 me shot mister sunil that I am cool currently assistant professor department of mechanical engineering vachan institute of technology solar per today I am going to deliver video session on coefficient of performance of an idea vapor absorption system at the end of this video session students will be able to derive an expression for COP of an ideal vapor absorption system contents veland  ninete  Punkt  fuego ictional  zodium 等  explosives 等   oversize  喔 The derivation of these video sessions is the derivation for the C O P of an ideal    अ बभ�  अ बब बब बल से� । nants of side system. In the case of v什麼 Очень System or desired if it is same it is there in a very Stockholm system. आपार �祂 मोँ प् andarbike alummaせः टी रसीक mnie ःी На कॉरहिइट म् dBbdh constrom मैं उ மोंगजना क skirtown काई भाब रावया कॉरह देखब nessa प् hasha pl說 iqay list p1 is earthquakes not . प् अड़ Mad해概une put kam छिहद और सब आप्झजना कpeas जिहचना याji misha j anyhow इतषको census . आप focusing j Wahl is given by qe upon wp plus qg so let us see in the following figure what are the various energy transfer that takes place in the vapor absorption system now if you look at this diagram this is the vapor absorption system which we have shown schematically so in this we have four main items one is the pump so pump is required for pumping the strong solution of ammonia and water in case of aqua-ammonia system let us consider we pump the strong solution of aqua-ammonia from the absorber to the generator so the first input is the pump work then in case of evaporator the heat is absorbed from the space in the evaporator so this we have shown as qe and it will be at temperature T evaporator we are assuming the temperature of the evaporator is T and heat absorbed is qe whereas in the generator temperature we are assuming is Tg and heat supplied in the generator is qg whereas in case of vapor absorption system the heat is rejected two items two components first is in the condenser and secondly in the absorber so total heat rejected is qe plus qc and which we will assume at temperature T0 or Tc both will be same so let us see what terminology we are going to use for in the derivation let us say qg is equal to heat supplied in the generator qp is equivalent to the work that is supplied so it is in the form of heat supplied equivalent to pump work to the refrigerant then qe is the heat absorbed by refrigerant in the evaporator and qc is here remember we are assuming qc is total heat rejected from condenser and absorber at temperature Tc ok we are not considering separate heat rejected total heat rejected to the atmosphere or cooling water from condenser and absorber at temperature Tc ok now we have seen that temperature in the evaporator generator is Tg temperature in the evaporator is T and temperature in the condenser is going to be Tc so let us see now what we are going to make while deriving this equation we are assuming that all the processes are reversible means without any losses there are no other heat transfer except consider whatever 4 heat transfers we have shown only those we are going to consider means there are no other heat transfers which will come into the picture and third important assumption is very valid assumption for vapor absorption system we are neglecting the pump work pump work being very small compared to the heat supplied in the generator usually it is neglected in case of vapor absorption system and it is negligible compared to heat transfer so we neglect the pump work now let us see how to start the derivation so according to first law of thermodynamics of conservation of energy we can write the total energy entering equal to total energy living now total energy living entering the system if we go back to this particular figure let us go back to this figure so that it will be clear to all of you let us see this figure the total energy entering into the vapor absorption system is summation of qg plus qe heat supplied in the generator whereas total heat coming out is qc which is sum of heat rejected to the cooling water in the condenser and absorber so by first law and because wp is neglected so we can write the first law equation and just now as we have seen here let us go back to that slide and let us see that equation so according to first law of thermodynamics we can write the energy balance as qg plus qe is equal to qc total energy entering equal to total energy living let us call this as equation number one now since we are assuming that this particular all processes are reversible this system is considered reversible according to the clauses equality for reversible system or reversible processes the net change in entropy must be equal to zero nothing but according to second law of thermodynamics therefore we can write by using the second law of thermodynamics change in entropy of generator delta hg plus change in entropy of operator delta hc plus change in entropy of condenser delta hc equal to zero so this net change in entropy has to be zero addition of net change in entropy so because we know that ds is equal to dq by t in differential form so s will be equal to q by t because temperature is constant we can write delta hg as qg upon tg by definition of entropy s is equal to q by t and temperature being constant we can directly write qg upon tg similarly for evaporator I can write qe upon tg similarly for condenser we can write qc upon tg but in condenser the heat is rejected qc is negative because heat is rejected so we get the equation qg upon tg plus qe upon tg minus qc upon tg equal to zero so by rearranging taking this term on the right hand side we can write the equation as qg upon tg less qe upon tg is equal to qc upon tc but as we have written by first law the equation number one that total energy entering is equal to total energy leaving so from equation one we already know that the qc that is total heat rejected in the condenser is sum of heat supplied in the generator plus heat absorbed in the evaporator so qc is equal to qg plus qe so if we put this qc equal to qg plus qe in this above equation then we will get the equation as the left hand side will remain same see here left hand side is same qg upon tg plus qe upon tg equal to qc we are replacing by qg plus qe so qc will be replaced by qg plus qe divided by temperature of qc now this term we can split further and we can write the equation as qg upon tg plus qe upon t is equal to qg by tc separation of the terms plus qe upon tc just we have separated the term now we will rearrange the equation to get the equation in required format so now what we have done we have brought the terms consisting of qe on one side and qg on other side so by rearranging the terms in the previous slide for example i will show you here in the previous slide what we are rearranging so we have already qe upon t over here so this qe upon tc i am taking on the left hand side so we will get qe upon t which will be equal to qg upon tc and this we take right hand side qg minus minus qg upon tg so by rearranging the term what we are getting is that qe upon tg minus qg upon tc is equal to qg upon tc minus qg upon tg it is simple rearrangement so taking qe and qg outside the bracket we will get qg into one upon minus one upon tc qg into one upon tc minus one upon tg now so when you simplify the bracketed term you get qg into tc minus te upon tc into qg into bracket tg minus tc upon tc into tg just simplification so finally we will get qe is equal to qg into te upon tc minus te into tg minus tc upon tg so finally we know that cop is what refrigerating effect upon its supply which is nothing but qe upon qg so if you put this qe upon qg in this equation I mean q term in this equation so qg will be there in the numerator and denominator that will get cancelled and we get the equation of cop as tc minus te into another bracket tg minus tc upon te remember these values are to be taken in degree kelvins so if you compare this then you will can see that this is the equation of Carnot refrigerator working between the temperature limit of te and tc whereas this is the equation of Carnot engine working between the temperature limit of te and tc so cop of vapor absorption system can be considered as product of cop of Carnot refrigerator into Carnot engine so let us see this in graphical form so vapor absorption system we can consider as combination of heat engine and refrigerator so heat engine operating between the temperature limit te and tc between te and te so this combination we can consider as vapor absorption system these are the references thank you