Power station 28

 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 station

» It is reasonable to assume that the maximum demand on the base load station is equal to the installed capacity ( 18 MW ) . It operates throughout the year for 8760 hours . 

» So , annual load factor = ( kWh output per annum ) ÷ ( installed capacity × hours in a years ) , 

» ( 101.35 × 10⁶ kWh ) ÷ ( 18 × 10³ kW × 8760 hours ) , = after calculate = 0.642 , 

» for % ( × 100 ) = 0.642 × 100 = 64.2 % , answer ✓ 

» as the base load station has no reserves above peak load and it is in continuous operation , therefore , its capacity factor is also 64.2 % . answer ✓ 

Comments

Post a Comment

Popular posts from this blog

Polyphase circuit part 6

Image
  Question = एक balanced load , 3 phase , star connected , resistance load में प्रति phase 1000 Ω का resistance लगा हुआ है। जो कि 230 volt/ phase 50c/s supply के साथ जुड़ा हुआ है। लाइन current , cos Ø और power factor मालूम करें ? Solution = is given.  R = 1000 Ω , V L  = 230 v/pH.. , f = 50 c/s , Formula = I L  = V L  ÷ R » 230 ÷ 1000 , » 0.23 ampere , answer ✓ » W= V×I » 230 × 0.23 » 52.9 watt, तो wattage in 3 phase = 3 × 52.9 =  158.7 watt , answer ✓  » Cos Ø = Input ÷ √3 × V L  × I L  ,  » 158.7÷ 1.732 × 230 × 0.23 , » 1.72 या 1(unity ) (क्योंकि power factor 1 से अधिक नहीं होता है)            (Circuit में resistance load हैं इसलिए PF unity होगा।) Answer ✓
Read more

Specific resistance की calculation

Image
  Question = एक 9.15 meter लंबी copper की bus-bar का resistance 0.000168Ω हैं। यदि इसका कटाक्ष क्षेत्रफल 7.62×1.27 cm² हो तो copper का specific resistance मालूम करें ? Solution= Length=9.15 meter, R=0.000168Ω, Area= 7.62×1.27cm²,  = spe... Resis... (rho) ? ,  Formula, Rho= R×A/L, =0.000168×7.62×1.27/9.15 Ω cm, (9.15meter ×100=915cm), और (μ=10⁶) =168×762×127×100/915×10⁶×100×100×100;  =1016127/ 571875×10⁶;  =1.77/10⁶;  = 1.77 μΩ cm,  answer ✓ Question =एक copper की wire का specific resistance (rho)1.79 μΩ cm हैं, यदि इसका व्यास या diameter 0.091 cm हैं और length 3.66 cm हैं। इसका resistance मालूम करें? Solution = rho=1.79 μΩ cm, dia= 0.091 cm, l=3.66 cm, R=?  Formula =R= ρL/A, =A=πd²/4; (π=3.14)  =A=3.14×0.091²/4;   =314×91×91/4×100×1000×1000, =2600234/4×10⁸ ; =650058.5/10⁸ ; =0.00650085 cm²,  इसलिए, R= 1.79×3.66/10⁶ ×0.0065;   =179×366×10000÷10⁶ ×65×100×100;  =1007.9÷10⁶: =0.001 Ω, answer ✓
Read more

Heating element की resistance और power निकालना,

Image
  Question= एक electrical water pot (जिसमें पानी गर्म किया जाता है) की efficiency 85% हैं। इसमें 2 pint (2×0.5682) पानी को 15°c से 98°c तक 1minute में गर्म करना पड़ता है। इसकी supply voltage 200 v हैं। (1)  तो water pot की heating eliment की resistance मालूम करें? (2) heating element की power भी निकाले,?  Solution= efficiency (η) =85%,  M=2 pint; (2×0.5682=1.1364 liter या 1.1364×1000 gram);  temperature difference= 98-15= 83°c, Time=1 minute या 60 second ;  V L  =200v,. R=?  Formula; Heat Required=Heat produced;                        = M.T. Calories = η ×0.24×v×i×t calories ;         = M.T. Calories=  η ×0.24×v×v/r×t calories;       =M.T. Calories=  η ×0.24×v²/r×t; calories;           =1136×83=85/100×0.24×200×200/r×60 calories ; So, R =    85×0.24×200×200×60      Ω;                                            1136×83×100 R =    85×24×200×200×60       Ω;                                       1136×83×100 ×100 = 489600÷94122 ,. =R=5.2Ω, answer ✓(1)  (2) v=200;  = w=v²/r = 20
Read more