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