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# Heat Engines - PowerPoint PPT Presentation

Heat Engines. Coal fired steam engines. Petrol engines Diesel engines Jet engines Power station turbines. DECC http://www.decc.gov.uk/assets/decc/statistics/publications/flow/193-energy-flow-chart-2009.pdf. Combined Cycle. THE LAWS OF THERMODYNAMICS

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• 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

1. You cannot win you can only break even.

2. You can only break even at absolute zero.

3. You can never achieve absolute zero.

p

p  1/V

1/V

V

V  T

T

p

p  T

T

p

p  N

Number of molecules, N

(constant temperature)

p

p

Isochors

(constant volume)

1/V

T

Isobars (constant pressure)

V

T

= constant

T

For ideal gases only

A gas that obeys Boyles law

In summary…

p  1/V

p  T

V  T

Most gases approximate ideal behaviour

Ideal gases assume:-

• No intermolecular forces

Not true - gases form liquids then solids as temperature decreases

• Volume of molecules is negligible

Not true - do have a size

= constant

T

p2V2

p1V1

T1

T2

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

=

pV = nRT

p = pressure, Pa

V = volume, m3

n = number of moles

R = Molar Gas constant (8.31 J K-1 mol-1 )

T = temperature, K

Macroscopic model of gases

pV = NkT

N = number of molecules

k = Boltzmann’s constant (1.38 x 10-23 J K-1)

v

y

x

Kinetic Theory

First there was a box and one molecule…

Molecule:- mass = m

velocity = v

-2mv

-v

v

mv - mu

pmol

Molecule

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

t

Molecule hits side of box…(elastic collision)

= -mv - mv = -2mv

pbox = -pmol = 2mv

Box

y

x

s

2x

=

=

v

v

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

Time between hitting same side, t

p

= p v

= 2mv v

= mv2

t

2x

2x

Force exerted on box

x

Time

Actual force during collision

Average force, exerted by 1 molecule on box

Average Force

-v6

z

v2

v5

v1

-v8

vN

v3

v4

-v7

y

x

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

Mean square velocity

x

Nmv2

Nmv2

Pressure = Force

=

=

Area

xyz

V

Nmv2

1

p=

V

3

Force created by N molecules hitting the box…

But, molecules move in 3D

Kinetic Theory equation

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

Nmv2

pV=

3

pV

= NkT

Nmv2

=

NkT

1

3

Microscopic

Macroscopic

(In terms of molecules)

(In terms of physical observations)

mv2

=3kT

(1)

Already commented that looks a bit like K.E.

K.E. = ½mv2

3

K.E. =

kT

2

Average K.E. of one molecule

Rearrange (and remove N)

Substitute into (1)

K.E.Total =

NkT

2

3

NkT

U =

2

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

This is translational energy only

- not rotational, or vibrational

And generally referred to as internal energy, U

NkT

U =

2

Physically hit molecules

Energy and gases

Internal 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

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

+ve when energy added

-ve when energy removed

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

i.e. change in volume

+ve when work done on gas - compression

-ve when work done by gas - expansion

Bonds between atoms a temperature difference

Jiggling around

(vibrational energy)

Einstein’s Model of a solid

Atom requires energy to break them

U  kT

Mechanical properties change with temperature a temperature difference

T = high

can break and make bonds quickly – atoms slide easily over each other

Liquid: less viscous

Solid: more ductile

T = low

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

Solid: more brittle

Liquid: more viscous

Activation energy, a temperature difference 

Can think of bonds as potential wells in which atoms live

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

The magic a temperature difference /kT ratio

 - energy needed to do something

kT - average energy of a molecule

/kT = 1

Already happened

/kT = 10 - 30

Probably will happen

/kT > 100

Won’t happen

Probability a temperature difference

1

Exponential

0

Energy

Probability of molecule having a specific energy

e a temperature difference -/kT

/kT

Boltzmann Factor

e-/kT

Probability of molecules achieving an event characterised by activation energy, 

1

0.37

10 - 30

4.5 x 10-6 - 9.36 x 10-14

3.7 x 10-44

> 100

Nb. 109 to 1013 opportunities per second to gain energy

Amongst particles a temperature difference

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

2 a temperature difference nd law of thermodynamics

S = k ln W

S = entropy

k = Boltzmann’s constant

W = number of ways

Δ a temperature difference S = ΔQ

T

At a thermal boundary a temperature difference

Energy will go from hot to cold

Hot

Number of ways decreases – a bit

Cold

Number of ways increases – significantly

Result - entropy increase

• Efficiency = W/Q a temperature difference H = (QH – QC ) / QH

• BUT Δ S = Q/T

• SOEfficiency =(TH – TC)/ TH

• = 1 – TC/TH

Symbol = c a temperature difference

Unit = J kg-1 K-1

Energy and solids (& liquids)

Supplying energy to a material causes an increase in temperature

Specific Thermal Capacity

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

E = mc a temperature difference 

E = Energy needed to change temperature of substance / J

m = Mass of substance / kg

c = Specific thermal capacity of substance

/ J kg-1 K-1

 = Change in temperature / K

Energy Units a temperature difference

From E = qV

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 = NAkT

Energy of 1 mole’s worth of particles

kJ mol-1

Latent Heat a temperature difference

Extra energy required to change phase

Solid liquid

Latent Heat of fusion

Latent Heat of vaporisation

Liquid gas

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

N a temperature difference A = 6.02 x 1023

Molar masswater = 18g

mass evaporated

NA

molar mass

energy used

no of molecules evaporated

1kg

NA Energy to vaporise one molecule

molar mass

SLHV - water

Calculate

1) Number of molecules of water lost

2) Energy used per molecule to evaporate

3) Energy used to vaporise 1kg of water