Transcript Electricity

Electricity
Part 4: Power Plants, Distribution
and Cooling,
Electrical Power Generation
Motor
External energy runs current through wire
loop in magnetic field.
Force on current-carrying wire causes
torque. Loop rotates.
Use rotating loop to move things.
Electrical Power Generation
Generator
Motor with input and output switched
External energy mechanically rotates wire
loop in magnetic field.
Electric charges (in wire) move in
magnetic field.
Magnetic force pushes charges through
wire (current).
Generators
Basically a motor used backwards.
Instead of running current through the loop
to get the shaft to rotate, rotate the shaft
with a heat engine, water wheel, or
windmill to get electrical current.
This is what is done in essentially all
power plants.
Electrical Power Generation in
the U.S.
Electrical Power Generation in
the U.S.
Fossil fuels 67%
Renewables 11.6% (Hydroelectric 7%)
Nuclear 19%
Typical Coal Fired Power Plant
Power Distribution
Long distances, use very
high voltage: up to
750,000 Volts.
For shorter distances use
~50,000 Volts
Locally step down to wall
voltage
Use TRANSFORMERS to
change voltages either up
or down.
Advantage to High Voltage
Transmission
Minimizes losses in power lines
Transmitted power is P = IV. If V is high,
then I can be low.
Losses in line are Ploss= I2R = V2/R
Lower losses high voltage transmission
Current for 3 W power line and
50,000 V transmission of 1 x 109 W
A. 200 A
B. 2000 A
C. 20,000 A
D. 200,000 A
0%
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0
20
0,
20
,0
00
A
A
0%
A
0%
20
00
20
0
A
0%
Loss for 3 W power line and
50,000 V transmission of 1 x 109 W
A. 6000 W
B. 6 x 104 W
C. 1.2 x 107 W
D. 1.2 x 109 W
1.
2
x1
09
W
0%
07
x1
2
1.
6
x1
04
W
60
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0%
W
0%
W
0%
Current for 3 W power line and
500,000 V transmission of 1 x 109 W
A. 200 A
B. 2000 A
C. 20,000 A
D. 200,000 A
0%
00
0
20
0,
20
,0
00
A
A
0%
A
0%
20
00
20
0
A
0%
Loss for 3 W power line and
500,000 V transmission of 1 x 109 W
A. 6000 W
B. 6 x 104 W
C. 1.2 x 107 W
D. 1.2 x 109 W
1.
2
x1
09
W
0%
07
x1
2
1.
6
x1
04
W
60
00
0%
W
0%
W
0%
Example: Assume Rline= 3 W, and
P =1GW
Case 1 V = 50,000 V
Case 2 V = 500,000 V
P
10 9 W
I

 20,000 A
V 5  10 4 V
Thus
P
10 9 W
I

 2,000 A
V 5  105V
Thus
Ploss  I 2 Rline  (2  10 4 A) 2 (3W )
Ploss  I 2 Rline  (2  10 3 A) 2 (3W )
 12
.  10 9 W
 12
.  10 7 W
This is more than we
have, i.e. all our power is
lost in transmission.
Here we only lose 1.2%
of total power in
transmission
We only want to go short distance without
much power at low voltages.
Transformers make it possible to raise and
lower voltages with essentially no power
loss.
This is the main reason we use AC power.
Why do power companies use high voltages to
transmit electric power over long distances?
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To
A. To increase current
in the power line
B. To decrease
resistance in the
power line
C. To reduce
transmission losses
D. To keep squirrels
and children away
from power lines.
Transformer
According to Faraday’s Law, we may induce a
voltage in a loop of wire immersed in a
magnetic field by
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A. Changing the area of
the loop
B. Changing the strength
of the magnetic field
passing through the
loop
C. Changing the
orientation of the loop
with respect to the
magnetic field
D. All of the above
What is the primary advantage of
AC electricity over DC electricity?
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It
A. It is safer
B. It is more reliable
C. We can easily
change the
voltage
D. It has more
groovitude
Cooling the plant.
We have to dispose of at least as much
energy as we generate. (Power plants do
not operate above 50% efficiency due to
the 2nd law.)
Usually remove the waste heat with a
continuous supply of water.
Question: How much water do we need?
Cooling the plant.
What rate of water flow per megawatt of
electricity generated do we need to cool a
power plant?
Volume/time or mass/time (l/s or kg/s)
We will allow the water temperature to
raise 7C
Cooling the plant.
Energy
Power is
Time
Q
P
t
Cooling the plant.
Energy
Power is
Time
Q
P
t
for heat Q  mcT
mcT
P
t
Cooling the plant.
m/t is mass flow rate
To convert to volume flow rate use the
density of water (kg/liter)
Rate of water flow per megawatt
A. 34.1 l/s
B. 341 l/s
C. 3410 l/s
D. 341000 l/s
l/s
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34
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l/s
0%
34
10
l/s
0%
34
1
34
.1
l/s
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Estimate the amount of water per
megawatt
Assume we only want the water to raise 14F
(7C).
Q m
P
 C T
t
t
Thus
m
P
10 6 W


t C T 
 o
J
 4186
o  7 C 
kg
C

 34.1kg / s
 34.1liter / s
 9 gal / s
For 1 MW this is easy, but a typical power
plant is 1000 MW.
To keep the temperature of the water from
rising more than 14F on a 1000 MW =
1GW plant we would need approximately
9000 gallons per second. (Almost 800
million gallons per day.) (Approximately
6000 swimming pools per day.)
To keep the temperature for rising only
~5F we would need ~3 time as much
water.
Total freshwater runoff in the US is
approximately 1200 billion gallons per day
(this includes floods)
Electrical plants need approximately 400
billions of gallons per day.
US Water Usage
Methods of Cooling
Once through cooling:
Requires a good size
river OR a good size
reservoir.
Least expensive
method
Infrared imaging of the
power plant near Joliet
Wet Cooling Tower
Extremely large towers
400 ft high and 400 ft
across.
~2% of water passing
through evaporates,
approximately 225
gallons per second.
Enough water to cover
1mi2, 1 in deep in water
every day.
Shape enhances natural air flow
Vortex engine powered by
excess heat
Dry Cooling
Used in places without
access to large supplies
of water.
Much more costly than
once through or wet
cooling.
Looses efficiency if
ambient temperature
much above 90F.
Sort of like a big car
radiator.
Since 1977 power plants have been
required to dispose of waste heat without
directly dumping it into the aquatic
environment.
Must construct cooling ponds or towers.
Effects of Increased Water
Temperature
Amount of oxygen
dissolved in water
decreases with
temperature
Most fish simply can't
stand warm water
and/or low levels of
dissolved oxygen.
Increased rate of chemical reactions
Changes in reproduction, behavior, and
growth patterns throughout the food chain.
Long term damage to natural bodies of
water.
Preferred Temperature Ranges for
Fish
Stratification of lake in the summer
In the winter the top layer becomes colder
and sinks to the bottom
This take oxygen rich water down and
brings up nutrient rich water.
A power plant doesn’t allow this mixing to
happen for as long.
Bottom dwelling animals have less oxygen
Discharge water from the bottom is rich in
nutrients (Nitrogen and phosphorus) that
stimulate algae and plant growth
This can cause a mat of algae (green
scum) on the surface that is toxic to many
species.
Dead algae sinks to bottom and is
decomposed using up even more of the
oxygen at the bottom.
This can eventually lead to the “death” of
the lake.
Eutrophication
The process in which a body of water is
enriched by the addition of extra nutrients ,
stimulating the growth of algae