This video provides a comprehensive preparation strategy for Thermal Engineering, the most important subject in Mechanical Engineering for TSPSC AEE, APPSC AEE, and AMVI exams. The instructor, Mr. Malakondaiah Sir from ACE Engineering Academy with over 25 years of experience, covers the three main parts of Thermal Engineering: Basic Thermodynamics, Power Plant Engineering and Internal Combustion Engines, and Refrigeration and Air Conditioning. The session focuses on Basic Thermodynamics, covering topics like First Law, Second Law, Entropy, and Properties of Pure Substances, with expected marks ranging from 1-3 marks per topic. The instructor emphasizes problem-solving strategies, including understanding that not all given data in problems is necessary for solutions, and demonstrates key concepts through solved examples such as ideal gas expansion, critical point properties, and thermodynamic processes.
TSPSC AEE Thermal Engineering: Preparation Strategy & Imp Qs
Added:foreign [Music] platform because of small technical issue it got late Lisa don't feel otherwise in this session today we are going to discuss the preparation strategy and important questions for the most important subject in the mechanical branch that is thermal engineering dear friends today's session is very very useful for those students who are preparing for tspsc a double e exam it is not only useful for the tspsc awe exam but also useful for pspsc AE exam and also useful for am v i assistant motor vehicle inspectors and this class is also important for appsc awe exam and AE exam okay dear friends I am faculty at Ace Engineering Academy dear friends my hard work more than 25 years of experience and blessings from parents and gurus elevated me to teach almost all thermal oriented subjects from the basic level 2 high end level having this introduction let us start the subject called thermal engineering dear friends we have divided the thermal engineering into three parts namely the first part is basic Dharma Dynamics the second part is power plant engineering and internal combustion engines and third part is Refrigeration under air conditioning today we are going to discuss the most expected questions for the very first part that is basic dharamodynamics the second part will be discussed 24th of this month the same timing 8 30 to 9 30.
and 26th of December we are going to discuss the third part of the thermal engineering that is refrigeration and air conditioning MO before going to start the most expected questions for the basic dharmodynamics let us see the syllabus for the basic thermodynamics the very first part that is basic the basics of thermodynamics similarly heat and work first law of thermodynamics second law of thermodynamics entropy availability and irreversibility properties of pure substances and the last one dharamore dynamic property relations so these are the topics we are going to discuss under the basic thermodynamics and the expected weight is is like this from the very first part that is basics of thermodynamics minimum one Mark can be expected maximum two marks can be expected similarly from the Heat and work also minimum one Mark can be expected maximum two marks can be expected first law of thermodynamics it is the most important unit further basic part of the thermodynamics from the first law of thermodynamics minimum two marks can be expected maximum three marks can be expected similarly the second law of thermodynamics is also equally important you can expect minimum two marks maximum three marks from the second law of thermodynamics similarly from entropy minimum one Mark can be expected maximum two marks availability and irreversibility one Mark can be expected properties of the pure substances that is also equally important minimum two marks can be expected and the last one Dharma Dynamic property relations one Mark can be expected that is a expected weight is for the basic Dharma Dynamics okay now let us see the weight is which were asked in different pspsc awe exam and a exam in the year 2015 a double g 16 marks were asked that is a weightage of the basic thermodynamics 16 marks were asked in the year 2015 for the aw exam in the same year for the exam a 18 marks were asked 18 marks similarly for the hmw SSB Hyderabad Metropolitan water supply seaways board in the year 2015 12 marks were asked look at the same boat asked in the year 2020 22 marks from this You can conclude that minimum 15 marks can be expected from the basic Dharma Dynamics only from the basic thermodynamics minimum 15 marks can be expected okay now let us see the material for the basic thermodynamics the best reference book is our Academy material Ace material it is equally useful for both tspsc and appsc similarly the other reference books are object to question Bank in mechanical engineering by a panel of experts Aroma Publishing House and other reference bookies object to type questions for the mechanical engineering by DS Kumar dear friends remember you need not to purchase all the three materials one material is sufficient the best material for both tspsc and appsces is engineering Academy material that is the best material okay having this introduction part now let us take the most expected questions for the tspsc awe exam and appsc awe exam look at that very first question try to read it clearly problem understanding is very very important try to read it clearly dear students remember by the time you are solving the problem it is not compulsory to use the entire data which is given in the problem it means Sometimes some extra data will be given you need not to consider that extra data for solving the problem the problem solving is very similar to like how the mother is preparing the chicken curry while our mother is preparing the chicken curry she does not take the all ingredients available in the kitchen room you know the dalda is available kerosis is available sugar also available in the kitchen room but these are not the ingredients required for preparing the delicious chicken curry similarly while you are solving the problem in the exam it is not compulsory to use the entire data which is given in the problem okay look at the very first one during an experiment an ideal gas is found to obey an additional law that is v p square equal to constant v p square is equal to constant that is the condition given for the problem the gas is initially at a temperature of t let the initial temperature of the gas is T1 that is T and volume is V means initial volume v 1 equal to V V1 is equal to V V 1 equal to V when it expands to a volume 2 V it is expanding to a volume 2 V means final volume B to equal to 2 V the temperature will become means you are asked to decide what is the final temperature different options are given you are asked you to decide the final temperature by the time the ideal gas is expanding to increase the volume from we want to V 2 V 1 is v v 2 is 2v okay now let us take the another slide it is given that v p square a minute it is given that v p square equal to constant but for this problem the temperature limits on the volume limits are given initial volume V1 given V final volume v two given 2 V initial temperature T 1 P final temperature T 2 equal to how much but look at the relation relation is given in terms of v p square no pressure limits are given in the problem but it is given an important condition that the material for this problem is ideal gas the material for this problem is ideal gas what is the ideal gas equation the ideal gas equation is PV equal to RT for 1 kg mass of gas from this I can write that P equal to RT by V or P Square because here we have V into P Square now I am going to write the P Square in terms of temperature and volume P Square equal to R square T Square by V Square now you put this value in the place of P Square let me write that value B into P Square is R square T Square dividing by V Square is equal to a constant from this one V is canceled look at the V finally it is T Square by V is equal to constant by R square but here R is the characteristic gas constant since R is the constant constant by square of the constant is another constant Let It Be C finally we got the relation between temperature and volume T Square by V is a constant C okay from that you can decide it T Square by V equal to a constant T Square by V equal to constant it means P 1 square by V 1 is equal to T 2 square by v v two but what is the T1 given t one equal to T V 1 given is V V 2 given is 2v okay let me put that values here it can be written T 1 square T Square by V 1 is V equal to T 2 square T2 to be found T 2 square dividing by V 2 equal to 2 V now let us cancel the V finally finally we got T 2 square by 2 T 2 square by 2 equal to T Square T 2 square equal to 2 T Square or t 2 equal to root 2 T that is a final temperature of the ideal gas by the time volume of the gas is increasing from V to 2 V I hope all of you clearly understood so T 2 equal to root 2 t look at the options look at the option b option b is absolutely correct okay shall I move to the next question yes yes look at them next question try to read it clearly at its critical point any substance will this problem connected to critical point option A exists in all the three phases option b change directly from solid to Vapor option C loss phase distinction between liquid and Vapor option b behaves as an ideal gas first of all what is the meaning of critical point critical point is the point where the liquid is directly flashed to Super critical material the state where the liquid is directly flashed to supercritical material without having the observable phase change that critical point with the temperature versus entropy coordinate system temperature versus entropy coordinate system can be marked like this that is a saturation Dome the Apex point is representing the critical point ok then the Karu a b b is marketed as a critical point the Karu a b is called saturated liquid curve similarly the Karu BC is called saturated Vapor Karu this curve is saturated Vapor Karu so these are saturated liquid curve these are saturated Vapor curve the saturated Vapor Karu and saturated liquid Karu both are meeting at a critical point then this region is a liquid region this region is the two phase region where liquid and Vapor both are coexisting under thermal equilibrium state and the region right to the saturated Vapor Karu is the superheated region superheated region okay that means the region left to the saturated liquid curve is the liquid region is also called subcode region the region right to the saturated Vapor Karu is the superheated region the region within the saturation Dome is the two phase region the saturated liquid carbon saturated Vapor Karu both are meeting only at a critical point dear friends remember at the critical point at the critical point this is very very important information at critical point the specific volume of the saturated vapor is equal to specific volume of the saturated liquid specific volume of the saturated vapor is equal to specific volume of the saturated liquid these are the steam table notations similarly the entropy of the saturated vapor is equal to entropy of the saturated liquid similarly enthal if you have the saturated vapor is equal to enthal if you have the saturated liquid therefore the latent heat of vaporization therefore the latent heat of vaporization h of G equal to 0 similarly increased entropy during the evaporation is zero similarly the increase in volume during the phase change is zero the change in volume during the phase change at critical point zero means the saturated liquid is directly flashed to supercritical material without having the observable phase change it means at the critical point at the critical point there is no phase distinction between the saturated liquid under saturated water once again exactly at critical point since the saturated liquid is directly flashed to supercritical material there is no phase distinction between the saturated liquid unsaturated Vapor exactly at a critical point okay now look at the options look at the option C loss face distinction between liquid and Vapor therefore option C is absolutely correct okay similarly the other important information regarding the critical point of water for water at a critical point for water at critical point this is very very important information the pressure is called critical pressure is equal to 221 bar and the corresponding temperature is called critical temperature is equal to 374.14 degrees integrate okay similarly at critical point the dryness fraction at critical point dryness fraction generally dryness fraction denoted with small X cannot be defined cannot be cannot be defined at critical point the dryness fraction cannot be defined and one more important information the surface tension Sigma is equal to zero surface tension equal to zero at the critical point means when you heat the water with a pressure is equal to 221 bar or above that critical pressure there is no bubble formation during the evaporation process why the bubble formation is zero by the time the water is evaporating either at critical pressure or above critical pressure is the surface tension of the water at critical point is zero I hope all of you clearly understood the information connected to the critical point now let us take the another important question try to read the question try to read it clearly it is asked like this in case of power failure a battery is used to light a bulb a battery is used for lightning one bulb run FM running a fan and heat an electric iron and for heating the iron box each of 100 words rating each consuming the power equal to 100 Watts for 10 minutes how long the three fellows are running the three fellows are running in 10 minutes in this process the work done and heat supplied by the battery are given by for that particular process you are asked to decide what is the heat transfer what is the heat transfer and what is the work transfer okay for that let us take the battery arrangement let us take the battery arrangement let that is your battery is your battery for this battery what are the elements connected for this battery a bulb is connected let the hear the bulb the energy consumed by the bulb egg minute the energy consumed by the bulb is 100 Watts a fan is connected the energy consumed by the fan also equal to 100 Watts and one iron box is connected one iron box is connected iron box heater the energy consumed by heat are also equal to 100 Watts 100 words now you tell me the energies transferring from Battery to bulb fan and heater either with the temperature gradient or other than temperature gradient why I am asking that question is as we all know only a kind of energy which is crossing the boundary of the system due to temperature gradient is called thermodynamic Heat thermodynamic heat is having a special meaning that only a kind of heat energy which is crossing the boundary of the system because of the temperature gradient like this let us take a system let that is your system assume your system is maintaining the temperature 80 degree centigrade surroundings also maintaining 80 degree centigrade for this case the heat transfer Q equal to 0 why the heat transfer Q equal to 0 for this particular arrangement because the system and surroundings both are maintaining same temperature but if I take the another system let me take the one more system for this system the temperature may be 40 degrees Centigrade and the surroundings may be maintaining 20 degree centigrade there is a temperature gradient because of the temperature gradient some energy is Crossing from system to some energies crossing the boundary of the system that energy which is crossing the boundary of the system due to temperature difference between the system and surroundings is the heat transfer from that the moral is the heat transfer is happening only because of the temperature gradient for example while you are keeping the hard chicken peas in your mouth the chicken piece may be maintaining around 150 degree centigrade that tongue may be maintaining 20 degree Centigrade there is a temperature gradient between the hot chicken peas and tongue because of that temperature difference some energy is transferring from hot chicken peas to your tongue that energy which is transferring from hot body to cold body is called heat energy or heat transfer from that the moral is the heat transfer is becoming zero when there is no temperature difference between the system and surroundings next to come to the work energy because the energies which are crossing the boundary of the system may be in the form of heat energy or in the form of work energy dear friends remember any energy which is crossing the boundary of the system other than the temperature gradient is coming under Dharma Dynamic work once again any energy which is crossing the boundary of the system other than the temperature gradient here gradient means difference is called thermodynamic work okay now come to our problem for this case nothing is mentioned about the temperature of the battery under temperature of the different equipments it means there is no thermodynamic heat because there is no temperature gradient between the system and surroundings for this particular problem the thermodynamic Heat Q equal to zero okay now come to the Dharma or dynamic work the heater is consuming the work energy the fan is consuming work energy bulb is also consuming work energy because energy is Crossing from Battery boundary to different equipment other than temperature gradient the energy which is crossing the boundary of the system other than temperature gradient is coming under Dharma Dynamic work only ok now what is the work energy right total work energy energy consumed by the heater plus energy consumed by the fan plus energy consumed by the bulb energy consumed by the bulb hundred plus hundred plus hundred is equal to 300 joules per second because the energy is given in terms of what what equal to Joule per second but look at the options all the options are given in terms of kilojoules for that we need to convert from Joule per second to joule for that right let us consider the time 10 minutes time is right used for driving the gray equipment 300 into 10 minutes 10 minute one minute equal to what actually let me write it clearly let me write it clearly these now 300 300 joules per second multiplied by 10 minutes 1 minute equal to 60 seconds the second second cancel finally it is equal to 300 300 into 10 into 60 it is equal to 1 8 1 8 0 0 0 one eight zero zero zero one more Zero D is in joules that is the work energy given by the battery since all options are given in kilojoules let us convert 82 kilo joules it is equal to 180 kilo joules the data work energy so the heat transfer is 0 work transfer is how much 180 kilo joules that is the answer got it yes now look at the options look at the option b work transfer is 180 kilojoules heat transfer is 0 option b is absolutely correct okay similarly let us take the another important question try to read it clearly in an adiabatic process in an adiabatic process 500 of work are performed on a system work done on the system negative okay in the non-adiabatic process by which the system returns to its original state thousand joules of heat are added to the system heat is added to the system heat added to the system should be taken as positive the work done during the non-adiabatic process is this problem connected to closure system because nothing is given regarding the mass transfer since or until unless nothing is given regarding the mass transfer that should be considered as a closed system okay for a closed system let us take the one to two process one is is the adiabatic process one two two is adiabatic process and you know after the adiability process there is another process called non-adiabatic process by which the system returns to original state by which the system returns to original state means the two to one process is the non-adiabatic process let me write that point two to one let me write the 2 to 1 2 2 1 is the non adiabatic process non-adiabatic process both are happening in the closure system okay now for adiabatic process the work interaction is given 5000 joules of work are performed on a system it means work for one two two is equal to minus 5000 minus 5000 joules I hope you understood why I took the negative sign because it is given work done on the system work done on the system should be taken as negative okay in the non-adiabatic process by which the system returns to its original state thousand zoules of heat is added to the system that means for the two to one process heat transfer Q two to one is thousand joules it is positive because heat supplied the system is thermodynamically taken as positive okay what you are asking to decide you are asked to decide what is the work done in the non-adiabatic process means work done in the non-adiabatic process is how much that is a problem okay now as you know adiabatic process means what the process in which the heat transfer is zero the process in which the heat transfer is zero is called adiabatic process that means for the 122 process heat transfer is zero okay then what is the energy balance equation for a closure system energy balanced equation for a closed system is equal to general formula first let me write the general formula the general formula is heat transfer Q equal to work done plus change in internal energy this is the energy balance equation for a closure system I can call as man flow energy equation okay now it can be written as Q one two two first let us take the 1 2 to adiabatic process is equal to work for one two two plus u 2 minus U1 ok but what is the heat transfer for one two two heat transfer for one two to zero then 0 equal to work transfer for 1 to 2 minus 5 000 plus u 2 minus U1 from that I can write that u 2 minus U1 is equal to you bring this quantity to other side minus 5000 becomes Plus 5000.
plus 5000 joules that is a change in internal energy for the one two two process now come to the non-adiabatic process for the non-adiabatic process Q two two one non-adiabatic process happening from two to one is equal to W two to one final internal energy minus initial internal energy for the two to one process one is the final State that's why final internal energy minus initial internal energy let me take the another slide Q two to one is equal to W 2 to 1 plus u 1 minus U2 but what is a q two to one thousand Q two to one is thousand is equal to W 2 to 1 is unknown Value Plus look at the u 2 minus U1 look at the U2 minus u 1 is 5000 then what is the U1 minus u 2 because we need u 1 minus u 2 here minus 5000 minus 5000 joules okay so U1 minus U2 minus 5000 joules from this work for 2 to 1 equal to 1000 minus of minus plus 5000 is equal to 6000 joules that is the answer for this particular problem so work for the non adiabatic process six thousand joules look at the options plus six thousand joules therefore option A is absolutely correct okay I hope all of you clearly understood now let us take the another important question try to read it clearly try to read it the net energy transferred as heat and as work to a closer non-flow system non flow system means closed system the closed system can also be called as non-flow system is also called fixed Mass system okay anyhow the net energy transferred as heat and as work to a closed system is equal to change in different options are given entropy enthalophy internal energy temperature Etc what is the non flow energy equation non-fly energy equation is Q equal to W plus d u only then what is the Q minus w d the net energy difference between heat transfer and work transfer is equal to d u what is the d u d u equal to change in internal Edge only this du is the changing changing internal energy genuine internal energy therefore the difference between the heat transfer and work transfer for a closer system process is equal to change in internal energy option C is absolutely correct okay very easy problem but important similarly look at the next question try to read it clearly try to read it clearly it is asked like this if a quasi equilibrium quasi equilibrium is also called quasi-static process if a quasi equilibrium process exist we have assumed different options are given like the pressure at any instant to be constant B an isothermal process option C the heat transfer to be small option D the boundary motion to be infinite smelly small now let us take the quasi-static process take the rigid cylinder the rigid cylinder is one which does not go for any deformation that rigid cylinder is connected with a boundary called piston this is the piston assume the Piston is moving without friction without friction the Piston is moving let us take some gas let us take some gas inside the cylinder above the piston above the piston let us take some sand sand means right you know that sand weight is acting that sand weight is acting vertically downwards whereas this gas pressure Force present inside the cylinder is maintaining upward force on the piston by the combined effect of upward gas pressure Force on the downward weight of the sand assume the system is in equilibrium by the combined effect of gas pressure force on the piston the vertical weight on the Piston which is acting vertically downwards azum the whole system is in equilibrium when the system is in equilibrium assume the pressure of the gas is p volume occupied by the gas is V under temperature of the gas is T okay now what I am doing is since we are placing the sand on the top of the piston among the total and I am removing means I am lifting one sand particle among the total send I am lifting one sand particle once I lift one sand particle a little bit downward weight decreases because of the decrease of a little bit downward weight the Piston tries to move upwards by the time the Piston is moving upwards along with the Piston definitely gas is expanding by the time the gas is expanding the volume of the gas increasing along with the volume the pressure and temperature are also changing it means by the time you remove one sand particle the properties of the gas may be changing from P2 P1 volume v to V1 and temperature T to T1 similarly when I lift another sand particle when I lift another sand particle again the Piston is moving up along with the piston the gas is also expanding by the time the gas is expanding definitely properties are changing from these properties to some other properties the pressure may be changing from P1 to P to volume may be changing from V1 to V2 similarly temperature may be changing from T1 to T2 like that the properties are keeping on changing by the time you are removing the sand particles one after other now you try to replace the second sand particle once you replace the second sand particle the downward weight increases once the downward weight increases the Piston comes down then the properties may be changing from P2 to P1 T2 to T1 V2 to V1 like that once I remove once I replace once I replace all sand particles the Piston can get back to its original state then the gas can maintain its original properties like P be the pressure VB the volume T be the temperature such a process is called quasi-static process now you tell me what is the movement of the boundary here piston is the boundary what is the movement of the boundary during the quasi-static process dear friends remember quasi quality means almost Cosi means almost C static means static means equilibrium equilibrium so cozy static process is also called Almost equilibrium process in the almost equilibrium process the boundary is moving very very slowly the Quasi static process is the best example for irreversible process got that point yes now look at the options look at the option D the boundary motion to be infinitely small infinite smell is small means very very slow okay therefore option D is absolutely correct okay are you following clearly yes now let us take the another important question try to read it clearly try to read it a heat Reservoir at 900 Kelvin is brought into contact with the ambient at 300 Kelvin for a short time a heat Reservoir is there maintaining 9000 Kelvin is open to Atmosphere maintaining 300 Kelvin during this period nine thousand kilojoules of heat is lost by the heat Reservoir it is like this let us take the heat Reservoir I can say the hard source hard Source given that hard source is maintaining the temperature 900 Kelvin it is exposure to atmosphere let the heeda atmosphere atmosphere is maintaining the temperature Let It Be t a given 300 Kelvin since the hard source is exposed to the atmosphere naturally heat is transferring from hot source to Atmosphere because the atmosphere care is maintaining low temperature how much amount of heat is transferred the amount of heat transferred is nine thousand kilo joules it is nine thousand kilo joules ok then the total loss of availability due to this process means we are asked you to decide the loss of available energy okay dear friends remember the loss of available energy loss of available energy is equal to T naught that is a atmospheric temperature into d s for the universe that is also called as net change in entropy that is a mathematical formula or general formula for deciding loss of available energy now here the atmospheric temperature itself is marked as dead State temperature right written with t naught I know the T naught but I don't know the net change in entropy that net change in entropy to be found ok now the net change in entropy listen carefully net change in entropy is equal to change in entropy of the hard source plus change in entropy of the surroundings means atmosphere okay it is equal to it is equal to what is the change in entropy otherwise let us write it separately what is the entropy of the heart source change in entropy of the heart source is it hard source losing energy gaining energy is it hard Source losing energy or gaining energy heart source is losing energy thermodynamically losing heat energy is negative therefore minus Q by minus Q by hard Source temperature why I took the minus sign because the hard source is losing the heat energy to atmosphere okay it is equal to minus minus 9000 dividing by 900 it is equal to minus 10 kilo joules per Kelvin okay similarly look at the change in entropy of the atmosphere receiving energy or losing energy the atmosphere is receiving energy receiving heat energy thermodynamically taken as positive therefore plus Q by the atmospheric temperature Q equal to 9000 9000 dividing by dividing by 300 it is equal to 30 kilo joules per Kelvin then what is the net change in entropy the net change in entropy is equal to entropy change of the hard Source plus entropy change of the atmosphere here entropy change of the hot Source look at the entropy change of the hard Source entropy change of the heart source is minus 10 plus entropy change of the atmosphere plus 30 is equal to 20 kilo joules per Kelvin I got the net change in entropy then the loss of available energy T not equal to what 300 Kelvin 300 Kelvin into net change in entropy 20 you simplify this value it is equal to six thousand six thousand kilo joules that is a loss of available energy now look at the options look at the options option C is absolutely correct these are all very very important problems shall I go for the next question yes look at them next question try to read it clearly try to read it clearly for two cycles coupled in series the topping cycle has an efficiency 30 percent and the bottoming cycle has an efficiency 20 percent the overall combined cycle efficiency the Top Cycle efficiency given Top Cycle efficiency given 30 percent and the bottom cycle efficiency given 20 percent both Cycles are connected in series then the overall efficiency if both Cycles are connected in series is equal to the Top Cycle efficiency plus bottom cycle efficiency minus Top Cycle efficiency into bottom cycle efficiency this is a mathematical formula try to remember this what is the top cycle efficiency 30 bottom cycle efficiency 20 minus minus 30 into 20 30 into 20 otherwise 0.3 plus Point these are all the percentages point to minus 0.3 into 0.2 point three plus point two point five minus 0.3 into point two point zero six you simplify this value 0.44 it means the overall efficiency of the system is 44 percentage okay look at the options look at the option b option b is absolutely correct this is very very important problem I can say that this is a five star problem most expected problem for the competitive exams okay similarly let us take the next question try to read it clearly try to read it clearly it is asked like this in a reversible cycle The Source temperature is 227 degree centigrade in a reversible cycle the hard Source temperature given two to seven degrees Centigrade and the sink temperature you know single temperature means cold Source temperature the sink temperature or lowest temperature is equal to 27 degrees Centigrade the maximum available work for a heat input of 100 kilo joules if the heat input let me write here the heat supplied equal to the heat supplied equal to 100 kilo joules 100 closures what is the work output of the cycle you are asked to decide the work output of the cycle ok now better to convert the temperatures in absolute form that is Kelvin form two to seven plus 273 it is equal to 500 Kelvin similarly the lowest temperature 27 plus 273 it is equal to 300 Kelvin okay then what is a reversible Heating and efficiency reversible Heating and efficiency is nothing but Carnot cycle efficiency that is equal to 1 minus lowest absolute temperature by highest absolute temperature okay it must be equal to network output by heat supplied okay from that the network output of the heat engine cycle equal to heat supplied into 1 minus lowest temperature by highest temperature but what is the heat input given heat input given 100 kilojoules 100 kilo joules into 1 minus lowest temperature is 300 Kelvin lowest temperature 300 Kelvin highest temperature 500 Kelvin 300 by 500 simplify this value 100 into 1 minus 300 by 500.26 1 minus 0.6.4 0.4 into 100 equal to 40 kilo joules that is a maximum expected network from the reversible heat engine okay look at the options look at the option C option C is absolutely correct okay similarly look at them another important question try to read it clearly the fundamental unit of enthalpy is you're asked you to decide the unit for the nth dear friends remember enthalophy is a kind of energy is a kind of energy what is the unit of energy joule unit of energy is Joule one joule equal to what Newton meter one joule equal Newton meter you know Newton is the unit of force Newton is the unit of force for shape equal to let me write here Force F equal to m a m is the mass dimensional right unit capital m l by T Square acceleration velocity by time simple mathematics School level mathematics m l t power minus 2 that is a unit of force that is a no unit of force or dimensional formula Newton e m l t power minus 2 into meter means distance unit l the unit is m l Square t power minus 2 that is the dimensional formula for the nth look at the option C option C is absolutely correct okay are you following yes similarly look at the another important question try to read it clearly try to read it clearly a Carnot engine receiving Heat at 400 Kelvin has an efficiency of 25 percent the cornered heat engine efficiency is equal to 25 percent means 0.25 of course it is given the temperature where it is receiving that may be called as hot body temperature 400 Kelvin hot Source temperature is 400 Kelvin refrigerator working between the same temperature limits you're asked her to decide the Carnot cycle refrigerate the cop of The diversity Carnot cycle the cop of The Riverside corner cycle is how much for the same temperature limits okay for that first let us take the cop of the heat pump cop of the heat pump is what 1 by efficiency of the heat engine cop of the heat pump is equal to one by efficiency of the heat engine it is equal to 1 by what is the efficiency of the heat engine 0.25.25 1 by 0.25 how much one by point two five four that is the cop of the heat pump but what it is asking it is asking the cop of the refrigerator not heat pump is asking cop of the refrigerator for that we have the relation between the cop of the heat pump on the coop of refrigerator cop of heat pump equal to cop of the refrigerator plus one that is the relation between the coop of the refrigerator and cop of the heat pump from this The Coop of the refrigerator for the same temperature limits dear friends these relations are valid as long as the temperature limits are remaining same for heat pump heat engine on the refrigerator the temperature limits are differing I cannot use the same relations anyhow c o p of the refrigerator equal to c o o p of the heat pump minus 1.
what is the cop of the heat pump four minus 1 is equal to 3 that is the cop of the refrigerator okay look at the option C option C is absolutely correct dear friends look at them hard Source temperature given for this problem we did not use that hard Source temperature anywhere for solving the problem that means this is the extra data or super polar data sometimes the extra data will be given in the problem you just ignore it okay similarly look at them another important question pressure reaches a value of absolute zero you know the absolute zero pressure is found only at Absolute Zero state absolute zero State means what the state at which the molecular momentum is 0 is called absolute zero state the state at which the molecular momentum is 0 is called absolute zero state that absolute zero state you can decide like this for that let us take the volume versus temperature coordinate system volume versus temperature coordinate system now let us take the gas in a container let us take the gas in a container assume the gas present in the container is maintaining the temperature 30 degrees Centigrade and volume of the gas present in the container is 5 meter cube these assumed value randomly has selected corresponding to this temperature and volume corresponding to temperature you try to project a vertical line representing 30 degree centigrade and corresponding to volume that is 5 meter cube try to project a horizontal plane representing 5 meter cube wherever the temperature line and volume line are meeting that is the initial state of the gas the initial state of the gas initial state of the gas initial state of the gas okay then what you do is you place this gas container in your domestic refrigerator you place this gas container in the domestic refrigerator assume your domestic refrigerator is maintaining 0 degree centigrade that means the gas temperature keep on dropping to zero degree Centigrade when you keep this container in a domestic refrigerator where the temperature is zero degree centigrade that means by the time the gas temperature is dropping the volume occupied about the gas due to Free Motion keeping on decreasing that means by the time gas temperature is dropping volume occupied by the gas is also dropping but even after the gas reaches to zero degree Centigrade this is the volume occupied by the gas due to free motion the volume occupied by the gas due to Free Motion at the zero degree centigrade is V naught okay then what you do is you extend the negative temperature coordinate you extend the negative temperature coordinate like this you try to extend the negative temperature coordinate and you tried the you try to place the gas container in the cryogenics dear friends remember the method of artificially providing and maintaining up to minus 100 degrees Centigrade can be called as Refrigeration the process of providing and maintaining up to minus 100 degree Centigrade can be called as Refrigeration below that minus hundred degree centigrade is called cryogenics okay generally the blood banks are working with the cryogenic refrigeration where the temperature is below minus 100 degrees Centigrade anyhow once I keep this gas container in the cryogenics where the temperature is below zero degree Centigrade then what is happening the gas temperature the gas temperature is further decreasing by the time the gas temperature further decreasing the volume occupied by the gas keep on decreasing then at a particular temperature at a particular temperature the volume occupied by the gas due to Free Motion is zero that means if we keep on cool the gas if I keep on cool the gas to a temperature minus 273.15 degree centigrade if you cool the gas to a temperature minus 273.15 degree Centigrade at that particular temperature no molecule present in the gas unable to move from one location to another location that particular State you can call as absolute zero state absolute zero state then what is the definition for absolute zero state the state at which the molecules present in the gas unable to move from one location to another location is called absolute zero state the pressure of the material at that state is called absolute zero temperature and the corresponding temperature is called the pressure corresponding to absolute zero state is called absolute zero pressure and the corresponding temperature is called absolute zero temperature okay once again if I cool the gas to minus 273.15 degree centigrade the molecules which are present in the ideal gas are any gas unable to move from one location to another location the state the the state where the molecules are unable to move from one location to another location is called absolute zero state the temperature of the material at that absolute zero state is called absolute zero temperature and a corresponding pressure is called absolute zero pressure and one more important information is at this absolute zero State the CP of the any material is equal to CV that means specificated constant pressure and specificated constant volume for any material for any material the CP and CV are identical only at Absolute Zero state then what is the CP by CV CB by c b equal to gamma since the CP and CV both are identical gamma equal to one that means for any material at Absolute Zero State the adiabatic index gamma equal to 1. the adiabatic index gamma equal to 1 at Absolute Zero State means the adiabatic process and isothermal process both are identical only at this absolute zero state once again at the absolute zero State the isothermal process and adiabatic process both are identical because you know the adiabatic index gamma equal 1 at Absolute Zero state is it clear understood yes now let us see the answer pressure reaches a value of absolute zero pressure when molecular momentum of the system is becoming zero option D is absolutely correct understood clearly I try to give the complete information for the each and every question though it is taking some time please wait okay similarly look at them another important question try to read it clearly two reversible engines operating between the thermal reservoirs at a 1200 Kelvin under T2 Kelvin under 300 Kelvin such that first engine receives heat from 1200 Kelvin Reservoir and rejects heat to Thermal Reservoir at T2 Kelvin while the second heat engine receives at T2 Kelvin and Rejects at 300 Kelvin the efficiency of both engines is equal what is the value of the temperature T2 OK it is like this there is a there are two cornered heat engines let this is a car not heat engine one engine 2.
okay in between the in between the two car not heat engines let us take the intermediate source intermediate source now it is given that it is given that for the first cornered heat engine heat is available from the hard source is maintaining t one equal to 1200 Kelvin that hard source is giving Q an amount of heat energy a path of heat energy is converted to work energy the remaining is dumped to the intermediate source while intermediate source is maintaining T2 temperature okay now here intermediate source is acting as the cold source with respect to First Corner heat engine ok now it is given that that temperature the you know the cold Source temperature with respect to first heat engine is equal to hard Source temperature with respect to Second heat engine because it is clearly given that first car not cycle is working between 1200 Kelvin under T2 and the second Corner cycle is ask it to work between T2 Kelvin and 300 Kelvin that means this intermediate source is acting as a cold source with respect to first car not heat engine the same heat the same intermediate source is acting as a hot source with respect to second car not engine that means the second cornered engine receives Q2 amount of heat energy from the intermediate Source a part of this Q2 is converted to work energy the remaining energy is dampered to the cold Source while the cold source is maintaining T3 temperature let that is not let that is given as 300 Kelvin once again let us read the problem two reversible engines operating between thermal laser wires at 1200 Kelvin T2 Kelvin and 300 Kelvin such that first engine receives heat from tolerant Kelvin first engine receives heat from 1200 Kelvin and rejects heat to the reservoir at T2 Kelvin the first heat engine is dumping the heat to the reservoir at T2 Kelvin similarly similarly the second engine receives at T2 Kelvin the second engine receives a t to Kelvin and dumping it to 300 Kelvin then what we are asking to find the efficiency of both engines equal what is the value of T2 it is like this condition is given efficiency of the first coordinate engine equal deficiency of the second coordinate engine but what is the there is a condition which is given clearly given the efficiency of both engines equal if efficiency of both engines equal what is the value of T2 ok now you know efficiency of the first heat engine is 1 minus T2 by T1 only 1 minus P two by T1 efficiency of the second heat engine 1 minus T3 by T2 1 minus lowest temperature dividing by highest temperature for that particular cycle 1 minus T3 by T2 let us simplify that part from this I can write that this one cancel negative sign is canceled T2 by T1 is equal to T three by P2 from this I can write that t 2 square equal to P 1 T 3 T 2 equal to root under T1 T3 so that is a intermediate Source temperature when the two heat engines are maintaining equal efficiency okay now T 2 equal to P two equal to root under T1 T3 T 1 T 3 it is equal to root under T1 is how much 1200 Kelvin 1200 Kelvin into T3 T3 is 300 Kelvin 300 you simplify this value for getting 600 Kelvin that is the 36 okay correct absolutely correct this is the absolutely correct answer is it clearly understood yes but if the same problem is given for equal work if if the work of the first engine is equal to work of the second engine what is the work of the first engine work out the first engine is q1 minus Q two Q one minus Q two work of the second engine Q 2 minus Q3 Q to be the heat received by second engine Q3 with the heat dumped by second engine therefore W2 equal to Q2 minus Q3 okay now if the work output of the both Cycles remain same work output of the first engine Q one minus two Q two is equal to Q2 minus Q3 from this you can also write 2 Q 2 equal to Q one plus Q3 or the Q two the heat dump to intermediate Source equal to q1 plus Q three dividing by 2.
okay but dear friends remember when a heat engine is constructed with all reversible process and with the two temperature limits for any heat engine or for any reversible heat engine for any reversible heat engine if it is constructed between the two temperature limits the heat transfer always proportional to temperature of the source since heat transfer is proportional to temperature for any reversible heat engine operating between the two temperature limits this Q2 I can write as Q2 is proportional to T2 Q2 is proportional to T2 equal to T1 plus T3 dividing by 2.
you try to remember this formula dear friends remember if the two reversible heat engines are connected in series for getting the equal work the intermediate Source temperature equal to automatic mean between the hot Source temperature and cold Source temperature but if they are constructed to have the equal efficiency to have the equal efficiency the you know ah intermediate Source temperature equal to geometric mean temperature between the hard source and cold source is it understood clearly yes now let us take the another important question try to read it clearly what are the properties of thermodynamic system whose value for the entire system is equal to some of their values for individual parts of the system you know the thermodynamic system is having two kinds of properties namely extensive property and intensive property the property which is depending on the size of the system or mass of the system is coming under extensive property the property which depends on size of the system or mass of the system is called extensive property the best example for extensive property is volume let the volume of the system is 5 meter cube but there are some properties like temperature does not depend on the size of the system the property which is not depending on size of the system is called intensive property the best example for intensive property is temperature let the temperature is 30 degree centigrade that means this system is maintaining 5 meter cube volume as extensive property under 30 degree Centigrade as intensive property okay now let us cut the system now let us cut the system into three blocks let us cut the system into three blocks these are block one block two block three though you cut the system into three blocks the temperature of the each block remains same that is 30 degree centigrade since temperature of the each block remains same that is 30 degree Centigrade that is equal to original temperature from this You can conclude that temperature is not depending on the size of the system but here volume V1 of the first system is not equal to 5 V 2 is not equal to 5 and V3 also not equal to 5.
volume is going to change with the change in size or mass of the system since the volume is changing with change in size of the system You can conclude that volume is coming under the size dependent quantity that kind of size dependent quantity thermodynamically called as extensive property but the total volume v equal to V1 plus V2 plus V3 that must be equal to 5 meter cube that means the summation of the extensive properties are the different blocks cut from the main block is equal to original original extensive property the summation of the V1 V2 and V3 definitely equal to original volume of the system that means the summation of the properties of the system is equal to original property that kind of property is coming under extensive property ok now look at the options look at the option b once again read the question what are the properties of thermodynamic system whose value for the entire system is equal to some of their values extensive property understood clearly yes now let us take the another important question try to read the another important question try to read it the throttling of certain gases may be used for getting the refrigeration effect the throttling process may be used for refrigeration effect the best example for the throttling process connected to refrigerator is capillary tube process capillary tube is constructed to work by throttling process okay what is the value of joule terms and coefficient for the cooling region of such throttling process you're asked to decide the Joule Thompson coefficient for the cooling region of the throttling process okay it is like this for that let us take the temperature versus pressure coordinate system let me draw a big diagram let us take the temperature versus pressure coordinate system for the throttling process ignit temperature versus pressure coordinate system temperature versus pressure coordinate system you know throttling process is coming under irreversible adiabatic steady flow process throttling process is the irreversible adiabatic steady flow process for the throttling process enthalpy remains constant since enthalpy remains constant for the throttling process throttling process can also be called as isentolithic process remains constant for throttling process the throttling process can also be called as ice enthalaphic process next to since it is irreversible process the Dharma Dynamic sense for a irreversible process irreversible process must be drawn by a broken line or discontinuous line and one more important information regarding the throttling process the throttling process is for dropping the pressure 32 the pressure of the refrigerant in the throttling process or in the capillary tube is dropping due to friction since the frictional pressure drop is happening since the frictional pressure drop is happening in the capillary tube process or throttling process that is a practical reason the throttling process is coming under irreversible process that means friction effect is not equal to zero for the throttling process anyhow the throttling process is for dropping the pressure that means pressure of the refrigerant keep on dropping more recently dual Thompson experimentally proved that though the pressure of the refrigerant is keeping on dropping during the throttling process initially the temperature of the refrigerant initially the temperature of refrigerant is increasing up to certain length of the process after certain length of the process the temperature starts to drop along with the pressure that is how the throttling process can be represented now one two two is the throttling process representation since throttling process maintaining constant enthalpy throughout this one to do process remains constant therefore the one two two process is maintaining constant enthal of a that means for the entire one to do process nth Alpha remains constant and Jolson has given a coefficient called Joule Thompson coefficient the Joule Thomson coefficient mu j t equal to dou T by dou P at constant enthalpy Joel Thompson has given a coefficient called Joule Thompson coefficient the Dual Thompson coefficient equal to dou T by dou P at constant enthal of a dear friends remember Joel Thompson coefficient is the slope of eisenthalophic curve with TP coordinate system the Joule Thompson coefficient is the slope of isenthalpic curve with the temperature versus pressure coordinate system ok now look at the very first half process let us Mark a point K here 1 2 K is the first half process k 2 2 is the second half process during the entire process pressure is keeping on dropping that means the change in pressure dou p for the entire process is negative why it is negative the pressure keeping on dropping final pressure minus initial pressure is the Dopey that is negative but look at the first half process for the first hour process though the pressure is dropping the temperature increasing the temperature increasing for the first half process let me write that point the dou t dou t temperature increasing means dou T is positive dou T is positive for the first half process dou p is negative dou p is negative dou T positive dou P negative 2 for the first half process therefore the Joule Thompson coefficient mu J T is negative the Joule Thompson coefficient is negative for the first half process that first half process I can call as warming process the first half process I can call as warming process why that process can be called as warming process the temperature increases thermodynamically during the warming process or heating process temperature is increasing since temperature is increasing in this particular process that's why I am calling these warming process then what is the Dual Thompson coefficient for the warming process for the warming process Joule Thompson coefficient is negative okay let me write that point mu J T Joule Thompson coefficient is negative for for what kind of process warming or cooling for warming process for warming process the Joule terms and coefficient is negative okay similarly look at the next half process for the next hour process now Dopey continuously negative but dirty look at the dirty in this region dou T also negative 2 because temperature is decreasing dou T negative dou P negative then what is the Joule Thompson coefficient the Joule Thompson coefficient mu j t equal to dou T by dou P at constant enthalpy since dou T negative dou P negative 2 both are negative Thompson coefficient is positive coefficient is positive since temperature is decreasing in this process that process You can conclude as the cooling process during the cooling process the temperature is decreasing since the temporary is decreasing for the next half process that process You can conclude as cooling process further cooling process what is the Joule compression coefficient positive yes so Joule Thompson coefficient mu JT is positive is positive is positive 2 let me write it clearly Joe Thompson coefficient is positive 2 for the cooling process this is very very important information coefficient is negative for the warming process whereas Joule Thompson coefficient is positive for the cooling process ok now exactly at the Apex point the Joule Thompson coefficient is 0 that particular point you can call as inversion point inversion point dear friends remember inversion Point means the point on the essential the point on the ice enthalpy curve where the Joule Thompson coefficient is 0 is called inversion point okay similarly the Joule Thompson coefficient mu j t equal to 0 for ideal gas for ideal gas okay having this information let us decide the answer the Joule Thompson coefficient for the cooling region for the cooling region for the cooling region Joel Thompson coefficient is positive Joule Thompson coefficient positive means it is greater than 0 therefore you can select the option D is it understood clearly yes dear friends I have selected concept oriented problems therefore you will get the better knowledge okay similarly look at the another important question try to read it clearly according to the Maxwell relation which of the following is or correct you are asked to decide the Maxwell relation dear friends remember there are four Maxwell relations there are four Maxwell relations all the four Maxwell relations connected to four properties namely entropy is pressure p temperature T volume v all the four Maxwell relations connected to four properties entropy is pressure p temperature T volume v okay you try to remember the same order by remembering a word called Spore two by remembering a word called sportive you try to remember that order s p t v okay then what you do is first take the equal symbol first take the equal symbol write the differential form of entropy in the left side numerator write the differential form of entropy in the left side numerator then you move in the same order in the clockwise Direction suppose p o we moved in the clockwise Direction since it is moved in the clock Direction right it should be right it should be written with a positive sign okay now since here temperature is there put air temperature volume is there put your volume this is the one of the Maxwell relations understood yes now you take the equal symbol write the differential form of entropy in the left side numerator move in the anticipation with the same order with negative sign move in the anticlockwise Direction with negative sign in the same order let me read SPO SPO p t o since it is written in the anti-clockwise direction take the negative sign that negative sign you can write in any side sincere p is there put your p t is there put here T this is another Maxwell relation these are one Maxwell relation these are another Maxwell relation understood similarly you take the differential form of entropy in the right side denominator write the differential form of entropy in the right side denominator with the same order move in the clockwise Direction with a positive sign for t o since it is written in the clockwise Direction it should be written with a positive sign then here pressure is there put air pressure entropy is there put your entropy this is the another Maxwell relation okay similarly the last maximal relation write the differential form of entropy in the right side denominator move in the anti-clockwise direction with the same order sptv for t o since it is written in the anti-clockwise direction you write the negative sign in any side I am writing here now here volume is there or if entropy is there put your entropy volume is there put here volume another Max violation that is the fourth maximum relation of course there is no order there are four Maxwell relations the four maximal relations are connected to four properties like sptv you try to remember this order by remembering a word called sport2 for easily remembering the Maxwell relations ok now let us take the options look at the option C option C how we have written we have taken differential form of entropy in the left side numerator Spore t o with positive positive sign because it is written with a positive direction milk clock by Direction since your temperature is there put your temperature volume is there put your volume this is one of the maximum relations there is no order there is no order so look at the third relation for two this is absolutely correct option C is absolutely correct is it clear understood you try to remember this Logic for easily remembering the four Maxwell relations is it understood clearly yes now let us take the another important question try to read it try to read that question for a gas that is allowed to expand reversible adiabatically there is no change in the process reversible reversible adiabatic the process is reversible adiabatic process as we all know adiabatic process means the process in which the heat transfer is zero reversible means all internal irreversibility is zero reversible means all internal irreversibility is zero if all internal irreversibility is zero definitely the entropy generation equal to zero okay now let us take the general formula to decide the entropy change general formula for deciding the entropy changes Del Q by T plus entropy generation okay now reversible process means all internal irreversibilities are zero if all internally reversibilities are zero entropy generation equal to zero adiabatic means q0 if Q is 0 the right the Del Q 0 if Del Q 0 Del Q by t equal to zero Del Q by t equal to zero means entropy transfer due to heat transfer is zero OK since Del Q by T is zero entropy generation 0 d s equal to zero DS means what change in entropy S 2 minus s 1 S 2 minus S one equal to 0 S 2 is equal to S1 means entropy remains constant for the reversible adiabatic process so at that point since entropy remains constant for the reversible adiabatic process this reversible adiabatic process you can also call as isentropic process okay look at that point for a gas that is allowed to expand reverse per diabetically there is no change in entropy change in entropy 0 for a reversible adiabatic process understood clearly yes similarly look at them please wait we'll have the some more important questions try to read it clearly the expression integral pdv can be used for obtaining work you know dear friends remember non-floor reversible work non-flow reversible work non-flow reversible work equal to integral one two two integral one two two p d v integral one two to p d v is the non flow reversible work similarly steady flow work steady flow work steady flow work equal to minus integral one to two VDP steady flow work formula is minus integral one two to v d p whereas non-flow reversible work formula is integral one two two pdv look at the options the expression integral pdv is for non flow reversible process therefore option C can be selected understood clearly yes similarly look at the another important question try to read it clearly please wait write some more important questions are there since we started late and we had taken most important questions right patience is important please wait some more time we'll complete the problems an ideal gas at 27 degrees Centigrade is heated at constant pressure at constant pressure till its volume becomes 3 times initial temperature of the gas T1 equal to 27 plus 273 is equal to 300 Kelvin the gas is heated at constant pressure the gas is heated at constant pressure until the volume becomes three times means final volume equal to 3 times the initial volume okay then what would be the final temperature of the gas you are asked to decide final temperature of the gas it is given for ideal gas for any ideal gas for any listen carefully for any ideal gas for any ideal gas at constant pressure at constant pressure V2 by V1 is equal to T2 by T1 ok but V two is given 3 V or 3 V 1 3 V 1 3 V 1 by B1 is equal to T 2 by T 2 by T1 for the T1 value 300 300 this V 1 is getting canceled V1 is canceled finally P two equal to 3 into 300 900 Kelvin but look at the options all options are given in degree Centigrade let me change it to degree Centigrade 900 minus 273 is equal to 620 620 seven eight minute 6 27 already it got lit but the better to complete all the problems 627 degrees Centigrade that is the final temperature of the gas understood these are you know according to constant pressure at constant pressure volume is directly proportional temperature from this I have written this particular relation understood yes so option is option b 627 degree Centigrade absolutely correct okay shall I go to the next problem try to read it clearly yes try to read it clearly the relation BS equal to d q by T where s is representing entropy Q represents heat T represents temperature absolute holds Good in which one of the following process you know for a reversible process for a reversible process reversible process means what internally reversible to I equal to zero if internal reversible I equal to 0 entropy generation equal to zero okay then what is the general formula for finding the change in entropy DS equal to Del Q by T plus s generation s generation the S generation equal to 0 for a reversible process since s generation equal to 0 for a reversible process d s equal to Del Q by T d s equal to Del Q by T only for a reversible process okay d s equal to Del Q by T only for a reversible process anyhow the d s equal Del Q by T only for a reversible process look at the option a reversible process for a reversible process since entropy generation equal to 0 d s equal to Del Q by T because d s equal to d s equal to what Del Q by T plus s generation for a reversible process s generation equal to zero this has generation equal to zero then d s equal to what Del Q by T only for a reversible process cut that point yes now let us take the another important question try to read it clearly at a triple point of a pure substance at a triple point of a pure substance look at the triple point state take the water in a container let us take the water in a container I had taken the water in a container with some empty space that empty space is occupied by atmosphere or atmospheric gas once the empty space is occupied by atmospheric gas the atmospheric pressure is acting on the surface of the water isn't it the atmosphere gas pressure is acting on the surface of the water now what I am doing is I am connecting a vacuum pump here this is the vacuum pump once I connect a vacuum pump for the system that vacuum pump can lift the gas that vacuum pump can lift the gas present on the surface of the water once the gas which is present on the surface of the water is lifted the pressure which is acting on the water the pressure of the gas which is acting on the surface of the water keeping on reducing once the pressure on the surface of the water reaches to point zero zero six one one bar once again by the time the vacuum pump is lifting the gas present on the surface of the water the pressure which is acting on the surface of the water keeping on reducing once the pressure on the surface of the water reaches two point zero zero six one one bar a part of the water a part of the water is becoming ice a part of the water is available in Steam the remaining part of the water is available in liquid state okay once the pressure on the surface of the water reaches to point zero zero six one one bar a part of the water is becoming ice that is solid a part of the water is becoming steam that is vapor a part of the water is available in the liquid state that means liquid phase Vapor phase and solid phase these three phases are coexisting under equilibrium State at this particular pressure that particular point is called a triple point state the definition is like this the state where the solid phase liquid phase and the gas phase of the same material are co-existing under Dharma dynamic equilibrium state is called triple point state the pressure at the triple point state is the PTP equal to point zero zero six one one bar and the temperature at the triple point state sorry temperature the triple point State point zero one five degree centigrade these are triple point pressure and this is a triple point State temperature anyhow triple point State means the state at which the three phases of the material are coexisting under equilibrium state is called triple point state got that point yes now look at the options look at the option D solid liquid and Vapor phase exist yes at the triple point State solid phase liquid phase and Vapor phase are co-existing got that point yes therefore option b is absolutely correct okay similarly look at the another important question try to read it try to read it clearly the process in which both low grade energy that is q and high grade energy W are identical dear friends remember only for isothermal process only for isothermal process both heat energy and work energy both are identical here heat is coming under low grade energy work is coming under high grade energy you know the meaning of high grade energy high grade energy means the energy which can be completely converted to useful energy the energy which can be completely converted to useful energy is called high grade energy example for high grade Energy electrical energy mechanical energy hydraulic energy wind energy tidal energy these are all coming under high grade Energies but heat energy cannot be completely converted to useful energy since heat energy cannot be completely converted to useful energy that's why heat energy is coming under low grade energy you may have doubt that sir how the heat energy cannot be completely converted to other form of energy you are supplying the petrol to your bike engine that petrol is firing inside the combustion chamber because of the burning of the petrol inside the combustion chamber some heat energy is liberated if the entire heat energy sorry if the entire heat energy which is released by the firing of the petrol is converted to work energy for a driving your bike the exhaust gases must be coming out from the bike Engine with atmospheric temperature but if you touch the exhaust gases while they are leaving from the bike engine they are maintaining very high temperature in fact your hand will be burning if you keep the Hand by the time the combustion gas are leaving from the engine of the bike that means there in your bike engine complete heat energy is not converted to work energy only a part of the heat energy is converted to work energy that kind of energy which cannot be completely converted other as is called low grade energy remember both low grade energy and high grade energy are identical only for isothermal process why you know the change in neurology d u equal to m c v into DT for isothermal process for ISO thermal process or thermal process isothermal process means what temperature remains constant if temperature is remaining constant then what is the DT value d t equal to 0 if this DT equal to 0 change in interlange d u equal to 0 the other point then what is the energy balance equation for a closed system Q equal to W plus d u for isothermal process the d u equal to 0 if d u equal to 0 heat transfer and work transfer heat transfer and work transfer both are identical for isothermal process here heat is coming under low grade energy work is coming under high grade energy anyhow both low grade energy and high grade energy both are identical for isothermal process therefore I can select the option C OK similarly look at the another important problem try to read it clearly hot coffee stored in a well insulated thermos flask is an example of isolated system the hot coffee present in the thermos flask is coming under isolated system that is a you know meaning of open system closure system isolated system it is like this if you want to call this system by open system with respect to boundary of the system Mass should be Crossing and energy should be Crossing that means with respect to boundary of the system if both mass and energy crossing the boundary that kind of system is coming under open system here example for open system is turbine turbine okay now come to the another system if you want to call this system by closed system here Mass interaction Mass interaction between the system and surroundings is zero but only energy is interacting between the system and surroundings if only energy is interacting between the system and surroundings in the options of mass interaction that kind of system you can call as closury system the best example for closure system is battery any battery I am giving the advertisement for the nippo battery anyhow any battery is coming under closed system once you connect the battery to the torch light what is happening the chemical energy of the battery is converted to light energy but mass of the battery remains constant before it is using and after it is using therefore battery is the best example for closure system similarly the Transformer is the closure system TV is the closure system Sachin Tendulkar bat is coming under closed system as long as he is playing you know as long as the Sachin Tendulkar is playing Just Energy is interacting between the ball and the bat no mass is interacting between the ball and bat therefore as long as the Sachin Tendulkar is playing the ball is coming under a closed system similarly battery is also coming under closed system okay but once he kept the once the Sachin Tendulkar is kept the uh bat in a particular location its mass remains constant energy also remaining constant the same bat is coming under isolated system the isolated system is like this if you want to call if you want to call this is the isolated system with respect to boundary of the system mass interacts with zero under energy interaction also zero example universe example is universe in the universe total energy remains constant and the total mass also remaining constant means with respect to boundary of the universe no mass is crossing no energy is crossing therefore universe is the best example for isolated system similarly this battery is coming under closer system as long as it is working but once the battery is kept at a particular location its mass remains same energy remains same the same object is coming under isolated system like how the Sachin Tendulkar bat is coming under closed system as long as he is playing but once he kept the bat in a particular location the mass remains same energy remains same that kind of system is coming under isolated system that means all non-living things are dead elements up to no external Force acts on them are coming under isolated systems only okay now come to our problem dharma's flask thermos flask is perfectly insulated since the dharmos flask is perfectly insulated perfectly insulated yeah the heat transfer is zero heat transfer Q equal to zero and there is no pressure gradient between the coffee present in the between the coffee present in the thermos flask and outside therefore thermodynamic work equal to zero because of the fourth gradient or pressure gradient some energy is crossing the boundary of the system that is coming under work energy here there is no pressure gradient between the there is no pressure gradient between the coffee present in the thermos flask and outside thermody since it is insulated heat transfer is zero and the coffee particles are not coming out from the system mass equal to zero Mass transfer is zero energy interaction is zero that's why the hard coffee present in the dharmas flask is the best example for isolated system gather point yes similarly look at them another important question try to read it clearly it is given like this a system of mass 100 kg undergoes a process in which its specific entropy increases from 0.3 kilojoules per kg Kelvin 2.35 kilosoles per kg Kelvin let me select a system this is my system the mass of my system is 100 kg 100 kg initial entropy 0.3 initial entropy 0.3 final entropy 0.35 S2 final entropy s two equal point three five both are given kilo joules per kg Kelvin okay and you know this is coming under surroundings surroundings at the same time the entropy of the surroundings decreases from 80 kilojoules per Kelvin to 75 kilojoules per Kelvin by the time the system entropy increasing from this value to this value surroundings entropy initial surrounding Center will let it be S1 Dash is initial entropy 80 final entropy 75 S1 Dash is the initial entropy of the surroundings 80 final entropy of the surroundings 75 but both are given in kilo joules per Kelvin surrounding Center be given in kilojoules per Kelvin system entropy given as specific entropy that is kilozoles per kg Kelvin then what is the name of the process whether the process is reversible or irreversible decided by entropy change of the universe if the entropy change of the universe is equal to zero that is irreversible process the entropy change of the universe is greater than zero irreversible process entropy change of the universe is less than zero that is impossible process other point the entropy change of the universe is zero that is reversible process then drop a change of the universe is greater than zero irreversal process the entropy change of the universe is less than zero impossible process okay for that let us find the entropy change for the universe you know entropy change for the universe entropy change for the universe equal to entropy change of the system entropy change of the system plus entropy change of the surroundings okay now what is the entropy change of the system entropy change of the system is equal to S2 minus S1 into Mass why I am multiplying with mass because given entropies are specific in trophies okay it is equal to 0.35 minus 0.3 into 100 it is equal to 5 kilo joules per Kelvin ok now come to the entropy change of the surroundings entropy change of the surroundings is equal to final entropy of the surroundings minus initial introop minus initial entropy let me write it clearly yes one dash S1 Dash S2 Dash how much 75 minus 80 is equal to Minus 5 kilo joules per Kelvin the change in entropy of the system file change in entropy of the surroundings minus five what is a plus 5 minus 5 0 that means entropy change of the universe is zero the entropy change of the universe is zero definitely that is reversible process look at the option A right dear friends remember let me write that point if the entropy change of the universe equal to 0 that is for reversible process reversible process entropy change of the universe greater than zero irreversible process irreversible process next if the entropy change of the universe less than zero is impossible is impossible I hope all of you clearly understood okay go for the next problem yes let us move to the next problem try to read it clearly change in anthology in a closed system is equal to heat transfer if the reversible process takes place at constant pressure dear friends remember heat transfer equal to change in internal energy for a for a constant volume process okay next heat transfer is equal to work transfer for constant temperature process next heat transfer is equal to change in enthalpy for constant pressure process okay next heat transfer equal to 0 for adiabatic process a diabetic process try to remember these four points heat transfer Q equal to change in internal energy for constant volume process heat transfer Q equal to work transfer that means low grade energy high grade energy identical for isothermal process heat transfer is equal to change in enthalpy for constant pressure process heat transfer equal to 0 for adiabatic process anyhow the change in enthalpy in a closure system is equal to heat transfer if the reversible process takes place at constant pressure therefore option C is absolutely correct understood yes similarly look at the another important question try to read it with increase in pressure the latent heat of steam it is like this let us take the pH coordinate system temperature versus enthalpy coordinate system the saturation Dome is looking like this that the saturation Dome now let me draw the constant pressure line we let P1 equal to 1 bar corresponding to one bar you know the latent heat corresponding to one bar this is a latent heat h of G these are latent heat right that means one two two is the evaporation process this heat equal to late and heat corresponding to P1 that is equal to one bar but when I in pressure when I increase the pressure from one bar to let the pressure p to equal hundred bar corresponding to 100 bar the evaporation is beginning corresponding to 100 bar Eva operation is beginning here evaporation is ending then these are latent heat transfer HF G2 now compare is it h of G2 these are the latent heat of vaporization corresponding to high pressure this latent heat of vaporization corresponding low pressure definitely I can say that h of G2 h of G two is less than h of G one hfg be the latent heat that means when I increase the pressure from 1 bar to 100 bar the latent heat is decreasing from h of g 1 to H of G2 similarly at critical point at critical point at a critical point the H of G h of G equal to zero there is no latent heat of vaporation at critical point the pressure at the critical point is called critical pressure is equal to 221 bar and the corresponding temperature is the critical temperature is the 374.14 degree centigrade anyhow when I increase surface pressure definitely latent heat decreases look at the options with increase in pressure the latent heat of steam decreases therefore option C is absolutely correct okay similarly look at the last question try to look at the last question try to read it the free expansion process dear friends remember there are two irreversible process irreversible process irreversible process for a closed system for a closure system is free expansion process free expansion process that means free expansion process the closed system irreversible process next irreversible process irreversible process for a steady flow system irreversible process for a steady flow system is the throttling process throttling process that means irreversible process connected to steady flow process the throttling process and irreversible process connected to closed system is the free expansion process but free expansion process free expansion process is irreversible adiabatic non-flow process map flow process okay next throttling process foreign process is irreversible adiabatic steady flow process steady flow process okay now dear friends remember here adiabatic means Del Q equal to zero but irreversible means I is not equal to 0. if I is not equal to 0 entropy generation always greater than 0 means entropy generation is positive then what is the change in entropy the change in entropy d s equal to Del Q by T plus s generation s generation here Del Q by T Del Q by t equal to 0 then d s equal to entropy generation that means d s d s for throttling process also equaled entropy generation anyhow conclusions are the free expansion process is the irreversible adiabatic process in the closure system okay look at that option a irreverse body process for open system no irreverse per diabetic process for a closed system option b is absolutely correct let me read the option C irreverse priceothermal process no irritable price on process no free expansion process the irreversible adiabatic non-flow process whereas throttling process the irreversible adiabatic steady flow process I hope all of you clearly understood with this my dear friends I am concluding once again on behalf of no uh like a class it is sorry for starting late because of the small technical issue right thank you have a great red uh success God bless you okay bye bye thank you is online learn anywhere anytime [Music] wherever maybe your present Journey 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