Determine motor pump power
Motor pump power is 2HP which is the same as 1492W, I want to design the system to be able to run for 8 hours during from morning the rising of the sun till evening when the sun is down
Determine servo motor power needed
>Mass of a single panel is 22kgs, the length of the panel is 1.5m and the width of the panel is 1m
>Power of the motor.
From the equation (newton 2nd law of motion)
>F=ma
>m = 22kgs and a = 9.81m/s2
>F=22kg x 9.81m/s2
>F = 215.82N
The average insolation over a year’s average, which is obtained from table above as 5.1 KW/m2/day.
>panel generation factor is 0.61
>Energy lost in the system is 1.3
Amount of Solar power needed = total watt-hour×1.3(energy lost in system) ÷ (0.62 ×5.1)
Solar power =((1492+0.525×8)×1.3)/(0⋅61×5.1)
=4989.5W ≈ 5000W
Selection of a light dependent resistor (LDR)
>LDR datasheet specifications
>Power dissipation at 30°C = 250mW
>Current = 75mA
>From voltage divider circuit
>𝑉𝑜𝑢𝑡=(R/(LDR+R))𝑋 𝑉𝑖𝑛
>Vout = 250mW/75 mA
3.33𝑉=(10k/(LDR+10k))𝑋 5𝑉
LDR = 4.99k ≈ 5kΩ
1 Comments
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