atm litres k n t c - eclecticon gases and heat engines.pdf · means a real gas is not infinitely...
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Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 1
Ideal Gas Equation
pV nRT1
constant
pV
pV
Boyles’s Law. At constant
temperature, gas pressure is
inversely proportional to volume.
constant
V T
V
T
Charles’ Law. At constant
pressure, gas volume is
proportional to temperature.
p
V
n
R
T
Pressure in Pascals (Pa)
Volume / cubic metres
Number of moles of gas
Molar gas constant
8.314 Jmol-1K-1
Temperature /Kelvin
Ideal Gases and Heat Engines
The operation, and theoretical efficiency, of combustion driven piston engines (e.g. diesel or petrol fuelled) can
be analysed by considering charges to pressure, temperature and volume of the gaseous components. This
proceeds by accounting for the energy changes in the gas as a result of heat applied and work done
combined with the ideal gas equation, which relates the physical properties of the gas.
An ideal gas assumes a large number of point
particles colliding elastically. It neglects any
short-range intermolecular forces resulting from
repulsion or attraction due to molecular
charges, and the fact that molecules have a
finite volume i.e. are not infinitely small! This
means a real gas is not infinitely compressible
whereas an ideal gas has no such limits.
Special cases of the ideal gas equation:
Jacques Charles
1746-1823
Robert Boyle
1627-1691
Temperature scales
273C
T T
9
532
FT T
/T K
-40o F = -40o C
Fahrenheit is a temperature scale, where 32o F is
the freezing point of water and 212o F is the boiling
point of water, defined at sea level at standard
atmospheric pressure (101,325Pa).
It was proposed in 1724 by Daniel Gabriel Fahrenheit.
0o F corresponded to the lowest temperature he could
cool brine (salt water) and 100o F was the average
human body temperature (37oC).
A more popular scale
is the Celsius scale,
with 0o C and 100o C
representing the
freezing and boiling
points of water at
standard atmospheric
pressure.
The Kelvin temperature scale (or
“absolute” scale) is proportional to the
mean kinetic energy of molecules.
1
2U nRT
Internal energy of n
moles of gas
Number of degrees of freedom of
molecular motion (e.g = 3 for
x,y,z translation)
Anders Celsius
1701-1744
Daniel Fahrenheit
1686-1736
1 mol = 6.02 x 1023 molecules. So energy of a molecule is
100o C is
equivalent to
212o F.
At 1atm = 101,325Pa, one mole of gas at
200C = 293K has volume
V = 2.40 x10-2 m3 = 24 litres
11000
o
101325 8.314atm litres K
/ 273/ atm
12.187 / litres
p V Tn
n T Cp
V
Ideal Gas Equation in practical units
William Thompson (Lord Kelvin)
1824-1907
o/ 273/ atm
12.187 / litres
n T Cp
V
o500T C
o20T C/ litresV
/ atmp
1
2 Bu nk T
Boltzmann's constant 23 23 1/ 6.02 10 1.38 10 JK
Bk R
Heat, work and internal energy
of an ideal gas
12
12
0
V
V
dV
dQ dU
C dT nRdT
C nR
V
V
dQC
dT
dQ C dT
Constant volume process
(isochoric)
1
2U nRT
1
2
pV nRT
U pV
12
1
2
U nRT
dU nRdT
dx
A
F
Consider a cylinder of gas
being compressed by a force F.
The work done on the gas by the force is:
x
dW Fdx
The pressure acting upon the gas is:
Fp
A
and the volume change is:
dV Adx
dVdW pA
A
dW pdV
If heat dQ is supplied to the gas
then the First Law of
Thermodynamics (that Energy in a
closed system is conserved) means
the internal energy change is
dU dQ pdV
dQ dU pdV
The internal energy for n moles of
an ideal gas is
The Ideal Gas Equation is
First Law
First Law
i.e. no work done on gas
Constant volume heat capacity for n moles
So for a constant volume
change, the heat capacity
is a constant for an ideal gas.
12V
Rc
M
Constant volume specific heat capacity.
M is the molar volume /kg
Gas M /gmol-1
Acetylene 26.04
Air 28.966
Ammonia 17.02
Argon 39.948
Benzene 78.11
Butane 58.12
Carbon dioxide 44.01
Carbon
Monoxide
28.011
Chlorine 70.906
Ethyl Alcohol 46.07
Fluorine 37.996
Helium 4.003
Hydrogen
Chloride
36.461
Hydrogen
Sulphide
34.076
Krypton 83.80
Methane 16.044
Natural Gas 19.00
Nitrogen 28.0134
Neon 20.179
Oxygen 31.9988
Ozone 47.998
Propane 44.097
Sulphur dioxide 64.06
Toluene 92.13
Xenon 131.30
Water vapour 18.02
http://www.engineeringtoolbox.com/molecular-weight-gas-vapor-d_1156.html
VQ mc T
Total amount of heat supplied to m kg of
gas is therefore:
Constant pressure process
(isobaric)
P
P
dQC
dT
dQ C dT
12
1
2
U nRT
dU nRdT
pV nRT
pdV nRdT
Since constantp
12
12
P
P
dQ dU pdV
C dT nRdT nRdT
C nR nR
12V
P V
C nR
C C nR
Experimentally it is very hard to maintain a
constant volume as heat is added, so
constant volume heat capacity is difficult to
measure directly.
However, constant pressure heat capacity
is much easier to measure, as one can
allow volumes to change in order to
maintain equilibrium with the ambient
pressure.
The Mayer relationship is therefore very
useful in working out the constant volume
heat capacity from the constant pressure
heat capacity.
Constant pressure heat capacity for n
moles of ideal gas
First Law
Ideal Gas Equation
Mayer Relationship
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 2
Constant temperature
(isothermal) process
Adiabatic (or isentropic) process
i.e. no heat added. Work done
is the sole cause of changes in
internal energy
0
0
0
constant
1
ln
V
V
dU
T
dQ pdV
nRTdQ dV
V
Q nRT dVV
VQ nRT
V
2
2
12
12
1
2
1 12 2
1 12 2
1 1
2 2
2
2
2
1
1
( )
0
1
1
1
1 ln ln
ln
constant
U nRT
pV nRT
U pV
dU d pV
dU Vdp pdV
dQ
dU pdV
dU pdV
Vdp pdV pdV
pdV Vdp
pdV Vdp
dV dp
V p
V p k
pV k
pV
1
2
12
2
1
1
1
P V
P
V V
V
P
V
P P
V V
C C nR
C nR
C C
C nR
C nR
C nR
C c
C c
0 0
constantpV
pV p V
Work done on gas
0ln
dW pdV
W pdV
W dQ
W Q
VW nRT
V
Work done on gas for an adiabatic change
00 0
1 10 0
0
1
0 0
0
2
10 02
0
2
010 02
1
11
1
1
V
V
dW pdV
W pdV
W p V V dV
p VW V V
p V VW
V
VW p V
V
VW p V
V
First Law
Using the Ideal Gas Equation
Heat supplied to gas
The Mayer Relationship and
2 21 1P
V
C
C
0 0
0 0
pV p V
p p V V
This defines an adiabatic change
Heat supplied and work done for an
isobaric process
P
p
dQ C dT
Q mc T
12
12
( 1)
VP
p
p
C nRC nR nRc
nM nM nM
Rc
M
0
0
V
VW pdV
W p V V
Since p is constant
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 3
1
0 0 011
p V VW
V
So work done by gas on the
surroundings is:
P V
VP
P V
C C nR
CC nR
Mn Mn Mn
Rc c
M
1
1
1
1
V V
V
V
P
Rc c
M
Rc
M
Rc
M
Rc
M
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 4
Heat Engines
Carnot Cycle
Positions 1 to 2
Isothermal expansion of an ideal gas at
the hot reservoir temperature. Since gas
temperature, and therefore internal
energy is constant, the work done by the
gas on the surroundings must exactly
equate to the heat absorbed by the gas.
Positions 2 to 3
Isentropic (i.e. adiabatic or ‘no heat
added or lost’) expansion of the gas. The
work done by the gas on the
surroundings is powered by the loss of
internal energy of the gas as it cools
from the temperature of the hot reservoir
to the temperature of the cold reservoir.
Positions 3 to 4
Isothermal compression of the gas. In
order for the temperature, and hence the
internal energy, to remain constant, the
heat lost by the gas to the cold reservoir
must equate to the work done on it by
the surroundings.
Positions 4 to 1
Isentropic compression of the gas,
heating it from the temperature of the
cold reservoir to the temperature of the
cold reservoir.
1
0 0 0
constant
11
pV
p V VW
V
0
lnV
Q nRTV
1
2
pV nRT
U nRT
constantpV
P
V
c
c
Assume we have n moles of ideal gas, none of which
are lost in the process. Input parameters are:
H
C
T
T
Hot reservoir temperature (Kelvin)
Cold reservoir temperature (Kelvin)
Volume of gas at position 1 in the cycle
Volume of gas at position 2 in the cycle
Volume of gas at position 3 in the cycle
1 1 HpV nRT
2 2 Hp V nRT
3 3 Cp V nRT
4 4 Cp V nRT
Between positions 1 and 2
2
in
1
2
1,2
1
ln
ln
H
H
VQ nRT
V
VW nRT
V
1 1 H
H
pV p V nRT
nRTp
V
Between positions 2 and 3
2
2 2 2
1
2 2 2
2,3
3
11
VpV p V p p
V
p V VW
V
Between positions 4 and 1
4
4 4 4
1
4 4 4
4,1
1
11
VpV p V p p
V
p V VW
V
Between positions 3 and 4
3
out
4
3
3,4
4
ln
ln
C
C
VQ nRT
V
VW nRT
V
3 3 C
C
pV p V nRT
nRTp
V
1
2
V V
V V
Volume of gas at position 4 in the cycle
3
4
V V
V V
To complete the cycle
4 4
1 4
1 1 4 1
1
11 1
4 1 4 1
CH
H H
C C
nRTV nRT Vp p
V V V V
T TV V V V
T T
2 3
2 3
2
3 2
3
2
3 2 3
1
11 1
3 2 3 2
, CH
C H
H H
C C
nRTnRTp p
V V
Vp p
V
nRT nRT V
V V V
T TV V V V
T T
Ideal gas equation Mayer relation
Nicolas Léonard
Sadi Carnot
(1796-1832)
Work done by the gas on the
surroundings is the area
enclosed by the cycle
W pdV
1( ,1)p
4
4
1
,V
pV
3
3
1
,V
pV
2
2
1
,V
pV
inQ
outQ
Work done by the ideal
gas on the surroundings
These two volumes are derived from the other inputs:
1
2
3
4
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 5
Carnot engine cont....
Total work done is:
1,2 2,3 3,4 4,1
1 1
32 2 2 2 4 4 4
1 3 4 1
11
12 2
1 1
ln 1 ln 11 1
ln 1 ln1
H C
CH
H C
H
W pdV W W W W
VV p V V p V VW nRT nRT
V V V V
TV nRT VW nRT nRT
V T V
11
1
2 2
1 1
2
1
2
1
11
ln 1 ln 11 1
ln1 1 1 1
ln
C H
C
C CH H
H C
H C
C CH H
H C
H C
nRT T
T
T nRTV nRT V TW nRT nRT
V T V T
nRT nRTV nRT nRTW nR T T
V
VW nR T T
V
1
1
3 2
H
C
TV V
T
1
1
4 1
H
C
TV V
T
1 1 H
pV nRT 2 2 Hp V nRT
3 3 Cp V nRT 4 4 C
p V nRT 2
in
1
lnH
VQ nRT
V
in
2
1
2
1
ln
ln
1
H C
H
H C
H
C
H
W
Q
VnR T T
V
VnRT
V
T T
T
T
T
1
12
3
C
H
TV
V T
1
14
1
H
C
V T
V T
3 2
4 1
V V
V V
Define heat engine efficiency as the ratio
of Work done by the gas to the heat input
The Carnot Engine efficiency
depends only on the reservoir
temperatures.
The efficiency of a Carnot heat engine can be more simply derived by
consideration of Entropy S. This is a measure of disorder in a substance.
The Second Law of Thermodynamics states for any change, the total
amount of Entropy in the Universe must increase.
If heat is added in a reversible process: dQ
dST
For the Carnot cycle, the isentropic stages have no heat change hence they are at constant
Entropy. (Note this applies to the ideal gas, the surroundings will change in entropy due to the
exchange of work with the ideal gas). We cannot create entropy in the cycle for the gas, as the
cycle returns to the original state (p1,V1, TH), and Entropy is a scalar function of state (i.e. like
potential energy – the path does not matter). The (S,T) curve for the cycle is therefore a rectangle.
In the isothermal stages, temperature is a constant, so in both cases outin 2
1
lnH C
QQ VS nR
T T V
From the First Law of Thermodynamics
dU dQ pdV dU TdS pdV
Over the whole cycle the internal energy doesn’t
change, so the work done by the gas is:
2
1
2
1
in 2
1
ln
ln
1
ln
H C H C
H C
C
H
H
W pdV Tds
VW T T S T T nR
V
VT T nR
V TW
Q TVnRT
V
Since the temperature
range cannot go beyond
the range of reservoir
temperatures, the Carnot
cycle represents the
most efficient way of
extracting work given an
amount of heat input.*
Any other process would
occupy less area in the
S,T diagram.
1 2
34
*There is a better justification on the next page!
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 6
General notes regarding the Second Law of Thermodynamics, and the maximum possible efficiency of heat engines
It is possible to bound the efficiency of any heat engine using (i) the law of conservation of energy and (ii) the Second Law of Thermodynamics. That the upper bound of
efficiency equates with the efficiency of the Carnot engine is perhaps an even stronger justification of the statement that a Carnot engine is the most efficient scheme
possible, if indeed it could be practically realized.
Any heat engine is essentially flow of heat from a hot reservoir to a colder one. By a reservoir we mean a thermal mass that is so large that it will not change
temperature when the heat we associate with our engine is taken from or added to it. The difference in heat taken from the hot reservoir, and the heat transferred to the
cold reservoir, is the maximum possible work W done by the engine. This must be true to satisfy the law of energy conservation, or the First Law of Thermodynamics.
Cold reservoir
Heat Engine
Hot reservoir
in outQ Q W
inQ
outQ
W
HT
CT
First Law:
The Second Law of Thermodynamics states that for every change there can never be an overall decrease
in Entropy. For our idealized system, this means the loss of entropy of the hot reservoir must at least be
compensated for by the gain in entropy of the cold reservoir.
0
0
in
H
H
out
C
C
in out
total H C
H C
total
in out
H C
QS
T
QS
T
Q QS S S
T T
S
Q Q
T T
Second Law:
Entropy changes of
hot and cold reservoirs
Combining with the
First Law expression:
0
0
11
0
1 0
in out
H C
out in
in in
H C
in
H C
C
H in
Q Q
T T
Q Q W
Q Q W
T T
W
Q
T T
T W
T Q
Define engine efficiency:
in
1 0
1
C
H
C
H
W
Q
T
T
T
T
So since the Carnot engine has
efficiency
1 C
H
T
T
this is as efficient as
thermodynamics allows, so the
Carnot cycle is (one example*) of
the most efficient heat engine
possible.
*A Brayton Engine (adiabatic compression, isobaric heating, adiabatic expansion, isobaric cooling) has a
similar theoretical efficiency as a Carnot Cycle.
Notes on reversibility
A reversible heat engine is one which is assumed to operate at thermodynamic
equilibrium at all times. The ideal gas equations, and associated relationships, hold
and there are no losses due to friction etc. In other words, the differential form of the
First Law holds at all times; i.e. where changes dU in internal energy are fully accounted
for by heat change dQ and work done dW = -pdV. This means that there is no net
internal energy and indeed entropy change over the complete cycle. This means the
‘ideal’ cycle could be run in reverse without breaking the Second Law, since a zero net
entropy change is permitted.
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 7
Heat Engines - Rectangular p,V cycle
Assume we have n moles of ideal gas, none of which
are lost in the process. Input parameters are:
1p Pressure at position 1 in the cycle
Pressure at position 3 in the cycle
Volume of gas at position 1,4 in the cycle
Volume of gas at position 2,3 in the cycle
Between positions 1 and 2
Between positions 2 and 3
1
2
V V
V V
2
1
1
1 1
1 1 1
1
1,2 2 1
1,2 1 2 1
2
1,2
1
( )
ln
p
T
p pT
p p
p Vp V nRT n
RT
Q nMc T T
W p V V
TdTS nMc nMc
T T
3
2
2
1 2
1 2 2 2
2,3 2 3
2,3
3
2,3
2
0
ln
V
T
V VT
V V
p Vp V nRT T
nR
Q nMc T T
W
TdTS nMc nMc
T T
Between positions 4 and 1
4
3
3
3 2
3 2 3 3
3,4 3 4
3,4 3 2 1
4
3,4
3
( )
ln
p
T
p pT
p p
p Vp V nRT T
nR
Q nMc T T
W p V V
TdTS nMc nMc
T T
1
4
1
3 1
3 1 4 4
4,1 1 4
4,1
1
4,1
4
0
ln
V
T
V VT
V V
p Vp V nRT T
nR
Q nMc T T
W
TdTS nMc nMc
T T
Between positions 3 and 4
Heat input to gas
Work done by gas
Heat output from gas
Work done by gas
Heat input to gas
Work done by gas
Heat output from gas
Work done by gas
Net heat input
in 1,2 4,1
in 2 1 1 4
in 1 2 1 1 1 1 3 1
in 1 2 1 1 1 3
1
in 1 2 1 1 3
( ) ( )
( ) ( )
( ) ( )
( ) ( )1 1
p V
p V
p V
Q Q Q
Q nMc T T nMc T T
nM nMQ c p V p V c p V p V
nR nR
MQ p c V V V c p p
R
VQ p V V p p
Net work done by gas
2 2
1 1
T V
T V
3 3
2 1
T p
T p
4 1
3 2
T V
T V
1 1
4 3
T p
T p
1,2 2,3 3,4 4,1
1 2 1 3 2 1
1 3 2 1
( - ) - ( - )
- -
W W W W W
W p V V p V V
W p p V V
21p
V
c
c
Engine efficiency
1 3 2 1
1in1 2 1 1 3
1
1 1
1 3 2 1
- -
( ) ( )1 1
1
1 - -
p p V VW
VQp V V p p
p V
p p V V
12V
Rc
M
1T Gas temperature at position 1 in the cycle
3p
1 1,p V
1 2,p V
3 2,p V 3 1
,p V
Note Carnot efficiency
for this heat engine
would be
13.5 2731 75%
873 273
1 1( , )p V
1 2( , )p V
3 2( , )p V
3 1( , )p V
1 2
34
1
2
3
4
1
1
1
V
P
Rc
M
Rc
M
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 8
Heat Engines – The Otto cycle
The Otto Cycle is the basis of spark-ignition
piston engine, which is essentially how a typical
petrol driven engine operates.
Positions 0-1:
Air is drawn into piston/cylinder arrangement at
constant pressure.
Process 1–2
Adiabatic (isentropic) compression of the air via a
piston.
Process 2–3
Constant-volume heat transfer to the working gas from
an external source while the piston is at maximum
compression. This process is intended to represent the
ignition of the fuel-air mixture and the subsequent rapid
burning.
Process 3–4
Adiabatic (isentropic) expansion (power stroke).
Process 4–1
Constant-volume process in which heat is rejected from
the air while the piston is at maximum expansion.
Process 1–0
Air is released to the atmosphere at constant pressure.
Assume we have n moles of ideal gas, none of which
are lost in the process. Input parameters are:
1p Pressure of gas at position 1 in the cycle
Volume of gas at position 1 in the cycle
Volume of gas at position 2 in the cycle
Between positions 1 and 2 Between positions 4 and 1
Between positions 3 and 4
Between positions 2 and 3
1
2
V
V
To complete the cycle
Nikolaus Otto
(1832-1891)
Work done by the
gas on the
surroundings is the
area enclosed by
the cycle W pdV
1 1( , )p V
inQ
outQ
1 1
1 1 1
1
1
1 1 1
1
1 1 1
1,2
2
1,2 1,2
11
0, 0
p Vp V nRT n
RT
VpV p V p p
V
p V VW
V
Q S
3
2
2
2 2
2 2 2 2
3 2
3 2 3 3
2,3 3 2
2,3
3
2,3
2
0
ln
V
T
V VT
V V
p Vp V nRT T
nR
p Vp V nRT T
nR
Q nMc T T
W
TdTS nMc nMc
T T
4 1
4 1 4 4
2
3 2 3
1
3 2 2
3,4
1
3,4 3,4
11
0, 0
p Vp V nRT T
nR
VpV p V p p
V
p V VW
V
Q S
1
4
1
4,1 4 1
4,1
1
4,1
4
0
ln
V
T
V VT
V V
Q nMc T T
W
TdTS nMc nMc
T T
2 2 1 1
1
2 1
2
p V p V
Vp p
V
3 2 4 1
2
4 3
1
p V p V
Vp p
V
3p Pressure of gas at position 3 in the cycle
1T Temperature of gas at position 1
in the cycle
-ve since work
is being done
on the gas
2 2( , )p V
3 2( , )p V
4 1( , )p V
1
2
3
4
Heat input to gas
via spark ignition
Work done
by gas
Heat output
via exhaust
in 1,2 3 2
out 4,1 4 1
V
V
Q Q nMc T T
Q Q nMc T T
21p
V
c
c
12V
Rc
M
1
1
1
V
P
Rc
M
Rc
M
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 9
Otto engine cont....
Total work done is:
in
2
3 11
2 3 1
1
11
1
1
11
W
Q
Vp p r
r
V p p r
r
Define heat engine efficiency as the ratio
of Work done by the gas to the heat input
So the Otto Engine efficiency
depends only on the compression
ratio, and the ratio of specific heats
1,2 3,4
1 1
3 21 1 1 2
2 1
12
1 3 1
1
12
1 31 1
2
31
1 11 1
11 1
1
11 1
1 1
11
1
W pdV
W W W
p Vp V V VW
V V
VW p r r p
r
V rW p r r p
r r
VW p
r
1p r
Work done by the
gas on the
surroundings is the
area enclosed by
the cycle W pdV
1 1( , )p V
inQ
outQ2 2
( , )p V
3 2( , )p V
4 1( , )p V
1
2
3
4
Define the
compression ratio
1
2
Vr
V
in 3 2
1
in 3 2 1 2
2
2 3 1
in
1
1
1
1
1
V
V
V
V
Q nMc T T
nMc VQ p V p V
nR V
Rc
M
Mc
R
V p p rQ
3 2
3
p VT
nR2 2
2
p VT
nR
1
2 1
2
Vp p
V
From the previous page:
1
2
3
4
21p
V
c
c
12V
Rc
M
1
1
1
V
P
Rc
M
Rc
M
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 10
Heat Engines – The Diesel cycle
The Diesel Cycle is the basis of a diesel engine, which is
ubiquitous in transport applications. Unlike petrol-driven
engines, diesel variants are more suited to heavy machinery.
They can be found powering most ships as well as trucks,
buses and cars.
Positions 0-1:
Air is drawn into piston/cylinder arrangement at constant
pressure.
Process 1–2
Adiabatic (isentropic) compression of the air via a piston.
Process 2–3
Constant-pressure (isobaric) heat transfer to the working gas
from an external source while the piston is at maximum
compression. This process is intended to represent the
ignition of the fuel-air mixture and the subsequent rapid
burning. This is different from the Otto cycle, which is constant
volume (isochoric) heating during this stage. In the Diesel
cycle, the heat generated from air compression is sufficient to
ignite introduced fuel vapours. In the Otto cycle a spark plug is
used instead to ignite the fuel.
Process 3–4
Adiabatic (isentropic) expansion (power stroke).
Process 4–1
Constant-volume process in which heat is rejected from the air
while the piston is at maximum expansion.
Process 1–0
Air is released to the atmosphere at constant pressure.
Assume we have n moles of ideal gas, none of which
are lost in the process. Input parameters are:
1p Pressure of gas at position 1 in the cycle
Volume of gas at position 1 in the cycle
Volume of gas at position 2 in the cycle
Between positions 1 and 2 Between positions 4 and 1
Between positions 3 and 4
Between positions 2 and 3
1
2
V
V
To complete the cycle
Work done by the gas
on the surroundings is
the area enclosed by
the cycle W pdV
1 1( , )p V
inQ
outQ
1 1
1 1 1
1
1
1 1 1
1
1 1 1
1,2
2
1,2 1,2
11
0, 0
p Vp V nRT n
RT
VpV p V p p
V
p V VW
V
Q S
3
2
2
2 2
2 2 2 2
2 3
2 3 3 3
2,3 3 2
2,3 2 3 2
3
2,3
2
ln
P
T
P PT
p p
p Vp V nRT T
nR
p Vp V nRT T
nR
Q nMc T T
W p V V
TdTS nMc nMc
T T
4 1
4 1 4 4
3
2 3 2
1
2 3 3
3,4
1
3,4 3,4
11
0, 0
p Vp V nRT T
nR
VpV p V p p
V
p V VW
V
Q S
1
4
1
4,1 4 1
4,1
1
4,1
4
0
ln
V
T
V VT
V V
Q nMc T T
W
TdTS nMc nMc
T T
2 2 1 1
1
2 1
2
p V p V
Vp p
V
2 3 4 1
3
4 2
1
p V p V
Vp p
V
1T Temperature of gas at position 1
in the cycle
-ve since work
is being done
on the gas
2 2( , )p V 2 3
( , )p V
4 1( , )p V
1
2 3
4
Heat input to
gas during
combustion
Work done
by gas
Heat output
via exhaust
in 1,2 3 2
out 4,1 4 1
P
V
Q Q nMc T T
Q Q nMc T T
Rudolf Diesel
(1858-1913)
3V Volume of gas at position 3 in the cycle
21p
V
c
c
12V
Rc
M
1
1
1
V
P
Rc
M
Rc
M
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 11
Diesel engine cont....
Total work done is:
in
1 2
1 2
1
(1 ) 11
11
(1 ) 1
1
11
1
1 11
1
W
Q
pVr s r s
r s pV
r s r s
r s
r s
r s
s
r s
Define heat engine efficiency
as the ratio of Work done by the
gas to the heat input
1,2 2,3 3,4
1 1
2 3 31 1 1
2 3 2
2 1
1
11 2 1 2
1 2
1
11 2
1 11 1
1 1 11 1
1 1 1 11
W pdV
W W W W
p V Vp V VW p V V
V V
p V p r V sW r r p r V s s
r
p V sW r r r s r s
r
1 2
1 2
1 2
1 11
(1 ) 1 1 ( 1)1
(1 ) 11
p VW r r r s r s s r
p VW r s r s r s
p VW r s r s
Define the
compression ratios
1
2
Vr
V
in 3 2
1
in 1 3 2
2
2 1 12 2
12
in 1 2
1
1
11
P
P
p
V P V
V
P
Q nMc T T
nMc VQ p V V
nR V
c R Rc c c
c M M
Mc
R
Q r s p V
2 3
3
p VT
nR 2 2
2
p VT
nR
1
2 1 1
2
Vp p p r
V
From the previous page:
3
2
Vs
V
Work done by the gas
on the surroundings is
the area enclosed by
the cycle W pdV
1 1( , )p V
inQ
outQ
2 2( , )p V 2 3
( , )p V
4 1( , )p V
1
2 3
4
The Diesel engine is typically
more efficient than a petrol (Otto)
engine since the former
works on the basis of self ignition
due to high compression. This
‘knocking’ is undesirable for petrol
engines, so a lower r value is
required.
1 3 2V V V
r s
1
2
3
4
Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 12
in
diesel 1
otto 1
1 11
1
11
W
Q
s
r s
r
Work done by the gas
on the surroundings is
the area enclosed by
the cycle W pdV
1 1( , )p V
inQ
outQ
2 2( , )p V
2 3( , )p V
4 1( , )p V
1
2 3
4
Work done by the
gas on the
surroundings is the
area enclosed by
the cycle W pdV
1 1( , )p V
inQ
outQ2 2
( , )p V
3 2( , )p V
4 1( , )p V
1
2
3
4
Comparing Otto and Diesel heat engines
1
2
Vr
V 3
2
Vs
V
1 3 2V V V
r s
1 2 (1 ) 11
pVW r s r s
in 1 21
1Q r s pV
2
3 11
11
1
VW p p r
r
2 3 1
in1
V p p rQ
Diesel
Otto
21p
V
c
c
12V
Rc
M
1
1
1
V
P
Rc
M
Rc
M