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

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%