Power station 29

 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 annual load duration curve = [ 1/2 ( 320 + 160 ) × 8760 ] MWH ,  = 2102400 MWH , or , 2102400 × 10⁶ WH , or , 2102.4 × 10⁶ kWh , 

» as the same plant , run - of - river plant (3) and hydro plante generated (1) units in the ratio of 7:4:1 , therefore units generated by each plant are given by ,  

» reservoir hydro plant (1) , steam (2) , and hydro generated (3) units ratio add = 7 +4 +1 = 12 , 

» reservoir  hydro Plant (1) = 2102.4 × 10⁶ × 1/12 = 175.2 × 10⁶ kWh , 

» steam plant (2) = 2102.4 × 10⁶ kWh × 7/12 = 1226.4 × 10⁶ kWh , 

» run of river plant (3) = 2102.4 × 10⁶ kWh × 4/12 ,= 700.8 ×10⁶ kWh , 

(1) maximum demand of run of river plant (2) = area OEBA/OA = ( 700.8 × 10⁶ ) ÷ 8760 = 80,000 kW , answer ✓ 

» Suppose the maximum demand of reservoir Plant is Y MW and it operates for X hours = then = Y ÷ 160 = X ÷ 8760 ,  » So , X = 8760 y/160  , 

» units generated per annum by reservoir hydro plant (1) = area DFG , 

» X value input 

» 1/2 (XY) , = 1/2 × ( 8760 y/160 ) × Y , 

» 8760 Y² /320 MW ,  

» But , the unit generated by reservoir hydro Plant (1) are 175.2 × 10⁶ kWh , 

» both value equal = 8760 Y² /320 MW = 175.2 × 10⁶ kWh , 

» So , Y² = ( 175.2 × 10⁶ kWh × 320 MW ) ÷ 8760 hours ,

» calculation details =  175.2 × 1000 = 175200 , WH ,

 MW = 10⁶ × W = MW , ( 175200 MW )

8760 hours , and , 320 MW , ( 0 cut )

175.2 kWh × 10⁶ and , 8760 hours , ( hours cut ) 

 below 👇

» Y² = ( 175200 MW × 32 MW ) ÷ 876 hours , 

»Y² = 6400 MW² , = Y = √ 6400 MW² , = 80 MW ,  (maximum demand of reservoir hydro plant (1) ) answer ✓ 

» maximum demand on steam station (2) EF = 320 - 80 - 80 = 160 MW , answer ✓ 

(2) » load factor run of river plant (3) is 100 % , answer ✓ 

» load factor reservoir hydro plant (1) = ( unit generated per annum ) ÷ ( maximum demand × 8760 )  , 

» ( 175.2 × 10⁶ ) ÷ ( 80 × 10³ × 8760 ) = 0.25 , 

» for % ( ×100 ) = 0.25 ×100 = 25 % , answer ✓ 

» load factor of steam plant (2) = ( unit generated per annum ) ÷ ( maximum demand × 8760 )  , 

» ( 1226.4 ×10⁶ ) ÷ ( 160 × 10³ ) × 8760 , = 0.875 , 

» for % ( ×100 ) = 0.875 × 100 = 87.5 % , answer ✓