Energy Systems & Climate Change – week 4 Winter – Zita solns – Friday 3 Feb. 2012
Wolfson Ch.10: Water, Wind, Biomaass
Wolfson Ch.10 Q 1, 3, 5 / Choose one AYC / Ex.# 1,4,6,8,10,14
Ex.# 1,4,6,8,10,14
1. It takes approximately 2.5 MJ to convert 1 kg of liquid water to vapor at typical atmospheric
temperatures (this is the latent heat introduced in Section 4.5). Compare this quantity with the
gravitational potential energy of 1 kg of water lifted to a typical cloud height of 3 km.
Solution: Gravitational potential energy U  mgh, where g  9.8 m/s2 is the acceleration due to
gravity.
U = 1 kg 9.8 m/s2  3103 m  2.9104 J  30 kJ.
energy to lift water
2.9 104 J

 1% . Vaporizing water takes a lot of energy.
energy to vaporize water 2.5 106 J
4. Estimate the water head at Hoover Dam shown in Figure 10.7b, given that it produces 2.1
dV
GW of electric power from a flow of
= 1,800 m3/s. Assume an energy conversion efficiency
dt
of 75%.
Solution: Electric power output  efficiency  dE/dt and E  mgh where mass  density 
volume and the density of water    103 kg/m3:
dE
dm
dm
dV
 egh
where

dt
dt
dt
dt
dV
.
P  egh 
. Solve for the water head h :
dt
2.1 109 J
P
s
h

 159 m
3
dV
kg
3
m
m
eg 
0.75

9.8

10

1,800
dt
s
s2
m3
Power  e
6. By what factor must the wind speed increase in order for the power carried in the wind to
double?
Solution: Recall the derivation on p.306, where s  distance and   density:
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1
1
mv 2
mv3 12 v3
Energy
Power =
 2
2

distance
time
s
s
v
Power 1 3
Since area = s 2 ,
 2 v .
area
mass
1 3
s 3  2 v density  1 v3 s 2 
2
1 3
1 2
.
s
s
Say we start with initial power P0 at initial speed v0: P0 =
1
2
 s 2v03 .
P  2 P0
To double the power,
1
2
 s 2v 3 = 2  12  s 2v03 
v 3  2v03
v   2  3 v0
1
We only have to increase the speed by a factor of  2 
1
3
 1.26 , an increase of just 26.
8. The density of air under normal conditions is about 1.2 kg/m3. For a wind speed of 10 m/s,
find
(a) the actual power/area carried in the wind;
(b) the maximum possible power (Betz limit) extractable by a wind turbine with a blade area of
10 m2; and
(c) the actual power extracted by a wind turbine with a blade area of 10 m2 and a power
coefficient 0.46.
Solution:
3
(a) Recall that the
Power 1 3 1 
kg  m 
 kg W 
 2  v  2 1.2 3  10   600  3 = 2  .
area
m 
 m  s 
s
Therefore the total wind power is
W
 Power 

2
Ptotal = 
 Area =  600 2 10m  6.0 kW.
m 
 area 

(b) The Betz maximum Pmax  0.59 Ptotal  0.596 kW  3.5 kW .
(c) The actual power extracted is Pactual  e Ptotal  0.46  6 kW  2.8 kW.
The actual power extracted is generally less than the Betz maximum.
10. (a) Estimate the total energy produced by a wind turbine with the power curve shown in
Figure 10.13 during a day when the wind blows at 2 m/s for six hours, at 10 m/s for six hours, at
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15 m/s for six hours, and at 25 m/s for six hours. (b) What’s the turbine’s average power output
over this day?
Solution: Let’s tabulate the power
produced in each time interval:
Speed (m/s)
2
10
15
25
Time (hours) Power (MW) Energy(MWhr)
6
0
0
6
0.25
1.50
6
1.00
6.00
6
0.85
5.10
(a) Total energy generated is about 12.6 MWhrs.
(b) The average power output over this day is: Power =
Energy 12.6 MW  hr

 0.5 MW .
time
24 hr
11. If you’ve had calculus, you know that you can find the maximum or minimum of a function
by differentiating and setting the derivative to zero. Do this for Equation 10.2, and show that the
resulting maximum occurs when a=1/3 and that the maximum is equal to 16/27 of the total wind
power 12 v 3 .
Solution: The maximum extractable wind power is P  2a 1  a   v3 .
2
dP
d
2 da
2

 2  v 3 1  a 
 a 1  a    0 when P is an extremum:
da
da
da


dP
d
2


 2  v 3 1  a  1  a 2 1  a  1  a  
da
da


 2  v 3 1  a   2a 1  a  1 


2
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dP
2
= 0 when 1  a   2a 1  a  or
da
1  a 
1  3a
2
 2a
1
 a .
3
2
1
1 1
22
8

Then Pmax  P  a    2 1    v 3     v 3 
 v3.
3
3 3
33
27

16 
1

Sure enough, this is
  total wind power  v3  .
27 
2

14. Figure 10.19 suggests biodiesel production of around 10 billion litres (L) in Europe in 2009,
which compares with a total diesel fuel consumption of about 150 billion L annually. European
biodiesel production yields about 1.23 kL per hectare (ha) of cropland.
(a) What fraction of Europe’s 49-Mha of total cropland is now used for biodiesel production?
(b) How much land would it take to replace all of Europe’s petrodiesel with biodiesel,
considering that the energy yield per litre of biodiesel is 10% lower than for petrodiesel?
Solution: (a) The amount of land Europe needed for biodiesel production in 2005:
1 ha cropland
10 109 L
 8.13 106 ha cropland
3
1.23 10 L
What fraction of the total cropland is this?
8.13 106 ha
 17% of Europe’s cropland required for biodiesel production
49 106 ha
(b) To replace petrodiesel, Europe would need 1.1150 109 L = 1.65 1011 L of biodiesel.
How much land would this take?
1 ha cropland
1.65 1011 L
 1.34 108 ha cropland = 134 million hectares
3
1.23 10 L
This is more than Europe’s total cropland.
Q 1, 3, 5 / Choose one AYC
Ch.10 Questions:
1 As used today, are the indirect solar energy sources of water, wind, and biomass truly
renewable? Answer separately for each.
Answer: Hydropower and wind: yes, in theory. Biomass: not really. It can be burned faster
than it grows.
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Hydropower is theoretically renewable as long as the hydrologic cycle keeps cycling. Where
hydropower relies on freshwater sources that people also use for other purposes (drinking,
agricultural irrigation, salmon runs), in some places we are drawing down water faster than it can
be replenished.
Wind is renewable as long as the Sun keeps driving temperature gradients in the Earth’s
atmosphere. Global warming will change global temperature gradients and may move locations
of windy places.
Biomass is renewable only if our consumption rates are lower than growth rates of the plants
used. If we emit biomass carbon faster than plant growth sequesters carbon, then it is not
renewable (and this is only one of several criteria).
3 Why can’t a wind turbine extract all of the wind’s kinetic energy?
Answer: If a wind turbine extracted all of the wind's kinetic energy, it would stop the air passing
through the turbine, deflect the air around it, and produce no power.
Betz analyzed the tradeoff and found a theoretical maximum extractable power of 59%.
Today’s best windmills have power coefficients of about 45%.
5 Tailpipe emissions of VOCs are lower for ethanol than
for gasoline, but overall VOC emissions are higher with
ethanol fuels. Why?
Answer: Ethanol is more volatile than gasoline. The highest VOC losses occur during handling,
transport, and dispensing of ethanol.
Ch.10 “Argue Your Case:”
1 One environmentalist argues that wind turbines belong in the Great Plains but not on New
England’s scenic mountain ridge-tops. Another claims that ridge-top wind turbines are preferable
to the nuclear and gas power plants that supply much of New England’s electricity. How do you
argue?
Response: CON: Nuclear power is already in place, and is generally safe. Putting wind power
in the mountains will disturb not only the scenery but also the environment. Trees will have to
be logged, roads cut through the forest, and the ground disturbance will also release CO2. Birds
and wildlife will be killed or displaced, and a place of natural beauty will be ruined.
PRO: Wind power is safer than nuclear power, which produces long-lived radioactive waste and
has the potential for catastrophic releases. We must also replace gas power plants with
electricity generators that do not emit GHGs. Turbines are best placed on windy ridges (or
offshore) for efficient wind capture. Beauty is in the eye of the beholder. Wind turbines look
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more beautiful when considering the nuclear waste and emissions they displace. The Audubon
Society has endorsed wind power, for global warming kills about 10,000 times more birds than
do turbines. The CO2 emissions displaced by the wind turbines will make up for the trees lost by
their installation. Without much more renewable power such as wind, our planet will be ruined
by continued GHG emissions.
2 Your university is planning to replace its oil-fired heating plant with a wood-fi red boiler
supplied by wood chips produced from logging waste. The university claims it will then be
“100% carbon neutral.” Formulate an argument supporting or opposing this claim.
Response: CON: Not true. Fossil fuels are used in logging and in transporting the waste to the
boiler. If the wood is burned, it emits greenhouse gases at faster rates than those at which they
were sequestered in the growth process. It is unlikely that this boiler will provide all the energy
needs of the university.
PRO: While not 100% carbon neutral, this is a good idea, especially in areas with active logging
already. The boiler will be much greener if the wood is gasified, the wood gas burnt cleanly,
and the waste sequestered as char which is returned to the soil as fertilizer. The logging
company should pile biomass close to logging roads (instead of in the middle of logging fields)
to minimize the fuel required to harvest and transport the waste to the biomass burner.
3 Is it ethical to use cropland for biofuel production in a world where some of the population still
does not have adequate food? Argue your view on this question.
Response: Yes. We should export less food, and more food production assistance. Enabling
more local food production in hungry nations will make them less dependent on our imports. It
is socially unjust in the long term to feed hungry nations (consider Haiti) instead of helping them
learn to feed themselves. Transporting food across oceans costs fuel and emits GHG. We
should produce domestic biofuels which have a high energy ratio, to reduce our fossil fuel
dependence.
No. Cropland should be used first for human food. Only when every person is well nourished
should cropland be used for livestock feed. Only after human needs for meat protein are met
should cropland be used for biofuels, and then only biofuels which have a high energy ratio.
Fossil fuel use should meanwhile be reduced by conservation, renewable electricity sources,
mass transport and alternative transport solutions.
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