Chapter 3 Pollution Energy for Human Usectaps.yu.edu.jo/physics/courses/phys645/PDFs/3... · Pollution ©Dr. Nidal M. Ershaidat Environmental Physics -Chapter 3 Energy for Human Use

1

© Dr. Nidal M. Ershaidat

Phys. 645: Environmental Physics

Physics Department

Yarmouk University

Chapter 3 Chapter 3 Energy for Human UseEnergy for Human Use

http://ctaps.yu.edu.jo/physics/Courses/Phys645/Chapter3

3-2-7 Energy Transport & Pollution

© Dr. Nidal M. Ershaidat Environmental Physics - Chapter 3 Energy for Human Use

3

Energy transport: The example of crude oil

� The cost of energy transport

It is largely dependent on the cost of transport of the

energy carrier.

� Pollution may occur in the transport process.

Leakages in the pumping system

� Transport of Crude oil :

Pipelines are used:

Increasing the transported volume per time

unit costs a lot of extra power!

This depends on “the distance” of transportation” and the

viscosity of the oil


4

Transport of crude oilSome Fluid Mechanics!

Consider a volume dV of a fluid, of density ρρρρ and viscosity µµµµ, that we want to transport through a

cylindrical pipe of radius R.

We shall compute the power needed to “transport”

a flux V per second.

5

Horizontal Motion of Fluids� Newton’s equation for the mass (ρρρρdV) can be written as

follows (u is the velocity):

.vispressgravityCoriolis FFFFdt

uddV

→→→→→→→→→→→→→→→→

++++++++++++====ρρρρ

where: and are, respectively, the forces due to pressure and to the viscosity

presF→→→→

.visF→→→→

The Coriolis term takes into account any eventual

rotational motion (which exists when treating fluids), or

the fact that calculations are done in a rotational frame.

The centrifugal terms are taken into account, partly by the

use of a local gravity g. And other terms are neglected

since we are considering a horizontal motion.

3-74

6

No acceleration, No gravity� There is no acceleration 0====ρρρρ

dt

uddV

� Motion is horizontal: (Neglecting the effect of gravitation on

a horizontal flow) 0====→→→→

gravityF

� Assuming there is no rotation, the Coriolis force is zero.

� Under these conditions the resultant force is simply:

0====++++→→→→→→→→

vispres FF

� and the pressure drop per meter that is pushing the fluid

is: pgradpG→→→→→→→→

−−−−====∇∇∇∇−−−−====

dVGdVpgraddVpF pres ====−−−−====∇∇∇∇−−−−====→→→→→→→→→→→→

and

3-75

2


7

Derivation of the viscous forces

jdVdz

udidV

dz

udF

yx.vis

2

2

2

2

µµµµ++++µµµµ====→→→→

� Newton’s assumption for viscous forces:

� If we neglect the z component since the motion

is horizontal, then:

3-76


8

What is the viscous force ?

→→→→→→→→

αααα−−−−==== uF .vis

The viscous force is a dissipative force just like

friction. A good approximation is to consider that:

αααα is the viscosity (unit N.s/m2 = J s m-3)

3-77


9

Viscous forces in a horizontal cylindrical pipe

Consider a cylindrical ring (radius r, thickness

δδδδr and length δδδδx). The viscosity forces on the inside and the outside of the ring are:

R

rδδδδr

u(r)u(r+dr)


10

Viscous forces in a horizontal cylindrical pipe

xrdr

duF

r

in.vis δδδδππππµµµµ−−−−==== 2

(((( )))) xdrrdr

duF

drr

out.vis δδδδ++++ππππµµµµ++++====

++++

2

Thus the net force is:

xdr

dur

dr

durF

rdrr

.vis δδδδππππ

−−−−µµµµ====

++++

2

µµµµ = is the viscosity per unit area (unit J s m-5)

3-78

3-79

3-80


11

dr

dr

dur

dr

dur

dr

dur

dr

d rdrr

r

−−−−

====

++++

Using the derivative:

xdrdr

dur

dr

dF .vis δδδδππππ

µµµµ==== 2

Fvis. can be written as follows

3-81

3-82


12

Equation of motionThe equation of motion of the considered fluid is then:

022 ====

δδδδµµµµππππ++++δδδδππππ−−−−dr

durdrxxdrr

dx

dP

0====−−−−

µµµµ rGdr

dur

dr

d

This 2nd order homogenous differential equation gives the volume passing through a cross-section per time unit.

3-83

3-84

3


13

(((( )))) BrlnArG

ru ++++++++µµµµ

−−−−====2

4

1

Solution of the equation of motionThe general solution of the former equation

0====−−−−

µµµµ rGdr

dur

dr

d

is:

and using the initial conditions:u(r=0) must be finite and u(r=R)=0

(((( )))) (((( ))))µµµµ−−−−

−−−−====4

22RrG

ru

3-85

3-86

3-87


14

Calculating the flux

The volume passing a cross-section per s is:

(((( )))) (((( ))))∫∫∫∫∫∫∫∫ µµµµ

−−−−====ππππ====

RR

drrrRG

drrruV0

22

0 42

µµµµππππ

====8

4RG

V

3-88

3-89


15

Calculating the power PThe power needed to transport the fluid from point 1 (pressure p1) to

point 2 (pressure p2) is:

(((( )))) LRG

VLGVppPµµµµ

ππππ========−−−−====

8

42

21

LR

VP µµµµππππ====

4

2

8

(((( )))) (((( ))))∫∫∫∫ ππππ−−−−====R

Force

rudrrppP0

21 2��

[[[[ ]]]]

µµµµππππ==== L

R

VP

4

2

8 WsJm

mmsJsm============ −−−−

−−−−−−−−1

4

526

Dimensional Analysis

3-90

3-91


16

Optimal conditions for this transport

The relation we found indicates that:

1) It’s better to keep the ratio V/R constant if the radius of

the pipe to be increased.

2) Decreasing viscosity µµµµ, would help.

For crude oil, decreasing µµµµ is possible if the crude oil is

heated

LR

VP µµµµππππ====

4

2

8

THIS IS DONE WITH THE OIL IN ALASKA AND

SIBERIA!!!

3-92


17

Electricity transport

Homework:Study the paragraph treating electricity transport (pages 105-106)


18

Reducing PollutionPollutants

• They may occur in air, water and soil.

Definition:

Pollutants are substances that when found in concentrations

bigger than a natural environment cause damage to plant,

animal and human life.

• They may be transported from one place to

another.

• Primary pollutants may react with each other or

with natural substances and produce secondary

pollutants which could be very dangerous for life.

4


19

Pollutants & the greenhouse :

• Interaction of “particles” with the natural

compounds occurring in the atmosphere are

responsible of the warming of our planet.

• Some other particles contribute to a

reverse effect (cooling)

• etc...


20

Pollutants & Chemistry :

• Atmospheric chemistry studies the interaction between primary pollutants & the atmosphere.

• Examples: - The Ozone hole,- The production of photochemical smog by

nitrogen oxides (NO & NO2)

• From a chemist point of view, the major pollutants are : sulfur (S), nitrogen (N), carbon (C) & halogens (Cl, F, …) compounds (oxides or carbonates)

Atmospheric chemistry

21

Nitrogen Oxides (Nox)*

• Major Sources:

3) Nitrogen is used in fertilizers and thus spread into the soil.

The reaction in which Nox is formed is:

N2 + O2 NOx

* x = 1 or 2

2) Nitrogen is present in the fuel itself. Some coal or fuel contain 1%of nitrogen.(Natural gas does not contain any nitrogen)

1) Combustion engines where air is used:N2 forms 78% of air, (21% of oxygen)

The NOx is formed in the flame at T = few thousands K. The reaction is endothermic, the Nox formation increases with temperature


22

Reducing NOx emission• Reduce the flame temperature

BUT this means lower efficiencies!

• Reduce the available oxygen for combustion

BUT this means the disappearance of the oxide

materials used to protect the combustion chamber,

and this solution means the search of new

materials for the combustion chambers.


23

Some technical solutions

• The fluidized bed: Adding Mg or CaCO3 to

crushed coal where air enters from below

(See textbook for further details)

• Spray NH3 into the exhaust gases of power

stations to reduce NO2 to N2


24

Sulphur (S) - The Acid Rain problem

• What is Acid Rain?

• Coal and oil contain considerable

concentrations of sulphur and burning coal and

fuel liberates Sulphur.

• Acid rain is caused by emissions of sulphur

oxides ( SO2) and nitrogen oxides (Nox).

• Sox and NO2 are converted in the atmosphere

to H2SO4 and HNO3 among other products.

5


25

Definition: The Acid Rain• Diluted forms of these acids can fall to earth as

rain, snow hail, or fog!.

• In non rainy days, these acids may interact

directly with soil, vegetation and water in a variety

of ways referred to as dry deposition.

• The term “acid rain” refers to rain fall with a pH

less than 5.6, which is the pH of distilled water

containing atmospheric carbon dioxide (CO2)


26

Acidity


27

Conditions for the damageEmissions cause damage to the ecosystem if 2 factors

prevail:

a) The ecosystem downwind the source must

be sensitive

b) The weather patterns must be such that they

will transport the pollutants to the sensitive

area allowing enough time and distance for

the chemical reactions to occur


28

Effects of acid rain1) Agricultural corps can be damaged directly or

by reduced soil fertility

3) Long exposure to acid rain can lower the pH of the water

of lakes and streams, making them more acidic, affecting all

people who use or drink this water. Marine life could also be

seriously affected.

4) Acid rain may also destroy roads and buildings (It erodes

stone, bricks and it corrodes metals)

2) Forest growth can be retarded.


29

Some technical solutions• The fluidized bed: Already mentioned

• The wet scrubbers:

Spray solutions of calcium carbonate (CaCO3) into

the pollutant gases to form calcium sulfate (CaSo3)

which is left behind and do not spread in the

atmosphere.

• Catalysts:

To remove sulphur from exhaust gases.

3-2-8 Nuclear Energy

6

31

Nuclear Energy - Introduction

FissionFission: Some nuclei simply fission spontaneously into other nuclei and energy is liberated in the process.

The order of magnitude of such processes is 100100 MeV.

The traditional way of producing energy:When burning fuel or coal, we are talking about chemical reactions. Energies involved are those of transitions of

atomic electrons. The order of magnitude is 1 eV.

FusionFusion: 2 nuclei may fuse producing a third one and energy is liberated in the process.

The order of magnitude of such processes is 100100 MeV.


32

Is Nuclear Energy cleaner?� The Factor of a million (the Factor Mega!)

The precedent figures show that a factor of a million is gained when using the energy liberated in the nuclear fission and fusion.

ProPro--nuclear energy people argue that this factor of a nuclear energy people argue that this factor of a million will be used for safety improvements.million will be used for safety improvements.

As we have said earlier, the capital cost of a nuclear As we have said earlier, the capital cost of a nuclear

power station is very high but the cost per power station is very high but the cost per kWhkWh is is rather moderaterather moderate


33

Major problem- Managing the “waste”

� Nuclear energy is not clean

The major problem in producing energy using fission is the management of the reaction waste!

As far as fusion is concerned, it is cleaner than all other means of energy production.

But the major obstacle is still technical.It is not easy to produce a controlled fusion.


34

Nuclear spectroscopists notationsThe following notations are used in Nuclear Spectroscopy:

EAN

AZ Eor EA

Z or

U23592 143

23592 Uor orU235

Example:

where E is the element (nucleus) symbolZ is the atomic number (number of protons)

A is the atomic mass numberN = A-Z = number of neutrons


35

Power from nuclear fission

What is Fission?What is Fission?

The most well-known fission is that of 236U:� A 235Uranium nucleus is bombarded with

slow neutrons. � The 235U nucleus “captures” a neutron,

becomes 236U in an excited state. This deformed nucleus, oscillates and like a “liquid drop” is “cut” spontaneously into two

fragments. � We say that that the 236U nucleus

“fissions” into two lighter nuclei and 2 to 3 fast neutrons are produced.


36

Thermal neutrons

When the energy of a neutron is around 0.025 eVwhich corresponds to a neutron energy in thermal

equilibrium with the surroundings (for T = 293 K,

E = kT = 1/40 eV), the neutron is called thermal neutron.

Fission

n (Thermal) +235U 236U X + Y+ ννννn +200 MeV

3-93

7


37

Thermal neutrons fission and fast neutrons fission

The number of the resulting fast neutrons (E = 2 MeV) is

νννν = 2.47 for 235U ( 2.89 for 239Pu)

Induced Fast neutrons

Some nuclei with odd atomic mass number A, such as 235U, 233U, 239Pu and 241Pu fission by “slow n capture”.

Other nuclei, such as the even-even 238U (Z=92, A=146) only suffer fragmentation by “fast n capture”.This is due to the so-called “fission barrier”

Chain Reaction


39

The multiplication factor, Chain reactionThe 2.47 induced Fast neutrons have themselves a probability to induce fission. From a generation to another their number is multiplied by a factor k called the multiplication factor.

This fact makes this reaction goes multiplying: fission and induced neutrons. This is the chain reactionchain reaction


40

k = 1: Stationary chain reaction

k > 1: divergent chain reaction

k < 1: decreasing chain reaction

Chain Reaction vs. k

Figure 3-14


41

What is a Nuclear Reactor?A Nuclear reactor is an installation where a controlled chain reaction takes places.

Nuclear reactors are classified according to:

1) The “fissile” fuel,

2) Type of heat exchanger,

3) Coolant used.

SEE the slide entitled

THE TYPES of NUCLEAR REACTORS

A controlled chain reaction is one which we can stop at reasonable limits.


42

The fission fragments

X and Y are called the fission fragments and the sum of their atomic numbers should be 92 (Z for Uranium).

The fission fragments mass distribution shows an asymmetry around A = 112. The Bt-Kr-Zr region (A near 95) and the complementary “elements” (with A near

140), i.e. I-Xe-Ba are privileged.

The resultant fragments are “rich” in neutrons and thus are generally ββββ- emitters.They can be identified by classical methods of nuclear chemistry.


8

43

Nuclear Binding Energies

Figure 3-15© Dr. Nidal M. Ershaidat Environmental Physics - Chapter 3 Energy for Human Use

44

Fission: resultant energyThe binding energy per nucleon for A=236 is EB=7.55

MeV and for A=118 (symmetric fission special case) this

energy is EB= 8.45 MeV.

The gain in energy in fission is:

∆∆∆∆E= (8.45 - 7.55) ×××× 236 = 210 MeV

For comparison the combustion of 1g of coal provides 3.3 ××××104 J.

The order of magnitude is 106 (The factor Mega)

If 236 g of 235U fission (1 mole of uranium), then the expected energy

is:

(6××××1023××××210 MeV)/235 = 8.6××××1010 J = 86 GJ = 20.5 Gcal.

Which is equivalent to the energy needed to “boil” 106 kg (1000

tons) of water from 0 to 100 0C


45

Power from nuclear fission

The other 17% of the energy available are

emitted as ββββ- and γγγγ radioactivity and as kinetic

energy of the induced neutrons (average 2 MeV

per emitted neutron)

Almost 83% of the available energy goes into

kinetic energy of the fission fragments who

have a small free path (< 1 mm). This energy is

rapidly converted into heat.

And this is the heat used in a nuclear (power

plant) heat engine.


46

Fission: some definitionsFor an initial number of nuclei N, the fraction dNsusceptible to undergo a fission interaction (or

having decayed) during an interval of time dt is

proportional to N and dt, i.e. dN ∝∝∝∝ - N dt.

dN = -λλλλ N dt

Which gives by integration N(t) = N0 e-λλλλt

where N0 is the number of nuclei present at some

reference instant t = 0.

3-94

The proportionality constant (λλλλ) is called the decay constant, i.e.


47

Half-life and mean lifetimeThe half-life ττττ is defined by

N(t) = N0/2 (e-λλλλt =2) or:

λλλλ====ττττ

2ln

The mean lifetime is defined by (the average):

(((( ))))

(((( ))))λλλλ

====λλλλ====λλλλ====

−−−−====−−−−

∫∫∫∫∫∫∫∫

∫∫∫∫∫∫∫∫

∞∞∞∞λλλλ−−−−

∞∞∞∞

∞∞∞∞∞∞∞∞

111

11

0

0

000

0000

dteNtN

dttNtN

dtdt

dNt

NdNt

N

t

3-96

3-95


48

equals ππππRU2 = π×π×π×π× (7.4 F)2 = 172 F2 = 1.72 barn

The fission cross-section

The radius R of a 235U nucleus (assumed to be spherical) is given by the empirical relation:

R = 1.2 ×××× A1/3, where A is the atomic mass number (235)

which gives RU = 7.4 F.

The cross-section of an interaction is the area of the target seen by the incident particle.

The experimental cross-section (582 barns) is 340 times the geometrical cross section which indicates that this reaction is strongly favoured

The geometrical cross-section for the reaction:


9


49

Total cross-sectionIn a reactor we deal with N per unit volume nuclei.

Each cross-section is related to the free path in each process

The relation between the mean free path λλλλi and ΣΣΣΣi is

simply: λλλλi = 1/ΣΣΣΣi, where i represents f, a or s.

Please refer to the textbook page 152.

Macroscopic cross-section is introduced and

defined by: ΣΣΣΣ = N σ σ σ σ(ΣΣΣΣf = N σσσσf, ΣΣΣΣa = N σσσσa, ΣΣΣΣs = N σσσσs)

Some nuclei fission (σσσσf), but others simply absorb (σσσσa) (radiative capture) or scatter (σσσσs) the fast neutrons.


50

Control Rods

Reflector

Core + Moderator

Water from condenser

Steam to Turbine

He

at E

xcha

ng

er

Coolant

A schematic Nuclear reactor

Figure 3-16


51

Components of a Nuclear reactorThe core = The part of the reactor which contains the nuclear fuel

(235U, 239Pu or 233U )

The moderator is used for thermalization of fast neutrons (i.e to

slow them enough to thermal energy so the fission can take place

again

The Control rods (Cadmium generally) are used for the absorption

of superfluous neutrons if any

The reflector’s task is to maintain a sufficient number of neutron projectiles

The coolant of the reactor core which generates steam for a turbine

by means of a heat exchanger (Sodium rods)


52

Types of Nuclear Reactors� BWR (Boiling Water Reactor)

Water is used as a moderator and a steam source

� CANDU (Canadian Deuterium Uranium)Heavy water is used as a moderator and a steam source. The cooling is done through 2, 3 or 4 cycles and pressure is maintained as in PWR, No enrichment needed.

� PWR (Pressurized Water Reactor)Water is used as a moderator and a steam source. The cooling is done through 2, 3 or 4 cycles and pressure is maintained. Needs enriched Uranium

53

The BWR� BWR (Boiling Water Reactor)

Water is used as a moderator and a steam source

Figure 3-17

54

The PWR

� PWR (Pressurized Water Reactor)Water is used as a moderator and a steam source. The cooling is done through 2, 3 or 4 cycles and pressure is maintained. Needs enriched Uranium

Figure 3-18: The PWR

10

55

� CANDU (Canadian Deuterium Uranium)Heavy water is used as a moderator and a steam source.

The CANDU

� The cooling is done through 2, 3 or 4 cycles and pressure is maintained as in PWR, No enrichment needed.

Figure 3-19: The CANDU reactor


56

The CANDU

� F = Fuel Loading Machine

F

� S = Steam Generator

S

R

� R = Reactor

W

� W= Circulating Water Generator

T

� T = Turbine

G

� G = Generator

3-2-9 Fission, FusionNuclear Energy & Pollution


58

Estimating ηηηη for a 235U Reactor

[[[[ ]]]] ννννσσσσ++++σσσσ++++σσσσ

σσσσ====ηηηη

238238235235235

235235

,c,c,f

,f

NN

N

007202999

710

238

235 ..

.

N

N========

b.,c 72238 ≅≅≅≅σσσσ

ηηηη

for an average number of neutrons νννν = 2.47

b,f 582235 ≅≅≅≅σσσσ b,c 100235 ≅≅≅≅σσσσ

3-97

238U Reactor Pure 235U Reactor

1.36 2.11

59

Uranium Enrichment - Increasing ηηηηIf the “crude” fuel (UO2) were to contain more fissile 235U, then ηηηη would increase and substantially.

For the same calculations done before if we replace the ratio

00720238

235 .N

N====

0260597

52

238

235 ..

.

N

N========

by 2.5% of 235U in the fissile fuel, i.e.

Then

ηηηη

2.5% Enriched uranium

1.83

20% Enriched uranium

2.11

60

The ModeratorIt is necessary to slow down the fast neutrons resulting from fission. A Moderator is usedModerator is used

�Water (H2O):available and cheap, can be used but it absorbs strongly the fast neutrons (σ(σ(σ(σ a= 664 mb).).).).

� Heavy water (D2O) is an excellent decelerator of fast neutrons. It is expensive but has a very week σσσσc

(σ(σ(σ(σ a= 1 mb).).).)..�Graphite (carbon) is a third option, cheap, and has a

relatively small σσσσa (σ(σ(σ(σ a= 4.5 b).).).).

Characteristics of a moderator:1) It should have a small Atomic mass number (A)To allow a fast thermalization (loss of (kinetic) energy of the

neutrons)2) It should not have a big cross-section of absorption.

11


61

The Reflector

A Reflector is used to reduce this factor lf.

The resultant number of fast neutrons to be slowed down is:

(ηηηηf εεεεf - ηηηηf εεεεf lf) n = ηηηηf εεεεf (1 - lf) n

Some of the fast neutrons may induce fission in 238U and 235U. Let εεεεf = the number of the fast neutrons induced

(called the fast fission factor). We thus have ηηηηf εεεεf fast neutrons.

There will also be some leakage. Let lf be the fraction of those fast neutrons which “leak”.(The corresponding number is ηηηηf εεεεf lf n)


62

The resonance escape probability

And after the slowdown a fraction ls may escape the reactor.

The absorption cross-section of (238U,n) has some absorption resonances not leading to fission. Some of the fast neutrons who have the required resonant energy would be absorbed.

A resonance escape probability p takes into account this “resonant absorption”


63

The thermal utilization factorFinally, only a fraction called the thermal

utilization factor f will induce new fission

The resultant number of useful neutrons:

k n = ηηηη εεεε p (1 - lf) (1 - ls) f n 3-98

64

The four factor formulae k is called the multiplication factor of the reactor:

For a theoretically infinite reactor, ls and lf may be neglected and thus:

The resultant number of useful neutrons:

The ratio moderator/fuel is chosen so as to get k=1 as close as possible

See discussion and Fig. 4.40 (pages 154-155)

Typically k = 1.1 or 1.2

Considering leakage factors, k could reach the value 1

necessary for a stationary reactor....

k∞∞∞∞ = ηηηη εεεε p f (The 4 factor formulae) 3-99

65

The Nuclear Fuel Cycle

Fuel Element Fabrication

Enrichment Plant

Reactor

Uranium mines

Tailings

Fuel ConditioningRadioactive Waste

Reprocessing


66

Homework : Study this figure

Figure 3-20

12


67

Fusion: The other source of nuclear energy

• Fusion = Isotopes of Hydrogen “fuse” under

particular condition to form a heavier nucleus

(Helium) and a large amount of energy is

released.

• Many different nuclear fusion reactions occur in

the sun and other stars, but only a few such

reactions are of practical value for potential

energy production on earth.


68

Physics of FusionFusion Reactions

Nuclei carry positive charges and thus they normally

repel one another. The higher the temperature, the

faster the atoms or nuclei move. When they collide at

these high speeds, they overcome the force of repulsion

of the positive charges, and the nuclei fuse.

In such collisions energy is released. The difficulty in

producing fusion energy has been to develop a device

which can heat the deuterium-tritium fuel to a

sufficiently high temperature and then confine it for a

long enough time so that more energy is released

through fusion reactions than is used for heating.


69

Fusion needs high temperatures

The products are energetic helium-4 (4He) the (alpha particle), and a more highly energetic free neutron (n). The helium nucleus carries one-fifth of the total energy released and the neutron carries the remaining four fifths.

To produce net power, fusion reactions must take place at high temperatures. The power production process which can occur at the lowest temperature and, hence, the most readily attainable fusion process on earth is the combination of a deuterium nucleus with one of tritium.


70

Fusion: Deuterium Tritium

Figure 3-21


71

Fusion: Power Plant

Figure 3-22

72

The Nuclear Fuel Cycle & Pollution

So

urce:

http

://

ccnr.o

rg

Figure 3-23

13

Appendix: Appendix: The Reactor EquationsThe Reactor Equations

Nuclear Energy & PollutionNuclear Energy & Pollution

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Chapter 3 Pollution Energy for Human Usectaps.yu.edu.jo/physics/courses/phys645/PDFs/3... · Pollution ©Dr. Nidal M. Ershaidat Environmental Physics -Chapter 3 Energy for Human Use