Ideal Gases and Heat Engines Ideal Gas Equation in practical units The operation, and theoretical efficiency, of combustion driven piston engines (e.g. diesel or petrol fuelled) can p  V1   T  be analysed by considering charges to pressure, temperature and volume of the gaseous components. This 101325  1000   8.314     n atm litres K 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. n T/o C 273  p / atm  12.187V / litres Ideal Gas Equation An ideal gas assumes a large number of point particles colliding elastically. It neglects any pV nRT short-range intermolecular forces resulting from Special cases of the ideal gas equation: 1 repulsion or attraction due to molecular Boyles’s Law. At constant p  p Pressure in Pascals (Pa) R Molar gas constant charges, and the fact that molecules have a temperature, gas pressure is V 8.314 Jmol-1K-1 finite volume i.e. are not infinitely small! This inversely proportional to volume. pV constant V Volume / cubic metres means a real gas is not infinitely compressible T Temperature /Kelvin n Number of moles of gas whereas an ideal gas has no such limits. Charles’ Law. At constant VT pressure, gas volume is proportional to temperature. V At 1atm = 101,325Pa, one mole of gas at constant Temperature scales T 200C = 293K has volume p / atm V = 2.40 x10-2 m3 = 24 litres TT9 32 FC5 n T/o C 273 100o C is The Kelvin temperature scale (or  p / atm  equivalent to TTC 273 “absolute” scale) is proportional to the 12.187V / litres 212o F. mean kinetic energy of molecules.

1 U 2  nRT

Internal energy of n Number of degrees of freedom of moles of gas molecular motion (e.g  = 3 for TC 500o -40o F = -40o C x,y,z translation)

o 1 mol = 6.02 x 1023 molecules. So energy of a molecule is u  1  nk T TC 20 2 B V / litres Boltzmann's constant 23 23 1 kRB / 6.02  10  1.38  10 JK

Fahrenheit is a temperature scale, where 32o F is A more popular scale the freezing point of water and 212o F is the boiling is the Celsius scale, point of water, defined at sea level at standard with 0o C and 100o C atmospheric pressure (101,325Pa). representing the It was proposed in 1724 by Daniel Gabriel Fahrenheit. freezing and boiling 0o F corresponded to the lowest temperature he could points of water at o cool brine (salt water) and 100 F was the average standard atmospheric Anders Celsius Daniel Fahrenheit William Thompson Robert Boyle Jacques Charles human body temperature (37oC). pressure. 1701-1744 1686-1736 (Lord Kelvin) 1627-1691 1746-1823 1824-1907 Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 1 Heat, work and internal energy The internal energy for n moles of Gas M /gmol-1 Constant pressure process of an ideal gas an ideal gas is (isobaric) x Acetylene 26.04 U 1  nRT 2 Air 28.966 dQ Constant pressure heat capacity for n CP  moles of ideal gas The Ideal Gas Equation is Ammonia 17.02 dT F dQ C dT pV nRT Argon 39.948 P U1  pV Benzene 78.11 A 2 Butane 58.12 dx Constant volume process Carbon dioxide 44.01 Consider a cylinder of gas (isochoric) Carbon 28.011 Since p  constant being compressed by a force F. Monoxide Constant volume heat capacity for n moles pV nRT Ideal Gas Equation The work done on the gas by the force is: Chlorine 70.906 dQ pdV nRdT dW Fdx C  V dT Ethyl Alcohol 46.07 Fluorine 37.996 The pressure acting upon the gas is: dQ CV dT dQ dU pdV First Law F Helium 4.003 1 p  1 CP dT 2  nRdT  nRdT U 2  nRT A Hydrogen 36.461 C 1  nR  nR dU1  nRdT Chloride P 2 and the volume change is: 2 Hydrogen 34.076 C 1  nR dV Adx dV  0 i.e. no work done on gas Sulphide V 2 dV dQ dU First Law CPV  C  nR Mayer Relationship dW  pA   Krypton 83.80 1 A CV dT2  nRdT Methane 16.044 Experimentally it is very hard to maintain a dW pdV 1 CV 2  nR Natural Gas 19.00 constant volume as heat is added, so So for a constant volume Nitrogen 28.0134 constant volume heat capacity is difficult to If heat dQ is supplied to the gas change, the heat capacity measure directly. then the First Law of is a constant for an ideal gas. Neon 20.179 Thermodynamics (that Energy in a Oxygen 31.9988 However, constant pressure heat capacity closed system is conserved) means 1 R is much easier to measure, as one can c  Ozone 47.998 the internal energy change is V 2 M allow volumes to change in order to Propane 44.097 dU dQ pdV Constant volume specific heat capacity. maintain equilibrium with the ambient M is the molar volume /kg pressure. dQ  dU  pdV First Law Sulphur dioxide 64.06

Total amount of heat supplied to m kg of Toluene 92.13 The Mayer relationship is therefore very gas is therefore: Xenon 131.30 useful in working out the constant volume

Q mcV T heat capacity from the constant pressure Water vapour 18.02 heat capacity. http://www.engineeringtoolbox.com/molecular-weight-gas-vapor-d_1156.html Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 2 Heat supplied and work done for an Adiabatic (or isentropic) process Work done on gas for an adiabatic change isobaric process i.e. no heat added. Work done is the sole cause of changes in dW pdV dQ C dT internal energy P W pdV 1 V  Q  mcp  T U  nRT  2 W p V V dV pV p00 V 00V 0  1 pV nRT p p00 V V CP CV  nR 2 nR nR  1  pV cp    U pV 00  11   2 WVV 0  nM nM nM 1  1 1 R(  1) dU2  d() pV c  2  1 p 11  M dU Vdp pdV  pV00 V 22 W  1 1  V0 V  W pdV V dQ  0 This defines an adiabatic change 2  0  V  C W p V V dU   pdV W1  p V 1 P 11 22     0  Since p is constant 2 00  V0 CV dU pdV  11 2 So work done by gas on the 22Vdp  pdV   pdV  The Mayer Relationship and   V0 surroundings is: 11 W1  p V 1  221 pdV   Vdp 2 00  C C nR V  1 PV p V V 12 pdV   Vdp  0 0 0 C nR    W 1  P  1V 1 dV dp  CC 2 VV 1    Vp C 1  nR V 2 Constant temperature Work done on gas 12 lnV   ln p  k C nR    (isothermal) process P dW pdV 1 1 2 C1  nR ln pV  k V 2   dU  0 W pdV Cc 1 2   PP  1  2 pV   constant T  constant Cc  W dQ VV dQ pdV First Law  R  WQ  cc pV  constant nRT VV dQ dV Using the Ideal Gas Equation M  V V0 pV p00 V W nRT ln CPV C nR R  c  1 V 1 V V   Q nRT dV CP CV nR M V  0 V Mn Mn Mn R 1 cV  V R M  1 Q nRT ln Heat supplied to gas cc V PVM R  0 c  P M  1

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 3 nRT nRT Heat Engines Ideal gas equation Mayer relation Between positions 1 and 2 Between positions 2 and 3 H C pp23, VV23 pV nRT cP pV11 nRTH p V nRT   22 H  Carnot Cycle  1 V cV 2 V U 2  nRT Q nRT ln  V2 2 in H  pV p V  p  p pp32  Positions 1 to 2 V 2 2 2  1 V V3 Isothermal expansion of an ideal gas at  1  the hot reservoir temperature. Since gas pV  constant V2  W nRT ln p V V nRTC nRTH  V2 temperature, and therefore internal 1,2 H  2 2 2  V V W2,3 1   energy is constant, the work done by the 1  VVV Q nRT ln   1 V3 3 2 3 gas on the surroundings must exactly  V0 1 equate to the heat absorbed by the gas. pV p11 V nRTH  1 11TTHH nRTH VVVV   Positions 2 to 3 p 3 2 3 2 TTCC Isentropic (i.e. adiabatic or ‘no heat  V  added or lost’) expansion of the gas. The pV  constant work done by the gas on the  1 Between positions 3 and 4 surroundings is powered by the loss of p0 V 0 V 0 W 1  internal energy of the gas as it cools  1V p33 V nRTC from the temperature of the hot reservoir  to the temperature of the cold reservoir. V3 Qout  nRTC ln  1 (p1 ,1) Positions 3 to 4 Work done by the ideal V4 Isothermal compression of the gas. In gas on the surroundings order for the temperature, and hence the V3 W3,4  nRTC ln  internal energy, to remain constant, the V W pdV heat lost by the gas to the cold reservoir 4  must equate to the work done on it by Work done by the gas on the pV p33 V nRTC the surroundings. Qin surroundings is the area enclosed by the cycle nRTC Positions 4 to 1 p 4 V Isentropic compression of the gas, V 2 V heating it from the temperature of the 4 p , 2 p4 , 2 cold reservoir to the temperature of the Between positions 4 and 1 V V1 1 cold reservoir. p V nRT V 44 C Qout 3  p3 , Assume we have n moles of ideal gas, none of which V V pV p V  p  p 4 1 are lost in the process. Input parameters are: 4 4 4  3 V TH Hot reservoir temperature (Kelvin)  1 T Cold reservoir temperature (Kelvin) p V V C 4 4 4 W4,1 1   Volume of gas at position 1 in the cycle  1 V1 VV 1   Volume of gas at position 2 in the cycle VV 2 V44  nRTH nRTC  V  pp14      VVVV These two volumes are derived from the other inputs: 1  1 4  1  1 Volume of gas at position 3 in the cycle Nicolas Léonard VV 3  1 11TTHH Sadi Carnot VVVV4 1  4   1 To complete the cycle VV 4 Volume of gas at position 4 in the cycle (1796-1832) TTCC

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 4 1 1 Carnot engine cont....  1  1   V2 TH TH p V nRT VV VV pV11 nRTH p22 V nRTH p33 V nRTC 44 C Qin  nRTH ln  32 41 V Total work done is: TC TC 1 1 1  1  1 V T VT VV32 W pdV  W  W  W  W 2  C 4  H   1,2 2,3 3,4 4,1 VV VT3 H VT1 C 41  1  1 V2  p 2 V 2 V 2 V3  p 4 V 4  V 4  W nRTHCln   1   nRT ln       1 VVVV11  1  3  4   1  The efficiency of a Carnot heat engine can be more simply derived by  1  1 consideration of Entropy S. This is a measure of disorder in a substance. 1 1  1  1 V22  nRTH  TC   V nRTC TH W nRTHCln   1     nRT ln 1 VTV 1   1 T The Second Law of Thermodynamics states for any change, the total 11  H   C amount of Entropy in the Universe must increase.   dQ V22  nRTHH TCC   V  nRT  T If heat is added in a reversible process: dS  W nRTHCln    1    nRT ln     1 T VTVT11 11  HC    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 V2 nRTHHnRTCC nRT nRT W nR THC  T ln      exchange of work with the ideal gas). We cannot create entropy in the cycle for the gas, as the V1 1   1   1   1 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 curve for the cycle is therefore a rectangle. V2 (S,T) W nR THC T ln    V1  QVinQout  2 In the isothermal stages, temperature is a constant, so in both cases S    nRln  TTVHC 1 Define heat engine efficiency as the ratio of Work done by the gas to the heat input From the First Law of Thermodynamics W   Since the temperature dU dQ  pdV  dU  TdS  pdV 1 2 Qin range cannot go beyond the range of reservoir Over the whole cycle the internal energy doesn’t V2 temperatures, the Carnot change, so the work done by the gas is: nR THC T ln  cycle represents the V1 most efficient way of   W pdV Tds V extracting work given an  2 amount of heat input.* nRTH ln  Any other process would V1 V2 occupy less area in the W THCHC  T  S  T  T nRln  V TTHC S,T diagram. 1   4 3 TH V2 THC T nRln  TC The Carnot Engine efficiency V T  1 W 1 C depends only on the reservoir    1  TH QTV temperatures. in 2 H nRTH ln  V1

*There is a better justification on the next page! Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 5 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.

First Law: QQWin out Hot reservoir T The Second Law of Thermodynamics states that for every change there can never be an overall decrease H 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. Q in Q in Entropy changes of Combining with the Define engine efficiency: SH   T hot and cold reservoirs First Law expression: Heat Engine W H W Q   out QQin out SC    0 Qin Q TC TT out HC T QQ C 10   in out QQWout in SSStotal   H   C    TH TTHC QQWin in     0 TC Cold reservoir S 0 Second Law:  1  total TTHC TH TC QQin out W    0 1 TTHC 1 Q  in  0 So since the Carnot engine has TTHC efficiency Notes on reversibility TWC T A reversible heat engine is one which is assumed to operate at thermodynamic  10    1 C TQH in equilibrium at all times. The ideal gas equations, and associated relationships, hold TH and there are no losses due to friction etc. In other words, the differential form of the this is as efficient as First Law holds at all times; i.e. where changes dU in internal energy are fully accounted thermodynamics allows, so the no net for by heat change dQ and work done dW = -pdV. This means that there is Carnot cycle is (one example*) of internal energy and indeed entropy change over the complete cycle. This means the the most efficient heat engine ‘ideal’ cycle could be run in reverse without breaking the Second Law, since a zero net possible. entropy change is permitted.

*A Brayton Engine (adiabatic compression, isobaric heating, adiabatic expansion, isobaric cooling) has a Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 6 similar theoretical efficiency as a Carnot Cycle. Heat Engines - Rectangular p,V cycle Between positions 1 and 2 Between positions 3 and 4

TV22 pp 1 pp 3  TV pV pV 11 p V nRT  n  11 p V nRT  T  32  pV, (,)pV (,)pV 1 1 1 3 2 3 3 1111 12 RT1 nR Tp 33 1 Heat output from gas  pV12, 2  Heat input to gas Q3,4 nMcp  T 3 T 4  Q1,2 nMcp  T 2 T 1  Tp21

W p() V V Work done by gas W3,4  p 3() V 2  V 1 Work done by gas 1,2 1 2 1 TV 41 T4 dT T T2  4 dT T2 S  nMc  nMc ln TV32 S  nMc  nMc ln 3,4 T pp 1,2 T pp  3 1 TT TT3 1 Tp 11 Tp Between positions 2 and 3 Between positions 4 and 1 43 4 3  pV31,   pV32,  (,)pV VV 2 VV (,)pV31 32 1 pV c 12 pV31 p 2 p1 V 2 nRT 2  T 2  p3 V 1 nRT 4  T 4    1   nR nR cV Q nMc T T Heat input to gas 2,3V  2 3  Heat output from gas Q4,1 nMcV  T 1 T 4  R c  1 Work done by gas Work done by gas V 2 W2,3  0 W4,1  0 M

T3 T T R 1 dT 3 1 dT T1 S  nMc  nMc ln S  nMc  nMc ln cV  2,3 T VV 4,1 VV 2 TT T4 M  1 2 TT4 R  c  Net heat input Net work done by gas P M  1

WWWWW1,2  2,3  3,4  4,1 QQQin 1,2 4,1 W p1(-)-(-) V 2 V 1 p 3 V 2 V 1 Qin nMcpV()() T 2  T 1  nMc T 1  T 4 W p-- p V V nM nM  1 3 2 1  Q cpVpV()()   cpVpV  innRpV 1 2 1 1 nR 1 1 3 1 M Engine efficiency Assume we have n moles of ideal gas, none of which Q p c()() V  V  V c p  p in 1pV 2 1 1 1 3  W  p-- p V V  are lost in the process. Input parameters are: R   1 3 2 1 Q  V1 Gas temperature at position 1 in the cycle  V1 in T1 Q p()() V  V  p  p p1()() V 2 V 1  p 1  p 3 Note Carnot efficiency in 1 2 1 1 3 11 p1 Pressure at position 1 in the cycle 11 for this heat engine 1 p3 Pressure at position 3 in the cycle would be  1  pV11 Volume of gas at position 1,4 in the cycle 13.5 273   VV 1  1   75%  1p - p V - V 873 273 1 3 2 1  VV 2 Volume of gas at position 2,3 in the cycle Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 7 Heat Engines – The Otto cycle Between positions 1 and 2 Between positions 4 and 1 To complete the cycle

pV   The Otto Cycle is the basis of spark-ignition 11 VV 1 p V nRT  n  p2 V 2 p 1 V 1 p3 V 2 p 4 V 1 1 1 1 Heat output piston engine, which is essentially how a typical RT1   Q4,1 nMcV  T 4 T 1  via exhaust petrol driven engine operates.  V1 V2 V pp21 pp43  1 W4,1  0 pV p1 V 1  p  p 1 V2 V1 Positions 0-1: V  T Air is drawn into piston/cylinder arrangement at 1 dT T1  1 S  nMc  nMc ln constant pressure.  4,1 T VV   4 p1 V 1 V 1 TT4 W1,2 1   1V Process 1–2 2 -ve since work Adiabatic (isentropic) compression of the air via a is being done piston. QS1,20,  1,2  0 on the gas

Process 2–3 Constant-volume heat transfer to the working gas from Between positions 2 and 3 an external source while the piston is at maximum VV 2 compression. This process is intended to represent the (,)pV32 ignition of the fuel-air mixture and the subsequent rapid pV p V nRT  T  22 3 burning. 2 2 2 2 nR

pV Process 3–4 p V nRT  T  32 Adiabatic (isentropic) expansion (power stroke). 3 2 3 3 nR

Q nMc T T Heat input to gas Qin Process 4–1 2,3V  3 2  Work done by the via spark ignition Constant-volume process in which heat is rejected from W  0 gas on the the air while the piston is at maximum expansion. 2,3 surroundings is the area enclosed by T (,)pV41 3 dT T3 Process 1–0 S  nMc  nMc ln 2 the cycle W pdV 2,3 T VV  2 4 Air is released to the atmosphere at constant pressure. TT2 (,)pV22 Qout 1 Assume we have n moles of ideal gas, none of which Between positions 3 and 4 (,)pV are lost in the process. Input parameters are: 11

Pressure of gas at position 1 in the cycle pV41 p1 p V nRT  T  4 1 4 4 nR p3 Pressure of gas at position 3 in the cycle V  V Volume of gas at position 1 in the cycle  2 1 pV p3 V 2  p  p 3  Q Q  nMc T  T V in 1,2V  3 2  Volume of gas at position 2 in the cycle  V2  1 Q Q  nMc T  T  pV  out 4,1V 4 1 T1 Temperature of gas at position 1 R 1 32 V2 Work done cV  W3,4 1  in the cycle M  1  1V by gas 1 c p 2 1 R R    1  cV  2 c  QS3,40,  3,4  0  M P Nikolaus Otto cV M  1 (1832-1891)

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 8 Otto engine cont.... Define the compression ratio Total work done is: V Define heat engine efficiency as the ratio r  1 of Work done by the gas to the heat input (,)pV W pdV V 32  2 3 WWW W 1,2 3,4   11Q     in p1 V 1 V 1 pV32  V 2  W 11          V 1 11VV    2  21    1 1 p31 p r       1 r Qin    Work done by the V 1  gas on the 2  1 V2 p 3 p 1 r  W p13 r11  r  p    1 surroundings is the  1r area enclosed by  1 (,)pV41  1 2 the cycle W pdV V 1  1 r 1  4 W2 11  p r  r  p  1 1113   1 (,)pV22 Qout  11rr r 1

V2 1  (,)pV11 Wp1  1  3  pr1  So the Otto Engine efficiency  1 r  depends only on the compression ratio, and the ratio of specific heats From the previous page:  V1 pV pV pp 22 32 21 T2  T3  V2 nR nR

Qin nMcV  T 3 T 2   nMcV V1 Qin p 3 V 2 p 1 V 2 nR V 2 R 1 c  V M  1 c p 2 1 R Mc 1   1  cV  V c  2 M R  1 V

 R 1 V2 p 3 p 1 r  cV  Q M  1 in  1 R  c  P M  1

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 9 Heat Engines – The Diesel cycle Between positions 1 and 2 Between positions 4 and 1 To complete the cycle The Diesel Cycle is the basis of a diesel engine, which is pV   ubiquitous in transport applications. Unlike petrol-driven 11 VV 1 p V p V p V p V p1 V 1 nRT 1  n  2 2 1 1 2 3 4 1 engines, diesel variants are more suited to heavy machinery. Heat output RT1 Q nMc T T   They can be found powering most ships as well as trucks, 4,1V  4 1  via exhaust  V1 V3 buses and cars. V pp21 pp42  1 W4,1  0 pV p1 V 1  p  p 1 V2 V1 V Positions 0-1:  T 1 dT T1 Air is drawn into piston/cylinder arrangement at constant  1 S  nMc  nMc ln  4,1 T VV pressure.   4 p1 V 1 V 1 TT4 W1,2 1   1V Process 1–2 2 -ve since work Adiabatic (isentropic) compression of the air via a piston. is being done QS1,20,  1,2  0 on the gas Process 2–3 Constant-pressure (isobaric) heat transfer to the working gas from an external source while the piston is at maximum Between positions 2 and 3 compression. This process is intended to represent the Qin ignition of the fuel-air mixture and the subsequent rapid (,)pV (,)pV23 pp 2 22 burning. This is different from the Otto cycle, which is constant pV 2 3 volume (isochoric) heating during this stage. In the Diesel p V nRT  T  22 cycle, the heat generated from air compression is sufficient to 2 2 2 2 nR ignite introduced fuel vapours. In the Otto cycle a spark plug is used instead to ignite the fuel. pV23 p2 V 3 nRT 3  T 3  Process 3–4 nR Adiabatic (isentropic) expansion (power stroke). Heat input to Q2,3 nMcP  T 3 T 2  gas during Work done by the gas Process 4–1 combustion W2,3 p 2 V 3 V 2  on the surroundings is Constant-volume process in which heat is rejected from the air the area enclosed by while the piston is at maximum expansion. T the cycle 3 dT T3 (,)pV S  nMc  nMc ln W pdV 41 2,3 T PP  2 Process 1–0 TT2 4 Air is released to the atmosphere at constant pressure. Qout 1

Assume we have n moles of ideal gas, none of which Between positions 3 and 4 (,)pV11 are lost in the process. Input parameters are: pV p V nRT  T  41 p1 Pressure of gas at position 1 in the cycle 4 1 4 4 nR Volume of gas at position 1 in the cycle  V1 V pV p V  p  p 3 Volume of gas at position 2 in the cycle 2 3 2  Qin Q 1,2  nMcP  T 3  T 2  V2 V

V Volume of gas at position 3 in the cycle  1 Qout Q 4,1  nMcV  T 4  T 1  3  R 1 p2 V 3 V 3 Work done T Temperature of gas at position 1 c  W3,4 1  1 V  1V by gas in the cycle M  1 1 R  cp 1 R QS0,   0   1  2 c  c  3,4 3,4 Rudolf Diesel  V 2 P M  1 cV M (1858-1913)

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 10 Diesel engine cont.... Define the From the previous page: compression ratios  V Total work done is: p p1 p r V V 2 1 1 1 3 VVV1 3 2 V r  s  2 Qin (,)pV V V (,)pV22 23 W pdV 2 2 rs pV23 pV22  T3  T2  2 3 nR nR WWWW1,2  2,3  3,4 11    p1 V 1 V 1 p2 V 3  V 3  W11     p2 V 3  V 2        Define heat engine efficiency 11VV    21       as the ratio of Work done by the gas to the heat input   1 Work done by the gas p V1 p r V s W1 2 r1  r  p r V s  1  1 2 s 1  on the surroundings is   12   11r W the area enclosed by   the cycle (,)pV  1 Q 41  in W pdV pV 1   s  W12 r1  r  r s  1  1  r s 1   4 Q      pV12  out  1r r(1 s )  r s  1    1 1   pV12      (,)pV W r  r  r s 11   r s  s r r  s1 pV 11  1  1 12 pV  W12 r(1  s )  r  s  1  1  r  ( s  1) r(1 s )  r s  1   1     rs 1 pV  W12 r(1  s )  r s  1  rs  1  1  1  rs  1 Q nMc T T 11s  inP  3 2   1  1   rs  1 nMcP  V1 Qin p 1 V 3 V 2  nR V2 c The Diesel engine is typically p 2 1RR 1  1  cVPV 22  c   c   more efficient than a petrol (Otto) cV M M engine since the former works on the basis of self ignition McP 1   2   due to high compression. This R  1 ‘knocking’ is undesirable for petrol  engines, so a lower r value is Q  r  s 1 p V in 1 1 2 required.

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 11 Comparing Otto and Diesel heat engines c p 2   1   V1 V3 (,)pV32 r  s  cV V V 2 2 R c  1 V 2 M

R 1 cV  M  1 Work done by the gas on the VVV1 3 2 R  surroundings is the cP  rs M  1 area enclosed by the cycle

Q (,)pV22 in (,)pV23 2 3

W Diesel   pV12  Qin W r(1  s )  r s  1  1 11s    Work done by the gas diesel 1  1  Q r s1 pV rs  1 in  1 2 on the surroundings is   1 the area enclosed by 1 the cycle (,)pV41 otto 1 Otto W pdV r 1  V 1 4 2  Qout W1  1  p31  p r   1r 1

 (,)pV11 V2 p 3 p 1 r  Q  in  1

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 12