 One of the things that we've been asked to solve for this example problem is the mass flow rate through the boiler and the way that we're going to about getting that is we are going to we also know the total power that the total power coming out of the system is I believe 120 megawatts so consequently what we're going to do is we're going to do a balance where we're going to find the amount of work that is the work that we're putting in through the pumps and then we'll find the work coming out through the turbines the difference of that would be the power out we can then put that in terms of per unit mass flow rate through the boiler and from that we can get mass flow rate through the boiler so that's what we're going to proceed and do now so I'm going to start by writing out the first law and in all the applications that we're doing here we'll neglect kinetic energy as well as potential energy and when we're looking at work in a pump we can also neglect heat transfer and I'm going to express it in terms of work in remember work in is a negative and that's why we get rid of the negative sign by doing that so what we're going to do we're going to start by looking at pump one and I'm going to non-dimensionalize the work and I'll be non-dimensionalizing all of the work terms here by the mass flow rate going through the boiler and that is m dot five now mass flow rate at one we can express that in terms of mass flow rate through the boiler minus the mass that is stripped off and sent through the open feed water here we talked about that earlier and with that we can re-express it in terms of the mass fraction that we saw earlier so that becomes the expression for work in one we can then substitute in values so that's pump one we'll proceed and do the same thing for pump two now m dot three and m dot five those that's mass flow rate into the boiler and out of the boiler if we go back and look at our schematic where was it here there we go m dot three and m dot five there's m dot five and m dot three is here all of those flow through the boiler and there's no stripping or removing there and so it's the same flow rate at both points consequently that ratio is one and so we can plug in values and so that's what we get for pump two continuing on let's take a look at the turbine so in a general sense the equation for a turbine from the first law now a turbine is a device that does work and consequently it will be a positive work so with that we can switch around the enthalpies we'll look at the high pressure turbine first and if you recall all of the mass flow is going through the high pressure turbine we haven't stripped anything off yet i'm now going to express that in terms of a per unit mass so dividing by m dot five plugging in values we get output for the high pressure turbine kilojoules per kilogram and now that's per kilogram mass through the boiler taking a look at the low pressure turbine and remember here we have stripped some of the mass off so we need to account for that and we will use the continuity equation in order to express mass flow rate at eight and that would be m dot five the mass through the boiler minus what we strip off and send through the open feed water heater non-dimensionalizing by that or dividing by that i should say and when we plug in the values we get work out for the low pressure 787.20 so the net work is going to be the work out of the turbine minus the work that we put into the pump we get that for the work net and we can now equate that to the total work out of the steam cycle or for a plant so that all needs to equal 120 megawatts and i will express that in terms of kilojoules per second and so here we've determined that we know that sets the power output so we can directly then solve for m dot five doing that we get the mass flow rate through the boiler 81.89 kilograms per second so that answers the first part of the problem the last thing that we're going to do in the next segment is we'll take a look at the thermal efficiency