 Let us move on to tutorial 6, we will see few problems because enough examples were worked out by Professor Babu. So, we will see some examples. The first one is a steam turbine is used to drive an air compressor like this. A steam turbine is used to drive an air compressor. The value of R for air and gamma for air is given. The heat loss from the compressor and turbine, each amount to 5 percent of their respective power. That means, there is a heat loss from the, this is not fully insulated, so there is a heat loss from this, I will say Q compressor, Q dot we will say and there will be heat loss Q dot turbine. This actually will be 5 percent of their respective power, ok. So, for example, it will be 5 by 100 into W dot T, whatever be the power turbine by the power done by the turbine, developed by the turbine. So, let the power developed by the turbine be W dot T, so 5 by 100 into W dot T will be Q dot T. Similarly, Q dot C will be equal to 5 by 100 into, here I am putting absolute of the power given to the compressor because it is negative, no power is consumed, so W dot C that is negative. So, absolute value of W dot C into 5 by 100 will be Q dot C, both are actually lost, so you can see the direction of the heat transfer. Assume steady state operation, neglect kinetic and potential changes. Compression process, see initial condition is given, air comes at 100 kB, 27 degree centigrade to the compressor at the volumetric flow rate of 8.66 meter cube per second and it goes out, but the compression obeys this. That means I can find the, apply this relationship for the compression. Now, the expansion process in the turbine is such that no liquid at the exit, there is no liquid at the exit, the final pressure is given 1 bar, ok. Here also initially a 30 bar, 600 degree centigrade steam enters at the volumetric flow rate of 1.363 meter cube per second. Compression with proper justification, highest possible compressor exit power. So, when highest possible compression can occur, when maximum power is delivered by the turbine to the compressor, ok. That means the condition at the exit of the turbine should be such that it should be saturated vapor, ok. Now, we will draw this and show, ok. So, now for the compressor, you can see that is higher 30 bar. So, let us say 30 bar, 600, sorry, so somewhere here, that is state for this, for the steam turbine. State 1 is this, ok. This is T V diagram. So, now this is 600 degree centigrade, this is 30 bar pressure. Now, here this is 1 bar, 1 bar. Now, the exit state if it is, see the condition given is there should be no liquid, ok. Now, that means the exit state can be superheated also. So, I can be here, here, anywhere. But what happens when we are here? The work delivered by the turbine will reduce. So, where the maximum work will be delivered with the condition of no liquid at the exit is this. When you are in the saturated liquid condition, then only maximum work will be delivered. If you go to superheated, then also no liquid condition will be satisfied, but the work delivered will decrease. So, state away I can fix the state 2 as, for the turbine has 1 bar x 2 equal to 1. See, like this is the justification. Why? For maximum compression, I need to deliver maximum work to the compressor. For maximum work to be delivered, the maximum work should be developed by the turbine for which the exit state should not be superheated, should be saturated paper, because no liquid should be there. I can also go to somewhere here, so that h decreases. I can also go somewhere here, but liquid will come in. So, I cannot take any excess here, I could only take this dot as the exit state. So, exit state is state away fixed, ok. So, that is the understanding you have to make. There are two important things here. There is a heat loss from the compressor as well as turbine. And for the compression, there is a PV power 1.35 equal to constant is given, from which you have to find the exit pressure. Similarly, for the exit of the turbine, the condition is no liquid should be there and the pressure should be 1 bar. That is the pressure ratio of the turbine is 30 bar, ok. Now, this is what is given. And let us see how to calculate the highest compression exit pressure. Solution from steam tables P i is 30 bar and T i equal to 600 degree centigrade for the steam turbine. And also volumetric flow rate V i is given as 0.0363 meter cube per second. So, now, from steam tables, what is the saturation temperature at 30 bar equal to 233.9 degree centigrade. So, which is actually less than T 2 T i, which is 600 degree centigrade. So, the state is superheated. So, I can go to the superheated tables now for 30 bar here, 30 bar superheated tables. Corresponding temperature is 600 degree centigrade. Now, from that I can take the value of V as 0.132 and H, because it is a control volume problem. Now, I need enthalpy. So, I go to H as 3682. So, now, V1 is 0.132 meter cube per kg and H1 is 3285 kilojoules per kg, and it is values for the state 1. Now, next is, from this I can find the mass flow rate of steam. Mass flow rate of steam is 1 volumetric flow rate of steam divided by the specific volume that will be equal to 0.363 divided by 0.132 equal to 2.75 kg per second, that is also fixed. Now, we have discussed for maximum power developed by the turbine, which can be used to get the maximum compression, correct, which can be used to get maximum compression, ok, compressor pressure. So, the exit state should be saturated paper. So, that means final state P2 equal to 1 bar, sorry, 1 bar and X2 equal to 1, that is it. Now, go to that and get the values for the exit state here. We go to the saturation, 1 bar, what is Hg, ok, that is 2675, 2675, V is 1.694, so I will say V2 equal to or existed, no, so I can say this is E output, so exited. Final state I have put i, so I have put here V equal to 1.694 meter cube per kg and He, the exit state for the steam is 2675, 2675 kilojoules per kg, that is the justification for what we are using. Now, for the turbine, I can write Q, sorry, Q dot minus Wt, worked out by a turbine, equal to m dot into He minus Hi, calculating the kinetic energy and potential energy changes, which is given. So, now what is Q dot as we discussed, it is minus 5 by 100 of Wt dot. Now what is this, minus 4, it is lost, so minus 10 comes and 5 percent of the turbine power minus Wt dot equal to m dot into He minus Hi, which implies Wt dot equal to m dot Hi minus He divided by 1.05, which is equal to 22637.4 kilowatts. That is the power developed by the turbine, maximum power developed by the turbine, imposing the condition that no liquid should be there in the exit of the turbine. Now, taking the compressor, for compressor, first we will calculate what is given data, Vi equal to 100 kilopascals, then Ti equal to 27 degrees centigrade, that is 300 Kelvin, then volumetric flow rate equal to 8.66 meter cube per second. Since volumetric flow rate is not conserved, we have to convert it to mass flow rate. So, what is mass flow rate of air will be equal to Pi divided by Rti, this is density into Vi dot. So, it is equal to 10.06 kg per second, ok. So, now I need to find the maximum compression pressure. I know that Pi Vi power 1.35 equal to Pi Vi power 1.35, that is given. I can go back and see in the problem. Pi Vi power 1.35 equal to constant. But first of all, how to get the values, the exit pressure, I do not know, correct? So, how to get that? So, first we go for the energy equation. I can say Q dot minus W dot C equal to m dot into H e minus H i, because again I am neglecting the kinetic energy potential changes. So, now he can, I can write for this as minus 5 by 100, minus is for Q is neglected, this is rejected by the compressor. I have to put absolute value, because this is actually negative, minus, minus of W c. I am again using absolute value, please understand this. So, that will be equal to m dot into H e minus H i, ok. Now, which implies W c magnitude will be equal to m dot into H e minus H i divided by 0.95. But what is this? This is nothing but the power developed by the turbine, that will be equal to 222637.4, correct? That is, we have already kilowatts. So, now, please see this. I do not know the exit temperature. So, I can now find exit temperature from this, T e minus T i divided by 0.95 equal to 22637.4 kilowatts. From that I can find T e as 548 Kelvin, ok. Now, I will find the pressure. How can I find pressure? P v power 1.35 equal to constant. So, and also I have P v equal to RT. So, I combine this and I get P 2 by P 1 will be equal to T 2 by T 1 power n divided by n minus 1, where n is 1.35, ok, n equal to 1.35. When I apply this, I will get P 2 as 1021.54 kilopascals. This is the answer. What is the compression? So, here you can see that the state 2 for the compressor is not fixed. But by using the energy equation, I found that temperature at the exit and from the temperature using P v power 1.35 equal to constant, combining that with the equation of state, we have found the final pressure. So, this is the answer. Next problem, consider a feed water heater that has two inlet and one exit. So, like this, a feed water heater. One exit, two inlet. So, let us say this is inlet 1, this is inlet 2 and exit we will say 3. Feed water at 800 kilopascals, 800 kilopascals and 50 degree centigrade and here steam at 800 kilopascals and 200 degree centigrade comes into the feed water heater and saturated liquid at 800 kilopascals goes out. That is 800 kilopascals, x equal to 0 goes out. Determine the ratio of mass flow rates of the feed water and the steam. So, let us say this is mass flow rate 1, this is mass flow rate 2, m dot 1 and m dot 2 and this is m dot 3. I want the ratio of m dot 1 divided by m dot 2 for steady operation. Neglect the kinetic energy, potential energy and no heat loss also. So, q is also 0, q dot equals 0, rate, steady flow problem. So, now, let us fix the states. State 1, because this is water, no, so state 1, that is state at the entry 1, we can say, is for since T sat at 800 kilopascals, what is that? That is, we will see these steam tables here. 800 kilopascals is 8 bar. The temperature is 170.4, equal to 170.4 is greater than 50 degrees. So, state 1 is subcooled liquid. Therefore, h1 will be equal to hf at 800, sorry, subcooled liquid. So, 50 degrees centigrade or we can say uf plus p into vf at 50 degree centigrade. So, go to the tables again. We can say h1 will be equal to 200, 209.3 into, sorry, plus 800 kilopascals into uf is 0.001012. So, that will be equal to 210.11. Better to use this. So, this kilo joules per kg, that is h1. Okay, now what is h2? State 2. State 2 is 800 kilopascals comma 200 degrees centigrade. Since T2, this is T2, this is T2, T2 is greater than T sat at 800 kilopascals, state is, state 2 is superheated. That means you will see that from the superheated tables, I can take h2, go to this. Superheated tables, 8 bar here, 8 bar, 200 degrees is the first line. What is h? 2839, 2839 kilo joules per kg. So, that is the state 2. State 3. State 3 is saturated liquid at 800 kilopascals. That means h3 will be equal to hf at 800 kilopascals. So, go to the table again, 800 kilopascals. What is the value? 721.1, 721.1 kilo joules per kg. So, now I got all the values. Now apply the energy equation. What is energy equation? q dot minus wx dot equal to m dot outgoing stream. That is m dot 3 h3 minus m dot 1 h1 plus m dot 2 h2. So, it is outgoing stream minus in-stream stream, but here q is 0. There is no work for this device also. And also mass conservation, m dot 3 will be equal to m dot 1 plus m dot 2. So, that we can apply here. So, I can say m dot 1 plus m dot 2 into h3 h3 equal to m dot 1 h1 plus m dot 2 h2. Now from that I can find the ratio. What is that? m dot 1 divided by m dot 2. That is the required ratio will be equal to h3 minus h2 divided by h1 minus h3 equal to 4.145. That is the required ratio for having a steady state operation. Now the third problem, a rigid vessel of volume 0.6 meter cube contains 8 kg of water at 20 bar. So, p1 is 20 bar. Liquid is allowed to exit. So, liquid alone is allowed to exit by having opening from the bottom of the vessel because liquid will be in the bottom and vapor will be in the top. So, now heat is transferred to the vessel in order to maintain the pressure constant at 20 bar. So, that is p2 equal to 20 bar. The problem is stopped when no more liquid is present. That is x2 will become 1. Only vapor will be there, saturated vapor. So, determine the mass that escapes and the heat which is transferred. What is the mass going out? So, we have to find this. Now state 1, v1 equal to v by m equal to m1 I will put because mass is going to change. So, v by m1 equal to 0.6 divided by 8 equal to 0.075 meter cube per kg. Then at the pressure is 20 bar. So, at 20 bar we will know from the we will take from the tables vf equal to 0.001177 meter cube per kg under vg equal to 0.1 meter cube per kg. You can see from the tables. So, from that I can calculate quality at state 1 as v1 minus vf divided by vg minus vf is equal to 0.747. Quality is gone. Now similarly, uf at 20 bar will be equal to 906.4 kilojoules per kg and ug equal to you can take from the tables as 2600 kilojoules per kg. Similarly, what is going out is liquid. So, I want hf also which is taken from the table as 908.8 kilojoules per kg. This is the data which I have retrieved. Why you want hf? Please see that the mass is going out. Liquid is only going out. That means it will take the enthalpy. What is the enthalpy going out? The liquid enthalpy. So, total enthalpy which is going out will be the mass which has gone out. Mass which has gone out say mass out into hf. This is total enthalpy which has gone out. So, for energy balance I need the value of hf. This is state 1. I fixed this, these values. So, from that I can get u1. u1 equal to uf plus x1 into ug minus uf which is equal to 2171.52 kilojoules per kg. So, that is the value of u1. Now, state 2 p2 equal to 20 bar and x2 equal to 1 where no more liquid is present. So, that means I can say v2 will be equal to what? vg at 20 bar equal to 0.1 meter cube per kg. Similarly, u2 will be equal to ug at 20 bar which is equal to 2600 kilojoules per kg. So, now what is mass at the state 2? That will be equal to the volume of the vessel divided by the v2 specific volume which is equal to 0.6 divided by 0.1 which is equal to 6 kgs. That is it. So, initially 8 kg of water was there, right? So, m1 equal to 8 kg, m2 equal to 6 kgs. That means mass that has left will be equal to m1 minus m2 equal to 2 kgs, 2 kg masses left. Now, control volume apply the first law q dot minus wx dot equal to de Cv by dt plus m dot e into he minus m dot i into hi. So, kinetic energy potential changes are neglected. Now, here you can see that there is no work involved here. So, that is 0. Similarly, e Cv is only u Cv. Then there is no mass coming in. So, this is also 0. Now, I can write this as q dot equal to du Cv by dt plus m dot e into he what is outgoing mass for it. Now, mass conservation de Cv by dt equal to m dot i minus m dot e, but here this is 0. So, I can say this as du Cv by dt minus de Cv by dt into he. What is the outgoing enthalpy? This is equal to hf at 20 bar. So, I can write this as de. Here I can write de m Cv u Cv divided by dt q dot equal to minus de m Cv by dt into this is constant. So, I can say hf. So, this is what I get. So, if you integrate this, I will get total q integrating. I will get total q will be equal to m 2 u 2 minus m 1 u 1. So, integrate 1 to 2 de m Cv u Cv divided by dt into dt. I will get m Cv u Cv state 1 to state 2 that is m 2 u 2 minus m 1 u 1. So, this is what I am playing here. So, that minus here if you integrate this, you will get m 2 minus m 1 into hf. So, I know m 1, m 2, u 1, u 2, hf also. So, I can get q as 45.44 kilo joules that is the answer. So, the mass that I have escaped is 2 kg. So, what we are doing here in this case is there is a constant volume of 0.6 meter cube. There is a state initially there is a water with a quality, water means what? It is a saturated mixture of liquid and vapor with a quality of 0.747 which we have determined. So, bottom will be having water and the top will be having vapor. So, when I remove this from the bottom, liquid around comes out. And after the q is transferred so that there is the precise maintained constant. Because when the liquid goes out, the pressure decreases, but the heat transfer will keep the pressure constant. So, that is the problem here. So, first we found the initial state and all the properties required for at the initial state is also calculated. Then final state was also fixed as per the problem from which we got the final mass. Initial mass minus final mass is the mass that I have left out by applying the energy equation and integrating that we found the total q. So, this is the problem 3. So, I will stop here. This is the completion of the tutorial 6.