Pratap technical experience

  Question = A hydro electric plant costs RS 3000 per kW of installed capacity . The total annual charges consist of 5 % (F) as interest , depreciation at 2% (F) , operation and maintenance at 2 % (R) and insurance , rent etc , 1.5 % (R). Determine a suitable two part tariff if the losses in transmission and distribution are 12.5% and diversity of loads is 1.25 . Assume that maximum demand on the station is 80% of the capacity and annual load factor is 40% . What is the overall cost of generation per kWh ? Solution = overall cost /kWh = total annual charges ÷ unit reaching the consumer ,  » Total annual charges = annual fixed charges + annual running charges ,  » annual fixed charges (A F.C ) = capital cost of plant × ( interest % + depreciation % ) /100 , » capital cost of plant = let installed capacity of the station kW × plant cost per kW , » 100 kW × 3000 RS = 3 × 10⁵ RS ,  » interest % + depreciation % ) /100 = (5 +2 ) ÷ 100  » So , A F.C. = 3 × 10⁵ RS ...

  Question = The annual working Cost of a power station is represented by the formula RS ( a+b kW+c kwh ) where the various terms have their usual meaning . Determine the value of a , b and c for a 60 MW station operating at annual load factor of 50% from the following data :  • F = fixed charges , R = running charges ,  (1) capital cost of building and equipment is RS 5 × 10⁶ ( 50 lakh ) ,  (2) the annual cost of fuel , oil , taxation and wages of operating staff is RS 9,00,000 ( 9 lakhs ) . (R) (3) the interest and depreciation on building and equipment are 10% per annum , (F) (4) annual cost of organisation and interest on cost of site etc is RS 5,00,000 ( 5 lakhs ) . (F)  Solution = is given ,  » annual operating cost = RS ( a + b kw + c kwh ) ,   » a = annual fixed cost , • b = b × kW = annual semi fixed cost , • c =  c × kWh =  annual running cost ,  (1) a = annual fixed cost = the annual fixed cost is due to the annual cost o...

  Question = At the end of a power distribution system , a certain feeder supplies three distribution transformers ,each one supplying a group of customer whose connected load are as under :  • transformer no. 1 load 10 kw , demand factor 0.65 , diversity of groups 1.5 , • transformer no. 2 load 12 kw , demand factor 0.6 , diversity of groups 3.5 , • transformer no. 3 load 15 kw , demand factor 0.7 , diversity of groups 1.5 , If the diversity factor among the transformer is 1.3 , find the maximum load on the feeder ?  Solution = Show a feeder supplying three distribution transformers .  » formula = maximum demand or , load  on feeder = sum of maximum demand transformer 1 + transf 2 +trans 3  ÷ diversity factor ,  » So , maximum demand trans.. no. 1 = ( connected load of trans. 1 × demand factor ) ÷ diversity factor  ,  = ( 10 kW × 0.65 ) ÷ 1.5 , = 4.33 kw ,  » maximum demand of trans . 2 =  ( 12 kw × 0.7 ) ÷ 3.5 , ...

  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 ...

  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 , =...

  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 , lubri...

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...

  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 ,  ...

  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 ge...

  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 ,  » ...

 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 ,...

  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 D...

  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 su...

  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...

  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...

  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 ...

  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 ,  »...

  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...

  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 ...

  Question = A power station has to meet the following demand : • group A - 200 kW between 8 a.m. and 6 p.m.  • group B - 100 kW between 6 a.m. and 10 a.m. • group C - 50 kW between 6 a.m. and 10 a.m. • group D - 100 kW between 10 a.m. and 6 pm and then between 6 pm and 6 am .  Plot the daily load curve and determine (1) diversity factor (2) unit generated/day (3) load factor ?  Solution = It is clear that Total load on power station is 100 kW for 0-6 hours , 150 kW for 6 - 8 hours , 350 kw for 8 - 10 hours , 300 kW for 10 - 18 hours and 100 kW for 18 - 24 hours . Plotting the load on power station versus time , we get the daiy load curve as shown in fig..  it is clear from the curve that maximum demand on the station is 350 kw and occurs from 8 am to 10 am . (1) diversity factor = sum of individual maximum demand of group ÷ max .. demand on station , = ( 200 +100 + 50 + 100 ) kw ÷ 350 kW , = 450 ÷ 350 = 1.286 answer ✓   (2) according to...

  Question = A generating station has the following daily load cycles : Draw the load curve and (1) maximum demand , (2) unit generated/day , (3) average load (4) load factor . ?  Solution = Daily load curve is drawn by taking the load a long y- axis time a long x- axis. For the given load cycle , the load curve is show in figure .  (1) it is clear from the load curve that maximum demand on the power station is 70 MW and occurs during the period 16 to 20 hours . So , maximum demand = 70 MW , answer ✓   (2) unit generated/day = MW × hours ,  »  [ 40 × 6 + 50 × 4 + 60 × 2 + 50 × 4 + 70 × 4 + 40 × 4 ] MWH ,  » [ 240 + 200 + 120 + 200 + 280 + 160 ] MWH , » 1200 MWH ,  » ÷ 1000 for kWh = 12 × 10⁵ kWh , answer ✓   (3) average load =unit generated/day ÷ 24 hours ,  » ( 12 × 10⁵ )  kWh ÷ 24 hours, = 50,000 kw , answer ✓   (4) load factor = average load ÷ maximum demand ,  » 70 MW or , 70 × 10³ kW ,...

  Question = It has been desired to install a diesel power station to supply power in a suburban . (1) 1000 houses with average connected load of 1.5 kW in each house . The demand factor 0.4 and diversity factor 2.5 respectively . (2) 10 factories having overall maximum demand of 90 kw .  (3) 7 tubewell of 7 kW each and operating together in the morning .  The diversity factor among above three types of consumers is 1.2 .what should be the minimum capacity of power station ? . Solution = formula = minimum capacity of station/power required = the sum of maximum demand of three types of loads ÷ diversity factor of three types of consumers ,  » maximum demand of three types of loads 1 = ( connected load × demand factor ) ÷ ( diversity factor of load 1 ) ,  » ( 1.5 kW × 0.4 ) × 1000  ÷ 2 .5 , = 240 kw ,  » maximum demand of factories = 90 kw ,  » maximum demand of tubewell = 7 × 7 = 49 kw ,  » So , the sum of maximum demand of three types of ...

  Question = A power supply is having the following loads : • Domestic load = maximum demand 1500 kW , diversity of group 1.2 , Demand factor 0.8 ,  • commercial load = 2000 kW , Diversity of group 1.1 , Demand factor 0.9 ,  • Industrial load = maximum demand 10,000 kW , diversity of group 1.25 , demand factor 1 , If the overall system diversity factor is 1.35 determine , (1) the maximum demand and (2) connected load of each type .  Solution = the sum of maximum demand of three type of loads is = 1500 + 2000 + 10,000 = 13,500 kw ,  As the system diversity factor is 1.35 ,  » maximum demand on supply system = 13,500/1.35 = 10,000 kW , answer ✓   (2) Each type of load has its own diversity factor among its consumers .  » sum of maximum demands of different domestic consumer = maximum domestic demand ×  diversity factor , = 1500 × 1.2/0.8 = 2250 kW , answer ✓ » connected commercial load = 2000 × 1.1/0.9 , = 2444 kW , answer ✓   » con...

  Question = A power station has a maximum demand of 15000 kW . The annual load factor is 50%  and plant capacity factor is 40% . Determine the reserve capacity of the plant ?  Solution = is given ,  » M.D. = 15,000 KW , L F. = 50% , Plant capacity factor = 40 % ,  » reserve capacity = plant capacity -  max demand ,  » plant capacity factor = ( unit generated/annum ) ÷ ( plant capacity × hours in a year ) ,  » unit or energy generated/annum = max .. demand × L.F. × hours in a year , = 15,000 × 0.5 × 8760 , = 65.7 ×10⁶ kWh ,  » plant capacity = ( 65.7 ×10⁶ kwh ) ÷ ( 0.4 × 8760 hour )  = 18,750 kW ,  » reserve capacity = plant capacity - max.. demand , = 18,750 - 15,000 = 3,750 kw , answer ✓  

  Question = A diesel station supplies the following loads to various consumer : industrial consumer = 1500 kw , commercial establishment  = 750 kw , domestic power = 100 kw , domestic light = 450 kw ,if the maximum demand on the station is 2500 kw and the number of kWh generated/year is 45 × 10⁵ , determine (1) the diversity factor and (2) annual load factor .  Solution = formula =  (1) diversity factor = sum of individual max. demand ÷ max. demand on Power station ,  » ( 1500 + 750 +100 +450 ) ÷ 2500 , = 2800 ÷2500 = 1.12 , answer ✓   » (2) annual load factor = average load /max. demand ,  » before , average load = kWh generated / hours in a year ,  » load factor = kWh/M D. × 24 H × 365 days ,   = 45 × 10⁵ kwh ÷ 8760 Hour , = 513.7 kw ,  » load factor = 513.7 kw ÷ 2500 kw = 0.205 , = 20.5 % , answer ✓ 

  Question = A generating station has a maximum demand of 25 MW a load factor of 60% , a plant capacity factor of 50% and a plant use factor of 72% . find (1) the reserve capacity of the plant , (2) the daily energy produce and (3) maximum energy that could be produced daily ? , if the plant while running as per schedule where ,  fully loaded .  • solution = is given ,  • generating station maximum demand = 25 MW ,  • L.F. = 60% , • plant capacity factor = 50 % , use factor = 72 % ,  » reserve capacity of the plant = plant capacity - maximum demand ,  » plant capacity = average demand ÷ plant capacity factor ,  » average demand = maximum demand  × load factor ,  = 25 MW × 0.60 = 15 MW ,  » plant capacity = a.d./p c.f. , = 15 MW/0.5 , = 30 MW ,  » reserve capacity of the plant = p.c. - m.d. , =  30 - 25 = 5 MW , answer ✓   (2) daily energy produce = average demand × 24 hours , = 15 MW × 24 H , = 360 MWH , answer ✓...

  Question = A 100 megawatt power station delivers 100 megawatt for 2 hour , 50 megawatt for 6 hour and is shut down for the rest of each day .  it is also shut down for maintenance for 45 days each year .  calculate the annual load factor ?  • Solution = • max. demand = 100 MW , • 1 st Watt hour = 100 MW × 2 H , mwh • 2 nd watt hour = 50 MW × 6 H , mwh •  Formula: » annual load factor = ( mwh supplied per annum ) ÷ ( max. demand in MW × working hours ) ×100 , » station operates for in a year = 365 - 45 = 320 days ,  » MWH/energy supplied per annum = energy supplied for each working day × day operates in a year ,  » energy supplied for each working day = (100 × 2 ) + ( 50 ×6 )  = 500 mwh ,  » MWH or , energy supplied/year = 500 mwh × 320 = 1,60,000 MWH ,  » annual load factor =( energy supplied/year  ) ÷ ( max. demand ) × ( 320 days × 24 hours ) *× 100 ( for % ) ,   » a.l.f. = ( 1,60,000 MWH ) ÷ ( 100 MW ) × ( 320 days ...

  Question = A generating station has a connected load of 43 megawatt and a maximum demand of 20 megawatt . the units generated being 61.5 × 10⁶ per annum . calculate (1)  the demand factor and (2) load factor  ?  • solution = connected load = 43 MW , • max.. demand = 20 MW , • unit generated/ annum = 61.5 × 10⁶ kWh ,  (1) demand factor = max. demand/ connected load , = 20 MW ÷ 43 MW , = 0.465 , answer ✓  (2) Load factor = average demand ÷ max.. demand.  » hours in a one year = 24 hours × 365 days ,= 8760 hours ,  » Average demand = ( unit generated/annum ) ÷ hours in a year , =( 61.5 ×10⁶ ) ÷ 8760 ,= 7020 kw ,  » L.F.= average demand ÷ maximum demand. = ( 7020 ) ÷ ( 20 × 10³ ) ,= 0.351 , or, 35.1 % , answer ✓  

  Question = the maximum demand on a power station is 100 megawatt . if the annual load factor is 40% . calculate the total energy generated in a year ?  Solution = » total energy generated in a year = max. demand × L.F. × hours , in a year , = ( 100 ×10³ kW ) × ( 0.4 ) × ( 24 hours × 365 days ) kWh , = 3504 × 10⁵ kWh , answer ✓ 

  Question = An atomic power reactor can deliver 300 MW . if due to fission of each atom of 92 u²³⁵ .  the energy released is 200 mev . Calculate the mass of Uranium fission  per hour .?  Solution = Uranium के कुछ परमाणु के नाभिक में 146 न्यूट्रॉन रहते हैं , जबकि अन्य परमाणु के नाभिक में 143 न्यूट्रॉन रहते हैं । अंत: यूरेनियम दो रूप में पाया जाता है । ये दोनों रूप यूरेनियम के समस्थानिक ( isotopes ) कहलाते हैं । इन दोनों की परमाणु संख्या समान , किंतु परमाणु द्रव्यमान अलग-अलग होते हैं । • परमाणु संख्या 92 + न्यूट्रॉन 146 = 238 , ( 92 u²³⁸ )  • परमाणु संख्या 92 + न्यूट्रॉन 143 = 235 , ( 92 u²³⁵ )  » energy deliver = 300 MW or, 3 × 10⁸ w , or, j/s ,  » isotopes type = 92 u²³⁵  » energy released/fission = 200 Mev ,  ( Joule को watt भी कहते हैं । Joule = V × A × Second , or, W = V × I × S , = or , W. S. , watt second  )  » mass of Uranium fissioned/hour = ( atomic mass unit (a.m.u. ) of Uranium ) ÷ ( avogadro number ) ,  » av...

  Question = A diesel engine power plant has 700 kilowatt and two 500 kilowatt generating units .  the fuel consumption is 0.28 kg/kwh and calorific value of fuel oil is 10,200 kcal/kg .  Estimate (1) the fuel oil required for a month of 30 days ? and (2) overall η .? plant capacity factor 40% .  Solution = »  first generating unit =  700 kW ,  and ,  »  second two generating units = 500 kw × 2 , » per unit fuel consumption = 0.28 kg/ kWh ,  » per kg calorific value of fuel oil = 10,200 kcal/kg ,  » plant capacity factor =40% ,  » 1 kWh fuel consumption = 0.28 kg ,  » x kwh fuel consumption = x × 0.28 kg ,  » x or , actual energy produced in a month , kWh = ?  Formula ,  » plant capacity factor = actual energy produced/month ÷ max . energy that could have been produced/month .  » maximum energy that could have been produce in a month = plant capacity × hour in a month , = (700 kw + 2 × 500 kw ) × (30...

  Question = A diesel power station has the following data : • fuel consumption per day = 1000 kg , • unit generated per day = 4000 kwh , • calorific value of fuel = 10,000 kcal/kg ,  • alternator efficiency= 96% , • engine mechanical efficiency = 95% , estimate (1) specific fuel consumption , (2)  η overall , (3) thermal efficiency of engine ?  Solution = (1)  specific fuel consumption = fuel consumption per day ÷ unit generated per day , = 1000 ÷ 4000 , = 0.25 kg/kWh , answer ✓  (2) η overall = electrical output in heat unit per day ÷ heat produced by fuel per day ,  • So first.  » heat produced by fuel per day = coal consumption/day × calorific value , = 1000 × 10,000 = 10⁷ kcal ,  » electrical output in heat unit per day = 4000 × 860 = 344 ×10⁴ kcal ,  »  η overall = ( 344 ×10⁴ kcal ÷ 10⁷ kcal ) × 100 , = 34.4 % , answer ✓  (3) η engine =  η overall ÷ η alternator ,= 34.4 ÷ 0.96 , = 35.83 % ,  So. Therm...

  Question= A diesel power station has fuel consumption of 0.28 kg/kwh the calorific value of fuel begin 10,000 kcal/kg . determine (1) the overall efficiency , and (2) efficiency of the engine if alternator efficiency is 95% .  Solution =•  Fuel consumption = 0.28 kg/kWh , • calorific value of fuel = 10,000 kcal/kg ,  (1) efficiency ( η ) overall = electrical output in heat unit ÷ heat of combustion ,  • So first ,  » electrical output in heat units or , heat equivalent of 1kwh =  860 kcal ,  » heat of combustion or , heat produced by 0.28 kg of oil = 10,000 × 0.28 = 2800  kcal ,  » η overall = 860 ÷ 2800 = 0.307 = 30.7 % , answer ✓ (2) engine η =  η overall ÷ η alternator , = 30.7 ÷ 0.95 = 32.3 % , answer ✓ 

  Question = A hydroelectric plant has a catchment area of 120 km² . the available run of is 50% with annual rainfall of 100 cm . A head of 250 m is available on the average . efficiency of the power plant is 70% find (1)  average power produce (2)  capacity of the power plant . ?  Assume a load factor of 0.6 . Solution = is given ,  • catchment area = 120 km² , or , 120 × 10³ × 10³  m² , or , 120 × 10⁶ m² , • yield factor or , run off = 50% , or, 0.5 , • annual rainfall = 100 cm , or , ÷ 100 = 1 meter , • head = 250 m , • efficiency ( η) = 70 % or, 0.7 , • load factor = 0.6 ,  ( 1)   average Power produced = kWh ÷ 8760 ,  »( yield factor or , run off  , is same ) » volume of water which can be utilised per annum =  catchment area × annual rainfall ×  yield factor , = 120 × 10⁶ m² × 1 m × 0.5 , = 60 × 10⁶ m³ ,  » weight of water available per annum is = W =  60 × 10⁶ × 9.81 × 1000 , = 588.6 ×10⁹ N ,  » el...

  Question = A hydroelectric power station is supplied from a reservoir having an area 50 km² and a head of 50 m . if overall efficiency of plant is 60% ,  find the rate of which the water level will fall ? , when the station is generated 30,000 kilowatt . Solution = Area of reservoir = 50km² or , 50 ×10⁶ m² ,  • H = 50 m. • η plant = 60% , or , 0.6 , • Water level fall = ? , Power generated = 30,000 kW ,  » let ,  » 1 hour or, 3600 sec.. = x metre fall Water ,  » in 1 sec.. = x ÷ 3600 fall Water ,  » average discharge/sec . = ( Area of reservoir ×x ) ÷ 3600 ,  » or ,  » 50 km²  or , 50 × 10⁶ = ( 50 × 10⁶ × x ) /3600 , = 1.388 × 10⁴ x m³  ,  » weight of water available/sec.  =W = 1.388 ×10⁴ x m³ × 1000 × 9.81 , N  » 13.624 × 10⁷ x N ,  » average Power produced = W × H × η plan , = 13.624x × 10⁷ × 50 m  × 0.6 , = 408.749x ×10⁷ Watt , or , 408.749x × 10⁴ kW ,  » but , kW produced = 30,000 kW ,  »...

  Question = it has in estimated that a minimum run - off  of approximately 94 m³/sec. will be available at a hydraulic project with a head of 39 m . determine the firm capacity and yearly gross output . ? Solution = weight of water available is W = volume of water × density . = 94 × 1000 , = 94,000 kg/sec.. ,  » water head = H = 39 m ,  » work done/sec.. = W × H , = 94000 ×9.81 ×39 watt , = 35,963 ×10³ watt , or , 35,963 kW , answer ✓  ( this is gross plant capacity ) » (2)  yearly gross output = firm capacity × hours in a year , = 35,963 × 8760 , = 312670680 kWh , or , 312 × 10⁶ kWh , answer ✓ 

  Question = a hydro-electric plant has a reservoir of area 2 km² and of capacity 5 million cubic meters . The net head of water at the turbine is 50 meters .  If the efficiency of turbine and generator are 85% and 95% respectively , calculate the total energy in kwh that can be generated from this station . If a load of 15000 kilowatt has been supplied for 4 hour , find the fall in reservoir .? Solution = (1) weight of water available , W = volume of reservoir × weight of 1 m³ of water , = 5 ×10⁶ × 1000 = 5 ×10⁹ × 9.81 N ,  » η overall= 0.85 × 0.95 = 0.8075 ,  » electrical energy that can be generated = W × H × η overall , watt sec... ,=   5 ×10⁹ × 9.81 N × 50 × 0.8075 /1000 ×3600 kWh , = 550109.375 kWh or , 5.5 × 10⁵ kWh , answer ✓   » (2) let , x meter be the fall in reservoir level in 4 hours ,  » 2 km²  or , 2 × 10⁶ m² ,  » average discharge/second = ( area of reservoir × x ) ÷ ( 4 ×3600 ) , =( 2 × 10⁶ × x ) ÷ ( 4 × 3600 ) , = ...

  Question = calculate the continuous power that will be available from hydroelectric plant having an available head of 300 meter ,  catchment area of 150 km² ,  annual rainfall 1.25 meter ,  yield factor 50% . assume penstock , turbine and generator efficiency to be 96% , 86% and 97% respectively . if the load factor is 40% what should be the rating of the generators installed ?  Solution =»  catchment area = 150 km² , or. 150 × 10³ × 10³ m² , or , 150 × 10⁶ m² ,  » head = 300 Meter , » annual rainfall = 1.25 meter , » Load factor = 40% , or , 0.4  , » yield factor ( K ) = 50% , or , 0.5 , » efficiency ( η ) overall = 96 % × 86 % ×97 % = 0.800832 % ,  » volume of water which can be utilised per annum =  catchment area × annual rainfall ×  yield factor , = 150 × 10⁶ × 1.25 × 0.5 , = 93.75 × 10⁶ m³ ,  » weight of water available per annum is W = 93.75 ×10⁶ × 1000 × 9.81 , = 919.6875 × 10⁹ N ,  » electrical energy availab...

  Question = a hydroelectric station has an average available head of 100 m and reservoir capacity of 50 million cubic metres .  calculate the total energy in kwh that can be generated , assuming hydraulic efficiency  of 85% and electric efficiency of 90% . Solution = weight of water available is W = volume of water × density , = 50 × 10⁶  ×1000 kg ,  » overall efficiency = 0.85 × 0.9 = 0.765 ,  » work = ( 500 × 10⁶ ) × ( 1000 ) kg × ( 9.81 N ) ,  » So , electrical energy available = W × H ×  η overall , watt - sec.. ,  » 50 × 10⁶ × 1000 × 9.81 × 100 × 0.765 / 3600 ×1000 , kWh ,  » 37523250 × 10⁶/3600 × 1000 , kWh ,  » 10.423 × 10⁶ kWh , answer ✓   (1 kWh = 1000 Watt × 3600 seconds , = 36 ×10⁵ joules , )  ( For , hour = ÷ 3600 , and , kilo = ÷ 1000 ) 

  Question = the weekly discharge of a typical hydroelectric plant is as under ,  • days/discharge ( m³/sec ) = sun/500 , mon/520 , Tues/850 , wed/800 , thur/875 , fri/900 , sat/546 ,  The plant has an effective head of 15 m and overall efficiency of 85% . if the plant operates on 40% load factor ,  estimate (1)  the average daily discharge , (2) pondage required and (3)  installed capacity of proposed plant .  Solution = is given ,  » Head (H) = 15 m ,  η overall = 85 % , load factor = 40 % ,  » show the plot of weekly discharge is taken along y- axis and day along x - axis . » (1) average daily discharge = ( 500 + 520 + 850 + 800 + 875 + 900 + 546 ) ÷ 7 days , = 4991 ÷ 7 = 713 m³/sec .. , answer ✓  (2) it is clear from graph that on three days (viz ,  Sunday , Monday and Saturday ) the discharge is less than the average discharge .  »  volume of water required on these 3 day = ( 500 + 520 + 546 ) × 24 × 3600 m³...

  Question = A run of river hydroelectric plant with pondage has the following data ;  installed capacity =  10 MW , water head , H =  20 meter , efficiency overhaul = 80% , load factor = 40% ,  (1) determine the river discharge in m³/sec.  required for the plant , (2)  if on a particular day ,  the river flow is 20 m³/sec. , What load factor can the plant supply ? Solution = consider the duration to be of one week unit generated / week = max demand × L.F. × hours in a week ,  » (10 × 10³) × (0.4) × (24×7) kWh = 67.2 ×10⁴ kWh. (1)  » let Q m³/sec. be the rever discharge required =  » weight of water available/sec. , W = Q × 9.81 × 1000 , = 9810 Q Newtown (N) ,  » average Power produced = W × H ×  η overall ,  = 9810 Q × 20 × 0.8 W , = 156960 Q watt , or , 156.96 Q Kw  » ( 24 hours × 7 days = 168 hours ) »  unit generated/week = 156.86 Q × 168 kWh , = 26369 Q kWh (2)  » equations (1) and (2) , We ge...

  Question = A factory is located near a water fall where the usable head for power generation is 25 m . The factory requires continuous power of 400 kilowatt throughout the year . The river flow in a year is (1) 10 m³/sec .  for 4 months , (2) 6 m³/sec.  for 2 months , and (c) 1.5 m³/sec .  for  6 months ,   (1) if the site is developed as a run of river type of plant without storage ,  determine the standby capacity to be provide .  assume that overall efficiency of the plant is 80% (2) if a reservoir is arranged upstream , will any standby unit be necessary ? what will be the excess power available ?  Solution = is given ,  • Water head = H = 25 m , • power requires continuous/ year = 400 kw , • river flow in months = 10 m³/sec. for 4 months , 6 m³/ sec. for 2 months , 1.5 m³/sec.  for 6 months ,  • efficiency ( η )  overall of plant = 80 % ,  (1) ran off rever plant =  in this type of plant ,  the ...