Economic of power generation 7

 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 RS , 

» depreciation charges of capital cost = 10% ×( 50 × 10⁶ ) = 5 × 10⁶ RS , or , 50 × 10⁵ RS , 

» charges of salaries , maintenance etc = 20% × 7,00000 = 1.4 × 10⁵ RS , 

»So , total annual fixed charges = 50 × 10⁵ RS + 1.4 × 10⁵ RS  = 51.4 × 10⁵ RS , 

» Cost per kW = ( 51.4 × 10⁵ RS ) ÷ ( 40 × 10³ ) +1 RS , = 128.5 +1 = 129.5 RS , answer ✓ 

» 2 nd part , Annual running charges , 

» formula , Cost per kWh = cost kWh due to running charges + royalty , 

»  cost per kWh due to running charges = RS of Salaries maintenance etc ÷ unit generated per annum , 

» royalty = given, 0.01 RS , 

» given , 20% fixed charges , extra = 100 - 20 = 80% , running charges , 

» Salaries maintenance etc = 80% × 7,00000 = 5.6 × 10⁵ RS , 

» unit generated per annum = M.D. × L.F. × hours in a years , = 40 ×10³ × 0.6 ×8760 = 210.24 × 10⁶ kWh , 

» So , Cost per kWh = ( 5.6 ×10⁵ ) ÷ ( 210.24 ×10⁶ ) + 0.01 , = 0.0027 + RS 0.01 = 0.0127 RS , answer ✓ 

» total generation cost in two part from is given by , ( 129.5 × kW + 0.0127 × kWh ) . 


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

Electrical bill निकालना,

  Question = किसी electrical circuit में 15 lights point प्रत्येक 60 watt, 8 light point प्रत्येक 40 watt , 6 fan प्रत्येक 40 watt और एक pump 1/2 HP का हैं।light और fan का 50% load प्रतिदिन 4 hours और pump प्रतिदिन 3 hours काम करता है। 33paisa/kWh unit की दर से एक महीने के लिए बिल बनाओ?  Solution = (1) 15 light point प्रत्येक 60w= 15×60= 900 watt,  (2) 8 light point प्रत्येक 40w = 8×40=320 watt, (3) 6 fan point प्रत्येक 40= 6×40= 240 watt,  (4) 1/2 hp का pump =746/2 = 373 watt, Lights और fan का Total watt =900+320+240= 1460 Watt,  Lights और fan का 50% load=1460×50÷100;  730 watt,       so, watthour= 730×4= 2920 wh, pump का watt hours =373×3= 1119 watt hours,   Total kWh= 2920+1119     = 4039                                                   1000                1000 =4.039 kWh;        = 30 दिन के kwh = 30×4.039,. =121.17 kWh,  खर्च = 121.17×33 = 3998.61 Paisa या 39.98 ₹, 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