🎯 Assumptions:
Laptop power usage: ~50 watts (moderate usage)
Time: 8 hours
Required energy:
50 W×8 hrs=400 Wh/day50 \text{ W} \times 8 \text{ hrs} = \boxed{400 \text{ Wh/day}}50 W×8 hrs=400 Wh/day
💧 Let’s reverse-engineer how much water is needed from a 30 ft (~9.14 m) drop:
We’ll use:
E=m⋅g⋅h⇒m=Eg⋅hE = m \cdot g \cdot h \Rightarrow m = \frac{E}{g \cdot h}E=m⋅g⋅h⇒m=g⋅hE
Where:
EEE = 400 Wh × 3600 = 1,440,000 joules
g=9.81g = 9.81g=9.81, h=9.14h = 9.14h=9.14
m=1,440,0009.81⋅9.14≈16,113 kg (liters)m = \frac{1,440,000}{9.81 \cdot 9.14} ≈ \boxed{16,113 \text{ kg (liters)}}m=9.81⋅9.141,440,000≈16,113 kg (liters)
⚠️ Adjust for 50% Efficiency:
16,113×2= 32,226 liters/day16,113 \times 2 = \boxed{~32,226 \text{ liters/day}}16,113×2= 32,226 liters/day
✅ Final Answer:
To power a 50W laptop for 8 hours/day using only water falling from 30 ft, you’d need:
~32,000 liters per day
That’s about 32 large water drums (200L) falling from that height daily
