 Welcome back. So, now we consider another problem with turbine and slightly more generic. Let me just read out the problem to you. Steam is supplied to a turbine with H i that is inlet enthalpy is equal to 3.2 mega joule per kg and it leaves with H e equal to 2.5 mega joule per kg. The entrance and exit velocities are 170 meters per second and 280 meters per second respectively. The heat loss is 50 kilo joule per kg. What is the work done? Now you realize that the mass flow rate has not been given. So, what we will do is we will find out the work done per kilogram basis and we just go ahead and write down the first law for steady systems again and let me write this down again. So, we have q dot minus w dot s is equal to m dot H e minus H i plus v e squared by 2 minus v i squared by 2 plus g z e minus z i. Now you notice that for H e and H i the numbers given are 3.2 mega joule per kg for H i which is in kilo joule per kg preferably written as 3200 kilo joule per kg where H e is 2.5 mega joules per kg which can be written as 2500 kilo joule per kg and you notice that these are numbers which I had mentioned in the previous snippet roughly around 3200, 3300, 3400 kilo joule per kg is a typical inlet to turbines and at the exit you can expect anywhere between 2500 to 2600 kilo joule per kg. So, these are typical numbers of course you know if the turbine had an inlet at much lesser pressures you would have much lower numbers here, but still the numbers would have been around 2900 or so. Now you notice what is being said about v e now v e we realize has been written down as 280 meters per second v e is 280 meters per second. Notice what we had mentioned in our previous snippet that only when you start getting velocities around 300 that is much more than 100 meters per second do those kinetic energy terms start getting significant. Now 280 meters per second is reasonably significant. So, if we calculate v e squared by 2 this will turn out to be 39200 joule per kg which is equal to 39.2 kilo joule per kg and if I look at vi, vi is also reasonably high it is around 170 meters per second. So, vi squared by 2 is equal to which you can see is still around 14.45 kilo joule per kg this is significant any number which is reasonably above 1 kilo joule per kg we will consider as reasonably significant. What has been mentioned about z e and z i nothing. So, we can make an assumption. So, let me put this down as assumption is nearly equal to 0. What about q dot? Now it seems like the turbine is not adiabatic and q dot is of the order of 50 kilo joule per kg and if you look at the delta H which is around 700 kilo joule per kg you realize that 50 is not a negligible number and definitely one must consider it. So, now we have all the numbers we do not have m dot here, but we will write down I mean we have actually been given q dot by m dot this has been provided. So, let me just write down the equation in terms of q dot by m dot and W s dot by m dot by just dividing the whole equation by m dot let us see how it turns out. So, we have the first law written as q dot by m dot this is because we have divided the entire equation by m dot the right hand side is no m dot it just has h e minus h i plus v e squared by 2 minus v i squared by 2 I am no longer writing down the potential energy term. So, we want W dot s. So, we just put everything else to the right we write W dot s by m dot is equal to q dot by m dot minus h e minus h i minus v e squared by 2 minus v i squared by 2. So, you realize that this number it is a negative number because you are losing heat this is minus 50 kilo joule per kg the units are correct h e minus h i we realize is 2500 minus 3200 units are correct it is kilo joule per kg minus v e squared by 2 is 39.2 kilo joule per kg minus v i squared by 2 is 14.45 kilo joule per kg. So, you realize this is equal to minus 50 this is a negative number and it turns out negative and negative this is positive plus 700 and we have minus 24.75 and all have the same units this is kilo joule per kg and we get this finally has 625.25 kilo joule per kg. So, this is W dot s by m dot it is specific work output from the turbine of the order of 625.25 kilo joule per kg. So, here we have neglected the potential energy term, but we have included the kinetic energy term specifically because they were reasonably big any velocity above 100 meters per second will lead to reasonably significant kinetic energy terms and the q dot was also not negligible there was a heat loss of 50 kilo joule per kg. So, that was a reasonably high number. So, we have now considered a slightly more general problem of course, in normal turbines you would not see such high velocities at inlet and exit, but it is worthwhile to consider if such a case exist and ensure that we do not neglect terms which cannot be neglected. So, next we will consider a slightly more generic problem that is it for now. Thank you.