Power Plant Engineering

NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION 111

Thermal energy input =

Electrical energy output

Plant efficiency

=

8760000

0.25

= 35040000 (mW hr)t

Thermal energy input to plant for one year = 35040000 (mW hr)t

Lignite input to plant for one year

=

(mW hr)

(Useful mW/kg) in lignite

t

Useful energy in fuel per kg = Available energy in fuel × Utility factor

= (14 mJ/kg) × 0.7 = 9.8 mJ/kg

Lignite input per year = 35040000 (mW hr)t

Conversion factor:

1 mW hr = 1 mW × 3600 s = 3600 mW.s = 3600 mJ

Lignite required per year = 35040000 ×

3600

9.8

= 1.287 × 10^10 kg = 12870 × 10^6 kg

Since (1 MT = 1000 kg) = 12870 × 10^3 MT.

Example 5. The incident beam of sunlight has power density of 0.9 kW/m^2 in the direction of the

beam. The angle of incidence θ is 60°. Calculate power collected by the surface having total flat area of

100 m^2.

A = 100 m^2

Ibn

= 0.9 kW/m^2

Sun

Ray

N

θ = 30°

S

Fig. 2.48.
Solution. Equivalent solar power falling on the surface S
PN (Watts) = IN (W/m^2 ) × A (m^2 )
IN (kW/m^2 ) = Ibn. cos θ
= (0.9 × 10^3 ) × 0.5 W/m^2 = 4.5 × 10^3 W/m^2
PN (kW) = (0.45 × 10^3 W/m^2 ) × 100 m^2 = 0.45 × 10^5 W = 0.045 mW.
Example 6. When a photovoltaic cell is exposed to solar insulation of 950 W/m^2 , the short circuit
current is 220 A/m^2 both based on a unit area of the exposed junction. The open circuit voltage is 0.60
V and the temperature is 300 K. Calculate