1. Topic 8: Energy, power and climate change
Can you read through the syllabus that Mr Porter is giving you?

2. Types of Energy Heat Chemical Light Gravitational Sound Elastic/strain
Kinetic Nuclear
Electric
Stored/potential

3. The Law of Conservation of Energy
Energy can be changed (transformed) from one type to another, but it can never be made or destroyed.

4. Energy Flow diagrams
We can write energy flow diagrams to show the energy changes that occur in a given situation.
For example, when a car brakes, its kinetic energy is transformed into heat energy in the brakes.
Kinetic heat
sound

5. Energy degradation!
In any process that involves energy transformations, the energy that is transferred to the surroundings (thermal energy) is no longer available to perform useful work.

6. Sankey Diagram
The thickness of each arrow is drawn to scale to show the amount of energy

7. Efficiency
Although the total energy out is the same, not all of it is useful.

8. Efficiency Efficiency is defined as Efficiency = useful energy output
total energy input

9. Topic 8 – Lesson 2 Workings of a generator Energy sources
Renewable and non-renewable
Energy density

10. Electromagnetic induction
If a magnet is moved inside a coil an electric current is induced (produced)

11. Electromagnetic induction
A electric current is induced because the magnetic field around the coil is changing.

12. Generator/dynamo
A generator works in this way by rotating a coil in a magnetic field (or rotating a magnet in a coil)

13. Non-renewable Finite (being depleted – will run out)
In general from a form of potential energy released by human action
Coal, oil, gas (fossil fuels), Uranium.

14. Renewable Mostly directly or indirectly linked with the sun
The exception is tidal energy
Photovoltaic cells, active solar devices, wind, wave, tidal, biomass.

15. World energy production
Fuel
% total energy production
CO2 emission g.MJ-1
Oil 40 70
Natural gas 23 50
Coal 90
Nuclear 7 -
Hydroelectric
Others
< 1

16. Energy Density
The energy that can be obtained from a unit mass of the fuel
J.kg-1
If the fuel is burnt the energy density is simply the heat of combustion

17. Energy density Coal - 30 MJ.kg-1 Wood - 16 MJ.kg-1
Gasoline – 47 MJ.kg-1
Uranium – 7 x 104 GJ.kg-1 ( MJ.kg-1)

18. Hydroelectric energy density?
Imagine 1 kg falling 100m.
Energy loss = mgh = 1x10x100 = 103 J
If all of this is turned into electrical energy it gives an “energy density” of the “fuel” of 103 J.kg-1

19. Electricity production
Generally (except for solar cells) a turbine is turned, which turns a generator, which makes electricity.

20. Fossil fuels
In electricity production they are burned, the heat is used to heat water to make steam, the moving steam turns a turbine etc.

21. Fossil fuels - Advantages
Relatively cheap
High energy density
Variety of engines and devices use them directly and easily
Extensive distribution network in place

22. Fossil fuels - Disadvantages
Will run out
Pollute the environment (during mining sulphur and heavy metal content can be washed by rain into the environment)
Oil spillages etc.
Contribute to the greenhouse effect by releasing greenhouse gases

23. Example question
A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%

24. Calculate the rate at which thermal energy is provided by the coal
A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%
Calculate the rate at which thermal energy is provided by the coal

25. Calculate the rate at which thermal energy is provided by the coal
A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%
Calculate the rate at which thermal energy is provided by the coal
Efficiency = useful power output/power input
Power input = output/efficiency
Power input = 400/0.35 = 1.1 x 103 MW

26. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%
Calculate the rate at which coal is burned (Coal energy density = 30 MJ.kg-1)

27. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%
Calculate the rate at which coal is burned (Coal energy density = 30 MJ.kg-1)
1 kg of coal burned per second would produce 30 MJ. The power station needs 1.1 x 103 MJ per second. So
Mass burned per second = 1.1 x 103/30 = 37 kg.s-1
Mass per year = 37x60x60x24x365 = 1.2 x 109 kg.yr-1

28. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%
The thermal energy produced by the power plant is removed by water. The temperature of the water must not increase by more than 5 °C. Calculate the rate of flow of water.

29. Rate of heat loss = 1.1 x 103 – 0.400 x 103 = 740 MW
A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%
The thermal energy produced by the power plant is removed by water. The temperature of the water must not increase by moe than 5 °C. Calculate the rate of flow of water.
Rate of heat loss = 1.1 x 103 – x 103 = 740 MW
In one second, Q = mcΔT
740 x 106 = m x 4200 x 5
m = 35 x 103 kg
So flow needs to be 35 x 103 kg.s-1

30. Controlled fission (see topic 7)
The chain reaction can be controlled using control rods and a moderator. The energy can then be used (normally to generate electricity).

31. Moderator
This slows the free neutrons down, making them easier to absorb by the uranium 235 nuclei. Graphite or water is normally used.
1 eV neutrons are ideal)

32. Control rods
These absorb excess neutrons,making sure that the reaction does not get out of control. Boron is normally used.

33. Heat
The moderator gets hot from the energy it absorbs from the neutrons.

34. Heat
This heat is used to heat water, to make steam, which turns a turbine, which turns a generator, which makes electricity.

35. Useful by-products
Uranium 238 can also absorb neutrons to produce plutonium 239 which is itself is highly useful as a nuclear fuel.
It makes more fuel!!!

36. Nuclear power - Advantages
High power output
Large reserves of nuclear fuels
No greenhouse gases

37. Nuclear power - disadvantages
Waste products dangerous and difficult to dispose of
Major health hazard if there is an accident
Problems associated with uranium mining
Nuclear weapons

38. The solar constant The sun’s total power output is 3.9 x 1026 W!
Only a fraction of this power actually reaches the earth, given by the formula
I (Power per unit area) = P/4πr2
For the earth this is 1400 W.m-2 and is called the solar constant

39. The solar constant
For the earth this is 1400 W.m-2 and is called the solar constant
This varies according to the power output of the sun (± 1.5%), distance from sun (± 4%), and angle of earth’s surface (tilt)

40. Solar power - advantages
“Free”
Renewable
Clean

41. Solar power - disadvantages
Only works during the day
Affected by cloudy weather
Low power output
Requires large areas
Initial costs are high

42. Hydroelectric power

43. Water storage in lakes
“High” water has GPE. AS it falls this urns to KE, turns a turbine etc.

44. Pumped storage
Excess electricity can be used to pump water up into a reservoir. It acts like a giant battery.

45. Tidal water storage
Tide trapped behind a tidal barrage. Water turns turbine etc.

46. Hydroelectric - Advantages
“Free”
Renewable
Clean

47. Hydroelectric - disadvantages
Very dependent on location
Drastic changes to environment (flooding)
Initial costs very high

48. Wind power
Calculating power

49. Wind moving at speed v, cross sectional area of turbines = A

50. Wind moving at speed v, cross sectional area of turbines = A
Volume of air going through per second = Av
Mass of air per second = Density x volume
Mass of air per second = ρAv A

51. Wind moving at speed v, cross sectional area of turbines = A
Mass of air per second = ρAv
If all kinetic energy of air is transformed by the turbine, the amount of energy produced per second = ½mv2 = ½ρAv3 A

52. Wind power - advantages
“Free”
Renewable
Clean
Ideal for remote locations

53. Wind power - disadvantages
Works only if there is wind!
Low power output
Unsightly (?) and noisy
Best located far from cities
High maintainance costs

54. Wave power

55. OWC
Oscillating
water column

56. Modeling waves
We can simplfy the mathematics by modeling square waves. λ L 2A

57. Modeling waves If the shaded part is moved down, the sea becomes flat.λ L 2A

58. Modeling waves
The mass of water in the shaded part = Volume x density = Ax(λ/2)xLxρ = AλLρ/2 λ L 2A

59. Modeling waves
Loss of Ep of this water = mgh = = (AλLρ)/2 x g x A = A2gLρ(λ/2) λ L 2A

60. Modeling waves Loss of Ep of this water = mgh= A2gLρ(λ/2)
# of waves passing per unit time = f = v/λ λ L 2A

61. Modeling waves Loss of Ep per unit time = A2gLρ(λ/2) x v/λ
= (1/2)A2Lρgv λ L 2A

62. Modeling waves
The maximum power then available per unit length is then equal to = (1/2)A2ρgv λ L 2A

63. Power per unit length
A water wave of amplitude A carries an amount of power per unit length of its wavefront equal to
P/L = (ρgA2v)/2
where ρ is the density of water and v stands for the speed of energy transfer of the wave

64. Wave power - Advantages
“Free”
Reasonable energy density
Renewable
Clean

65. Wave power - disadvantages
Only in areas with large waves
Waves are irregular
Low frequency waves with high frequency turbine motion
Maintainance and installation costs high
Transporting power
Must withstand storms/hurricanes

66. Global Warming

67. The Sun
The sun emits electromagnetic waves (gamma X-rays, ultra-violet, visible light, infra-red, microwaves and radio waves) in all directions.

68. The earth
Some of these waves will reach the earth

69. Reflected
Around 30% will be reflected by the earth’s atmosphere. This is called the earth’s albedo (0.30). (The moon’s albedo is 0.12) Albedo is the ratio of reflected light to incident light.
30%

70. Albedo
The albedo depends on the ground covering (ice = high, ocean = low), cloud cover etc.

71. Absorbed by the earth
Around 70% reaches the ground and is absorbed by the earth’s surface.
70%

72. Absorbed by the earth
This absorbed solar energy is re-radiated at longer wavelengths (in the infrared region of the spectrum)
Infrared
70%

73. In order to understand some of these processes we need to know a little physics.

74. Black-body radiation
Need to “learn” this!

75. Black-body radiation
Black Body - any object that is a perfect emitter and a perfect absorber of radiation
object does not have to appear "black"
sun and earth's surface behave approximately as black bodies

76. Black-body radiation
The amount of energy per second (power) radiated from a body depends on its surface area and absolute temperature according to
P = eσAT4
where σ is the Stefan-Boltzmann constant (5.67 x 10-8 W.m-2.K-4) and e is the emissivity of the surface (=1 for a black object)

77. Wien’s law
λmaxT = constant (2.9 x 10-3 mK)

78. Example
By what factor does the power emitted by a body increase when its temperature is increased from 100ºC to 200ºC?

79. Example
By what factor does the power emitted by a body increase when its temperature is increased from 100ºC to 200ºC?
Emitted power is proportional to the fourth power of the Kelvin temperature, so will increase by a factor of 4734/3734 = 2.59

80. Surface heat capacitance Cs
Surface heat capacitance is defined as the energy required to increase the temperature of 1 m2 of a surface by 1 K. Cs is measured in J.m-2.K-1.
Q = ACsΔT

81. Example
Radiation of intensity 340 W.m-2 is incident on the surace of a lake of surface heat capacitance Cs = 4.2 x 108 J.m-2.K-1. Calculate the time to increase the temperature by 2 K. Comment on your answer.

82. Example Each 1m2 of lake receives 340 J.s-1
Radiation of intensity 340 W.m-2 is incident on the surface of a lake of surface heat capacitance Cs = 4.2 x 108 J.m-2.K-1. Calculate the time to increase the temperature by 2 K. Comment on your answer.
Each 1m2 of lake receives 340 J.s-1
Energy needed to raise 1m2 by 2 K = Q = ACsΔT = 1 x 4.2 x 108 x 2 = 8.4 x 108 J
Time = Energy/power = 8.4 x 108/340 = seconds = 29 days
Sun only shines approx 12 hours a day so would take at least twice as long

83. Absorbed by the earth
This absorbed solar energy is re-radiated at longer wavelengths (in the infrared region of the spectrum)
Infrared
70%

84. Absorbed
Various gases in the atmosphere can absorb radiation at this longer wavelength (resonance) H O
They vibrate more (become hotter) O C H H C H H H O

85. Greenhouse gases
These gases are known as “Greenhouse” gases. They include carbon dioxide, methane, water and N2O. O H O H H C H H C H O

86. Balance
There exists a balance between the energy absorbed by the earth (and its atmosphere) and the energy emitted.
Energy in
Energy out

87. Balance
This means that normally the earth has a fairly constant average temperature (although there have been big changes over thousands of years)
Energy in
Energy out

88. Balance
Without this normal “greenhouse effect” the earth would be too cold to live on.
Energy in
Energy out

89. Greenhouse gases
Most scientists believe that we are producing more of the gases that absorb the infra-red radiation, thus upsetting the balance and producing a higher equilibrium earth temperature. This is called the enhanced greenhouse effect.

90. What might happen?
Polar ice caps melt

91. What might happen?
Higher sea levels and flooding of low lying areas as a result of non-sea ice melting and expansion of water

92. Coefficient of volume expansion
Coefficient of volume expansion is defined as the fractional change in volume per unit temperature change

93. Coefficient of volume expansion
Given a volume V0 at temperature θ0, the volume after temperature increase of Δθ will increase by ΔV given by
ΔV = γV0Δθ

94. Example
The area of the earth’s oceans is about 3.6 x 108 km2 and the average depth is 3.7 km. Using γ = 2 x 10-4 K-1, estimate the rise in sea level for a temperature increase of 2K. Comment on your answer.

95. Evaporation? Greater area cos of flooding? Uniform expansion?
Example
The area of the earth’s oceans is about 3.6 x 108 km2 and the average depth is 3.7 km. Using γ = 2 x 10-4 K-1, estimate the rise in sea level for a temperature increase of 2K. Comment on your answer.
Volume of water = approx depth x area
= 3.6 x 108 x 3.7
= 1.33 x 109 km3 = 1.33 x 1018 m3
ΔV = γV0Δθ
ΔV = 2 x 10-4 x 1.33 x 1018 x 2 = 5.3 x 1014 m3
Δh = ΔV/A = 5.3 x 1014/3.6 x 1014 = 1.5 m
Evaporation? Greater area cos of flooding? Uniform expansion?

96. What else might happen?
More extreme weather (heatwaves, droughts, hurricanes, torrential rain)

97. What might happen?
Long term climate change

98. Evidence? Ice core research Weather records
Remote sensing by satellites
Measurement!