RENEWABLE AND CLEAN ENERGY
Hydro-electric Dam
Problem Set
Problem 1: Rainfall
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?
Problem 1 Solutions: Rainfall
A) A typical heavy rain storm in Vancouver produces 30 mm of rain in 6 hours, or 5 mm an hour. The
roof area of my two car garage is 49m2 (7 m by 7 m).
volume flow rate of rain, Q  50 m 2  0.005 m/hour  0.25 m 3 /hour  250 L/hour
Each hour, 250 L of rain falls on the roof, flows into the gutter and is carried down the downspout
towards the ground.
What we know:
ρ = 1000 kg/m3
g = 10 m/s2
h=3m
Q = 250 L/h = 7 x 10-5 m3/s
If the water simply runs off the roof and into the downspout, the power available during the rainstorm
is:
P  ghQ
 (1000 kg/m 3 )(10 m/s 2 )(3 m)(7 x 10 -5 m 3 /s)
 2W
This is enough to light a single small LED.
B) According to the Weather Network, the annual precipitation at Vancouver International Airport is
1100 mm.
volume flow rate of rain  (50 m 2 )(1.1 m/year)(
1 year
)  1.8 x 10 -6 m 3 /s
7
3x10 s
P  ghQ
 (1000 kg/m 3 )(10 m/s 2 )(3 m)(2 x 10 -6 m 3 /s)
 0.06 W
When averaged over an entire year, only 0.06 W is available which is of no use to anyone.
©University of British Columbia Department of Physics and Astronomy
Problem 2: Brandywine Creek Run-of the-River Project
Run-of-the-river projects harness power from the natural flow and elevation drop of a river. In the
Brandywine project, water is diverted from the river into a pipe (r=0.75 m) in which it flows down the
hill, through the turbine, and returns to the river. The project at Brandywine Creek produces 40,000
MWh of electricity per year, i.e. a mean power of 4.6 MW. The diverted water drops 280 m in elevation
between where it leaves the river and where it reaches the turbines. Estimate the flow rate and speed
of the water going into the turbine.
Problem 2 Solution: Brandywine Creek Run-of the-River Project
What we know:
P ≈ 4.6 MW
ρ = 1000 kg/m3
g = 10 m/s2
h = 280 m
Q=?
After rearranging the equations for power, it is easy to solve for the flow rate.
P  ghQ
Q
P
gh
4.6 x 10 6 W

(1000 kg/m 3 )(10 m/s 2 )(280 m)
 1.6 m 3 /s
The volume flow rate is 1.6 m3/s and is defined to be the area the water is flowing through multiplied by
the speed of the water. The area is the cross-sectional area of the pipe A = 𝛑r2 = 𝛑(0.75)2 = 1.8 m2.
Therefore, the approximate speed of the water is
Q
A
1.6 m 3 /s

1.8 m 2
 0.9 m/s
v
Real flow rates and speeds will be higher than our calculated ones because of losses in the system.
References:
The Weather Network. Vancouver, British Columbia (online). http://www.theweathernetwork.com [22
June 2010].
Villholth Jensen & Associates Ltd. Brandywine Hydro Project (online).
http://www.vja.bc.ca/Brandywine.pdf . [23 June 2010].
©University of British Columbia Department of Physics and Astronomy
Run of River Power. Brandywine (online). http://www.runofriverpower.com/id/14 [22 June 2010]/
Brittany Tymos 2010/06/25
©University of British Columbia Department of Physics and Astronomy