Sustainable Energy Technologies
Homework #2
1.
Micro-Hydropower. You have a cabin in the Colorado wilderness far away from civilization and want to install a small hydro generator to provide electricity. A stream runs down a hill behind your cabin where you can construct a simple run-of-river hydro installation. You calculate that the effective head of the installation is 25 feet and that water would flow through the system at a rate of 50 cubic feet per minute. The simple inexpensive hydro turbo-generator (turbine mated with electric generator) that you intend to buy has a working efficiency of 60%. Answer the following (show your work):
H

25 ft

25 ft

1 ft
3 .
28 m

7 .
62 m
Q

50 ft
3 min

50 ft
3 min
 min
60 s
1 m
3

35 .
31 ft
3

0 .
0236 m
3 s a.
How much power can you generate with this system?
P
 g

QH

10 m

0 .
60
 s 2
0 .
0236 m
3 s

7 .
62 m

1 .
079 kW b.
How many 100 watt light bulbs can you power with this hydro setup?
About 10 light 100 watt light bulbs (10×100 W = 1000 W = 1 kW) c.
What type of turbine would you expect to find in your turbo-generator? Why?
Either a Kaplan or propeller turbine (or perhaps a cross-flow turbine) is appropriate for low flow, low head applications. For an inexpensive hydroturbine, a propeller turbine would probably be used because of cost and ease of maintenance (rigid veins).

Sustainable Energy Technologies
2.
Macro-Hydropower. The Glen Canyon Dam in Arizona dams the Colorado River creating
Lake Powell behind it. The dam has a height of 710 feet. Total nameplate (maximum continuous) generating capacity for the power plant at the dam is about 1 GW (1 million kW).
Assume an effective head of 700 feet and generating efficiency of 95%.
H

700 ft

700 ft

1 ft
3 .
28 m

213 .
41 m
P

1

10
6 kW a.
At maximum capacity, what flow is required through the dam’s turbines to generate its nameplate power? Express flow in cubic meters.
Q

P g

H

10

1

10 6
0 .
95

213 .
41

493 .
2 m 3 s

500 m 3 s b.
If the dam operates at 60% of capacity on average, how much energy does it produce in a year? Express your answer in watt-hours ( e.g., Wh, kWh, MWh, GWh, etc.).
E

10
6 
0 .
6

24

365

5 .
256

10
9 kWh

5 .
256

10
12
Wh

5 .
26 TWh c.
How many people should this energy support, assuming average community consumption of 3000 kWh per year per person?
N people

5 .
256

10
9 kWh
3000 kWh / person

1 .
75

10
6 
1 , 750 , 000 people d.
How much coal would be required in a coal-burning power plant to generate the equivalent amount of energy as the Glen Canyon Dam generates in a year? Assume that the dam generates 5 billion kWh hours (5×10 9 kWh or 5 terra-watt hours) of energy annually. How many coal cars would be required to carry this coal? How long would a train of these cars be?

The energy content of coal is 8,142 kWh/tonne (1 tonne = 2,206 pounds)

The energy efficiency of a typical coal-burning power plant is about 38%

A railroad coal car carries about 100 tons (90.72 tonnes) of coal and is about 75 feet (22.86 m) in length
E

5

10
9 kWh

1 tonne
8142 kWh

1
0 .
38

1 , 616 , 052 tonnes coal
N coalcars

1 , 616 , 052
90 .
72

17 , 814
L coalcars

17 , 814 coal cars cars

75 ft car

1 , 366 , 022 ft

253 miles

250 miles

Sustainable Energy Technologies e.
How much natural gas would be required in a gas-fired power plant to generate the equivalent amount of electrical energy as the Glen Canyon Dam generates in a year?
How many Goodyear blimps of natural gas does this represent?


The energy content of natural gas is about 0.95 kWh/m 3
The efficiency of a modern gas-fired power plant is about 55%

The Goodyear blimp holds about 200,000 ft 3 (5,663 m 3 ) of gas.
E

5

10
9 kWh

1
0 .
95 m
3 kWh

1
0 .
55

9 .
569

10
9 m
3 natural gas
N b lim ps

9 .
569

10
9 m
3
5 , 663 m
3

1 .
69

10
6 
1 .
7 million b lim ps

Sustainable Energy Technologies
3.
The total electrical energy consumption of the US in 2003 is estimated to be 3.66×10 12 kWh or
3.66 PWh ( CIA World Factbook - http://www.cia.gov/cia/publications/factbook/rankorder/2042rank.html
).
Hints

PWh = peta-watt-hrs = 1 quadrillion (10 15 ) watt-hrs = 1 trillion (10 12 ) kWh

The average depth of Lake Powell behind the Glen Canyon dam is 40 meters.

1 square mile = 2.59 million square meters (m 2 ) or 2.59 square kilometers (km 2 ) a.
How many Glen Canyon sized dams would be required to supply all of the electrical needs of the United States?
 Assume that the dam generates 5 billion kWh hours (5×10 9 kWh or 5 terra-watt hours) of energy annually.
N dams

E
E
US
GC

3 .
66

10 12 kWh
5

10
9 kWh

0 .
732

10
3 
732 dams b.
How much land area (square miles) would be inundated by these dams? What US state(s) haves land areas about equivalent to this inundated area?

Assume that that water flows through the dam at a rate of 500m 3 /sec when it is generating at its rated power.

The total area of states can be found at http://www.infoplease.com/ipa/A0108355.html
V yr

500 m
3 s

60 min s

60 min hr

24 hr day

365 days yr

15 .
77

10
9 m
3
/ year
A dam

Volume
Depth

15 .
77

10 9 m 3
40 m

394 .
2

10
6 m
2 
394 km
2
A dam

394 km
2 
1 sq
2 .
59 mi km
2

152 sq mi / dam
A total

N dams

A dam

732

152

111 , 411 sq mi
The total reservoir area required is 111,411 sq miles, about the area of Nevada or
Arizona. The total volume of water required would flood all of Nevada or Arizona to a depth of 40 meters (about 130 feet).

Sustainable Energy Technologies
4.
Pumped Storage. Consider a pumped hydro system that is used for energy storage. During off-peak hours, excess electricity is used to pump water up to a storage reservoir. During peak demand periods the flow is reversed and electric power is generated by flowing the water through a hydraulic turbine. The storage reservoir is 100 m above the hydro turbine/pump units and the hydraulic turbine can accommodate up to 650 l/s (liters per second) of flow.
Hints

 density
  of water is 1000 kg/m 3 (or 1.0 kg/liter) acceleration of gravity g is 9.8 m/s 2

1 acre = 4,047 m 2

1 J = 1 kg m 2 /s 2 = 2.78
-7 kWh a.
How much area would the pumped storage reservoir need to store 1000 MWh of hydro energy if its average depth were 10 m? Express your answer in acres. Is this reasonable?

The potential energy of water in the reservoir can be calculated as
 gH , where

is the density of water, g is gravitational acceleration, and H is the effective head of the system.

First calculate the total volume of water required in cubic meters, then the surface area of the reservoir if it is 10m deep on average. Then convert to acres.
E
E vol vol


 gH
10 m
6
3
J


1000 kg
 m
3
2 .
78

10

7
J
9 .
8 m s
2
 kWh
100 m


0 .
278 m
9
3
.
8

10
5 kWh m kg
3 m / s
2

1

10
6 kg m
3 m
2
/ s
2

10 m
6
3
J
V total

E total
E vol

1

10
6 kWh
0 .
278 kWh / m
3

3 .
7

10
6 m
3
A total

V total
Depth

3 .
7

10
6
10 m m
3

0 .
37

10
6 m
2
A total

0 .
37

10
6 m
2  acre

91 .
43 acres
4047 m
2
Yes, this appears to be a reasonable amount of land area to flood for the dam. b.
Estimate the theoretical power (assume 100% efficiency) produced during the times that electricity is generated.
Q

650 l s
P
 g

QH

0 .
65 m
3 s

10 m s 2

1 .
0

0 .
65 m
3 s

100 m

650 kW

Sustainable Energy Technologies c.
If the round-trip efficiency of the pumped storage unit is 75% ( i.e., energy out / energy in), how much water (m 3 ) must the reservoir hold to store the same 1000 MWh of energy?
V rqd

V
100 %


3 .
7

10
6
0 .
75 m
3

4 .
93

10
6 m
3