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കുറഞ്ഞ ചേരുവകൾ കൊണ്ടൊരു simple മുട്ടക്കറി l Egg Curry - Less Ingredients l W Eng Sub.l Recipe No # 7 - Duration: 6:02.
Assalamu Alaikkum.
Welcome to Calicut Flavours,
Today we have Special Flavor.
Today we are going to prepare a special Egg Curry.
For preparing this recipe, only few ingredients required and within very little time we can prepare it.
We normally serve it with Neeriya Pathiri (Rice Pathiri).
But more people have it with our special Kannu Vacha Pathiri. its a good combination and very tasty.
So we will discussed other details later on.
Now, we going to prepare the special egg curry.
Bismillah.
Lets start our cooking, regarding ingredients and measurements are given in the description box, kindly check it.
First we need to grind the coconut.
Here, I took 3/4 cup of Coconut for grinding.
Much better If you could use the fresh coconut.
Add 2 small green chilies.
This can be add at the time of sorting
But, I like and recommend for doing grinding in the first stage itself.
For kids, they like this curry very much. If we do not grind it well, then they spend time for picking the green chilies from the curry.
Add 1/2 Tsp or less Fennel Seed Powder.
Add 1/4 Tsp Turmeric Powder.
Add 1/4 Tsp Red Chili Powder.
Now, with altogether crush the mixture in a small mixer grinder.
Add required water and grind it.
If we do not crush it well, then the green chili will not grind well.
Here, I add 1/2 cup of water to grind well. its not necessary to grind very loose type.
For normal curry making, no need for very loose paste. normal grinding is well enough.
It should be like this format as shown in the video.
We need 4 small Shallot.
Required Curry leafs.
Let start to make our curry.
Start the stove and place a pan and make it hot.
Add 1 Tbsp coconut oil or as needed.
Remember, for making such recipe, kindly use coconut oil always, it will give the right flavor.
You will get the correct flavor and taste.
Now, the oil got little hotter, we need to add chopped small shallots.
We need to sort it well until color change.
Add Curry leafs.
By adding more curry leafs you get more flavor and taste. No worries for adding more curry leafs.
As I said earlier, at this stage you may add the green chilies or spices according to your need.
Also, if you feel and smell the flavor as fresh, then you may add more spices.
Here, I do not feel such things.
Its all I leave to you, if you want to add more spices or Green Chilies up to you.
Now, our curry leaf and small onion get well cooked. it start forming like paste.
It starts smell the good flavor.
Now, add 1/2 cup of water.
Add enough Salt.
Add only little salt for this curry, suddenly the salt flavor increase, so please add very little salt as shown in the video.
Next, we are going to add egg, for that no need for much salt.
So, earlier said now, here you may add little salt and spicy spices.
Here, our curry is boiled well, we need to add the egg.
Now, am adding 2 eggs into this curry like this. Here you can beat or mix well the eggs and add to the curry, its all your choice.
Here, am adding the eggs to the curry. its has to boil very well.
After, boiling only you need to stir the curry with spoon.
If you are using organic or farmed eggs, then no need for more boiling time. we can add the coconut paste for final preparation
If you manage to get normal and organic or farmed eggs, the egg become very fluffy and soft also taste itself very good with flavor.
It taste special taste.
If you can get farmed or organic eggs, it is very good and taste well.
Anyway, I cannot manage to get it here, what I got here is a broiler chicken eggs, so it is very important to cook very well.
Now, our curry is boiled very well, the eggs are cooked well. slowly stir it well.
Now, add the grinded coconut paste with the curry.
Stir the curry well, so that it mix well.
At this point you can taste the salt.
Here, I found salt is less, so am adding a little more salt.
The curry had enough spicy. so no need to add any more spicy spices.
Now the curry got boiled up, no need for boiling too much time.
Now, we can close the flame.
Our curry consistency should be like this, shown as per the video.
After get cool down the curry will get little more thickness.
By this measurement you can prepare for 4 adults.
While you prepare the recipe, please do not reduce any ingredients, because only few ingredients are there, each one very important.
Specially, Curry leafs - curry leafs are not only added for flavor but its a herbal ingredient for health too.
While serving and eating, please have the curry leaf together with egg and gravy, here the combination give a different flavor and taste of feeling.
Here I would like to share one of my opinion that (which I try for my self), When ever you try to make any new recipes, kindly make it in less quantity, so that you will get the right flavor and taste, afterwards you may increase the volume of quantity.
Finally, our simple and delicious Egg Curry is ready for serving.
There is an important thing to be shared that, this recipe was thought to me by my mother, at the time of learning cooking.
The reason that, it is very simple and you can manage the few ingredient very well.
Another reason that, this recipe never fail you among other and very delicious, tasty..
Therefore, who ever learning cooking, they can easily make this egg curry.
Same for Bachelors, who wish to make a nice tasty curry with less ingredient, this is a best option.
Therefore, please try to make this recipe.
Before leave for this time, would like to request you to LIKE this video.
Share this video among your contacts
And encourage them to subscribe my channel.
Your valuable comments, encouragement and support highly appreciated.
In shaa Allah, will see you with another best recipe.
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Simple Macaroni Recipe | How to make macaroni | Masala veg macaroni recipe without chicken - Duration: 4:20.
macaroni
simple macaroni recipe
fms bawarchi khana
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5 Simple Masculine Qualities That Make All Women Fall In Love - Duration: 3:17.
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Суп с чечевицей - супер простой рецепт - Duration: 2:36.
List of ingredients for 4 servings: • lentils - 125 g; • champignons - 100 g; • onion - 1 pc .; • carrots - 1 pc .; • Chicken fillet - 1 pc .; • vegetable oil - for frying; • bay leaf - 1 pc .; • salt; • ground black pepper to taste.
Hello everyone! TodayPriprava Club prepares soup with lentils, chicken and mushrooms
The first thing to do is boil the chicken broth.
For broth, I use chicken fillet, half carrots, half onions, black pepper and bay leaf
The second half of the carrot is finely chopped or grated
Crush the second half of the onion
Fry the onion and carrot
Cut the mushrooms and add to the onions
Bouillon remove from heat, remove the onions and carrots
Slice Chicken Fillet
Add the fillet to the broth
Add onions, carrots and mushrooms
Bring to a boil and add lentils.
The cooking time of lentils depends on its grade and can last from 10 to 40 minutes.
Some varieties need to be pre-soaked.
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How To Make Simple And Professional LOGO ( In Easy Method ) Photoshop Tutorial - Duration: 5:14.
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One Simple Advice for You to Succeed in Math - Duration: 0:54.
So, after teaching mathematics over the past several years to a number of students,
I have observed one character, one trait, one habit that separates the
students who succeed in math, get excellent grades and the students who
get just so-so grades - is the following: the students who succeed do the problems
step by step they usually don't skip the steps, but the ones who get so-so grades
usually try to find a shortcut do it quickly and skip the steps
Because if you skip the steps, then there is a very high likelihood
that you will get the question wrong and the second thing is you will not get the
marks for the steps. Because IB and iGCSE exams will give you marks for the steps
so one simple advice here from me to you guys is to do the questions
step by step. Don't skip the steps and you will do well in mathematics
All the best!
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EVDE DİNAMOMETRE NASIL YAPILIR , Simple-How to make DYNAMOMETER at home - Duration: 14:31.
hi , my friends
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Thermodynamics: Brayton cycle with intercooling and reheating, Ideal simple Rankine cycle (33 of 51) - Duration: 1:06:05.
>> All right, well good morning everybody. Pardon me. So just a comment or two
on the homework that's being returned to you now. Well first off, if you haven't
picked yours up it's just laying up here at the front. You know once again, most
of you did just fine. You know the first several problems were review problems.
A lot of students solved problem 675 instead of 678 and you know at first I just
didn't understand, but then I realized I probably didn't write it down as neatly
as I should have on the board. Many of you did 78, but many of you did 75; so I
wasn't prepared to grade number 75; so I just put 78 with a little question mark
by it. But nonetheless I think the only other error of significance was that
many of you forgot that I had asked you to do particular problems using variable
specific heat, not constant specific heat as it mentioned in the book. And then
also several of you on those last couple of problems, was it 38 and 52, the book
specifically says accounting for variations in specific heat with temperature,
which his the authors way of saying solve using variable specific heats. Again,
many of you solved using constant specific heat; so you'll see a big red X
through those problems and you know, it could very well have affected your grade
getting a check instead of a plus. Remember plus is three points out of three,
check is two out of three; so if you didn't do those last two problems
correctly, I mean those two problems are really the longest problems in the
entire homework set. Again, it could have cost you a point on the homework, but
nonetheless in general I'm pleased with the homework and again the homework
solutions are posted outside my office. One other thing I wanted to note on
problem 9-90 from your current homework set it may not be entirely obvious how
to solve it. I would note that on that particular problem you're given the
actual temperature at 4A and what are you trying to find? I guess you're trying
to find the temperature of .3. You know you've never really been given a problem
exactly like that before. And all I would really note is that basically the
problem has to be solved iteratively. I'm not going to give you too may hints on
that but you actually have to guess the value of T3 and then look up the
relative pressure and the [Inaudible] at that point [Inaudible] then have to
plug it into the equation for pressure ratio equals the ratio of the relative
pressures and that's going to give you your relative pressure of .4S then you
can find the [Inaudible] at 4S and then you can apply that to the efficiency
equation and the unknown there will give you the [Inaudible] at 4A and then see
if the temperature at that point 4A is the same as the one given. If it's not
then you pick the wrong value of T3, so start over again. Pick a new T3, go
through the exact same procedure, 3-4, 4S to 4A using efficiency and then check
again to see if .4A has the same temperature as the one that's given; so that's
how you would approach that problem. So this is given and it's an iterative
solution, you can call it trial and error if you prefer. Same thing. So any
questions on that? All right, good. So with that let's just go back to where we
were on Wednesday. Remember that we're talking about the Brayton Cycle and we're
now looking at variations to the Brayton Cycle, so it's not the simple Brayton
Cycle anymore. I had mentioned that there's three variations that we're going to
look at, all of which are designed to improve the thermodynamic efficiency and
indeed they all will. First we could utilize some of the waste heat, exhausted
from the turbine in that cycle rather than just dumping it into the environment
as heat output. We're going to send it through a regenerator and we're going to
pick up some of that heat as we move out of the compressor such that when we
finally go into the combustion chamber we have a higher temperature, and
therefore we add less heat as we move from the point of discharge from the
regenerator that is .5 up to .3; s our heat input is lower and our work is going
to be the same, so we end up with a higher efficiency. The second thing was
using intercooling along with compression and stages and that's really what
we're talking about now. This is all in your notes from last time; at least I
think it is. A schematic diagram as well as a TS diagram that represents the
intercooling process, so we're going to compress in stages, so two stages,
right? One to two and three to four, those compression processes take place at
the same pressure ratio. The inlet to each compressor is going to be at the same
temperature. We also note that the intercooling process from two to three is
done at constant pressure. Nonetheless if we do this compression in stages with
cooling in between each stage then we should minimize the compressor work input
and therefore with less work input we've got more network output. And again, the
efficiency should increase. So that's what we'll continue talking about now and
then the third one is where we actually allow the work process, the work output
process to be done in stages through multiple turbine stages and we have some
form of reheating in between. So we'll get to that in just a little bit and then
I'm going to give you an example problem that basically throws all of this
together. And so last time again we talked about the intercooling process, we
understand the basic process now. The equations are not really any different
than they were before so we still have our thermal efficiency, which is still
the network over the heat input. The network is still just the difference
between the turbine and the compressor work, divided by the heat input and again
I'm writing this on a per unit mass basis, all lower case letters. But it could
be capital letters, it could be dots. It doesn't really matter but this is the
basic thermal efficiency equation. What about the work input then? I mean the
work input which is the work associated with the compressors is really the only
thing that we're addressing here, right? We'll talk about the turbine work later
and heat input later as we combine regeneration, intercooling and reheat. But
for now it's really just the work input that's changed by using the intercooler
and that work input is just going to be the sum of the work input required by
each of the two compressors, so it's just going to be H2 minus H1, plus H4 minus
H3 and of course these are magnitudes, right? So this is the magnitude of the
compressed work input. But let's also remember that we're treating the working
fluid here as air as an ideal gas and as such the [Inaudible] are only functions
of temperatures. We know that just by looking into our table A17. But
nonetheless H1 and H3 are the same because T1 and T3 are the same, right? H2 and
H4 are the same because T2 and T4 are the same; so really this just becomes two
times H2 minus H1. And again since H2 equals H4 and H1 equals H3. So this is
really relatively straightforward as far as being able to calculate the
compressor work input. Now I might note further that in reality if we had an
infinite number of compression stages that would truly maximize compressor work.
And we can't do that, right? Well I guess one could try but it will take you
forever to build such a device, right? And plus it will be so expensive it
wouldn't' really be worth it; so having anything more than one intercooler with
two compression stages really is about the limits of our economics if you will.
So we're not going to look at other situations. But certainly you could have two
intercoolers or three intercoolers. The only thing that would be different would
be that the pressure ratio across each stage would be the Nth route, that is N
being the number of compressors rather than square root. So we know that the
total pressure ratio when we take the square root will give us the pressure
ratio across each individual stage. If we had two intercoolers with three
compressors we'll still know the total pressure ratio. Well we take its cube
root and that will give us the pressure ratio across three stages. If we had
four stages, the fourth root; so just keep that in mind, although you really
never have to use that fact, but certainly you want to remember that in going
through the analysis of compression from one to two or from three to four we
have to use a pressure ratio per stage, right? We're still going to solve this
like any other compressor. We're still going to say that the ratio of pressures
for that stage is equal to the ratio of relative pressures or we're going to use
the fact that the temperature ratio from one to two equals the pressure ratio to
the K minus one over K power. Either way you do it that's how you would go from
.1 to .2. You don't even have to do three to four because one and two is the
same as three and four. So just go through one compression process using the
pressure ratio across one stage and then just multiply by the number of stages,
which in your case is always going to be two. Anyway are there any questions on
this particular process or why it's relevant or important to have the extra
compression stage? All right. So this now brings us to our third and final
modification, which deals with reheat. Now here when we talk about reheat we're
really referring to the expansion in turbine. And I do call it turbine
expansion. It is an expansion process, right? The pressure is dropping as the
air moves through it. The gas should expand some, so lower temperatures, it's
got lower density; so it is an expansion process. But nonetheless like the
compressor where we can compress in stages in order to minimize a compressor
work if we do our turbine work, that is work output in stages we can maximize
the work output, that is maximize the turbine work. Now between each stage you
have to add heat, essentially just the opposite of what we talked about in the
compressor, right? You compress and then you remove heat if we have inner
cooling with reheat you expand turbine and you reheat back up to the temperature
that you had going into that first turbine. So how does reheat work? Well before
I write all these words let me just illustrate. So we have our turbine and I'll
put a big T for turbine. And then we're going to have some sort of heat
exchanger where we're going to have some heat input and then we're going to
expand through the next turbine. I'm not going to use one, two, three and four
here. I'm going to use six, seven, eight and nine. Although frankly the numbers
don't really matter. I mean you can use A, B, C, D or pick an alphabet anywhere
in the world. You just have to keep track of what point is what, right? Anyway
this heat exchange in the middle here this is the reheater. And if we were to
look at this on a TS diagram rather than a schematic you know, we still want to
show our minimum and maximum pressure lines. We're going to start at an elevated
temperature, right? I mean we've added heat already. We've got the high
temperature air, combustion gasses in the real world but high temperature air
for us. And now we're going to expand in stages so let's show the intermediate
pressure line too. So we go from six and we expand down to seven. We then reheat
along the constant pressure line up to .8 and then we expand a second time, do
more work output as we finally get to .9. So the turbine work is really going to
be the sum of the work from each of the two compressors, I'm sorry, each of the
two turbines. Okay? So this is what's happening. Now again I should note that
this is really the ideal case because we're assuming in this particular case
that the turbine is ideal. That is isentropic. But it doesn't have to be, right?
We can certainly solve a problem like this or like a compressor problem for the
non-ideal case too, right? We do have to use efficiencies, isentropic
efficiencies associated with the compressor and with the turbine. So let's just
move ahead - so how do we do this. What's the importance? So I'll just say
similar to compression in stages with inter-cooling - We can maximize the
turbine work output if we expand in stages. And to do this properly you expand
in stages and again, like the compressor with equal pressure ratios - Across
each stage - So with equal pressure ratios across each stage and reheat in
between the turbines up to the same temperature that we had going into the first
turbine. And that's exactly what I've illustrated over here on the TS diagram,
right? We have the same temperature at eight and six. We have the same
temperature, therefore at seven and nine. I mean after all the pressure ratios
have to be the same. And therefore, you're temperature ratios have to be the
same; and therefore we simply would find that if we had the same inlet at six
and eight, we have to have the same outlet temperature at seven and nine; so
again very, very similar to what we saw in the compressor. So this is how we
would maximize the turbine work. Again if we expand in stages with equal
pressure ratios across each stage and reheat in between the turbines up to the
temperature that we had entering the first turbine. I should probably note for
completeness that the reheat has to be done at constant pressure. So with all
this in mind then first of all what would be the pressure ratio across each
stage and also what would be the equations we would use? So once again similar
to what we saw for the compressor. I'm not going to actually go through the
derivation, but if we know the total pressure ratio, that is Pmax over Pmin then
we just have to take the Nth route where N is the number of turbine stages.
Again in this class two is going to be the maximum number of turbine stages
we're going to deal with and therefore just a single reheater in between. Again
it doesn't have to be that way, but it certainly would generally be that way in
the real world. So I'll just say we can find that pressure ratio across each
stage, it's just going to equal the square root of the total pressure ratio and
of course we know that the total pressure ratio is just the ratio of the maximum
to the minimum pressure within the cycle. So that's what we would have as far as
the pressure ratio that we would be required to use. Again, just to make sure
it's in your notes, we would note that the temperatures at the inlet are going
to be the same. And the temperatures at the outlet of each turbine stage are
also going to be the same. So what would the work be? Well the turbine work is
just going to be the Nth will be changed as we go from six to seven. Plus the
Nth will be changed as we go from eight to nine. And furthermore since the
temperatures are the same then the Nth will be the same so H6 and 8 are the same
and H7 and H9 are the same. And therefore - well clearly the turbine work is
just going to equal two times the Nth will be changed across one of the turbine
stages. Okay? Now again how you would actually calculate those would certainly
depend on whether we're using constant or variable specific heats. Certainly if
we're using variable specific heats then we'll just use the [Inaudible] data
directly out of our air tables, table A17. On the other hand if we're using
constant specific heat then all of our [Inaudible] changes just become CP times
temperature changes, ultimately when we plug everything into our equation for
thermodynamic efficiency the CP's are going to cancel and everything will just
be in terms of temperatures. Now rather than illustrate all this for just this
one specific case of reheat, what I like to do now is combine everything. This
would really be the most complete non-ideal cycle, well I should say most
complete cycle and we'll talk later about whether it's ideal or not ideal and
how that changes our mathematics. So any questions on this then? Okay, so
remember again our purpose - the purpose is to maximize the work output. We do
that with reheat and of course now we're going to add to it, minimizing the work
input by compression in stages and minimizing the heat input. Now let me know
one other thing here before I move on pass reheat, when we talk about the heat
input to the cycle we do have heat input now from an external source, right?
It's not like well even the inner cooler had heat rejection to an external sink,
but it never appeared in our calculations. At least the way I use thermodynamic
efficiency with network over heat input we don't really care about how much heat
output there is. It's not something we have to pay for. It's not something we're
concerned with. But on the other hand with regards to heat input we definitely
have to add heat from external source; so one should really go back here and
recognize that there is additional heat input. There's so little space in here.
I guess I'll do this, heat input. The arrow should point towards that line from
seven to eight. So we do have to include that in thermal efficiency. Now you may
wonder, well wait a minute. If the whole purpose is to improve the efficiency
why are we adding extra heat? I mean adding extra heat is going to actually
lower the thermodynamic efficiency, right? I mean we're increasing the
denominator of our efficiency equation, efficiency is going to have to drop but
fortunately the added work you get by this expansion in stages will more than
make up for the added heat that we have to pay for during the heat input process
and reheater; so the efficiency still should go up. All right. So let's just
make that final note - Heat input and our thermal efficiency equation will
include the reheater and really as well as the heat that's added in the
combustion chamber. So please don't forget to include that when you're looking
at the various equations. And again I'm going to bring all this together now by
looking at all three of our improvements. So any questions before I move on?
Good. So let's just call this Brayton Cycle with everything. That's not really
the official name of this Brayton Cycle, but its good enough. So for the
schematic diagram, so we have state point one and from state point one we're
going to go into a compressor. And we come out of the compressor at .2 and we
have to provide our inner cooling. So here's our Q out. We cool down to .3 the
same as T1 and we go through our second compressor. And now we come out of this
compressor and instead of running this air directly into the combustion chamber,
we run it through our regenerator. So here's out big heat exchanger, we call
regenerator. And let's see we come out of the compressor at four, we go into our
regenerator and we come out of regenerator at five. Now we go into our
combustion chamber so CC for combustion chamber. This is our heat input. We come
out of the combustion chamber and we go into our turbine. However this is just
our first turbine stage; so let's say we have six coming out of the combustion
chamber. Now we go into turbine number one. We do some work, we then have to
reheat - By the way I really should have labeled this intercooler over here for
completeness. So we go through our reheater where we have more heat input. And
we go through the second turbine; so let's see seven coming out of the first
turbine, eight out of the reheater, nine coming out of our second turbine. And
now instead of letting this exhaust into the environment or instead of rejecting
heat from our cycle we go right back into the regenerator, we remove heat from
this exhaust at - to .10 and then we finally have our giant heat exchanger in
the sky, the environment and that brings us right back down to .1. Okay so this
is our heat output - Now for completeness here as well let's also show the work
input to the two compressors. And let's also show the work output from the two
turbines, okay? So there's your full schematic diagram of a very complete cycle
with inner cooling and reheat and regeneration. Now that might look a little
complicated but it really isn't. I mean separately we've already looked at the
regenerators; separately we've looked at the inner cooler with compressor and
the reheater with the turbines; so we have all this already in our notes. We've
thought about it, it's been discussed. You might think that the TS diagram is
going to look incredibly complicated, but it really isn't. I mean keep in mind
that everything occurs either as isentropic work processes or as constant
pressure heat exchange processes. So we're only going to have vertical lines and
slope lines representing pressure. So let's just show our cycle again with our
minimum and maximum pressure. But let's also show in here the intermediate
pressure that we know is going to be associated with both the inner cooler and
the reheater. So we'll start down here at .1, so this is not going to be simple
but I will note that it is still the ideal case we're looking at. We're not yet
going to be talking about isentropic efficiencies, although we will be doing
that very shortly. So we start at .1, we compressed to .2. We inner cool to .3
at the same temperate as one. We compress up to .4 at the same temperature as
two. We exchange heat in our regenerator up to .5. We add more heat in our
combustion chamber up to .6. We expand first down to .7, we then reheat up to
eight. We expand a second time down to .9. We now move through the regenerator
and we exchange heat until we get to .10. And then lastly we just end up here at
.1. So the cycle isn't too bad. Certainly you can see now why I chose to discuss
the turbine and reheater from .6, .7, .8 and .9 because that's consistent with
this particular cycle. Now what would be the thermodynamic efficiency? Well it's
the same. It's just the equation; it's just the network of heat input so that
doesn't change. It's still turbine minus compressor work over the heat input.
What does change however is the fact that we now have more than one work term
and more than one well I should say more than one work output term for the
turbine? We have more than one work input term to the compressor and we have
more than one heat input term now. So using the same numbering scheme that's
here on these diagrams, what would our efficiency be? Well let's see, the
turbine work output is going to be the sum of the work output from the two
turbines, but it's the same, right? Just like we saw. So it's really two times
eight, six minus eight, seven. And then we subtract from this the required work
input and that's going to be twice the compressors work, so that's two times H2
minus H1. And then in the denominator it's our heat input and again we have heat
input twice, right? We have heat input from five to six, which is going to be
our combustion chamber. But then we also have heat input from seven to eight,
which is our reheater. So H8 minus H7. And this then is our thermodynamic
efficiency if we have - well I mean this is the good general equation but this
would apply specifically to the case where we would have variable specific
heats. Again all our [Inaudible] data is going to simply come from table A17.
We're going to treat the compression process and the expansion process exactly
as we've already done, so it really doesn't require more discussion at this
point. There is one other thing though that we do have to be aware of, which is
that since regenerator we still have a regenerator effectiveness. And
regenerator effectiveness is still defined the same way, right? The actual heat
input to the air coming out of the compressor going through the regenerator,
divided by the maximum possible. The problem though is that our state points are
different. That is they're numbered differently than when we talked about this
previously. Again the numbers shouldn't matter to anybody, it's the processes
that you need to understand. So what is the actual amount of heat exchange? Well
that's as we go from .4 to .5 so this is just H5 minus H4 and then what would be
the maximum possible? Well it's as if we could heat from .4 all the way up to
the temperature of the exhaust coming out of the last turbine stage. That is all
the way up to .9; so H9 minus H4. The definition of the regenerator
effectiveness hasn't changed, the only thing I've done here is used my updated
thermodynamic state points. And that would be that. So any questions on any of
this? Yeah it should be pretty straightforward. Once again I do want to note
that if we had variable specific heats we use these equations presented
directly. And if we have constant specific heat then we just replace the
[Inaudible] change with CP times temperature change and of course the CP's will
appear in each term in both numerator and denominator so they cancel and your
equation will only be in terms of temperature. So quite frankly you can just
take this equation and just replace the H's with T's and that will be the
equation that you would use if we have constant specific heat. And the same
thing would apply to the regenerator effectiveness, right? You've [Inaudible]
will be numerator and denominator, those equal CP times temperature changes. The
CP's cancel and you end up with just the temperature. So again you could just
replace the H's with T's. T for temperature and that's going to then give you
the equation that you would require if you had constant specific heat. Okay, now
the only other thing then that needs a little bit more clarity is what if we
have the non-ideal cycle? Okay, so we're still going to be looking at this
ranking - I'm sorry Brayton Cycle with everything. But how the equations now
being different and how would our problem solving techniques be different? Well
quite frankly the equations don't really change much at all. Again it's just the
analytics that is the math of the problem that's going to be a little bit
different. The equations will change ever so slightly. I mean quite frankly and
maybe I'll redraw the TS diagram. Quite frankly the only real differences that
we're now going to have to use two different state points at two and four, 2S
and a 2A. We'll have a 4S and a 4A and we'll have two different state points at
7 and 9. We'll have a 7S and a 7A and a 9S and a 9A. So the basic diagram is
going to be the same but we'll go from one to 2S in an isentropic way. We'll
then go from 1 to 2A utilizing well the efficiency, right? The compression
isentropic efficiency. We'll then - inner cool back to .3, the same temperature
as 1. So again, these are going to be the same. We'll then compress to point -
well we're really compressing to .2 - I'm sorry 4S and then 4A. So again the
dashed line represents the process we have to analyze, but not the processes
occurring, right? We assume it's isentropic. That gives me my [Inaudible] at 4S
or temperature of 4S if we're using constant specific heat. We apply the
isentropic efficiency to get to 4A. Then we move through our regenerator again
up to 5. To our combustion chamber at 6. And then we do the same thing, we
expand to 7S, but really we're going to 7A. We then reheat up to 8. Once again
we expand to 9S but really we're expanding to 9A. Again use the efficiency of
the turbine and then we have heat rejection - I'm sorry, then we go back into
the regenerator. And we come out at .10 and then finally we have our heat
rejection back down to .1; so it's the same diagram, right? It's only different
between 1, 2, 3 and 4. And then again 6, 7, 8 and 9. Now the basic equations are
not going to be any different except that we have to make sure that we use the A
terms as applicable. So in other word the thermal efficiency is not going to
just be written the same way, right? We're expanding from 6 to 7A. Not 6 to 7
anymore; so we really need to show two times eight six minus eight 7A. We
compress from one to 2A now, not just 2S. So let's do times H2A minus H1. And as
a denominator, well again we have no differences until we get over here to .7
and then at .7 well again we have to use 7A. So we still have the heat input in
the regenerator from 5 to 6, I'm sorry, not in the regenerator, in the reheater.
I'm sorry I take that back. 5 to 6 is combustion chambers; so we still have heat
input in the combustion chamber from 5 to 6; so that doesn't change and then we
have heat input in the regenerator - heat input in the reheater. Okay and that's
taking us from 7 to 8. So we have H8 minus H7A. All right, so ever so slightly
different, but not that much different, right? Anyway and then of course the
regenerator effectiveness is also going to be effected - we're going to have to
use 4A's and 9A's instead of just 4's and 9's, so H5 minus H4A over H9 minus -
I'm sorry H9A minus H4A and this is going to be our regenerator effectiveness
equation. Now of course, when you see the A's that implies that we have to do
the isentropic analysis first. You know we know that we have to use the relative
pressures to go through both the compression and the expansion process in order
to get points 2S and 4S, or 7S and 9S, right? Then we use the isentropic
efficiency of the turbine. And the compressor to go from well 2A, I'm sorry 2S
to 2A or 4S to 4A or 7S to 7A, 9S to 9A. We've talked about this before. We
haven't used the same numbering scheme again, but again that's really
immaterial, right? It's all about understanding the processes and how to go
through them and you know maybe as a suggestion maybe you want to solve some
problems with different numbers, just kind of swap all the points and just make
sure you can go through something understanding the individual processes and not
just memorizing equations with numbers on them. I mean what if I gave you a
problem on an exam that had an inner cooler but no reheater, or a regenerator
and a reheater but no inner cooler. I wouldn't be using the same numbering
scheme. You know for every device we remove we're going to have one less or
maybe two less state points. So you know understand the processes, don't just
memorize equations. So let us now go through an example problem that will pull
all this together. Yeah it's a little bit long because we have a lot of
processes to analyze, but if you can do this example problem or at least
understand the example problem you should be able to do any problem I would
throw at you. So this is from your textbook with a slight revision. It's chapter
nine, problem 121. And turn this on. So this is problem 9-121. I'll put E8 for
the eighth edition of the book. I'm starting to do it this way because some of
the examples I'm going to use moving forward are from the seventh edition or the
sixth edition. I mean I've got some going all the way back to the second
edition, but that's only because I've been here 32 years. But nonetheless we're
going to modify this a little bit. This is going to be with regenerate
effectiveness of 75%, a compressor efficiency of 80%, and turbine efficiency of
85%. Okay and I guess technically we would call this part B of this problem.
Because part B already tells you that you have a regenerator with effectiveness,
but it doesn't give you anything with regards to the isentropic efficiencies. In
fact, if you read it, it says consider it an ideal gas turbine cycle. But this
is not ideal, okay? So that's why I say with these so please note again, this is
not an ideal cycle. Nonetheless we have our gas turbine cycle, not ideal and
there's two compression stages and two expansion stages. In other words, we have
inner cooling and reheat. We're given the pressure ration across each stage of
the compressor, which often is not given, sometimes they'll give you the total
pressure ratio or say what the maximum and minimum pressures are. And then you'd
have to use the square root of the total pressure ratio to get the compression
ratio across each stage. But it does give you the stage pressure ratios now, so
just keep that in mind. Read your problems carefully, it's very easy to make a
mistake at the very first step of the problem and use the wrong pressure ratios
and then everything is wrong. All right, they've also given you the temperature
entering each stage of the compressor and each stage of the turbine as 300 and
1200 kelvin respectively. Now again they don't really have to say each stage,
because we know that the temperature is the same going into each stage, so they
could have just said that the inlet temperature of a compressor is particular
value or turbine particular value and you just know that they're the same for
each stage. Anyway they want us to bind them [Inaudible] ratio as well as the
thermal dynamic efficiency of the cycle. And again in part B, with my
modification we have this other data. So let's just write out what we have. We
know that the pressure ratio across each stage is given as three. We're given
the temperature into the compressor so that's 300 kelvin, so that's T1. And by
the way I'm going to use exactly the same numbering scheme they have over here
on the far right hand side of the board. This is a non-ideal Brayton Cycle with
everything on it. So T1 is what's been given. We're also given the temperature
at .6 which is 1200 kelvin. And of course T1 and T3 are the same; T6 and T8 are
the same so you can note that here as well. We're trying to find the back work
ratio. And remember the back work ratio is just a ratio of the work output from
the compressor, divided by the work input - I said that wrong. The work input to
the compressor divided by the work output from the turbine. So that's what we're
going to have to find as well. And then of course we'll find the thermal
efficiency; so there's your problem. The question really is how do we approach
the problem. I think one thing I want to do is see if I can raise the screen up
just a little bit, just so we can see the rest of the schematic diagram over
there. I mean I can't adjust where the light is so yeah, do the best you can. So
maybe what I'll do is I'll just kind of leave this up here for now. I can't even
do that, I don't have enough room. That's the problem with these new classrooms
with these jumbo size screens. It would be nice if the screens were over on the
edge somewhere so I could use the white board. But it takes up half the space
and then I can't really do anything; so I guess I will have to start erasing
this. It's in your notes anyway. So what's the appropriate approach for this
problem? Well it's the same approach that we've always used, right? We'll start
at the entrance of the compressor; we'll go through isentropic compression
process. That will give me the information at 2S and now use the efficiency to
get to our results 2A and the have to move over to the turbine. I can't deal
with the regenerator yet because I need to know what the discharge is from the
turbine, so then I go over to the turbine stages and I'll do the same thing,
right. I'll find my data at .6 that is at T6, I'll find the relative pressure at
.6. I'll go through the expansion to 7S to get well my data at 7S. I'll then use
the isentropic efficiency to get the data at 7A. And note 7A is the same as 9A
so I know what's coming out of the turbine. And then I go back to the
regenerator. I do the regenerator effectiveness last. The regenerator
effectiveness will allow me to find the [Inaudible] leaving the regenerator,
right? Over here at .5 from this equation. I've already done the compressor so I
have 4A and I've done the turbine, 9A so this gives me H5. And that's all I
need. Once I have that then I just apply it to the isentropic efficiency - the
thermal dynamic efficiency equation. I have all my thermodynamic states and
problem is solved. So there's nothing different here, we've handled each of
these processes individually, now we're going to do it collectively. Now we're
going to do it collectively. So let's begin. I'm only going to do this using
variable specific heats. Certainly you could do is using constant specific
heats. It would not be as accurate and you're not going to get exactly the same
results but still it's worth noting that this is the only method that I'm going
to use. So let's just begin. So first of all at T1, we have to go into table A17
and we need two things. We need [Inaudible] and the relative pressure. So the
[Inaudible] at 1 is 300.19, [Inaudible] per kilogram, that would be T1 at kelvin
and the relative pressure is going to be 1.386. Now we have to go through the
isentropic compression so we note here that the pressure ratio across each stage
is really just going to be P2 over P1, which is PR2 over PR1. Again this is not
entirely correct because this only applies to isentropic process, right? So
don't forget your S. This will then allow me to find PR2S, so P over P1 is equal
to three. We have PR1 from above. So this gives me PR2S and it's going to equal
4.158. Can you guys see these numbers over in the far corner? Anyway I can't
hold it there but it's 4.158. So I'll just move it. So now we have that number,
so now we go back in at PR2S that is back in table A17 and this gives me H2S.
and of course we have to find this by [Inaudible] and we get 411.26, kilojoules
per kilogram. And now we apply the isentropic efficiency of the compressor which
is given as .8. And of course this is just going to be the ideal over the actual
work. And the work terms are just [Inaudible] terms, we have H1, we have H2S. So
we get H2A then and this is going to equal 439.02 kilojoules per kilogram. And
this is the same as H4A. So we know H1 and H3 are the same. I probably should
have also noted that this is equal to H3 up here above. But nonetheless we now
have all of our [Inaudible] from one to four. And we do exactly the same thing
for the turbine. Really I should say for the turbine stages. So again same
thing; so now we're going to start at 8.6, so at T6, into table A17. We need the
[Inaudible] and the relative pressure at six. So eight six is 1277.79 kilojoules
per kilogram. PR6 is 238.0. Now we go through the expansion so the pressure
ratio per stage is still equal to three and you know P6 over P7, but it's also
PR6 over PR7S. So again we can go through our math, just plug in PR6 from above.
And we can get the relative pressure at 7S, which is 79.33 and then by
interpellation go back into table A17 we get H7S and this [Inaudible] is going
to be 946.36 kilojoules per kilogram. And again now we just deal with the
isentropic efficiency of the turbine. This is given as 85%, we know that the
definition is inverse from that of the compressor, so it's the actual level of
the ideal work. So H7A - I'm sorry H6 minus H7A over H6 minus H7S and then what
we want to do is we simply want to plug in the numbers from above and solve and
we get the actual [Inaudible] leaving at turbine that was H7A of 996.07. And
again kilojoules per kilogram. Now once again we know that each turbine has the
same inlet so up above you can just note that H6 and H8 are the same. And down
below H7A and H9A are the same. So the next step it looks like I'll have to
erase the TF diagram now. The next step is to solve for the regenerator
effectiveness in order to get the temperature - well I'm sorry the [Inaudible]
that leaves regenerator. So again there's all the single unknown, right? In fact
I just erased the equation that we're going to use. I guess I didn't really have
to erase it. So H5 minus H4A over H9A minus H4A. Once again everything we
already know from above, so I'm not going to bother to plug in the numbers. You
can do that at home. We end up with H5 of 856.81 kilojoules per kilogram. That's
the hard part; the rest is just finding the data that we need. So we need the
compressor work, we need the turbine work so that we can find the back work
ratio and we need the heat input additional to find the thermal dynamic
efficiency. So the compressor work is just two times H2A minus H1, again I'm not
going to bother to plug and chug, but you can plug all the numbers in yourself
so we get 277.66 kilojoules per kilogram for the compressor work input. We have
two times H6 minus H7A as the total turbine work output. So again if we go
through the math this is 563.44 kilojoules per kilogram. The heat input is from
our combustion chamber as well as the reheater so H6 minus H5 is the combustion
chamber. And then H8 minus H7A represents the reheater. So again we have all the
data already here on the board, plug it all in and we get 702.7 kilojoules per
kilogram. We find out back work ratio, the compressor work input divided by the
turbine work output. So just take the ratio of the two terms above and we get
.493. That might sound like a lot, I mean fully half of the work being produced
by the turbine is being used up by the compressor. In other words only half of
the turbines work output is actually network that could be used to generate
electricity or thrust or whatever your work output is. But that's not unusual
for these types of systems. It's very common for a significant portion of the
turbines work to be required by the compressor; so there's one result. And then
lastly and the thermodynamic efficiency, network over the heat input we'll find
this is .407 or 40.7% so there's the results for hi particular problem. Now some
of you might think that a 40% thermodynamic efficiency is pretty lousy, it's
actually excellent. If we look at a typical steam power plant that exists out
there in the world and fully 80% or so of our electricity is generated from
steam electric power plants the best of them, the very best of them maybe 34%
efficiency. A gas turbine like this using the Brayton Cycle has efficiencies in
the 40% range. It's significantly higher. So these are actually very good
numbers. The nice thing about having achievable efficiencies in the 30-40% range
is that there's always going to be work for mechanical engineers. I mean you're
always going to have to try to improve the efficiency to do whatever you can
whether it's through materials or processes or controls or thermal dynamics or
fluid flow, whatever - you're always trying to make improvements, right? I mean
you know the company that come up with the next latest and greatest product is
going to be the one that wins if you will, in the business world. And that's a
good thing, right? Of course it's also good from an environmental point of view,
the higher our efficiency than the less heat output, the less thermal pollution
if you will and presumably the less fuel we would have to burn to produce the
work that we need. So that's good or pollution; so for all sorts of reasons it's
good to keep working on improving efficiencies. Now just as a matter of note
here if one were to solve this problem without a regenerator - Assuming that we
still have the compression and the turbine in stages, what we would find is that
the thermal efficiency is going to drop to about .368. The back work ratio is
going to go down significantly to about .335. And you know you might think well
that's good, right? If we're using less work for compression than perhaps we
have more network output. But that's not true, okay? Without the generator we
have more heat input, yeah we might have less work input but we also are going
to require quite a bit more heat input and the efficiency is going to drop,
okay? And again it's not just dropping by four 100's or if you will it's not
just dropping by 4%, it's dropping by a full 10%, right? We compare what is
achievable at almost 41% to what's producible now at 37%, I mean that's a 10%
drop; that's a significant number. So think about it in relevant terms and
you'll see this is a pretty good improvement by adding the generator. Anyway oh
- Without a regenerator I should have also said and with assumed compressor and
turbine efficiencies of 100% -- in other words that would be part A of the
problem as it was actually presented in the textbook. Again we have significant
differences. So nonetheless are there any questions then at all because this is
really our end of the discussion of Brayton Cycles. Are there any questions at
all on this particular type of cycle? Good, so now we will move on and I have
just the right amount of time to introduce you to my favorite cycle, certainly
the one you'll see most in the world, the Rankine Cycle, which is our steam
power plant cycle. Again vast majority of the energy, electricity that's
produced in this world, not just the US, but the entire world comes from steam
power plants, so we really need to understand how to analyze such a device. So,
this brings us the to the Rankine Cycle. And this is going to be the last of the
heat engine cycles that we're going to talk about. It's not going to be the last
cycle, after we talk about the Rankine Cycle we'll then get into the kind of
cycles that we use for air conditioning, refrigeration processes. So you know
that's going to be not the heat engine cycle, but the refrigeration cycle and
we'll get to that eventually but right now the Rankine Cycle. So this is used
for steam power plants. Okay. Now there's significant similarities between the
Rankine and the Brayton Cycle not necessarily in the way that we utilize our
data or how we use that in our equations but just like the Brayton Cycle we're
still going to have constant pressure heat exchange, so the heat input and the
heat rejection is still going to be done at constant pressure. So that's
certainly the same as the Brayton Cycle, also at least for the ideal case we're
going to have isentropic work. So the work input is going to be isentropic, the
work output is going to be the isentropic; so again just like the Brayton Cycle
and of course we use the non-ideal cycle, which we'll certainly do then we'll
still have to assume first isentropic, but then we'll have to apply the
isentropic efficiency in order to get the actual work associated with this
cycle. But nonetheless these are some similarities. The names of the devices
that we use are certainly going to be a little bit different than we do in the
Brayton Cycle. So let me just show a simple schematic. So this is going to be
for the simple Rankine Cycle. So first of all we'll start at some state point,
this is called one. And we'll note that this is used for steam power plants. So
we're not talking about air cycles anymore at all, you can just completely
remove that from your mind at this point. We're using water, right? Steam is a
form of water so we're actually going to take liquid water at .1 and rather than
send it through a compressor for our work input, we're going to send it into a
pump. Now again a pump and a compressor perform exactly the same function, we
just call it a pump because it moves a liquid rather than a compressor which
moves a vapor, but it's the same process. We're going to leave the pump state .2
and we're going to go into a heat exchanger of some sort. Now granted this is
not going to be a combustion chamber anymore, here is where we're going to turn
our liquid water into steam; so this is an actual boiler. So this is where we
have our heat input to the cycle, so we come out of the boiler state .3, so this
is our high temperature, high pressure of the cycle. We're now going to go
through a turbine, we still call this a turbine. So this is where our work
output is going to be. So work output is turbine work. By the way over here when
I put work input I probably should have put WP as well just for pump, but
nonetheless so we're back to the turbine. We have our work output and we leave
at .4 and then we have to reject heat into the environment. Now this is water,
right? It's not air. We're not going to just in the real world, we're not going
to just dump the water out into the environment, so we have to take that water
and cool it down. And in fact, as it comes out of the boiler it's steam, as it
leaves the turbine it's either steam or a two phase mix. But nonetheless we have
to send it through a device that's going to condense that steam back into liquid
water again and of course we use the word condenser, right? It's condensing the
steam, we call this condenser. Okay, now in the condenser there's going to be
some heat rejection, so there's the heat output. How this is done is not
something we're going to talk about immediately. It can be done through a
variety of means, but typically this is a big heat exchanger, you've got the
steam on one side of some tubes, inside a shell. On the other side of the tubes
you have cooling water from the local ocean or a river or a lake or it could be
from a device, like a cooling tower which we'll actually talk about near the end
of this quarter. But it's a heat exchanger again and we just condense back into
a liquid. So this is the basic Rankine Cycle and shown on a schematic diagram.
No more complicated than a Brayton Cycle for sure. In fact, in many ways the
analysis is going to be a little bit easier. We're not going to have to deal
with relative pressures and all that. We're just simply going to take our
[Inaudible] data and our steam tables, table A4 through A6. So this is the basic
process. It would certainly be appropriate to show this on a TS diagram. Now
here I would like to show it relative to the saturation lines. So there's my
saturation curve, my critical point up at the top. State point one without
exception is going to be a saturated liquid, okay so always a saturated liquid.
This represents the low pressure of the cycle, so let me also just show a couple
constant pressure lines, again show the maximum and minimum, but of course this
is different than our Brayton Cycle and our Brayton Cycle who was just air as a
gas, it would occur over here in this area and that's why we always had these
angled lines. Now we have the saturation region, we have constant pressure at a
uniform temperature as we go through the saturation region. So that's why I have
the curved lines that become flat for a while and they curve back up again.
Nonetheless we compress isentropically to .2, right? We then add heat and
constant pressure until we get up to 7.3. We then expand isentropically until we
get back to .4 and then we get right back over to .1 again. So we've got our
pump work input, our turbine work output, our heat input and our heat output and
there's your basic, simple and ideal - so I better write that both, simple and
ideal Rankine Cycle. Okay by the way we'll certainly adjust this later. We'll
look at non-ideal cycles, we'll look at cycles that are not simple. We'll look
at a form of regeneration to try and improve the efficiency, but that will be
later. So this is the basic Rankine Cycle and that's really all the time I have.
Question?
>> Are those from one to two and three to four, streamed down?
>> Yes, they're isentropic and I don't know, maybe my curves aren't as vertical
as they should be, but these are clearly isentropic processes. So anyway here is
our basic cycle. We'll get into this in a lot more detail next week; so have a
nice weekend everybody. Once again don't forget to pick your homework up at the
front. If any of you missed my hint on problem 9-90 then just talk to one of
your classmates and I will see you all then.
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Simple Present Story 1 - Duration: 1:02.
For more infomation >> Simple Present Story 1 - Duration: 1:02. -------------------------------------------
Thermodynamics: Stirling and Ericsson cycles, Ideal and non-ideal simple Brayton cycle (31 of 51) - Duration: 1:06:07.
>> 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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Simple Border Line Design 4 Embroidery - Hand Embroidery Work - Duration: 4:53.
Simple Border Line Design 4 Embroidery - Hand Embroidery Work
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