Calculating Power from rain flow

  • Thread starter Thread starter unknown1991
  • Start date Start date
  • Tags Tags
    Flow Power Rain
AI Thread Summary
Energy can be extracted from rainwater flowing from downspouts, with calculations showing that a typical heavy rain could yield approximately 0.47 watts at peak flow. Over a year, this could amount to 170 watt-days if it rains every day, but realistically, if it rains only half the year, the output would be about 85.2 watts. The calculations assume 100% efficiency and do not account for evaporation losses. A key point raised is the need to use average rainfall data specifically for rainy days rather than annual averages. Accurate estimates depend on these factors for realistic power generation projections.
unknown1991
Messages
2
Reaction score
0

Homework Statement



As we have seen, energy can be extracted from moving water. When it rains, water that lands on roofs is collected in gutters and carried to the ground via downspouts. In a typical heavy rain in your city, how much power could be harnessed from the rain flowing out of your downspouts? How much power could you extract averaged over the course of the entire year?

Homework Equations





The Attempt at a Solution


Mean rainfall in this area: 600 mm per year = 0.6 m/year
Height: 10 m
Q: (250 m2)(0.6m/year)/ (3.15 × 107s/y) = 0.000004761 m3/s
Power Generated
η= efficiency= assume its 1
ρ= 1000kg/m3
g= 9.80 m/s2
P =ηρghQ = (0.000004761 m3/s) (1)(1000kg/m3) (9.80 m/s2)(10m)
P= 0.47 Watts
Year: 0.47 Watts x 365 days= 170 watts
We could extract about 170 watts over the course of the year assuming it rains everyday. This
calculation assumes that η is100% efficient and there is no loss due to evaporation..
If it only rains half the year then we would extract 170 watts/ 2= 85.2 Watts.


I don't know what I did. Please help me! :)

 
Physics news on Phys.org
unknown1991 said:
Year: 0.47 Watts x 365 days= 170 watts
0.47 Watts x 365 days= 170 Watt-days. Convert that to Joules, or kWh.
assuming it rains everyday
You used the average rainfall over a year, not the average rainfall on those days when it rains.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top