[مجموعه ویدیویی آموزشی] - ترمودینامیک Thermodynamics - به زبان انگلیسی

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Lecture 6: Evaluation of Work > Evaluation of Work

Lecture 6: Evaluation of Work > Evaluation of Work

Simple and Complex Systems

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[FONT=comic sans ms]We now define what we mean by simple systems and complex systems.[/FONT]
[FONT=comic sans ms]The main issue here is the modes of work for a system.[/FONT]
[FONT=comic sans ms]We know that a given system can have any number of modes of work.[/FONT]
[FONT=comic sans ms]Thermodynamics does not put any restriction on this.[/FONT]
[FONT=comic sans ms]We also know that some of these modes of work are two way.[/FONT]
[FONT=comic sans ms]For example, expansion of a gas.[/FONT]
[FONT=comic sans ms]Some others could be one way.[/FONT]
[FONT=comic sans ms]For example, stirring a liquid.[/FONT]
[FONT=comic sans ms]And important number for us is the number of two way work modes.[/FONT]
[FONT=comic sans ms]Let's call this N2w, it represents the number of two way work modes possible for a system.[/FONT]
[FONT=comic sans ms]As an example, let's consider system as a simple gas, say air.[/FONT]
[FONT=comic sans ms]In this particular case the number of two way work modes is one - that is the expansion work mode.[/FONT]
[FONT=comic sans ms]We consider another example in which we take the system as say a dielectric fluid.[/FONT]
[FONT=comic sans ms]In which case the number of two way work modes could be two.[/FONT]
[FONT=comic sans ms]One of them would be expansion-compression.[/FONT]
[FONT=comic sans ms]The second one would be charging-discharging.[/FONT]
[FONT=comic sans ms]Is also possible for a system to have zero number of two way work modes.[/FONT]
[FONT=comic sans ms]For example, if you take an incompressible liquid.[/FONT]
[FONT=comic sans ms]In this case we can say that the number of two way work modes is zero.[/FONT]
[FONT=comic sans ms]Why? No expansion-compression possible, it's not a dielectric so charging discharging is out of question.[/FONT]
[FONT=comic sans ms]Can be stirred, but that is a one way mode of work.[/FONT]
[FONT=comic sans ms]So here we have seen illustrations of systems where you can have a single two way work mode, or system with two two way work modes and even a system with zero two way work modes.[/FONT]
[FONT=comic sans ms]Using this number, we are now going to classify the systems and this classification will be useful later.[/FONT]
[FONT=comic sans ms]For example, if the number of two way work modes is one.[/FONT]
[FONT=comic sans ms]We will called that system a simple system.[/FONT]
[FONT=comic sans ms]If the number of two way work modes is greater than one, two, three, four, we call that system a complex system.[/FONT]
[FONT=comic sans ms]And if it turns out that the number of two way work modes is zero, or the system is constrained such that it can not execute any two way work mode, we will call such a system a rudimentary system.[/FONT]
[FONT=comic sans ms]Rudimentary systems, we will see, will be useful when we consider thermometry and the measurement of temperature.[/FONT]
[FONT=comic sans ms]Most of our illustrations and many of our engineering systems will be simple systems.[/FONT]
[FONT=comic sans ms]And because we come across a large number of different simple systems, it is often possible to classify the simple systems into some subclasses.[/FONT]
[FONT=comic sans ms]For example, a single work mode - we mean a simple system.[/FONT]
[FONT=comic sans ms]But in that, that work mode if the single work mode is that of expansion compression, as that of a gas,[/FONT]
[FONT=comic sans ms]we will call it a simple compressible system - like that of a gas.[/FONT]
[FONT=comic sans ms]If that single work mode is that of electrical charging or discharging as in a rechargeable battery,[/FONT]
[FONT=comic sans ms]we may call it a simple electrical system, Ok?[/FONT]
[FONT=comic sans ms]Take for example a spring, a rubber band.[/FONT]
[FONT=comic sans ms]You can extend it or you can compress it as in case of a spring.[/FONT]
[FONT=comic sans ms]This will be a simple elastic system.[/FONT]
[FONT=comic sans ms]We should notice that only simple systems can be described thus using a single adjective.[/FONT]
[FONT=comic sans ms]Like a simple compressible system, or a simple electrical system, or a simple elastic system.[/FONT]
[FONT=comic sans ms]Complex systems will need more than two, or two or more adjectives.[/FONT]
[FONT=comic sans ms]Thank You.[/FONT]
[FONT=comic sans ms]


 

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Lecture 6: Evaluation of Work > Evaluation of Work

Lecture 6: Evaluation of Work > Evaluation of Work

Exercises

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[FONT=comic sans ms]Welcome back.[/FONT]
[FONT=comic sans ms]Now we are in a position to exercise ourselves in thermodynamics.[/FONT]
[FONT=comic sans ms]We will soon give you a set of exercises, the exercise set A, and you will be able to solve this.[/FONT]
[FONT=comic sans ms]But, before you do to any exercises in thermodynamics.[/FONT]
[FONT=comic sans ms]There are some preliminary ideas and preliminary issues that one should be aware of.[/FONT]
[FONT=comic sans ms]And that is we should be conscious of the formal method of solution of problems in thermodynamics.[/FONT]
[FONT=comic sans ms]We should know how to solve a problem.[/FONT]
[FONT=comic sans ms]There is a procedure or a formalism involved in it.[/FONT]
[FONT=comic sans ms]First thing we have to do is read the specification and[/FONT]
[FONT=comic sans ms]I have underlined the word read indicating that - what we want is not just reading, we want one to read, appreciate,[/FONT]
[FONT=comic sans ms]and understand - what is the problem specifications is all about.[/FONT]
[FONT=comic sans ms]Then as we read, and as we proceed with the solution,[/FONT]
[FONT=comic sans ms]we will sketch diagrams, diagrams of the system, or systems involved, and the processes that the system or systems execute.[/FONT]
[FONT=comic sans ms]Sometimes the specification - actually quite often - will not be complete.[/FONT]
[FONT=comic sans ms]So, we should read things or appreciate things which have not specified.[/FONT]
[FONT=comic sans ms]We can say we should be able to read in-between lines or in-between words.[/FONT]
[FONT=comic sans ms]And because of this we will have to make appropriate assumptions to take care of things which are not specified.[/FONT]
[FONT=comic sans ms]As good engineers, these assumption we will need to justify or if some data is available, we may even be able to check the validity of these assumptions.[/FONT]
[FONT=comic sans ms]Now when we come to the numericals, we must treat with respect the numbers, and the dimensions, and the units of the quantities associated with those numbers.[/FONT]
[FONT=comic sans ms]Now the moment we come to units, we will have to combine numbers of different units, so we will have to handle conversion factors properly.[/FONT]
[FONT=comic sans ms]And once we obtain the final answer the exercise is not over.[/FONT]
[FONT=comic sans ms]Quite often we will need to comment on the answers.[/FONT]
[FONT=comic sans ms]Because the purpose of solving a problem, doing an exercise, is not just doing it; it is to demonstrate that we have learned something.[/FONT]
[FONT=comic sans ms]In the next slot we will be looking at an illustrative example, where we will be going through this process.[/FONT]
[FONT=comic sans ms]Thank You.[/FONT]
[FONT=comic sans ms]


 

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Lecture 6: Evaluation of Work > Evaluation of Work

Lecture 6: Evaluation of Work > Evaluation of Work

Illustrative Example 1

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Welcome back.[/FONT]
[FONT=comic sans ms]So we are now in a position to look at an illustrative example.[/FONT]
[FONT=comic sans ms]Example problem concerns a balloon.[/FONT]
[FONT=comic sans ms]It has a initial diameter of 0.2 metres 20 centimetres and it is inflated using some gas in the initial pressure is 1.1 bar.[/FONT]
[FONT=comic sans ms]It rises through the atmosphere and as its rises it's expands.[/FONT]
[FONT=comic sans ms]At the top of its flight is diameter is 0.4 metres, so it has doubled.[/FONT]
[FONT=comic sans ms]The pressure of the gas is now 0.7 bar.[/FONT]
[FONT=comic sans ms]Notice that the pressure has reduced.[/FONT]
[FONT=comic sans ms]Assume that the pressure-volume relation is linear, that is a straight line on the pressure-volume diagram.[/FONT]
[FONT=comic sans ms]Determine the work done by the gas.[/FONT]
[FONT=comic sans ms]So that's the specification of the problem.[/FONT]
[FONT=comic sans ms]So the first thing we do is to sketch.[/FONT]
[FONT=comic sans ms]Actually the sketching can be done as we read the problem.[/FONT]
[FONT=comic sans ms]First we have a balloon, since a diameter is given, there is a volume associated with, we assume that it's a spherical balloon.[/FONT]
[FONT=comic sans ms]So, the simple system diagram will be something like this - a spherical balloon.[/FONT]
[FONT=comic sans ms]The gas inside the balloon with which we are concerned is our system.[/FONT]
[FONT=comic sans ms]So, this is the gas and that is our system.[/FONT]
[FONT=comic sans ms]The initial state, state 1 is given to us as diameter of 0.2 m pressure of the gas as 1.1 bar.[/FONT]
[FONT=comic sans ms]The final state, let's call it state 2 is given to us as diameter 0.4 m and pressure of 0.7 bar.[/FONT]
[FONT=comic sans ms]We are also told that the pressure-volume relationship is linear.[/FONT]
[FONT=comic sans ms]So, if you sketch the pressure-volume relationship, so the volume will go from V1 to V2 the pressure will go down from P1 to P2.[/FONT]
[FONT=comic sans ms]So the initial state will be this state 1, the final state will be this state 2.[/FONT]
[FONT=comic sans ms]And we are told that the pressure-volume relationship is a straight line.[/FONT]
[FONT=comic sans ms]So, let me sketch the process as a straight line from state 1 to state 2.[/FONT]
[FONT=comic sans ms]And from this diagram it's clear to us that it's a quasi-static process, and hence this area is going to represent the pressure volume work, the expansion work.[/FONT]
[FONT=comic sans ms]So this is the first step in the problem.[/FONT]
[FONT=comic sans ms]Thank You.[/FONT]



 

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Lecture 6: Evaluation of Work > Evaluation of Work

Lecture 6: Evaluation of Work > Evaluation of Work

Illustrative Example 2

کد:
[FONT=comic sans ms]
[/FONT]
[FONT=comic sans ms]Welcome back.[/FONT]
[FONT=comic sans ms]Continuing with the solution of the illustrative example.[/FONT]
[FONT=comic sans ms]The first thing we notice, is that we will to now determine the initial volume V1 and the final volume V2,[/FONT]
[FONT=comic sans ms]so that just the way we can write the pressures here initial pressure of 1.1 bar and the final pressure of 0.7 bar.[/FONT]
[FONT=comic sans ms]We can quantify the values of the volumes.[/FONT]
[FONT=comic sans ms]We know that the volume of any sphere equals pi by 6 d cube, where d is the diameter of this sphere.[/FONT]
[FONT=comic sans ms]So, we can say V1 is pi by 6 d1 cube, so this becomes pi by 6 (times) 0.2 cube, the units will be meter cube.[/FONT]
[FONT=comic sans ms]Now, here we have to be careful because if you put this into a calculator, you will notice that we will get a figure something like this,[/FONT]
[FONT=comic sans ms]calculator will not tell us that it is meter cube but we should not forget that it a meter cube.[/FONT]
[FONT=comic sans ms]The first thing you should notice is that all that this thing tells me is that your calculator has a very wide display.[/FONT]
[FONT=comic sans ms]More expensive a calculator the larger number of digits will be on display.[/FONT]
[FONT=comic sans ms]Quite often, we don't need all those displays and hence we will round this off to typically four digits.[/FONT]
[FONT=comic sans ms]In this course most of the numbers will be good to four digits or at most five digits.[/FONT]
[FONT=comic sans ms]Anything more than that is not worth it, and quite often we will not even be able to major to that accuracy and work with that accuracy.[/FONT]
[FONT=comic sans ms]So instead of this we will simply say that this is 4.189 into 10 to the minus 3 meter cube.[/FONT]
[FONT=comic sans ms]In a similar fashion V2 will be pi by 6 d2 cube, which will be pi by 6 into 0.4 cube meter cube and which we will note down 0.03351 meter cube.[/FONT]
[FONT=comic sans ms]So we know the final volume, we know the initial volume.[/FONT]
[FONT=comic sans ms]We want we can write this as 4.189 into 10 to the minus 3 and this is 0.03351 both values in meter cube.[/FONT]
[FONT=comic sans ms]The next thing is to note that the pressure volume relation is indicated to be linear.[/FONT]
[FONT=comic sans ms]That means the locus is known and that means this is an indication that the process is a quasi-static-process,[/FONT]
[FONT=comic sans ms]not just because it is linear but because there is a neat relationship provided.[/FONT]
[FONT=comic sans ms]And because it's a quasi-static-process expansion work can be evaluated as integral 1 to 2 p dV,[/FONT]
[FONT=comic sans ms]and since the relationship between p and V is given to be a straight line, this integral p dV represent this area.[/FONT]
[FONT=comic sans ms]And because it's a trapezoid, the value of this area will simply be the average height multiplied by the width, or the base.[/FONT]
[FONT=comic sans ms]Average height will be p1 plus p2 by 2 and the width of the base will be V2 minus V1.[/FONT]
[FONT=comic sans ms]And hence, we can write this as p1 plus p2 by 2 into V2 minus V1. Ok.[/FONT]
[FONT=comic sans ms]In this particular case because the geometry is that of a trapezoid, we don't have the go through the mathematical process of integration.[/FONT]
[FONT=comic sans ms]And now what we do is just substitute numbers.[/FONT]
[FONT=comic sans ms]Let us first substitute for pressure, the values of pressure are 1.1 bar for p1, 0.7 bar for p2.[/FONT]
[FONT=comic sans ms]It's a good idea to write down the units just after that, multiplied by V2 is 0.03351 minus 4.189 into 10 raised to minus 3, this is volume, so that is meter cube.[/FONT]
[FONT=comic sans ms]And when we multiply this out you will get a value which is 0.02644 and the units will be bar meter cube.[/FONT]
[FONT=comic sans ms]Now although this is good enough an answer, we notice that, this bar meter cube is not a proper unit, it's not one of the standard units for work.[/FONT]
[FONT=comic sans ms]Standard units for work are joule, kilojoule, megajoule and things like that.[/FONT]
[FONT=comic sans ms]And hence, we will have to convert this into the appropriate set of units. And hence, we will have to use a conversion factor.[/FONT]
[FONT=comic sans ms]Before we proceed, let's look at the conversion factor in some more detailed.[/FONT]
[FONT=comic sans ms]The traditional way of looking at conversion factors is something like this,[/FONT]
[FONT=comic sans ms]we know for example 1 bar is 10 raised to 5 pascal or we know one bar is 10 raised to 2 kilonewton per meter square.[/FONT]
[FONT=comic sans ms]This is the traditional way of writing conversion factors.[/FONT]
[FONT=comic sans ms]However, this may sometimes lead to confusion, so what we do is we converts this conversion factors into a ratio, whose value is 1.[/FONT]
[FONT=comic sans ms]Value is 1, it doesn't have any dimensions but it will have units.[/FONT]
[FONT=comic sans ms]For example, this conversion factor will be converted to 10 raised to 5 pascal per bar.[/FONT]
[FONT=comic sans ms]What is its value? One.[/FONT]
[FONT=comic sans ms]What are its unit? Pascal per bar.[/FONT]
[FONT=comic sans ms]What is its dimension? It is dimensionless.[/FONT]
[FONT=comic sans ms]Ok, or we can convert it in other way, we can call it 10 raised to minus 5 bar per pascal,[/FONT]
[FONT=comic sans ms]this can be converted to 10 raised to 2 kilonewton per meters square per bar[/FONT]
[FONT=comic sans ms]or the other way around - 10 raised to minus 2 bar per kilonewton per meter square.[/FONT]
[FONT=comic sans ms]Now, let's go back to our earlier solution. We had determined that the expansion work was 0.02644 bar meter cube.[/FONT]
[FONT=comic sans ms]Since, we want to go to, say, kilojoules, so we will use a conversion factor, which will allow us to get rid off this bar[/FONT]
[FONT=comic sans ms]and convert it into something like kilonewton or newton, and we will see that this is the conversion factor, which is useful.[/FONT]
[FONT=comic sans ms]So all that we do is multiply this by this conversion factor whose value is one;[/FONT]
[FONT=comic sans ms]because the value is one and it is dimensionless, we are not changing anything.[/FONT]
[FONT=comic sans ms]But when we do this - now consider this bar, meter cube, here and the bar and the kilo newton, meter square here as algebraic symbols.[/FONT]
[FONT=comic sans ms]So, I can cancel out bar and bar and there is a meter square which get rids off this meter cube and converts it into a meter.[/FONT]
[FONT=comic sans ms]So now, I end up with 10 raised to 2 into 0.02644, so 2.644, the units would be kilonewton meter.[/FONT]
[FONT=comic sans ms]Now we know a newton meter is known as a joule so a kilonewton meter will be known as a kilojoule. So you can write this as 2.644 kilojoule.[/FONT]
[FONT=comic sans ms]If you want to be really perfect you could say that I will use a conversion factor here which says[/FONT]
[FONT=comic sans ms]multiplied by one - conversion factor is kilo joule per kilonewton meter[/FONT]
[FONT=comic sans ms]and then this kilonewton cancels with kilonewton, meter cancels with meter and we end up with this answer.[/FONT]
[FONT=comic sans ms]Now, is the exercise over - no, because all that we have done is determine W expansion,[/FONT]
[FONT=comic sans ms]what we have to do determine the work done by the system, which is the gas.[/FONT]
[FONT=comic sans ms]Now here, what we say is: we assume, because there is no other hint, that the net work done or the total work done is the expansion work,[/FONT]
[FONT=comic sans ms]there is no other mode of work is in action, this is an assumption.[/FONT]
[FONT=comic sans ms]And hence, we can say, finally, that the work done by the gas which is our system is 2.644 kilo joule, this is the final answer.[/FONT]
[FONT=comic sans ms]In thermodynamics the answer is important but the procedure by which you have obtained the answer is perhaps equally, quite often more, important.[/FONT]
[FONT=comic sans ms]Thank You![/FONT]

 

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