heat engines coal fired steam engines. petrol engines diesel engines jet engines power station...

Heat Engines

• Coal fired steam engines.

• Petrol engines

• Diesel engines

• Jet engines

• Power station turbines

DECChttp://www.decc.gov.uk/assets/decc/statistics/publications/

flow/193-energy-flow-chart-2009.pdf

Combined Cycle

THE LAWS OF THERMODYNAMICS1. You cannot win you can only break even.2. You can only break even at absolute zero.3. You can never achieve absolute zero.

S = k log W

• Atoms don’t care.

• What happens most ways happens most often

p

Boyle’s Law

p 1/V

1/V

T

V

Charles’s Law

V T

T

p

Pressure Law

p T

Number of molecules, N

p

Common sense Law

p N

Isotherms(constant temperature)

1/V

p

T

V

Isobars (constant pressure)

Isochors(constant volume)

T

p

p 1/V V T p T

In summary…

pV

T= constant

For ideal gases only

A gas that obeys Boyles law

Ideal gas?

Most gases approximate ideal behaviour

Ideal gases assume:-

• No intermolecular forces

• Volume of molecules is negligible

Not true - gases form liquids then solids as temperature decreases

Not true - do have a size

p1V1

T1

p2V2

T2

=

pV

T= constant

Only useful if dealing with same gas before (1) and after (2) an event

Ideal Gas Law

pV = nRT

p = pressure, PaV = volume, m3

n = number of molesR = Molar Gas constant (8.31 J K-1 mol-1 )T = temperature, K

Macroscopic model of gases

pV = NkT

N = number of moleculesk = Boltzmann’s constant (1.38 x 10-23 J K-1)

Which can also be written as …

First there was a box and one molecule…

Molecule:- mass = m velocity = v

x

y

zv

Kinetic Theory

Molecule hits side of box…(elastic collision)

v

-v

pmolMolecule

pbox = -pmol = 2mvBox

mv - mu

= -mv - mv = -2mv

2mv-2mv

Remember p = F so a force is felt by the box t

Molecule collides with side of box, rebounds, hits other side and rebounds back again.

Time between hitting same side, tvs= v= 2xx

y

z

Average force, exerted by 1 molecule on box

F = pt

= p v

2x

= 2mv v

2x

= mv2

x

Force exerted on box

Time

Average Force

Actual force during collision

x

y

zv1

Consider more molecules

v4

v2 v5

vN

-v6

-v7

All molecules travelling at slightly different velocities so v2 varies - take mean - v2

v3

-v8

Pressure = Force

Area

Force created by N molecules hitting the box…

F = Nmv2

x

= Nmv2

xyz

= Nmv2

V

But, molecules move in 3D

p = Nmv2

V

1

3

Mean square velocity

Kinetic Theory equation

Brownian Motion

Why does it support the Kinetic Theory?

• confirms pressure of a gas is the result of randomly moving molecules bombarding container walls

• rate of movement of molecules increases with temperature

• confirms a range of speeds of molecules

• continual motion - justifies elastic collision

Microscopic Macroscopic

pV = Nmv2 13

pV = NkT

(In terms of molecules) (In terms of physical observations)

=Nmv2 13

NkT

Already commented that looks a bit like K.E.

K.E. = ½mv2

Rearrange (and remove N)

Substitute into (1)

= 3kTmv2 (1)

K.E. = 32

kTAverage K.E. of

one molecule

Total K.E. of gas (with N molecules)

K.E.Total = 32

NkT

This is translational energy only

- not rotational, or vibrational

And generally referred to as internal energy, U

U = 32

NkT

U = 32

NkTInternal Energy of a gas

Sum of the K.E. of all molecules

How can the internal energy (K.E.) of a gas be increased?

1) Heat it - K.E. T

2) Do work on the gas

Physically hit molecules

Energy and gases

Change in Internal Energy

Work done on material

Energy transferred thermally

= +

U = W + Q

Basically conservation of energy

Also known as the First Law of Thermodynamics

Heat, Q – energy transferred between two areas because of a temperature difference

Work, W – energy transferred by means that is independent of temperature

i.e. change in volume

+ve when energy added-ve when energy removed

+ve when work done on gas - compression-ve when work done by gas - expansion

Einstein’s Model of a solid

Bonds between atoms

Atom requires energy to break them

U kT

Jiggling around(vibrational energy)

Mechanical properties change with temperature

T = highcan break and make bonds quickly – atoms slide easily over each other

T = low difficult to break bonds – atoms don’t slide over each other easily

Liquid: less viscous Solid: more ductile

Liquid: more viscous Solid: more brittle

Activation energy, - energy required for an event to happen i.e. get out of a potential well

Activation energy,

Can think of bonds as potential wells in which atoms live

The magic /kT ratio

- energy needed to do something

kT - average energy of a molecule

/kT = 1

/kT = 10 - 30

/kT > 100

Already happened

Probably will happen

Won’t happen

Probability of molecule having a specific energy

Exponential

Energy

Probability 1

0

Boltzmann Factor

e-/kT

Probability of molecules achieving an event characterised by activation energy,

1

10 - 30

> 100

0.37

4.5 x 10-6 - 9.36 x 10-14

3.7 x 10-44

e-/kT/kT

Nb. 109 to 1013 opportunities per second to gain energy

Entropy

Number of ways quanta of energy can be distributed in a system

Lots of energy – lots of ways

Not much energy – very few ways

An “event” will only happen if entropy increases or remains constant

Amongst particles

S = k ln W

2nd law of thermodynamics

S = entropyk = Boltzmann’s constantW = number of ways

ΔS = ΔQ

T

Energy will go from hot to cold

At a thermal boundary

Hot

Cold

Number of ways decreases – a bit

Number of ways increases – significantly

Result - entropy increase

• Efficiency = W/QH = (QH – QC ) / QH

• BUT Δ S = Q/T

• SO Efficiency = (TH – TC)/ TH

• = 1 – TC/TH

• Atoms don’t care.

• What happens most ways happens most often

Specific Thermal Capacity

Energy required to raise 1kg of a material by 1K

Symbol = c Unit = J kg-1 K-1

Energy and solids (& liquids)

Supplying energy to a material causes an increase in temperature

E = mc

E = Energy needed to change temperature of substance / J

m = Mass of substance / kgc = Specific thermal capacity of substance

/ J kg-1 K-1

= Change in temperature / K

Energy gained by an electron when accelerated by a 1V potential difference

E = 1.6 x 10-19 x 1 = 1.6 x 10-19J = 1eV

From E = qV

Energy Units

From E = NAkT

Energy of 1 mole’s worth of particles

kJ mol-1

Latent Heat

Extra energy required to change phase

Solid liquid

Latent Heat of vaporisation Liquid gas

At a phase boundary there is no change in temperature - energy used just to break bonds

Latent Heat of fusion

SLHV - waterCalculate

1) Number of molecules of water lost

2) Energy used per molecule to evaporate

3) Energy used to vaporise 1kg of water

mass evaporatedmolar mass

NA

energy usedno of molecules evaporated

1kgmolar mass

NA Energy to vaporise one molecule

NA = 6.02 x 1023

Molar masswater = 18g