assuming these are ideal processes. The work is done without any heat
loss. And if we assume no appreciable kinetic potential energy change, well, the
work terms are also just equal to internal energy changes. So, the work out,
that's going to be between three and four. So just U3 minus U4. And the work in
is going to be that between point one and two. So U2 minus U1. And that's how we
would find the work for these particular processes. Now again, if we have ideal
gas with variable specific heat, we're just taking internal energy data out of
our tables, we don't have to modify these equations at all. If we're using
constant specific heats, we have to make a slight change. So let's just briefly
look at that. So if we have constant specific heat, then we would just note that
an internal energy change is equal to CV times a temperature change, right. So
we can go all the way back to these various equations. I mean, we wouldn't be
able to show that the thermal efficiency then is just one minus. And then in the
numerator, we're just going to have a CV times T4 minus T1. And the denominator
will have a CV times T3 minus T2. You can see that the CVs cancel. And we'll
find here that thermodynamic efficiency is only going to be a function of
temperature, okay. Now I can actually modify that equation a little bit further,
and I'll get to that in just a moment. But also, I just wanted to write out the
net work equation, right. Again, we're specifically look at constant specific
heat here, so the internal energy change becomes CV times a temperature change.
And we just get CV and then times T3 minus T4. And then minus T2 minus T1. I've
just factored the CV entirely out and moved it over to the far left-hand side of
the right-hand side of the equation. So, this is something we would note. Now, I
can go back to the efficiency. One thing I'd like to show is that we can
calculate the efficiency based only on the compression ratio. Now unfortunately,
I've just run out of time, so we'll continue this on Wednesday. Certainly,
you're not at a point where you can start doing your homework problems. I mean
some of you could certainly start if you want to read ahead a little bit. But,
you know, we're really not quite there yet. So, we'll continue this next time,
talk about autocycle, go through some example problems, move on to the diesel
cycle. Anyway, see you then. Reminder, don't forget to show proof of
prerequisite. If you have it today, you can bring it up now. Or just bring it up
any time within the next week or week and a half really. So see you all
Wednesday.
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