CHAPTER 7
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Dear friends hope that you are practicing well.
Now I am providing you with the solutions of chapter 7(Entropy)
Before I start providing you the solutions I would like to give a brief introduction of chapter.
In this chapter you have studied about how the concept of entropy came into existence. You have studied Clausius inequality and developed the notion of isentropic process. You have studied about the Principle of increase of Entropy and how entropy increase principle be helpful to know whether a certain process is possible or not. The very important concept of this chapter is the entropy generation so you have also studied about with respect to both closed and open system. You have also studied the equations when first and second law is combined. At last you have studied about reversible adiabatic work in a steady flow system.
Dear friends for a quick review of what I said above you can click the videos below where I am giving a quick review of Second law of thermodynamics.
This is the problem which kept me occupied for many years but last to last year when I was teaching in class, all of a sudden an idea strikes me that why not solve this problem with a different approach and the result of that idea is below. please be careful this is one the toughest problem of this book.
This is all I have to say on this chapter. These problems are of 4th edition.
One more thing I want to say ( which I have also mentioned in previous post) that please do not follow the solutions blindly. Firstly try yourself and if find difficulty in solving, then only look the solutions.
After few days I will be publishing next chapter solutions so stay connected.
your comments are valuable to me, so please do write them. If any problem in any solution then also let me know.
Thankyou
All the Best.
what will happen to entropy for a close system and open system while decreasing the pressure?
ReplyDeleteentropy is relateble to heat.for a closed system u cam relate heat with pressure through balancing the energy ..but for open system u have to take care of flow process energy balance equation .hence u can get heat from there amd then put in the formula of entropy
DeleteWhen we talk of closed system and consider it to be insulated then as pressure decreases it's volume will increase and that is because of decriment in temperature, now you look 2 effects are taking place simultaneously first is decrease in tem. and second is increase in volume now if the process is reversible then these 2 opposing effects will balance the net entropy and make its change zero but if this is irreversible then decriment in entropy because of decriment in temperature will be less than increment in entropy because of volume so net result will be increase in entropy of gas, mind it in this case the system i.e. gas is isolated from universe.
ReplyDeleteNow consider second case when the system is not insulated then in that case the decriment in pressure will be because of heat loss to environment and if free to move then also because of increase in volume now in this case the entropy of system will decrease but a greater amount of entropy increase occurs in surrounding so net entropy change of universe is positive.
Now for open system just by saying that the pressure is reduced in flow passage we cannot predict about entropy as we do not have any information regarding temperature.
Now consider a case of fluid flowing in an insulated duct such that in the flow passage pressure reduces but temperature increases then the change in entropy will be positive ( one of the above problem just demonstrate this)
Abhishek I hope that this will clear some of you doubts, if still have any please do ask.
Thankyou
okay sir i get it..but sir if we talk about the example that you illustrated above in open system ( say throtling process ), its temperature as well as pressure decreases but entropy inceases slightly ?
ReplyDeleteDear Abhishek,
Deletein the example above mentioned you might have forgotten about the velocity at the exit(which make makes the problem of different class than throttling),now consider throttling: first thing is that in case of throttling the temperature may decrease or increase it will depend where you are with respect it inversion curve. but still if you consider the case where the temperature decreases i.e. when you are on left side of inversion curve then the entropy will certainly decrease but this decrease of entropy will be of gas only and when you consider both gas and surrounding then you will notice that surrounding entropy increase is of greater amount than that of system( because of heat transfer to the surrounding) hence the entropy of universe will be positive and the throttling process will be possible and irreversible. and if you consider the throttling process to be adiabatic then also the because of dissipation the intermolecular energy will increase(of a general gas) which will lead to increase in entropy of gas.
basically any system whether open or closed will suffer a positive change of entropy when either heat is transferred to it or dissipative effect is present there and in case of throttling because of huge dissipation( friciton) the entropy increases.
Let us consider the second approach:
we know that Tds=Cp.dT-v.dP
now if the decrease in pressure is more that the decrease in temperature than also the change in entropy will be positive. and never ever try to make dS=0 because throttling is not a friction free process.
I hope this will help you
Thankyou
thanks sir
ReplyDeleteits really helpful for me as well as other students.
Sir i have a doubt that
ReplyDeleteIn the formula of isentropic efficiency and second law efficiency, the numerators are same W(actual) but what is the difference between denominators that are W(max).
Ans. to the qno.40 is incorrect at the end. the entropy generated is not correct.it comes out as -0.31kj/k.
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ReplyDeleteSir
ReplyDeleteIn Question 32 why you use mcpdt for vapour and on the other hand for water you use mcln(Tf/Ti). I think we either use mcln formulae or mcpdt please explain it creating confusion
Sir, l have problem in question 7.36 in chapter 7 (edition 5) and in the solution provided it is not there.
ReplyDeleteSo please provide solution of it.
I have tried it but unable to solve.
Durgeshwar
ReplyDeleteQuestion 50. Please
ReplyDeleteDear Pankaj sir,
ReplyDeleteIn Q-13 case-1 when the temp of the body changes from T1 to T2 by a series of reservoirs ranging from T1 to T2 you have assumed ∆Sbody=0. However since it is a reversible isothermal process ∆Sbody should be mCv*ln(T2/T1). Even though ∆S of all reservoirs will be zero.
Kindly tell me whether this is right or wrong.
Waiting for your reply.
look into qs no 13.heat has been delivered by series of reservoir.check ex 7.2.
ReplyDeletepk nag THERMODYNAMICS, 4th ADITION CHAPTET 15 - PSYCOMETRICS SOLUTIONS
ReplyDeleteSir there is 50th question and iIam not sure about the solution.can you please upload it
ReplyDeletesir in q39 the value of ∆s/sytm is differencr btw the temps not ln because in the integral T gets cancelled
ReplyDeletesir new edition has more problems ..will u plz solve them
ReplyDeleteHow to download all the solutions
ReplyDeleteSir
ReplyDeleteyou have not done q no 50
sir i cant solve it
sir please solve it
or anybody reading this please help
Sir , question no 13 is not going to understand , the problem is that if process is reversible ( not mention in problem) then how should be possible it is greater than 0. Ie dS total
ReplyDeleteSir please upload the solution of chapter 15 of pk nag
ReplyDelete