1. Ask any difficulties about Heat Transfer, Thermodynamics, Fluid Mechanics & Applications here.. Let help others & get helped from others also. What do you think?

      1. Check out the NPTEL video lectures on Thermal Engineering by PK Das. The video lectures are really very good giving out basic knowledge of Thermal Engg.
        Here are the links:
        Lecture 1 : Introduction & Definition

        Lecture 2 : First Law of Thermodynamics (Open System)

        Lecture 3 : First Law of Thermodynamics (Closed System)

        Lecture 4 : Second Law of Thermodynamics

        Lecture 5 : Second Law & carnot Principle

        Lecture 6 : Properties of Pure Substances

        (Guys do use InI community regularly)

      1. Inspire N Ignite

        Thanks Chetankumar this really helpful.

      1. @I n I : Its my Pleasure…

      1. Ritesh Das

        @chetan…for a closed system constant volume process work done will “always” be zero!!…is that correct!!!???

      1. @Ritesh: Absolutely it is always be zero. Work = Integration of P*dV. Now as Volume is constant it means change in volume is zero i.e. dV is zero, which results in zero work done.
        Lets took it in this way. Consider the standard “Piston-Cylinder” example. Whatever u do in the system (either change pressure or Temp.) but its volume is going to be same as it is Constant Volume process. So if u remember the definition of “Work” ; it is Force*Displacement. So it requires “Displacement”
        In this process there is no displacement of the piston because it will change the Volume of the cylinder so it will violet our essential requirement of “Constant Volume”. And also as it is “Closed System” no mass can cross the boundary of system so that there is no chance of Work to be done on the mass so it could cross the boundary.
        So moral of the story is “For a Closed System Constant Volume process work done will ALWAYS be zero”

      1. Ritesh Das

        thanks yaar…that was a good explanation!!!…

      1. Thanks!!!! I know I am not so good to Explain something but still I tried my best.

      1. Ritesh Das

        TdS= dU+pdV…according to pk nag this equation holds good for any process(reversible or irreversible)…but how is that possible??…the term “pdV” is only used for quasi-static process or a reversible process!!!!!

      1. @Ritesh, Can you please give the full sentence exactly given by PK Nag in the book.
        Also give in what concern this equation comes in the book?
        I am waiting for your reply…

      1. @all,i am finding convection part of heat transfer difficult bcos of various formulae for various conditions.Any suggestions??

      1. Ritesh Das

        @chetan…its from PK nag(4th edition)…page number 180…chapter name-entropy..sub-topic-“first law and second law combined”..

      1. @Ganesh, Yes I agree, Heat Transfer is difficult subject. Up to conduction it is quite easy but after that is becomes more difficult because of very long & complex formulas & conditions as you said.
        But we have to remember those formulas exactly; we dont have any choice for this 🙁
        Only thing we can do is write those formulas in systematic way & try to remember them.
        Also if you solve many problems on it you are getting used to remembering those formulas.
        All the best..

      1. @chetan,thanx a lot.

      1. @Ritesh,
        According to PK Nag,
        “TdS= dU+pdV this equation can be applied in all the processes undergone by closed system” is true provided that there is no other forms of major energies (like KE, PE, viscuos, Electical) involved in the process.
        You will get the answer after you can understands how actually this equation comes.
        The original equation is,
        dQ-dW=dE …………1 … (from First law of thermodynamics)
        where dQ=small change in heat
        dW=small change in work
        dE=small change in total energy
        now, dE=dU+dKE+dPE+dE+…..
        where, dU,dKE,dPE,dE are small change in internal energy, K.E., P.E., Electrical energy resp.
        we no for many processes we can neglect dKE,dPE&dE & other minor energies.
        Also we know dS=dQ/T
        & dW=pdV…. this is the work done on the system or by the system
        so equ 1 becomes,
        TdS= dU+pdV

        So the term “pdV” represents the work done on the system or by the system & it have no mean whether the process is reversible or irreversible.

        Hope you understands it. 🙂

      1. Rahul Ranjan

        @RITESH you should go for lectures by PROF. S.K.SOM(basic thermodynamics) for clearification of doubts like one u have asked

      1. Ritesh Das

        thank you chetan for ur reply…and sorry for being late…actually i read pk nag again…cleared my doubt…only for reversible process we integrtae the term pdv and get p(v2-v1)…whereas for irreversible process we cant do the same…but we can use the term pdv for both the processes…and yeah nptel videos of prof. s k som were very helpful too..:)

      1. @Ritesh, there are many more very good NPTEL videos to clear our concepts but unfortunately as now we don’t have that much time. May be after exam we could watch them!!!

You must be logged in to reply to this topic.