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The yearly load duration curve of power plant is considered as straight line from maximum 340 MW to minimum 90 MW. Power is supplied by one units of 200 MW and two units of 120 MW each.

Determine- i) installed capacity ii) Plant factor iii) Load factor iv) Utilisation factor.

Mumbai University > Mechanical Engineering > Sem 7 > Power Plant Engineering

Marks : 10M

Year: Dec 2015

1 Answer
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Data:

Straight line load duration curve Maximum demand = 340 x 103kW , Minimum demand = 90 x 103 kW,

Installed capacity = 200 x 103 kW of 01Nos, & 120 x 103 kW of 02Nos,

To find :i) Installed capacity ii) Plant Factor iii) Load Factor iv) Utilization factor.

Plant Installed capacity = 200 x 103 x 01 + 120 x 103 x 02

= 200 x 103 + 240 x 103 = 440 x 103MW

Total Energy Generated/year = Area Under the curve AB

= Area of Triangle ABC + Area of Rectangular EDBC

= (90 x 8760) + (0.5 x 8760 x (340 – 90))

= 1.8834 x 109 kWh

Average load = $\frac{(Total energy generated per year)}{(Time duration of year )}$

= $\frac{(1.8834 ×10^9)}{8760}$

= 215000 kW

Plant factor = (Total energy generated per year)/(Time duration of year ×Plant installed capacity) = Load factor x Utilization factor

= $\frac{(1.8834 × 10^9)}{(440 × 10^3×8760)}$

= 0.4886

OR

48.86 %

Load Factor = $\frac{(Average load)}{(Maximum load demand)}$

= $\frac{215000}{340000}$

= 0.6323

OR

63.23%

Utilization Factor = $\frac{(Maximum demand on Plant)}{(Plant Installed capacity)}$

= $\frac{(340 × 10^3)}{(440 × 10^3 )}$

= 0.7727

OR

77.27%

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