 Hello everybody, myself Dr. Sunil Lathain Kulkarni, Assistant Professor, Department of Mechanical Engineering, Vulture Institute of Technology, Solapur. In continuation to my previous video session on methods of Terip, today I am going to conduct a video session on some numerical problems on Terip methods. Let us see at the end of this session, student will be able to solve the numerical problems on Terip, that is nothing but charging for electricity to the consumer. The contents of the video session are some numerical problems on Terip. Now let us see the first problem. In problem one, it is given that a consumer has a maximum demand of 200 kW at 40% load factor. So maximum demand is given as 200 kW and load factor is 0.4. If the Terip is Rs 100 per kW of maximum demand plus 10 per kWh, find the overall cost per kWh. So we need to find out per kWh what is the charges of electrical energy solution. Let us see now. Now here the term load factor is given. What is load factor? We have discussed earlier. Let us think for a while and how we can write the equation for load factor. So we know that load factor is equal to average load upon peak load. So by using this definition, we can calculate the average load by multiplying the load factor with the peak demand. And by using the same, we can also find out the electrical energy that is produced. Therefore the units consumed per year will be equal to maximum demand that is 200 kW into load factor 0.4 into hours in the year. Now this is annual load factor we are considering. So 200 into 0.4 load factor into hours in the year is 8760. So this we get as the units consumed in a year. Now we can calculate the annual charges by using the given tariff N, the annual maximum demand charges which will be equal to Rs 100 per maximum demand into 200 kW plus it is given in the problem Rs 100 per kW of maximum demand plus 10 paise per kWh. Now we have calculated total number of units generated therefore the tariff will be equal to Rs 100 per maximum demand into 200 kW plus 0.1 rupees into total number of units generated. So the annual charges comes out to be Rs 90,080 rupees and when we divide by total number of units generated that is 7,800, we get the overall cost per unit of electrical energy as Rs 1 to 85 that is 12 rupees and 85 paise. Now let us see now the second problem. In this problem it is given that the maximum demand of a consumer is 20 ampere at 220 volts and its total energy consumption is 8760 kWh. If the energy is charged at the rate of 20 paise per unit for 500 hours use of the maximum demand plus per annum plus 10 paise per unit for additional units calculate annual bill and equivalent flat rate. Now for this particular maximum demand we will assume that the load factor and power factor to be in unity therefore the maximum demand will be equal to V into I 220 into 20 ampere 220 volts into 20 ampere that is V into I power factor we are assuming 1. So it will be equal to and we are dividing by 1000 to get it into the kW. So 220 into 20 that will divide by 1000 will be 4.4 kW. Now units consumed in 500 hours will be equal to because it is given that okay the total consumption is 8760 and for first 500 hours the charges are 20 paise and for remaining it is 10 paise for additional unit. Therefore the units consumed in for 500 hours will be 4.4 into 500 it will be equal to 2200 kWh. Therefore the charges for 2200 kWh will be equal to 20 paise so rupees 0.2 into 2200 so rupees 440 and now the remaining units will be 8760 total units minus the first 500 hours units are 2200 so it comes out to be 6560 kWh and for these units the tariff is given as 0.1 rupees that is 10 paise per unit. So charges for 6560 kWh units will be 0.1 into 6560 it comes out to be 656. Therefore total annual bill will be equal to charges for these units 440 rupees plus charges for 6560 that is 656 so it comes out to be rupees 1096 rupees. So equivalent flat rate if we want to calculate it will be equal to total annual bill divided by total number of units so 1096 divided by 8760 so it comes out to be rupees 125 that is 12.5 paise. Now let us see the third problem the following two parts tariffs are offered rupees 100 plus 50 paise per unit and a flat rate of 30 paise per unit so at what consumption the first tariff is economical. Now for this problem let us assume that x be the number of units at which the charges due to both tariffs will be equal. So at x number of units both the tariffs will have the same charges so first tariff we can write it as rupees 100 plus number of units are x into 15 paise means 0.15 rupees so 100 plus 0.15 into x is equal to flat rate of 30 paise per unit so 0.3 into x okay converting into rupees. So by solving this we get 0.3 x minus 0.15 x will be 0.15 x is equal to 100 so x will be equal to 100 by 0.15 so it will be equal to 666.67 units so at this particular number of units okay the first tariff will be economical. Now let us come to the another problem a supply is offered on the basis of fixed charges of rupees 30 per annum plus 3 paise per unit or alternatively at rate of 6 paise per unit for the first 400 units per annum and 5 paise per unit for all the additional unit. Find the number of units taken per annum for which the cost under the two tariff becomes same okay. So here there are two tariff methods which are given and we want to find out the number of units okay number of units per year for which the cost under the two tariffs will be same. Now let us assume let x be the greater than 400 units okay because be the number of units taken per annum for which the annual charges due to both tariffs become equal. So annual charges due to first tariff will be equal to rupees 30 plus 3 paise means 0.03 into x these are the annual charges for first tariff. Now annual charges due to second tariff will be equal to second tariff is given as just we can have once again here have the loop and second tariff is at the rate of 6 paise per unit for first 400 units per annum okay for first 400 6 paise and for additional unit 5 paise. So for second tariff we can write it as for first 400 units 6 paise means 0.06 rupees plus x are the number of units above 400 so x minus 400 into 5 paise. So second tariff equation we get as rupees 4 rupees 4 plus 0.05 into x after simplifying this term. So as the charges in both cases are to be equal we can write the first tariff equation is 30 plus 0.03 into x is equal to 4 plus 0.05 into x therefore x will be equal to if you simplify this okay x will be equal to 30 minus 4 divided by 0.05 minus 0.03. So number of units to be taken are coming to be x is equal to 1300 kilowatt hour this is the number of units to be taken per annum for which the charges will become equal. So these are the references thank you.