  SAYO VRIVAN రిగా కరంరడం కనిథకింరి ట౰ిలునాటిలోనం మారమ్రక్నికాకంత్న్ మర్లోన్నికి. పనింకాలునకికి. పచలడు సిహాసికికి. మర్లు సస్రనని Is. We are going to take sole two problem on various types of load factors and other parameters related to power plant performance. Now Let us see the problem number one statement. A generation station has a maximum demand of 20 MW, a load factor of 60%, plant capacity factor of 48% and plant use factor of 80%. So here we are given with the maximum load on the power station or generation station as 20 MW and the various types of load factors are given like load factor of 60%, plant capacity of 48% and plant use factor of 80%. And we are going to find the daily energy produce, the reserve capacity of the plant, the number of operating hours per day and the maximum energy that could be produced daily. If the generation station was running all the time means if generating station is running all the time. Now let us see how to find out first of all the daily energy produce. Now we know that the load factor is given by an equation average load upon maximum load. So load factor given in the numerical is 0.6 which will be equal to average load upon maximum load of 20 MW which is given and load factor is 60% that is 0.6. So from this we can get the average load equal to 0.6 into 20 MW which is equal to 12 MW. Now average load is nothing but the area under the load curve in Kilowatt hour divided by number of hours for which the plant is in operation. From this we can find the time for daily energy produce. Now we already know that the time for daily energy produce is 24 hours for a particular day. Then area under the load curve which is nothing but the energy produce in Kilowatt hour that is number of units produce which we can obtain from this equation as area under the load curve is nothing but energy produce which will be equal to average load into number of hours. So this equation I have written average load into number of hours. Now average load we have obtained is 12 MW and the number of hours for which plant is operated throughout the day is 24. So from this we get that the total energy produce in a day is 288 MW hour or 288 into 10 to the power 3 Kilowatt hours number of units are being produced. Now let us see how to find out the reserve capacity of the plant. The plant capacity factor is nothing but the average demand upon installed capacity. We are given with plant capacity factor and we also know the average demand. So from this we can find out the installed capacity by rearranging this equation as installed capacity will be equal to then average demand upon plant capacity factor which we have taken from here to denominator. So we get installed capacity as average demand upon plant capacity factor. So now putting the value average demand we have calculated as 12 MW divided by plant capacity factor given in the numerical is 0.48. So from this we will be able to get the installed capacity of 25 MW. But we know that the maximum demand already we know and installed capacity know. So reserve capacity will be equal to install capacity minus maximum demand. The installed capacity is 25 MW minus maximum demand is 20 MW. So reserve capacity for the plant is 5 MW. Now let us see the number of operating hours per day. How to find out the number of operating hours for which the plant is operated. Now we know the equation plant use factor is given by average demand into time divided by installed capacity into number of actual operating hours. So here by putting the value of plant use factor as 0.8 average demand is 12 MW multiplied by 24 hours per day divided by installed capacity is 25 MW multiplied by number of operating hours. So by rearranging this equation we will get the number of operating hours will be equal to 12 into 24 divided by 25 into 0.8. So number of operating hours we get it as 14.4 hours. Now here we should understand that the plant is in operation only for 14.4 hours in a particular day. Now let us see what is the maximum energy that could be produced if the generation station was running all the time. The maximum energy that could be produced daily will be equal to install capacity into 24 hours a day. So install capacity is 25 MW multiplied by 24 hours. So we can obtain 600 MW per day if the plant is operated for 24 hours or we can also write it as 600 into 1000 KW that is number of units 600 into 1000 KW that is number of units those can be produced. Now let us see the second numerical that is problem number 2 statement. A 60 MW turbo generator say has an overall efficiency of 25%. The calorific value of coal usage is 24 MJ per kg. Estimate the consumption of coal per kilowatt hour and also per day of 24 hours if the load factor is 30%. So important things to be noted are the capacity of the turbo generator say is 60 MW overall efficiency is 25%. Coal calorific value given is 24 MJ per kg. We want to find out coal consumption per kilowatt hour per unit of electrical energy produced and also per day of 20 hours if the load factor is 30%. Now let us see how we can find out first of all the the consumption of coal per kilowatt hour. Now we know the equation in general overall efficiency will be equal to output upon input. Now we want to find out the coal consumption per kilowatt hour. So therefore what output we will consider let us think what can be the output that we should consider when we want to need to find out the coal consumption per kilowatt hour. So here we will take output as 1 kilowatt hour so that we can find out the input required from the equation of overall efficiency. So input power will be given by output power divided by overall efficiency. Output power given is taken is 1 kilowatt hour which we have taken divided by overall efficiency 0.25. We get input power as 4 kilowatt hour. Now let us see which we can write this 4 kilowatt hour into kilo joules. So 4 kilowatt hour we have expressed as 4 kilo joules per second multiplied by 1 hour equal to 3600 second. Which comes out to be 14400 kilo joules. Now let us see how to find out then coal consumption per kilowatt hour. Heat required per kilowatt hour divided by calorific value of coal. Now heat required we have calculated earlier as that is input power as 14400 kilo joules divided by this is the heat required per kilowatt hour. Divide by calorific value of coal is 24000 kilo joules per kg of coal. From this we get the coal consumption required is 0.6 kg per kilowatt hour. Means for producing 1 unit of electricity we require 0.6 kg of coal. Now let us see we know that the load factor is given by average load upon peak load. Therefore average load will be equal to load factor into peak load. The load factor given is 0.3 whereas the peak load is 60. From this we will be getting the average load of 18 megawatt. Now from this we can find out the coal consumption per day. The energy generated in 24 hours will be equal to average load is 18 multiplied by 24. Which we get as 423 megawatt hour and that in terms of kilowatt hour when we multiply by 1000 we get 423000 kilowatt hours. Now for obtaining the coal consumption per day we can write the equation as energy generated per day in kilowatt hour multiplied by coal required per kilowatt hour. Therefore energy generated just now we have calculated at 423000 kilowatt hour. Multiplied by coal required per kilowatt hour we have calculated in our step 1 as 0.6 kg per kilowatt hour. So which comes out to 259.2 tons. So the coal consumed per day is 259.2 tons. These are the references. Thank you.