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

Topic 8: Energy, power and climate change

Can you read through the syllabus that Mr Porter is

giving you?

Types of Energy

Heat Chemical

Light Gravitational

Sound Elastic/strain

Kinetic Nuclear

Electric

Stored/potential

The Law of Conservation of Energy

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

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

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.

Sankey Diagram

The thickness of each arrow is drawn to scale to show the amount of energy

Efficiency

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

Efficiency

Efficiency is defined as

Efficiency = useful energy output

total energy input

Topic 8 – Lesson 2

• Workings of a generator

• Energy sources

• Renewable and non-renewable

• Energy density

Electromagnetic induction

If a magnet is moved inside a coil an electric current is induced (produced)

Electromagnetic induction

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

Generator/dynamo

A generator works in this way by rotating a coil in a magnetic field (or rotating a magnet in a coil)

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.

Renewable

• Mostly directly or indirectly linked with the sun

• The exception is tidal energy

• Photovoltaic cells, active solar devices, wind, wave, tidal, biomass.

World energy production

Fuel % total energy production

CO2 emission g.MJ-1

Oil 40 70

Natural gas 23 50

Coal 23 90

Nuclear 7 -

Hydroelectric 7 -

Others < 1 -

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

Energy density

• Coal - 30 MJ.kg-1

• Wood - 16 MJ.kg-1

• Gasoline – 47 MJ.kg-1

• Uranium – 7 x 104 GJ.kg-1 (70000000 MJ.kg-1)

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

Electricity production

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

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.

Fossil fuels - Advantages

• Relatively cheap

• High energy density

• Variety of engines and devices use them directly and easily

• Extensive distribution network in place

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

Example question

• A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%

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

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

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)

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

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.

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 – 0.400 x 103 = 740 MWIn one second, Q = mcΔT

740 x 106 = m x 4200 x 5m = 35 x 103 kg

So flow needs to be 35 x 103 kg.s-1

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).

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)

Control rods

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

Heat

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

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

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!!!

Nuclear power - Advantages

• High power output

• Large reserves of nuclear fuels

• No greenhouse gases

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

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

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)

Solar power - advantages

• “Free”

• Renewable

• Clean

Solar power - disadvantages

• Only works during the day

• Affected by cloudy weather

• Low power output

• Requires large areas

• Initial costs are high

Hydroelectric power

Water storage in lakes

“High” water has GPE. AS it falls this urns to KE, turns a turbine etc.

Pumped storage

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

Tidal water storage

• Tide trapped behind a tidal barrage. Water turns turbine etc.

Hydroelectric - Advantages

• “Free”

• Renewable

• Clean

Hydroelectric - disadvantages

• Very dependent on location

• Drastic changes to environment (flooding)

• Initial costs very high

Wind power

Calculating power

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

V

A

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

V

AVolume of air going through per second = Av

Mass of air per second = Density x volume

Mass of air per second = ρAv

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

V

AMass 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

Wind power - advantages

• “Free”

• Renewable

• Clean

• Ideal for remote locations

Wind power - disadvantages

• Works only if there is wind!

• Low power output

• Unsightly (?) and noisy

• Best located far from cities

• High maintainance costs

Wave power

OWC

Oscillating

water column

Modeling waves

• We can simplfy the mathematics by modeling square waves.

λ

L

2A

Modeling waves

• If the shaded part is moved down, the sea becomes flat.

λ

L

2A

Modeling waves

• The mass of water in the shaded part = Volume x density = Ax(λ/2)xLxρ = AλLρ/2

λ

L

2A

Modeling waves

• Loss of Ep of this water = mgh = = (AλLρ)/2 x g x A = A2gLρ(λ/2)

λ

L

2A

Modeling waves

• Loss of Ep of this water = mgh= A2gLρ(λ/2)

• # of waves passing per unit time = f = v/λ

λ

L

2A

Modeling waves

• Loss of Ep per unit time = A2gLρ(λ/2) x v/λ

• = (1/2)A2Lρgv

λ

L

2A

Modeling waves

• The maximum power then available per unit length is then equal to = (1/2)A2ρgv

λ

L

2A

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

Wave power - Advantages

• “Free”

• Reasonable energy density

• Renewable

• Clean

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

Global Warming

The Sun

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

The earth

Some of these waves will reach the earth

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%

Albedo

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

Absorbed by the earth

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

70%

Absorbed by the earth

70%Infrared

This absorbed solar energy is re-radiated at longer wavelengths (in the infrared region of the spectrum)

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

Black-body radiationNeed to “learn” this!

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

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)

Wien’s law

• λmaxT = constant (2.9 x 10-3 mK)

Example

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

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

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

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.

Example

• 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 = 2500000 seconds = 29 days

• Sun only shines approx 12 hours a day so would take at least twice as long

Absorbed by the earth

70%Infrared

This absorbed solar energy is re-radiated at longer wavelengths (in the infrared region of the spectrum)

Absorbed

• Various gases in the atmosphere can absorb radiation at this longer wavelength (resonance)

C

O

O

C

H

H

H

HThey vibrate more (become hotter)

HH

O

Greenhouse gases

• These gases are known as “Greenhouse” gases. They include carbon dioxide, methane, water and N2O.

C

O

O

C

H

H

H

HHH

O

Balance

There exists a balance between the energy absorbed by the earth (and its atmosphere) and the energy emitted.

Energy in Energy out

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

Balance

Without this normal “greenhouse effect” the earth would be too cold to live on.

Energy in Energy out

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.

What might happen?

• Polar ice caps melt

What might happen?

• Higher sea levels and flooding of low lying areas as a result of non-sea ice melting and expansion of water

Coefficient of volume expansion

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

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

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.

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?

What else might happen?

• More extreme weather (heatwaves, droughts, hurricanes, torrential rain)

What might happen?

• Long term climate change

Evidence?

• Ice core research

• Weather records

• Remote sensing by satellites

• Measurement!