Pratap technical experience

  Question = The capital cost of a hydro - power station of 50 MW capacity , is RS 1000 per kW. The annual depreciation charges are 10% (F) of the capital cost. A royalty of re 1 per kW per year and re 0.01 per kWh generated is to be paid for using the river water for generation of power. The maximum demand on the power station is 40 MW and annual load factor is 60 . Annual cost of salaries, maintenance charges etc is RS 700000. If 20% of the expense is also chargeable as fixed charges , calculate the generation cost in two part form ? Solution = 1 st part, annual running charges = formula ,  » Cost per kW = cost per kW due to fixed charges + royalty ,  » given, royalty = 1 RS ,  » cost per kW due to fixed charges  = total annual fixed charges ÷ M.D.  » is M.D. = 40 × 10³ kW ,  » Total annual fixed charges = depreciation charges of capital cost + charges of salaries , maintenance etc ,  » capital cost of plant = 50 × 10³ × 1000 , = 50 × 10⁶ RS ,  » depreciation charges of capital

  Question = A generating station has the following data : installed capacity = 300 MW , capacity factor = 50% or 0.5 , annual load factor = 60% or 0.6 , annual cost of fuel , oil etc (R) =  RS 9×10⁷ , capital cost (F) = RS 10⁹ , annual interest and depreciation (F) = 10% ,  • Calculate (1) the minimum reserve capacity of the station and (2) the cost per kWh generated ? Solution = reserve capacity = installed capacity - maximum demand ,  » is given , Installed capacity = 300 MW ,  » M.D. = ( capacity factor × installed capacity ) ÷ Load factor ,  » capacity factor ( c f. ) = Average demand (A.D.) ÷ Installed capacity (I C.) ,  equations (1) AND ,  Load factor = average demand ( A.D.) ÷ maximum demand ( M.D.) , equations (2) » Dividing 1 and 2 we get ,  » C.F. ÷ L.F. = ( A.D. ÷ I.C. ) ÷ ( A.D. ÷ M.D.) , » after calculate = C.F. ÷ L.F. = M.D. ÷ I.C. , » So , Maximum demand = ( C.F. × I.C. ) ÷ L.F. ,  » M D.= ( 0.5 × 300 MW ) ÷ 0.6 , = 250 MW ,  » So , reserve capacity = 300 - 250 =

  Question =Estimate the generating cost per kWh delivered from a generating station from the following data :  » F = fixed charge , R = running charge ,  Plant capacity = 50 MW , annual load factor = 40% , capital cost (F) = 1.2 crores , annual cost of wages , taxation etc (F) = RS 4 lakhs , Cost of fuel , lubrication , maintenance etc (R) = 1 paise/kWh generated . Interest 5% per annum (F) per annum , depreciation 6% per annum of initial value .  Solution = is given ,  » maximum demand and plant capacity = 50 MW , or 50 × 10³ kWh , • capital cost = 1.2 crores , or 120 × 10⁵ ,  • annual cost of wages , taxation etc (F) = RS 4 lakhs or RS 4 × 10⁵  , • convert % to point = 5% = 0.05 , 6% = 0.06 , • L F. = 40 % = 0.4 , • 1 paise = 1÷ 100 = 0.01 RS ,  » cost per unit = total annual charges ÷ unit generated per annum ,  » Total annual charges = cost of fuel , lubrication etc (R) + total annual fixed charges ,  »  cost of fuel , lubrication etc (R) = unit generated per annum × rate of co

Question =  A generating Plant has a maximum capacity of 100 kW and cost RS 1,60,000 . The annual fixed charges are 12% consisting of 5% intererst , 5% depreciation and 2% taxes . Find the fixed charges per kWh if the load factor is (1) 100% (2) 50% .  Solution = (1) load factor is 100% ,  » fixed charges per kWh = annual fixed charge ÷ unit generated per annum ,  » annual fixed charges = 1,60,000 × RS 0.12 = 19,200 RS ,  » unit generated per annum = maximum demand × L.F. × hours in a years , = 100 kW × 1 × 8760 ,= 8,76,000  kWh ,  » So, fixed charges per kWh = RS 19,200 ÷ 8,76,000 = 0.0219 RS or 2.19 paise , answer ✓  (2) load factor is 50% ,  »fixed charges per kWh = annual fixed charge ÷ unit generated per annum ,  » annual fixed charges = 1,60,000 × RS 0.12 = 19,200 RS ,  » unit generated per annum = maximum demand × L.F. × hours in a years , = 100 kW × 0.5 × 8760 , = 4,38,000 kWh ,  » So, fixed charges per kWh = RS 19,200 ÷ 4,38,8000 = RS 0.0438 or 4.38 paise , answer ✓ It i

  Question = A generating station has an installed capacity of 50,000 kW and delivers 220 × 10⁶ units per annum . If the annual fixed charges are RS 160 per kW installed capacity and running charges are 4 paise per kWh , determine the Cost per unit generated ? Solution = Cost per unit generated = total annual charges ÷ unit per annum ,  » is given , unit per annum = 220 × 10⁶ kWh ,  » Total annual charges = annual fixed charges + annual running charge ,  » annual fixed charges = 160 × plant capacity , = 160 RS × 50,000 kW = 80 × 10⁵ RS ,  » annual running charge = ( 0.04 RS ) × ( 220  × 10⁶ ) kWh = 88 × 10⁵ RS ,  » So , Total annual charges = ( 80 × 10⁵ ) + ( 88 × 10⁵ ) = 168 × 10⁵ RS ,  » now , Cost per unit generated = ( 168 × 10⁵ RS ) ÷ ( 220 × 10⁶ ) = 0.0764 RS or 7.64 paise , answer ✓  

  Question = A generating station has a maximum demand of 50,000 KW . Calculate the cost per unit generated from the following data : capital cost = RS 95 × 10⁶ , annual load factor = 40 % , annual cost of fuel and oil (R) = RS 9 × 10⁶ , interest and depreciation (F) = 12 % , taxes , wages and salaries etc (R) = RS 7.5 ×10⁶ ,  Where , F = fix charge , R = running charge ,  Solution = Cost per unit generated = total annual fix charge ÷ unit generated per annum ,  » unit generated per annum = max..  demand × load factor × hours in a years , = 50,000  kw × 40 % × 8760 hours = 17.52 × 10⁷ kWh ,  » annual fixed charge = (1) annual interest and depreciation = 12% of capital cost = 95 × 10⁶ × 0.12 = 11.4 × 10⁶ RS ,  ( 2nd ) annual running charge = total annual running charge  = annual Cost of  oil and fule + taxes wages etc , = ( 9 × 10⁶ + 7.5 × 10⁶ ) = 16.5 ×10⁶ RS ,   » add charge = ( 11.4 × 10⁶ + 16.5 ×10⁶ ) = 27.9 ×10⁶ ,  » So , Cost per unit generated = ( 27.9 ×10⁶ ) ÷ ( 17.52 × 10⁷ )

  Question = A transformer costing RS 90,000 has a useful life of  20 years . Determine the annual depreciation ( मूल्यहास ) charge  ( RS )using straight line mathod . ? Assume the salvage value of the equipment to be 10,000 . Solution = is given , » Initial cost of transformer , P = 90,000 RS ,  » useful life n = 20 years ,  » salvage value , s = 10,000 rs ,  » annual depreciation charge = ( p- s )÷ n ,  » So , ( 90,000 - 10,000 ) ÷ 20 , = 80,000 ÷ 20 , = 4000 , RS answer ✓

  Question = The annual load duration curve for typical heavy load being served by a steam station , a ran - of - river station and a reservoir hydro - electric station is as shown in figure . The ratio of number of unit supplied by these  station is as follows : Steam : run - of - river : reservoir : 7:4:1   The run - of - river station is capable of generating power continuously and work as a base load station . The reservoir station works as a peak load station . Determine (1) the maximum demand on each station ? And (2)  load factor of each station ?  » solution =  » ODCA is the annual load duration curve for the system as shown in figure . The energy supplied by the reservoir Plant (1)  is represented by area DFG , steam station (2) by area FGCBE and run - of - river ( part 3 ) by area OEBA . The maximum and minimum loads on the system are 320 MW and 160 MW respectively . » 24 hours × 365 days = 8760 hours ,  » Unit generated per annum = area ( in kWh ) under an

  Question = A base load station having a capacity of 18 megawatt and a stand by station having a capacity of 20 megawatt . Share a common load . Find the annual load factor and plant capacity factor of two power station from the following data :  Annual stand by station output = 7.35 × 10⁶ kWh ,   Annual base load station output = 101.35 ×10⁶ kWh.  Peak load on station = 12 MW , hours of use by standby station per year  = 2190 hours  » Solution = Standby station :  » annual load factor = ( unit generated per annum ) ÷ ( maximum demand × annual working hours ) ,  » peak loads एक तरह का maximum demand होता है ।  » So , A.L.F. = ( 7.35 × 10⁶ kWh ) ÷ ( 12 × 10³ × 2190 hours ) =0.28 ,  » for % ( × 100 ) = 0.28 × 100 = 28 % , answer ✓   » Annual plant capacity factor = ( unit generated per annum ) ÷ ( installed capacity × hours in a years ) ,  » A.P.L.F.= ( 7.35 × 10⁶ kWh ) ÷ ( 20 × 10³ × 8760 hours ) , = 0.0419 ,  » for % ( × 100 ) = 0.0419 × 100 = 4.19 % , answer ✓  (2) Base load st

 Question = A generating station is to supply four regions of load whose peak loads are 10 megawatt , 5 megawatt , 8 megawatt and 7 megawatt  . the diversity factor at the station is 1.5 and the average annual load factor is 60% .  (1) Calculate the maximum demand on the station . ? (2) Annual energy supply by the station ? (3) Suggest the installed capacity and the number of unit ? » Solution = (1) maximum demand = sum of maximum demand of the regions ÷ Diversity factor ,  » M D. = ( 10 + 5 + 8 +7 ) ÷ 1.5 = 20 MW , answer ✓  (2) unit generated per annum = M.D. × L.F. × hours in a years ,  » unit generated = 20 × 10³ × 0.6 × 8760 kWh , = 105.12 × 10⁶ kWh , answer ✓ (3) The installed capacity of the station should be 15 % to 20 % more than the maximum demand in order to meet the future growth of load . Tracking install capacity to be 20% more than the maximum demand . » So , installed capacity = maximum demand का 20 % ,  = 20 MW × 20% , = 4 MW ,  » Capacity = 20 MW + 4 = 24 MW ,

  Question = A proposed station has the following daily load cycle :  Draw the load curve and select suitable generator units from the 10000 , 20000 , 25000 , 30000 kva . Prepare the operation schedule for the machines selected and determine the load factor from the curve .  Solution = Load factor = average load ÷ maximum demand ,  » maximum demand by load curve = 70 MW or , 70 × 10³ kW ,  » average load = units generated per day ÷ 24 hours ,  » units generated per day = [ ( 20 × 8 ) + ( 40 × 3 ) + ( 50 × 5 ) + ( 35 × 3 ) + ( 70 × 3 ) + ( 40 × 2 ) ] ,  » after calculate = 925 MWH or , 925 × 10³ kWh ,  » average load = ( 925 × 10³ ) ÷ 24 hours = 38541.7  kW ,  » Load factor = ( 38541.7 kW ) ÷ ( 70 × 10³ kW ) , = 0.5506 ,  » for % ( × 100 ) = 0.5506 × 100 = 55.06 % , answer ✓  » Generators operation schedule : » for 10,000 kva  » let ,  cos Ø = 0.8 . Maximum demand = 70 MW , (common all DG sets ) » kW = kva × cos Ø = 10,000 × 0.8 = 8000 kW or , 8 MW ,  = 7

  Question = A Central station is supplying energy to a community through two substations . Each substation feeds 4 feeders . The maximum daily recorded demands are : •  power station is maximum demand 12,000 kW .  Sub Station A = maximum demand 6000 kW ,  Feeder no. 1 , 1700 kW , feeder no 2, 1800 kW , feeder no 3 , 2800 kW , feeder no 4 600 kW ,  • Sub Station B = maximum demand 9000 kW ,  Feeder no. 1 , 2820 kW , feeder no 2 , 1500 kW , feeder no 3 , 4000 kW, feeder no 4 , 2900 kW ,  Calculate the diversity factor between (1) Central substations , (2) feeder on substation A  , (3) feeders on substation B . ? Solution =  (1) Diversity factor of central station = ( maximum demand on substation A + maximum demand on substation B ) ÷ maximum demand of central station ,  » ( 6000 kW + 9000 kW )  ÷ 12,000 kW , = 15,000 ÷ 12,000 = 1.25 , answer ✓   (2) Diversity factor of substation A = sum of individual load substation A ÷ maximum demand substation A ,  » sum of indivi

  Question = A substation supplies power by four feeders to its consumers . Feeder no. 1 supplies 6 consumers whose individual daily maximum demand are 70 kW , 90 kW , 20 kW , 50 kW , 10 kW and 20 kW . While the maximum demand on the feeder is 200 kW .  Feeder no. 2 supplies 4 consumers whose daily maximum demand are 60 kW , 40 kW , 70 kW , and 30 kW . While the maximum demand on the feeder is 160 kW .  Feeder no.3 and 4 have a daily maximum demand of 150 kW and 200 kW respectively , while the maximum demand on the station is 600 kW .  Determine the diversity factor for feeder no. 1 , feeder no 2 and 4 feeder . ?  Solution =   feeder no 1 , following loads = 70 + 90 + 20 + 50 + 10 + 20 , = 260 kW ,  » maximum demand = 200 kW , » Diversity factor = sum of individual maximum demand ÷ max.. demand on feeder ,  » 260 ÷ 200 = 1.3 , answer ✓   Feeder no 2 following loads = 60 + 40 +70 +30 .= 200 kW , » maximum demand = 160 kW ,  » diversity factor = 200 ÷ 160 = 1.25 a

  Question = A power station has to meet the following load demand:  (1) Diversity factor (2) unit generated per day (3) Load factor .  Solution = the given load cycle can be tabulated as under :  From this  table it's clear that total load on power station is 20 kW for 0 to 10 hours , 50 kW for 10 to 4 pm hours , 70 kW for 4 pm to 6 pm hours , 50 kW for 6 pm to 10 pm hours and 20 kW for 10 pm to 12 pm , hours .  Plotting the load on power station versus time , it is clear from the curve that maximum demand on the station is 70 kW and occurs from 4 pm to 6 pm .  (1) Diversity factor = sum of individual maximum demand of groups ÷ M.D ,  » where , sum of individual maximum demand of groups = 50 + 30 + 20 = 100 kW , » maximum demand show in figure = 70 kW ,  » D.F. = 100 ÷ 70 = 1.42 answer ✓  (2) unit generated per day = kW × hours ,= [ ( 20 × 10 ) + ( 50 × 6 ) + ( 70 × 2 ) + ( 50 × 4 ) + ( 20 × 2 ) ,= 880 kWh , answer ✓   (3) » Load factor

  Question =  A generating station has the following daily load cycle. Solution = Daily curve is drawn by taking the load y-axis and time along x-axis. For the given load cycle , the load curve is show in figure .  (1) It is clear the load curve that maximum demand on the power station is 35 MW and occurs during the period 16 to 20 hours in maximum demand = 35 MW , answer ✓   (2) unit generated per day = [ ( 20 × 6 ) + ( 25 × 4 ) + ( 30 × 2 ) + ( 25 × 4 ) + ( 35 × 4 ) + ( 20 × 4 ) , MWH ,  » after calculate we found = 600 MWH or , 600 ×10³ kWh , answer ✓   (3) average load = unit generated per day  ÷ 24 hours , = ( 600 ×10³ kWh ) ÷ ( 24 hours ) , = 25 × 10³ kW , answer ✓   (4) Load factor = average load ÷ maximum demand , = ( 25 × 10³ kW  ) ÷ ( 35 × 10³ kW ) , = 0.714 or,  » for %  ( × 100 ) = 0.714 × 100 = 71.4 % . answer ✓ 

  Question =  A generating station has a maximum demand ( M.D. ) of 20 MW , a load factor ( L.F. ) 60 % , a plant capacity factor ( P.C.F.) of 60%  and a plant use factor ( P.U.F. ) of 80 % , find  (1) the daily every produced , (2) the reserve capacity of the plant , (3) the maximum energy that could be produced daily , if the plant was running all the time , (4) the maximum energy that could be produced daily , if the plant was running fully loaded and operating as per schedule . ?  Solution = is given , »  M D. = 20 MW or 20,000 kW , » L.F. = 60 % , P.C.F. = 60 % , P.U.F. = 80 % ,  (1) energy produced  = M.D. × L.F. × hours in a one day ,  » 20,000 kW × 0.6 × 24 h , = 288 × 10³  kWh , answer ✓   (2) the reserve capacity of the plant = plant capacity - maximum demand ,  » M.D. = 20 MW , ( is given )  » Plant capacity =average demand ÷ plant capacity factor ,  » average demand = 20 × 0.6 = 12 MW , » P.C. = 12 ÷ 0.6 = 20 MW ,  » So , reserve capacity of the plant = 20 MW - 20 MW

  Question = A generating station supplies the following loads : 15000 kW , 12000 kW , 8500 kW , 6000 kW ,and 450 kW . The station has a maximum demand of 22000 kW . The annual load factor of the station is 48 % . Calculate (1) the number of units supplied annually (2) the diversity factor and , (3) the demand factor .  Solution = is given ,  » according to question total load = 15000 + 12000 + 8500 + 6000 + 450 kW = 41951.90 kW ,  » maximum demand = 22,000 kW , » annual load factor = 48% ,  (1) the number of units supplied annually = M.D. × L.F. × Hours in a years ,  » units = 22,000 × 0.48 × 8760 = 92505600 kWh , or , 925 × 10⁵ kWh , answer ✓  (2) Diversity factor = total following loads ÷ maximum demand ,  » ( 15000 + 12000 + 8500 + 6000 + 450 ) ÷ ( 22,000 )  = 41951.90 kW ÷ 22,000 = 1.9 answer ✓   (3) Demand factor = maximum demand ÷ the following loads = 22, 000 kW ÷ 41951.90 kW ,= 0.5244 , = for % ( × 100 )= 52.44 % answer ✓ 

  Question = A power station is to supply 4 regions of loads whose peak value are 10,000 kW , 5000 kW , 8000 kW and 7000 kW . The diversity factor of the load at the station is 1.5 and the average annual factor is 60% . Calculate the ( maximum demand ) on the station and ( annual energy ) supplied from the station . ?  Solution = according to question We are found = 10,000 + 5000 + 8000 + 7000 kW = 30,000 kW ,  » diversity factor of the load = 1.5 ,  » L.F. = 60 % , or 0.6 ,  (1) maximum demand = total load ÷ diversity factor , = 30,000 ÷ 1.5 = 20,000 kW , answer ✓   (2) energy generated per annum = M.D. × L.F. × Hours in a years ,  » ( 24 hours × 365 days = 8760 hours )  » 20,000 MW × 0.6 × 8760 h = 105120000 kWh , or ,   105.12 × 10⁶ kWh, answer ✓

  Question = A 100 MW power station delivers 100 MW for  2 hours , 50 MW for 8 hours and is shut down for the rest of each day . It is also shut down for maintenance for 60 days each year . Calculate it's annual load factor .?  Solution = is given ,  » 1 st load = 100 MW × 2 hours = 200 MW ,  » 2 nd load = 50 MW × 8 hours = 400 MW , » station operate days = 365 - 60 = 305 days ,in a year ,  • annual load factor ( a.l.f. ) =( MWH supplied per annum ) ÷ ( maximum demand in MW × working hours ) ,  » ( Watt × hours )  = kWh or MWH ,  » MWH supplied per annum  = ( 100 × 2 ) + ( 50 × 8 ) , = 200 + 400 = 600 MWH ,  » energy supply on station operate days = 600 MWH × 305 days = 1,83,000 MWH ,  » So , a.l.f. = ( 1,83,000 MWH ) ÷ ( 100 MWH ) × ( 305 days × 24 hour ) ,  » 0.25 , » for % = × 100 , = 0.25 × 100 , = 25 % , answer ✓ 

  Question = A generating station has a connected load of 40 MW and a maximum demand of 20 MW , the units generated being 60 × 10⁶ kWh . (1) the demand factor (2) the load factor . ?  Solution = is given ,  »  connected load = 40 MW , » maximum demand = 20 MW or , 20,000 kW ,  » unit generated per annum = 60 × 10⁶ kWh ,  (1) Demand factor = maximum demand ÷ connected load. = 20 MW ÷ 40 MW = 0.5 . Answer ✓  (2) load factor = average demand ÷ maximum demand ,  » So , average demand = unit generated per annum ÷ hours in a years ,  » 24 hours  × 365 days= 8760 hours ,  » AV. D. = 60 × 10⁶ kWh ÷ 8760 hours , = 6849.315 kW ,  » L.F. = 6849.315 kW ÷ 20 × 10³ kW , = 0.3424 ,  » for % = × 100 , = 0.3424 × 100 = 34.24% , answer ✓ 

  Question =  the annual load duration curve of a certain power station can be considered as a straight line from 20 MW to 4 MW . To meet this load three turbine generator units , two rated at 10 MW each and one rated at 5 MW are installed determine (1) installed capacity (2) plant factor , (3) units generated per annum , (4) load factor , (5) utilisation factor .  Solution = (1) installed or plant capacity = 10 + 10 + 5 = 25 MW , answer ✓  (2) plant factor = average demand ÷ plant capacity ,  » Referring to the load duration curve , = average demand per unit = 1/2 (20 +4 ) , = 12 MW per unit ,  » so , P.F. = 12 MW ÷ 25 MW , = 0.48 , for % = × 100 , = 0.48 × 100 , = 48 % ,   answer ✓ (3) unit generated per annum = area ( in kWh ) under load duration curve , = 1/2 ( 4000 kW+ 20,000 kW ) × 8760 hours , = 105.12 ×10⁶ kWh , answer ✓   (4) load factor = average demand ÷ maximum demand ,  » L F. = 12 MW ÷ 20 MW = 0.6 ,  » for % = × 100 , = 0.6 × 100 = 60 % , answer ✓  (5) u

  Question = A power station has the following daily load cycle : Plot the load curve and load duration curve . Also calculate the energy generated per day . ?  Solution = figure 1 , show the daily load curve , whereas figure 2 , show the daily load duration curve . It can be readily seen that area under two load curves it the same . Note that load duration curve is drawn by arranging the loads in the order of descending magnitudes .  • According to figure 1 , ( under daily load curve ) ,  • Unit generated per day = watt × hours ,  = [ ( 20 × 8 )  + ( 40 × 4 ) + ( 60 × 4 ) + ( 20 × 4 ) + ( 50 × 4 ) ] , = 840 MWH , or , 840 × 10³ kWh , answer ✓ (2) According to figure 1 , ( under daily load duration curve ) ,  • Unit generated per day = watt × hours ,  = [ ( 60 × 4 ) + ( 50 × 4 ) + ( 40 × 4 ) + ( 20 × 12 ) ] , = 840 MWH , or , 840 × 10³ kWh ,  Which is the same as above . Answer ✓  

  Question = A power station has a daily load cycle as under: 260 MW for 6 hours , 200 MW for 8 hours , 160 MW for 4 hours , 100 MW for 6 hours . If the power station is equipped with 4 set of 75 MW each , calculate (1) daily load factor , (2) plant capacity factor , (3) daily requirement if the calorific value of oil used were 10,000 kcal/kg and the average heat reat of station were 2860 kcal/kWh . Solution = (1) load factor = average load ÷ maximum demand ,  » so, maximum demand = 260 MW ,  » average load = unit generated per day ÷ 24 hours ,  = Unit generated per day = [ (260 × 6 ) + ( 200 × 8 ) + ( 160 × 4 ) + ( 100 × 6 ) ] MWH ,  »4400 MWH , or , 4400 × 10³ kWh ,  » average load = 4400 ÷ 24 ,=  183.333 MW ,  » load factor = 183.333 MW ÷ 260 MW , = 0.70512 ,  » for % = × 100 = 0.70512 × 100 = 70.5 % , answer ✓  (2) plant capacity factor = average demand per day ÷ station capacity ,  » average demand per day =Unit generated per day  ÷ 24 hour , = 4400 MWH ÷ 24 H  = 1,83,333 KW

  Question = A daily load curve which exhibited a 15 minute peak of 3000 kW is drawn to scale of 1 cm = 2 hours and 1 cm = 1000 kW . The total area under the load curve is measured by planimeter and is found to be 12 cm² . Calculate the load factor based on 15 minute peak . ?  Solution = load factor = average load ÷ max.. demand ,  » So , average load = area under daily load curve ÷ 24 hours ,  » 1 cm² of load curve represent = 1000 W × 2 H  = 2000 kWh ,  » 12 cm² में होगा = 2000 × 12  = 24,000 kWh ,  » average demand/load = 24,000 kWh ÷ 24 H = 1000 kW ,  » load factor = 1000 ÷ 3000 ,= 0.3333 ,  » for % = × 100 करेंगे , = 0.333 × 100 = 33.33 % , answer ✓ 

  Question = The daily demands of three consumer are given below .  Plot the daily load curve and find (1) maximum demand of individual consumer (2) load factor of individual consumer (3) diversity factor (4) load factor of the station . ? ??  Solution =  load curve को जब हम horizontal addition करते हैं तो = 200 + 800 + 2400 + 800 + 400 ,watt पाते हैं । इसमें 2400 Watt maximum demand पर है ।  » (1) maximum demand of individual consumer , or consumer 1 = 800 Watt , answer ✓  » consumer 2 = 1000 watt , answer ✓  » consumer 3 = 1200 Watt , answer ✓  (2) Load factor = average load ÷ maximum demand , or ,  Load factor = ( energy consumed/day ) ÷ ( maximum demand × hours in a day ) ,  » So , load factor of consumer 1 =  ( 600 W× 6 H ) + ( 200 W ×2 H ) + ( 800 W × 6 H ) ÷ ( 800 W × 24 H ) ,  » 8800 WH ÷ 19200 WH = 0.45833 ,  » for % × 100 = 0.45833 × 100 = 45.833 % , answer ✓   • Load factor in consumer 2 = ( 200 W × 8 H ) + ( 1000 W × 2 H ) + ( 200 W