Pratap technical experience

  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 load curve , units generated

  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 ,  » 50,000 kW ÷ ( 70 × 10³ ) kW , = 0.714 or , 71.4 % , answer ✓ 

  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 loads = 240 +90 + 49 , = 379 kw ,  » So ,  m

  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 ✓   » connected industrial load = 10,000 × 1.25/1 , = 12,5

  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 ✓  (3) maximum energy that could be produced = actual energy produced

  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 × 24 hours )  × 100 (for % ) = 20.8 % , answer ✓ 

  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 ) ,  » avogadro number ( N or NA ) =  पदार्थ के एक मोल  में उसके  सत्ताओ

  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 ) × (30days × 24 hours ) , = 1700 × 720 kWh ,  » actual energy produced in a month = 0.4 ×1

  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. Thermal efficiency of engine = efficiency engine ÷ mechanical efficiency o

  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 ,  » electrical energy available per annum = W × H ×  efficency overall ,  Watt - seco

  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 ,  » So , 408.749x × 10⁴ kw = 30,000 kw ,  » x = ( 30,000 kw ) ÷ ( 408.749 × 10⁴ kw ) = 0.007

  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 ) , = 138.888 x m³ ,  » weight of water available per s

  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 available per annum = W × H ×  η overall , = 919.6875 ×10⁹ N × 300 m ×

  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 , = 1566 × 24 × 3600 m³/sec . ,  » ( 1 hour = 3600 sec.. , 24 hours

  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 get ,  » 26,369 Q = 67.2 × 10⁴ , = Q = 67.2 × 10⁴ /26369 , = 25.48 m³/sec , answer ✓

  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 whole water of stream is allowed to pass through the turbine for power generati

  Question =A hydroe-electrical power station has a reservoir of area 2.4 km² and capacity 5 × 10⁶ m³ . The effective head of water is 100 meter .  the pen stocks , turbine and generation efficiency are respectively 95% ,  90% and 85% .  (1) calculate the total electrical energy that can be generated from the power station .   (2) if a load of 15000 kilowatt has been supplied for 3 hour , find the fall in reservoir level .  Solution = is given ,  • Reservoir area = 2.4 km² ,or , 2.4 × 10⁶ m²  • water head , H = 100 meter , •  volume of reservoir = 5 × 10⁶ m³ ,  • efficiency ( η ) penstock = 95% ,  •  η turbine = 90% , •  η generation = 85% ,  »( 1) electrical energy that can be generated =  W× H × η  over all ,  » w , or weight of water available = volume of reservoir × weight of 1m³ water = 5 ×10⁶ m³ × 1000 kg , = 5 × 10⁶ m³ × 1000 kg , or 5 × 10⁹ kg , or , 5 × 10⁹ × 9.81 N ,  » overall η = 0.95 × 0.9 × 0.85 , = 0.726 ,  » energy generated = (  5 × 10⁹ × 9.81 N × 100 m × 0.726 )

  Question = calculate the average Power in kilowatt that can be generated in a hydroelectric project from the following data catchment area =  5 × 10⁹ m² , mean head = H = 30 m , annual rain fall , F = 1.25 m , yield factor. K = 80% , overall efficiency = 70% , if the load factor is 40% so what is the rating of generator installed ?  Solution = average Power = electrical energy available per annum = W × H ×  η overall , watt sec .  » w , or weight of water available per annum is = volume of water which can be utilise per annum × 9.81 × 1000 ,  » volume of water which can be utilised per annum = catchment area × annual rainfall × yield factor , = 5 ×10⁹ m² × 1.25 m × 0.8 = 5× 10⁹ m³ ,  » So , W = 5 × 10⁹ m³ × 9.81 ×1000 = 49.05 ×10¹² N ,  » electrical energy available per annum = 49.05 × 10¹² N × 30 m ×0.7 watt sec.. ,  (For kWh = ÷ 1000 and , for hour = ÷3600 , ) » ( 49.05 × 10¹² N × 30 m ×0.7 watt sec.. ) ÷ ( 1000 × 3600 ) , kWh ,  » 2.86 ×10⁸ kWh , and , average Power = 2.86 × 1

  Question = water for a hydroelectric station is obtained from a reservoir with a head of 100 metres . calculate the electrical energy generated per hour per cubic meter of water if the hydraulic efficiency be 0.86 and electrical efficiency 0.92 . Solution = water head = H = 100 m , discharge, Q = 1 m³/sec.. , η overall = 0.86 × 0.92 = 0.79 , energy generated/hour = ? ,  (Mass of 1m³ of water is 1000 kg )  » energy generated/hour = power produced × hour ,  » Power produced = w × H ×  η overall ,  » w or, weight of water available/sec. = Q × 1000 × 9.81 = 9810 N ,  » power produced = 9810 ×100 × 0.79 = 775 ×10³ watt ,  or ÷ 1000 = 775 kw ,  » energy generated/hour = 775 ×1 = 775 kWh , answer ✓ 

  Question = it has been estimate that a minimum run of approximately 94 m³/sec.  will be available at a hydraulic project with head of 39 m . determine (1) firm capacity (2)  yearly gross output .  asume the efficiency of plant to be 80% . Solution = weight of water available ,  w = 94 m³/sec. , Head of water = H = 39 metre , plant efficiency = 80% ,  » formula ,  Firm capacity = plant efficiency × gross plant capacity ,  » gross plant capacity = work done/sec ( m³ × 1000 = kgs ) ,  » w = 94 × 1000 = 94000 kg/sec ,  » work done/sec = w × h , = 94000 × 9.81 × 39 watts. = 35963 ×10³ watt , or, 35963 kw ,  » (1) firm capacity = 0.80 × 35963 = 28770 kw , answer ✓ » (2)  yearly gross output = firm capacity × hours in a year , = 28770 × 8760 = 252 × 10⁶ kWh , answer ✓ 

  Question = A hydroelectric generating station is supplied from a reservoir of capacity 5 × 10⁶ cubic metre at a head of 200 metres . find the total energy available in kWh, if the overall efficiency is 75% ? Solution = weight of water available is w = volume of water × density ,  » 5 × 10⁶ × 1000 ( mass of 1m³ of water is 1000 kgs )  » 5 × 10⁹ kg , = 5 × 10⁹ × 9.81 N ,  » electrical energy available = w × h × overall efficiency ,  = 5 × 10⁹ × 9.81 × 200 × 0.75 watt second ,  » for hour = ÷ 3600  , and for kilo  = ÷ 1000 ,  »  5 × 10⁹ × 9.81 × 200 × 0.75 ÷ 3600 × 1000 ,  » 2.044 × 10⁶ kWh , answer ✓ 

  Question = A thermal station has an efficiency of 15% and 1 kg of coal burnt for Every kWh generated . Determine the calorific value of coal ? Solution = let,  » X kcal/ kg be the calorific value of fuel .  » heat produced by 1 kg of coal = 1 kg × x kcal ,  » heat equivalent of 1 kwh = 860 kcal ,  » thermal efficiency = 860 kcal / 1 kg × x kcal/ kg ,  » 15% = 860 kcal / 1 kg × x kcal/kg ,  » x kcal/ kg = 860 kcal /1 kg × 0.15 ,  = 5733 kcal/ kg , answer ✓ 

  Question = A 65000 kW steam power station used coal of calorific value 15000 kcal/ kg . If the coal consumption per kWh is 0.5 kg and the load factor of the station is 40 % . Calculate (1) coal consumption per day ?  Solution = steam power station = 65000 kW , coal calorific value = 15000 kcal/ kg , per kWh coal consumption = 05 kg , load factor = 40 % ,  Formula ,  » coal consumption per day = unit generated/day = maximum demand × L.F × hour in a day ,   » 65000 × 0.4 × 24 = 624000 kWh ,  » 1 kWh = 0.5 kg ,  = तो 624000 kwh में होगा 624000 kWh × 0.5 kg = 312000 kg , or , 312 tons , answer ✓  

  Question = A 75 MW steam power station used of calorific value of 6400 kcal/kg . Thermal efficiency of the station is 30% while electrical efficiency is 80% . Calculate the coal consumption per hour when the station is delivering it's full output ?  Solution =formula ,   coal consumption/hour = H/calorific value ,( H = heat produced/hour ) ,  steam power station = 75 MW , coal calorific value = 6400 kcal/kg , thermal efficiency= 30% , electrical efficiency = 80 % ,   » coal consumption/hour = ? .  » heat produced/hour , H = electrical output in heat unit ÷ η overall ,  » η overall =  η thermal ×  η electrical , = 0.30 × 0.80 = 0.24 .  » now , electrical output in heat unit or unit generated/hour = 75 MW × 1 = 75 MWH or , 75 × 10⁶ wh ÷ 1000 = 75 × 10³ kWh ,  » 1 kWh = 860 kcal ,  » 75 × 10³ kWh = 75 × 10³ × 860 ,  » H = 75 × 10³ × 860 / 0.24 , = 268750000 kcal ,or , 26875 × 10⁴ kcal ,   » So , coal consumption/ hour = 26875 × 10⁴ kcal ÷ 6400 kcal/kg = 41992.18 kg or, 42 tons

  Question   =  A generating station has an overall η of 15%  and 0.75 kg of coal is burnt per kwh by the station . Determine the calorific value of coal in kilocalories per kg ?  Solution = let ,  » 1 kg be the calorific value of coal = x kcal ,  » 0.75 kg of coal by heat produced = 0.75 x kcal/ kg ,  » heat equivalent of 1 kwh = 860 kcal ,  » details of 860 kcal ,  ( 1 kwh = 1000 watts × 3600 seconds = 36 ×10⁵ joules ,  » 4.18 joules = 1 calories ,   » 1 joules = 1/4.18 calories ,   » 36 ×10⁵ joules = ( 36 × 10⁵ j × 1c ) ÷ 4.18 j , = 860 ×10³ calories , or 860 kcal , ) »  η overall = electrical output heat unit / heat of combustion ,  » 0.15 = 860 kcal / 0.75 x kcal , = x = 860 /0.15 × 0.75 , = 7644.44 kcal/ kg , answer ✓

  Question = A 100 MW  steam station uses coal of calorific value 6400 kcal/kg . Thermal efficiency of station is 30% and electrical efficiency is 92% . Calculate the coal consumption per hour when the station is delivering it's full output .  Solution = calorific value =  6400 kcal/kg ,  » heat produced/hours = H = electrical output in heat units /  η overall ,  » η overall =  η thermal × η electrical ,  = 0.30 × 0.82 = 0.276 ,  » 100 MW or , 100 MW × 1hour , = 100 MWH ,  Or , 100 × 10⁶ wh ÷ 1000 , = 100 × 10³ kWh or , 10⁵ kWh ,  » now. electrical output in heat unit or , unit generated/hour = ( 100 × 10³ ) = 10⁵ kWh or , 10⁵ × 860 kcal ,  » details of 860 kcal  ,   ( 1 kwh = 1000 watts × 3600 seconds = 36 ×10⁵ joules ,  » 4.18 joules = 1 calories ,   » 1 joules = 1/4.18 calories ,   » 36 ×10⁵ joules =  ( 36 × 10⁵ j × 1c )  ÷ 4.18 j , =  860 ×10³ calories , or 860 kcal , ) » heat produced/hours = H = 10⁵ ×860 / 0.276 ,= 311.6 × 10⁶  kcal ,  And,  » coal consumption/hour = 311.6

  Question = the relation between water evaporated (w kg) , coal consumption ( c kg) and kWh generated per 8 hours shift for a steam generating station is as follows :  W = 13500 + 7.5 kWh  (1)  C = 5000 + 2.9 kWh (2)  (1) To what limiting value dose the water evaporating per kg of coal consumed approach as the station output increases ?  (2) how much coal per hour would be required to keep the station running on no load ?  Solution = (1) For an 8 hours shift , weight of water evaporated per kg of coal consumed is = w/c = 13500 +7.5 kWh / 5000 + 2.9 kWh ,  » as the station output ( kWh) increases towards infinity , the limiting value of w/c approaches 7.5/2.9 = 2.6 . Therefore, the weight of water evaporated per kg of coal consumed approaches a limiting value of 2.6 kg as the kWh output increases .  (2) At  no load , the station output is zero . kWh = 0 , therefore , from expression (2) we get , coal consumption at no Load = 5000 + 2.9  × 0 = 5000 kgs , So , consumption/hour = 500

  Question = A steam power station spends rs 30 lakhs annum for coal used in the station . The coal has a calorific value of 5000 kcal/kg and costs rs 300/ton . If the station has thermal efficiency of 33% and electrical efficiency of 90 % , find .the average load on the station .?  Solution = formula , Average load on the station = unit generated/annum ÷ hours in a year ,  » hours in a year = 365 × 24 = 8760 hours ,  » unit generated/annum = η overall × heat of combustion ,  »  η overall = thermal efficiency × electrical efficiency ,  = 0.33 ×0.9 = 0.297  » heat of combustion = coal use/annum × calorific value ,   » So , coal use / annum = 300 rs –––> 1ton ,  1ton––> 1/300 ,  30×10⁵ –––( 30 ×10⁵ ) ÷ 300 = 10⁴ tons , या 10⁷ kgs ,.  » calorific value of coal = 5000 kcal/kg ,  So , heat of combustion = 10⁷ × 5000 , = 5 × 10¹⁰ kcal ,  » unit generated/annum या heat output =  0.297 × 5× 10¹⁰ = 1487 × 10⁷ kcal ,  » 860 kcal –– 1 kwh ,  » 1 kcal–– 1/860 kWh , » 1485 × 10⁷ kcal ·––

  Question = A thermal station has the following data = maximum demand = 20000 kilowatt , load factor = 40% ,  boiler efficiency =  85% ,  turbine efficiency =  90% ,  coal consumption =  0.9 kg per kwh , cost of one ton of coal = RS 300 ,  determine (1)  thermal efficiency (2)   coal bill per annum .( 1ton = 1000 kg ) . Solution = thermal efficiency = efficiency boiler × efficiency turbine ,  = 0.85 × 0.9 = 0.765 or 76.5 % , answer ✓ (2) formula ,   coal bill / annum =  coal consumption × unit generated/annum ,  » unit generated/annum = maximum demand × load factor ×hour in a year ,  » 20,000 × 0.4 × 8760 = 7008 × 10⁴ kWh ,  » coal consumption in 1 kwh  = 0.9 kg ,  » 7008 ×10⁴ kWh = 0.9 ×7008 ×10⁴ = 63072000 kg ,  » for ton = ÷ 1000 = 63072 tons ,  » annual coal bill = RS 300 × 63072 , = RS 1,89,21,600 , answer ✓ 

  Question = A steam power station has an overall of 20% and 0.6 kg of coal is burnt per kwh of electrical energy generated . Calculate the calorific value of fuel ? Solution =  let , 1 kg be the calorific value of fuel =  x kcal ,  » 0.6 kg of coal by heat produce   = 0.6 x kcal ,  » heat equivalent of 1 kwh = 860 kcal ,   » 1 kwh = 1000 watts × 3600 seconds = 36 ×10⁵ joules ,  » 4.18 joules = 1 calories ,   » 1 joules = 1/4.18 calories ,   » 36 ×10⁵ joules =  ( 36 × 10⁵ j × 1c )  ÷ 4.18 j , =  860 ×10³ calories , or 860 kcal ,  » efficiency (η) overall= electrical output in heat unit ÷ heat of combustion ,  » 0.2 = 860 kcal/0.6x kcal , » x = 860/0.6 × 0.2 , = 7166.67 kcal/kg , answer ✓