Generating stations 16

 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 generation . The plant utilise the water as and when available .  Consequently more power can be generated in a rainy season than in dry season . 

(1) (a) when discharge = 10 m³/sec. ,  Weight of water available/sec. , W = 10 × 1000 kg = 10⁴ × 9.81 N , 

» power developed = W × H ×  η overall , = 10⁴ × 9.81 ×25 ×0.8 watts , = 1962 × 10³ watts , = 1962 kW ,

(b) 10 m³/sec power developed in = 1962 kW, » 1 m³ in power developed in = 1962 kw/10m³ sec. , 

» 6 m³/sec.  in  power developed = ( 1962 kw × 6 m³/sec. ) ÷ 10m³/sec.. , = 1177.2 kw , 

(C) 10 m³/sec power developed in = 1962 kW , » 1 m³ in power developed in = 1962 kw/10m³ sec. , 

» 1.5 m³/sec. in power developed = ( 1962 kw/1.5m³/sec. ) ÷ 10m³/sec. = 294 kw , 

• it is clear that when discharge is 10 m³/sec. or , 6 m³/sec. , Power developed by the plant is more than 400 kilowatt required by the factory .

»  however when the discharge is 1.5m³/sec. , Power developed falls short and consequently stand by unit is required during this period . 

» capacity of standby unit = 400 - 294 = 106 kw, answer ✓ 

(2) with reservior - when reservoir  is arranged upstream , we can store water .  This permits regulated supply of water to the turbine so that power output is constant through out the year . 

» average discharge = ( 10 × 4 ) + ( 2 × 6 ) + (1.5 × 6 ) /12 months ,  = 5.08 m³/sec.  , 

»10 m³/sec power developed in = 1962 kW ,   » 1 m³ in power developed in = 1962 kw/10m³ sec. , 

» 5.08 m³/sec. in power developed =  (1962 kw × 5.08 m³/sec ) / 10m³/sec .  =  996.7 kw , 

Since , power developed is more than required by the factory , no standby  unit is needed . Excess power available =996.7 kw - 400 kw = 596.7 kw , answer ✓

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