Power Plant Engineering

112 POWER PLANT ENGINEERING

(a) reversed saturation current

(b) the voltage that maximizes the power

(c) the load current that maximizes the power

(d) the maximum power

(e) the maximum conversion efficiency

(f) the cell area for an output of 1 kW at the condition of maximum power.

Solution.

/

/

s

o

IA

IA

= exp eVac

kT







• 1 = 1.193 × 10^10 o

I

A

= 10

220

1.193 10×

= 1.8 × 10–8 A/m^2

The voltage Vmp that maximizes the power is given by

exp mp

eV

kT







1

eVmp

kT



+



= 1 +

/

/

g

o

IA

IA

Vmp = 0.52 V

Imp

A

=

/

1/

mp

mp

eV kT

−eV kT

so

II

AA

−





= 210 A/m^2

max

A

ρ

= mp

I

A







Vmp = 109 W/m^2

ηmax = max

/

in/

PA

PA

=

109

950

= 11.5%

A =

out required

max/

P

PA

=

1000

109

= 9.17 m^2.

Example 7. Estimate the average daily global radiation on a horizontal surface at Ahmedabad
(22°00′ N 73°10′E) during the month of April. If the average sunshine hours per day are 10. Assume a
= 0.28 and b=0.48.

Solution. Let

Isc = Solar constant = 4870.8 kJ/m^2 .hr

Ho = Daily global radiation on a m^2 horizontal surface at the location on a clear sky day in the

month Ho is calculated from as

Ho =

24

π

Isc

1 0.033 cos^360 (sin. sin cos. cos. sin )

365

+αns ss+α



Hg = Daily global radiation (monthly average) for a horizontal surface at the location is calcu-

lated by using value of Ho, a, b as

Hg = Hoa + b







h

m

L

L

kJ/m^2. day