Lecture 11 More on Heat Engines Last time we stated Carnot s Theorem which says that for a given temperature difference the most efficient heat engine possible is a reversible one To prove this let s start by asking what happens if Carnot is wrong That means I can make a Varnes engine which is not reversible but has better efficiency than a Carnot engine Now let s say I use my engine to run a Carnot engine backwards I can do this since the Carnot engine is reversible this really means I turn the Carnot engine into a refrigerator Both engines obey the 1st Law of Thermodynamics WC Qh C Qc C eC Qh C WV Qh V Qc V eV Qh V Let s say I set this up so that the Varnes engine provides the power for the Carnot refrigerator Then WC WV eC Qh C eV Qh V Qh C Qh V eV 1 eC Since we assumed at the start that eV eC Looking at the entire Varnes Carnot apparatus then one sees no net work input or output a net flow of heat into the high temperature reservoir In other words this violates the 2nd Law of Thermodynamics that means I can t build an engine better than Carnot s The Carnot Engine 1 2 3 4 In our idealized model of a heat engine we do the following Asborb energy from high temperature reservoir isothermal at Th Do work i e gas expands without any heat transfer adiabatic Exhaust energy to low temperature reservoir isothermal at Tc Compress gas to return to initial state without heat transfer adiabatic In the Carnot engine all of these steps are done reversibly Carnot Efficiency Heat input VB Qh nRTh ln VA Work done by the engine As heat is input VB W1 nRTh ln VA During adiabatic expansion 1 W2 PBVB PCVC 1 as heat is exhausted VD W3 nRT ln VC as gas is compressed 1 W4 PDVD PAV A 1 So the total work done is W W1 W2 W3 W4 VB 1 nRTh ln PBVB PCVC V A 1 VD 1 nRTc ln PDVD PAV A VC 1 VB VD W nRTh ln nRTc ln VA VC 1 nRTh nRTc nRTc nRTh 1 VB VD nR Th ln Tc ln VA VC This means the efficiency is VB VD nR Th ln Tc ln VA VC W e VB Qh nRTh ln V A 1 VD Tc ln V C VB Th ln V A 1 VD Tc ln V C VA Th ln V B Using the properties of adiabatic and isothermal expansions we have PAV A PBVB PCVC PDVD PAV A PDVD PBVB PCVC PBVBV A 1 PDVD PBVB PDVDVC 1 V A 1 V 1 B VD 1 VC 1 V A VD VB VC So the efficiency is VD Tc ln V VA Tc ln V Tc e 1 1 1 VA VA Th Th ln Th ln VB VB C B The Internal Combustion Engine The heat engine that you re most familiar with is probably the automotive gasoline engine heat from burning gasoline is converted to work in the form of motion of a car We ll study a somewhat idealized version of this engine The engine goes through the following steps in each cycle 1 Intake stroke Piston moves down drawing air gas mixture into cylinder at 1 atm Volume of gas increases from V2 to V1 P V2 A V1 V 2 Compression stroke Intake valve closes and piston moves upward air gas compressed from V1 to V2 temperature increases from TA to TB This happens quickly close to adiabatic since there s little time for heat transfer 3 Combustion Spark plug fires causing fuel to burn P This is very quick can assume the piston doesn t move during combustion Energy Qh enters the system Temperature increases from TB to TC B A V2 BANG V1 V C P B A V2 V1 V 4 Power stroke Piston pushed downward by expanding gas Pretty fast can treat as adiabatic Volume increases from V2 to V1 Temperature drops from TC to TD C P D B A V2 P D B A Pressure drops quickly 6 Exhaust stroke Piston moves upward pushing out burned fuel and air This 4 stroke sequence is known as the Otto cycle V C 5 Exhaust valve opens V1 V2 V1 C P V D B A V2 V1 V Example Efficiency of the Otto cycle In the Otto cycle heat enters along path BC This is at constant volume so we have Eint Qh nCV T nCV TC TB PCVC PBVB CV V2 nCV PC PB nR R nR The total work done is the sum of the positive work done by the gas on path CD and the negative work done on path AB Using the equation for work in an adiabatic process from Problem 21 53 in the text we have 1 W PCVC PDVD PAV A PBVB 1 In terms of V1 and V2 this is 1 W PCV2 PDV1 PBV2 PAV1 1 This means the efficiency is 1 PCV2 PDV1 PBV2 PAV1 1 W e CV V2 Qh PC PB R We can simplify this first by noting that CV CV 1 1 1 CP C P CV R 1 CV so the efficiency becomes CV PCV2 PDV1 PBV2 PAV1 R W e CV V2 Qh PC PB R PCV2 PDV1 PBV2 PAV1 V2 PC PB From the properties of adiabatic expansion we know that PCV2 PDV1 PbV2 PAV1 Substituting for PB and PC we find V1 V1 PD 1 PDV1 PA 1 PAV1 V2 V2 1 e V2 V1 V1 PD PA V2 V2 V1 V1 V1 V1 PD PA V V V V 2 2 2 2 V1 V1 PD PA V2 V2 V1 V1 1 PD PA V2 V2 V2 1 V1 PD PA V1 V2