 Ok, good morning. So, let us continue with the engineering thermodynamics. Now, we will be in the tutorial 5. This is on first law involving steam. The first problem, a two phase mixture of water with the quality of 0.25 is contained in a piston cylinder assembly as shown. The mass of the piston is 40 kgs and its diameter is 10 centimeters. The initial and final position of the piston are shown in the figure. So, initially it is at the height of 1 centimeter. Finally, it hits the stops and it is at the height of 4.5 centimeters. The water is hit until the pressure reaches 300 kPa. So, initial solution, initial quality is 0.25. Ok, now diameter of the piston is 10 centimeters. Mass of the piston is 40 kgs, then final pressure is 300 kPa. So, area of cross section of the piston will be equal to 5 by 4 into 10 into 10 power minus 2 to convert centimeter into meter square. Then volume 1 will be equal to 1 centimeter height into area. Similarly, volume 2, when it touch the the piston touches the stops will be equal to 4.5 centimeters minus 1 centimeter. Ok. So, now what is the pressure? Pressure at state 1 will be equal to atmospheric pressure which is given as 100 kPa plus the pressure imported by the piston that is mass of the piston into G divided by area that will be equal to 150 kPa. So, you can this is basically 40 into 9.81 divided by area that is 5 by 4 0.1 square. So, that gives the initial pressure 150 kPa. So, now using see the first state is quality is given as 0.25 that is state 1 is state 1 is saturated mixture of liquid and vapor. So, using pressure based because pressure is known pressure based saturation table steam tables. So, I will show you the steam table now here. This is the saturation tables based upon temperature. You can see the temperature is the first column. If you go down, there will be another table is called saturation table based upon pressure. So, pressure is the first column. Now, it is given a pressure is given in bar here. We want to convert this into kilopascals 150 kilopascals. So, 150 kilopascals will be 1.5 bar. So, this is the entry for us 1.5 bar. So, in 1.5 bar you will see that there is a Vf value that is the specific volume of the liquid, similarly specific volume of the vapor. So, this is 1.053 into 10 power minus 3 meter cube per kg and this is 1.159 meter cube per kg. Similarly, the Uf is specific volume of the liquid. Similarly, Ug is the specific volume of the, sorry, Uf is the specific internal energy of the liquid and Ug is the specific internal energy of the vapor. So, you can take these values and find the state 1 specific volume and specific internal energy state 1. So, here U at 150 kilopascals as I showed Vf will be equal to 0. here 0.001053 meter cube per kg. Similarly, Vg equal to 1.159 meter cube per kg. So, V1 will be equal to Vf plus x1 times Vg minus Vf which is equal to 0.2906 meter cube per kg. So, that is the specific volume of the mixture at 150 kilopascals. Now, we can also find U1, U1 equal to Uf plus x1 into Ug minus Uf which is equal to tables. Uf is at 150 kilopascals 469.9, 469.9 plus 0.25 into sorry 466.9 it is 2520. So, this is 466.92520 minus 466.9. So, that will give you U1 as 980.175 kilo joule per kg. So, the properties are taken state 1 is fixed and properties are taken. Now, what is the mass in the system? Mass m is equal to the total volume divided by the specific volume which is given by 2.70268 into 10 power minus 4 kg. So, that is the mass. So, V1 you have found here you can see that V1 is you know the area of cross section area of cross section into 1 centimeter that is the height of the piston at the particular state 1. So, using that I can find the volume area of cross section into the height that is the volume of the mixture which is entrapped here. Now, that from that total volume by specific volume will give you the mass. So, the state 1 and mass are got now. The state 2 volume from this you can find state 2 volume total volume V2 equal to 3.53 into 10 power minus 4 meter cube from this equation. So, this is actually 4.5 only because in total volume you have to take. So, 4.5 into 10 power minus 2 into AP. So, that will give you this volume. So, the delta V only is 4.5 minus 1. The total volume at state 2 will be 4.5 the total height into the area of cross section that will be this 3.53 into 10 power minus 4. So, now the state 2 state 2 I have to fix V2 is equal to V2 by m. Now, mass is known V2 also is known which is equal to 1.3077 meter cube per kg. Now, the final volume is final state 2 volume is now calculated final pressure is given as 300 kilo Pascal. So, based upon that the state 2 can be fixed. So, you can see how to fix the state you know the specific volume is 1.3077 at pressure of 300 kilo Pascal. So, going forward to steam tables for 300 kilo Pascal this line Vg is 0.606, but the specific volume at state 2 was calculated as 1.3077. So, that is greater than this to understand. So, Vg at P2 equal to 300 kilo Pascal is equal to 0.606 is less than V2 which is equal to 1.3077. So, that means state 2 is superheated state 2 is superheated. So, we have to use the superheated tables. So, you will understand here when you draw this diagram temperature versus volume we can say. So, here you are talking about this 300 kilo Pascal's isobar in a TV diagram and this temperature and this volume this volume here is Vg. So, this is Vg this Vg is 0.606, but the volume given is here somewhere that is 1.3077. So, it should be in the superheated state in the same diagram the superheated state ok. So, use super so using superheated tables for the pressure of 300 kilo Pascal's. So, we will go to the tables now. So, this is the superheated steam tables. So, as in the saturation tables then the superheated steam tables here different pressures have different entries. So, we want 3 bar pressure. So, this 3 bar pressure basically and our specific volume what we calculated at the state 2 is 1.3077 meter cube per kg. So, here we can see this in this 3 bar see the entries for the V we can see that for 500 degrees centigrade it is 1.187 meter cube per kg and 600 it is 1.347. So, we have to interpolate between these two temperatures. So, what we should do is here we want to interpolate for the value of internal energy. So, internal energy at the state 2 what we want is 2 is equal to we can say 3130 plus this is the u at 500 plus we will say du by so that is 3301 minus 3130. So, this du by dv dv is this 1.341 minus 1.187. So, this is nothing but du by dv into dv dv is what now the required given value for state 2 is 1.3077 minus here the previous value is 1.187. So, if you do this then you will get the value of u2. So, calculate so this is a simple linear interpolation what we understand here is between two rows of this table the properties vary linearly. So, with that assumption we make this interpolation. So, we also have entries for 50 degrees the temperature interval etcetera, but this is enough for this to understand how to use the tables. So, if you see this we have got the pressure and volume for the state 2 based upon the pressure we have added the superheated tables. Because we find that for the pressure 3 bar this volume is more than vg at 3 bar. So, we have come to the superheated tables. In the superheated tables we find that these are the two entries which actually form the boundary of this value. So, u is interpreted by using this particular. So, this is actually a simple linear interpolation. So, coming to the problem for v2 equal to 1.3077 meter cube per kg, you can see that we have calculated u2 as 3, 2, 6, 4 kilo joules per kg from the superheated tables by linear interpolation. So, by interpolation ok. So, now you know the two states what is q1 to 2 you have to calculate that that will be equal to w1 to 2 plus u2 minus u1. So, that means you have w1 to 2 you have already calculated that will be m into p into constant pressure 1.3077 minus 0.2906. So, then plus again m into u2 minus u1 this is actually v2 minus v1 ok. So, using this we can calculate the q as 2.70268 into 10 power minus 4 into 150 into 1.3077 minus 0.2906 plus I will say here I put a bracket plus 3, 2, 6, 4 that is the u2 value minus u1 value was calculated as 980.175, 980.175. So, this will give you q1 to 2 as 658.47 into 10 power minus 3 kilo joules or q1 to 2 equal to 658.47 joules. So, that is the value. So, now we can see the process. Process if you go back you can see that u1 is 980.175 that corresponds to that was calculated using the x1 value. Initially we had a mixture of water that is liquid water and steam with a quality of 0.25. So, you can see this is the state 1, 0.25 in 150 isobar line this is corresponding to 0.25. Then there is a constant pressure process where it reaches the state 2 dash. Then a constant volume process occurs. So, after it basically it lifts up from 1 centimeter to 4.5 centimeters there the pressure remains constant at 150. Then after it hits the stops what happens volume remains constant. So, the intermediate state 2 dash you can say somewhere it is occurring. So, that to this the pressure increases volume remains constant no work is done here that is 150 to 300 kilo pascals the pressure increases. So, this is the problem where the total work is basically done during 1 to 2 dash only here. Then after that since delta v is 0 no work is done here. Finally, the intermediate changes from this state final state to initial state whatever be the u2 minus u1 that is the value we used ok. So, this is about the first problem. So, first problem is done second problem where 2 phase mixture of water the quality of 0.25 is contained in a piston cylinder assembly as shown the mass of the piston is 100 kgs and its area is 0.02 initial and final positions are shown again similar to the previous problem. But here there is a spring as shown in the figure. Initially the spring when it is in this position spring just touches touches the top surface of the piston without any pre compression. So, it is not going to exert any force in the initial state. The stiffness is 125 kilo Newton per meter water is heated again like in the previous problem. So, the piston moves up and goes to a position from the bottom it is about 5 centimeters away where the spring is sufficiently compressed. The spring is also compressed by 5 minus 1 4 centimeters. So, atmospheric pressure is given as 100 kilo basketball and G is given as 10. Solution from this problem initially what is P1? P1 will be equal to P atmosphere plus mass of the piston into G divided by A. Okay, there is no since spring is not compressed here. So, that will be equal to 100 plus 100 kg into 10 is the given actually divided by area is 0.02. So, which will be equal to 150 kilo Pascal's or 1.5 bar that is the P1. Then we can calculate the area of cross section is given as 0.02 meter square. So, we can calculate V1 equal to 1 into 10 power minus 2 into 0.02 which is equal to 0.0002 meter cube. Similarly, V2 equal to final position 0, height is 5 into 10 power minus 2 into 0.02 which is equal to 0.001 meter cube. So, the finishing and final volumes are known. Now, for the state 1 we can find state 1 is fixed by P1 equal to 1.5 bar x1 is given quality is given as 0.25. So, similar to the previous problem so saturated mixture saturated mixture of the state 1. So, from pressure based saturation tables for 1.5 bar I can take Vf equal to like in the previous problem 0.001053 meter cube per kg. Similarly, Vg will be equal to 1.159 meter cube per kg. Uf also can write Uf equal to 466.9 kilo joule per kg under Ug equal to 250. So, from this we can calculate U1, U1 equal to Uf plus x1 into Ug minus Uf. Similarly, V1 equal to Vf plus x1 into Vg minus Vf. So, we will get the values of U1 equal to 980.18 kilo joule per kg and V1 equal to 0.29054 exactly like the previous problem state 1 is fixed. Now, what is the mass of water? Mass of water will be equal to V1 by small v1 which is equal to 0.0002 meter cube divided by 0.29054 meter cube per kg. So, it is equal to 6.883 into 10 to the power minus 4 kg. So, that will be the mass. Spring is compressed by 5 minus 1 equal to 4 centimeters as the piston moves from state 1 to state 2 ok. So, here from 1 centimeter to 4 centimeter, 5 centimeters it compresses the spring by 4 centimeters. So, now final pressure can be found as PATM plus P applied by the piston plus P applied by the spring. So, which is equal to 100 plus we have already seen this 50 plus the spring constant K of spring is 125 kilo Newton per meter spring constant. So, K into that is 125 ok into significantly it be kilo Newton into 4 centimeters is compressed. So, 0.04 divided by the area, area is given as 0.02. So, this is nothing but K into X divided by A that is the pressure given by this. So, that will give you 400 kilopascals or 4 bar that will be the final pressure. Do you understand? So, now you can see finally what happens here go to the problem. So, the what is it until the piston moves the final position ok determine the total amount of heat transfer and the final state of the water final state we need to know. So, now we can see that the state 2 is fixed by the pressure which we have calculated and the volume volume also is known to me mass is known to me. So, specific volume of state 2 can be calculated ok. So, that means we need two properties what is specific volume at state 2 that will be V2 divided by M and V2 is nothing but we have calculated here here straight away V2 is 0.001 meter cube. So, 0.001 meter cube divided by mass we have calculated here 6.883 into 10 to the power minus 4 kg. So, which is equal to 1.452 meter cube per kg that is V2. So, now pressure is 4 bar in the final state and volume is 1.452. So, what is this state? We have to find this state. So, for that we have to see the tables go back to the table. So, 4 bar. So, first you go to saturation based pressure tables here this is saturation based pressure tables. So, here we can see that 4 bar the volume is Vg is 0.463 ok. So, Vg at 4 bar equal to 0.463. Now, V2 is 1.452 greater than Vg at 4 bar. So, that means state 2 is superheated. So, we have to go to the superheated tables now superheated table corresponding to 4 bar this is the table now 4 bar table. And again the volume 1 point the volume is 1.59452 1.452. So, here you can see that this will be in these two lines. So, we have to interpolate basically here these two lines ok. So, 1.452 will be in this between these two lines correct. So, this these two. So, you have to consider this and interpolate interpolate to get the value of u. So, again u2 is interpreted using this basically. Same way what we have done here u2 will be equal to here u at v1 plus du by dv into dv plus ok that is it. So, here it is u this is actually u v2 and this is 1 ok. So, now this if you find I can get the value ok. Interpreting the tables I got u2 equal to 4023 kilojoule per kg 4023 kilojoule per kg. So, now we can find the answer what is the work interactions w atmosphere first we will find that is p atm into v2 minus v1 ok. So, that will be equal to minus 100 into 0.02 into 0.04 this is area this is delta h it is change in volume. So, delta v. So, that will be equal to minus 80 joules yeah. So, here into 10 power 3 you can put for this kilo Pascal's to this. So, in joules we will write. Similarly, for a spring this will be equal to minus k x square by 2 x is actually delta h basically. So, that will be equal to minus 125 into 1000 into 0.04 x delta h this is or x divided by area area sorry divided by 2 which is equal to minus 100 joules. Similarly, work for the piston piston equal to minus mass of the piston into g into x or delta h delta h. So, which is equal to minus 50 into 10 into 4 into 10 power minus 2 which is equal to minus 20 joules. So, all the works are calculated. Now, what is work done by the the steam work all these are done by steam. So, work done by the steam will be equal to minus of work involved in atmosphere plus work involved by the piston plus work involved by the spring is equal to 200 joules delta u of the steam will be equal to m into u 2 minus u 1 which is equal to 6.883 into 10 power minus 4 into 4023 minus 980.18 which is equal to 2079.15 joules. So, q equal to w plus delta u first law delta ke equal to delta pe equal to 0. So, now we can say this is equal to 2279.15 joules. So, here we try to draw this TV diagram. So, here the first state is what first state is 150 kilopascals. So, 150 kilopascal isobar here this is x equal to x equal to 0.25. Then it goes to 4 bar. So, 4 bar let us say something like this something like this. So, now what happens the pressure actually linearly increases because of the spring. So, this is state 2 is superheated state. So, somewhere here. So, this is state 2. So, now we can see that if we go back we will see approximately the temperature is in between these two. So, around say 1000 degrees centigrade. So, 1000 degrees centigrade is the temperature. So, it will be very high here. I can show you the plot. This is 4 bar pressure. So, somewhere here 1000. So, this is where the process will be somewhat like this. Pressure increases linearly after this to this. So, crossing a state saturated state at a particular pressure then it goes back. So, it goes to the superheated state. So, this is the value of total Q.