1. Hydropower introduction
TIM 155
Week 4, Spring 2017
Hydropower introduction

2. Hoover Dam, Colorado River

3. Hydropower Practice Problems - Part 1
Shasta Dam, Sacramento River, Northern California

4. Shasta Dam Power Station

6. I
A_Sluice Gate
B_Spillway
C_Penstocks
D_Power Station
E_Freeboard
F_Gross (Static) Head
G_Tailrace
H. Turbines
I. Water Level E B A F C D H G

7. Shasta Dam, Sacramento River, Northern California
Reservoir water level
Gross (Static) Head
Turbines

8. Shasta Dam, Sacramento River, Northern California
Freeboard
Spillway
Penstocks
Sluice gates
Power station
Tailrace

12. How much power is generated by hydropower?

13. The key equation:
E = m x g x h

14. E = m x g x h
E Energy produced (in this case, Joules) =
m mass of water passing through the turbines (kg) x
g acceleration of gravity (a constant, 9.8 meters/sec2) x
h how far the water is falling (meters), measured from the
reservoir water level to the turbines

15. E = m x g x h
kg . m2 / s2 = kg x m/s2 x m

16. “mass”
E = m x g x h
kg . m2 / s2 = kg x m/s2 x m
“meters”

17. E = m x g x h
kg . m2 / s2 = kg x m/s2 x m
This is the unit of energy called a Joule, or J.

18. E = m x g x h
kg . m2 / s2 = kg x m/s2 x m
This is the unit of energy called a Joule, or J.
A flow of energy, 1 Joule per second, is called a Watt, or W

19. E = m x g x h
kg . m2 / s2 = kg x m/s2 x m
This is the unit of energy called a Joule, or J.
A flow of energy, 1 Joule per second, is called a Watt, or W.
1 J/s = kg . m2 / s2 / s = kg . m2 / s3 = 1 W

20. E = m x g x h
kg . m2 / s3 = kg/s x m/s2 x m
To calculate Watts; change kg to kg/s, since a flow of water will produce a flow of power.
1 J/s = kg . m2 / s2 / s = kg . m2 / s3 = 1 W

21. E = m x g x h
kg . m2 / s2 = kg x m/s2 x m
Question: How much energy is produced each year by a
dam with a height from turbine to reservoir level of
20 meters through which one million tonnes of water flows?

22. E = m x g x h
kg . m2 / s2 = kg x m/s2 x m
Question: How much energy is produced each year by a
dam with a height from turbine to reservoir level of
20 meters through which one million tonnes of water flows?
E = 1,000,000,000 kg x 9.8 m/s2 x 20 m
= 2 x 1011 Joules

23. Adding an efficiency factor
If the system is only 90% efficient at generating power, multiply m x g x h by .9
E = m x g x h x .9
Question: How much energy is produced each year by a
dam with a height from turbine to reservoir level of
20 meters through which one million tonnes of water flows? Assume the hydropower system has an efficiency factor of 90%.
E = 2 x 1011 Joules x = 1.8 x 1011 Joules

24. Hydropower quantitative question guidance…
Additional guidance…
E = m x g x h
Always use meters (you may have to convert other units to meters)
Always use kilograms. It’s easy to convert from cubic meters since for water 1,000 kg = 1m3

25. Hydropower quantitative question guidance…
There are ~3.1 x 107 seconds in one year!

26. Hydropower quantitative question guidance…
10,000,000 W = MW
You can convert W to MW by dividing the number of Watts by 1,000,000 and adding the M for “Mega” to Watts in your answer (see above).
The “M” in MW is equivalent to six zeros.