 So, now we consider a still more generic steady state problem. It is not a turbine, but a system which involves nearly every term that we thought of. Let me read out to you a steady flow system receives 60 kilogram per minute of gas at 2 bar 90 degree centigrade with negligible velocity and discharges it at a point 20 meters above the entrance section at a temperature of 300 degree centigrade with a velocity of 2200 meters per minute. During this process, 2 kilowatt of heat is supplied from external sources and the increase in enthalpy is 7.8 kilojoule per kg. Determine the power output. Now, this has all terms that we can think of. There is a Q dot, there is W dot S, there is delta H, there is delta kinetic energy and there is delta potential energy. Let us see how if we put each of these terms what we get? So, of course, this is a steady state problem. So, we start with our simple steady state first law equation. Before writing down the equation, let us just draw a diagram. It is no longer a turbine. I will draw an arbitrary shape here. This is the inlet section here i, this is the exit section e and this height difference has been given as 20 meters. There is a Q dot. The Q dot is given as 2 kilowatt. There is a W dot S which we are expected to find out and M dot has been given. This is 60 kilogram per minute which of course translates as just 1 kilogram per second. The problem has been invented to wake up the students because the units are in minutes. Typically, we would always work in seconds. Vi, it is mentioned is negligible. So, we will just put this as equal to 0. Ve, if you see, is equal to 2200 meters per minute. Again, this is non-standard. Let me convert it into standard units and this is what is expected. You just have to divide it by 60 and you will get meters per second. So, if we look at Ve squared by 2 minus Vi squared by 2, that would just be equal to Ve squared by 2 because Vi is negligible. This will turn out to be 36.667 squared by 2 and this would be just 672.2 joules per kg and that would be of course a reasonably small number. It is 0.672 kilojoule per kg and you would have noticed that Ve was even lesser than 50 meters per second. Hence, we do not expect this term to be big. What about GZe minus Zi? We have been given that this is 20 meters. 20 meters is a reasonably large number but GZe minus Zi would still be a reasonably small number. Let me just calculate it. This is 9.81 meter per second square multiplied by 20 meters. So, this would be 196.2 joule per kg. It is a very small number, 0.196 kilojoule per kg. He minus HI has been given. It is a reasonably large number. It is 7.8 kilojoule per kg. So, it is not as large as in turbines. It is a reasonably small number. It is not of the order of 100 or 200 or 900. Hence, the Ve squared by 2 term is at least comparable to this. You can say it is at least 10 percent of it and hence, we cannot neglect this. Of course, GZe minus Zi is much smaller but still compared to 7.8. It is not that negligible and hence, we consider it. Q dot of course, is of the same order as He minus HI. It is 2 kilojoule per second and if we divided by 1 kilochrom per second, you realize that it is of the same order. Let us just write down the first law now. So, we know all the quantities. This is a positive quantity. It is 2 kilowatt. This has to be found out. This is 1 kilogram per second. This He minus HI is 7.8 kilojoule per kg. Here, this is negligible but so, Ve squared by 2 minus Vi squared by 2 is nothing but Ve squared by 2 and we found that this is 0.672 kilojoule per kg and GZe by Zi, we found out was 0.196 kilojoule per kg. Luckily, we are just multiplying this by 1 here. So, W dot S, it will turn out is now, this entire thing kilogram per second multiplied by kilojoule per kg. So, this will turn out to be 7.8 kilowatt when you multiplied by m dot is 0.67 kilowatt, 0.196 kilowatt. So, we would end up with W dot S is 2 minus 7.8 minus 0.672 minus 0.196 and all the units you realize are in kilowatts. You will realize that this will turn out to be a negative quantity and to ensure that there is significant delta H increase in kinetic energy, increase in potential energy, you have to do both work input into the system and also input heat energy into the system. So, this will be a negative quantity which means work is done on the system and it will turn out hopefully as minus 6.668 kilowatt. So, this is a problem where we consider every term and we neglected none of the changes in potential and kinetic energies and that was reasonable since the terms were comparable. But typically in a turbine and compressor problems, you will realize that these terms will be negligible. Right now, that is it. Thank you.