Agriculture Reference
In-Depth Information
Solution
We know, Peak energy demand,
P
e
=
Q
×
9.81
×
H
/
E
P
Given,
Discharge rate
0.009921 m
3
/s [Assuming daily bright sunshine
(solar) hour, i.e., pumping hour
250 m
3
/d
=
=
=
7]
Head,
H
10 m
Assuming efficiency of the pump
=
=
70%
=
0.07
Putting the values,
P
e
=
1.3903 kW
Now, the number of PV array,
N
PV
=
P
e
/
(
R
S
×
A
PV
×
E
PV
)
6MJ/m
2
/d
0.23809 kJ/m
2
/s
Here, Assured solar irradiance,
R
S
=
=
0.23809 kW/m
2
[considering one (1) solar day
=
=
7 h, as mentioned above]
3m
2
Assuming efficiency of the PV cells,
E
PV
=
Area of single PV unit,
A
PV
=
30%
=
0.30
Putting the values,
N
PV
=
6.488
≈
7 nos (Ans.)
Example 13.3
Determine the increase in power from a solar panel if the efficiency of PV cells
increase from 30 to 50%.
Solution
Consider a PV cell of 1 m
2
, and solar irradiance
R
s
MJ/m
2
/d
=
Power to be generated by PV cell,
P
PV
=
E
PV
As the power is directly related to the efficiency of the PV cell, the increase in
power due to increase in efficiency will be (50-30%)
R
S
×
1
×
=
20%.
=
×
=
Relative power increase
(20/30)
100
66.66% (Ans.)
13.4.5 Uses of Solar System Other than Irrigation Pumping
In this context, “solar energy” refers to energy that is collected from sunlight. Solar
energy can be applied in many ways, including, to
•
generate electricity using photovoltaic solar cells.
•
generate electricity using concentrated solar power.
•
generate electricity by heating trapped air which rotates turbines in a solar updraft
tower.
•
heat buildings, directly, through passive solar building design.
•
heat foodstuffs, through solar ovens.
•
heat water or air for domestic hot water and space heating needs using solar-
thermal panels.
•
heat and cool air through use of solar chimneys.
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