Second Law Thermo 4 451 Fall 2011 - University Of Colorado Boulder

EnginesMuch of the thinking in the 2nd Law developed from theinvestigation of the properties of heat engines.While this will now be useful for us, we must first developthe concept of an engine! An engine is a device (system) that converts energy towork. We care about a specific type of engine: a HEAT engine. A heat engine draws heat from a hot reservoir, convertssome heat to work, and releases some heat to a coldreservoir. The engine itself is a system that undergoes a cyclicalprocess.

Experiments that suggested 2nd Law Clausius: An engine does not exist whose sole effect is totransfer heat from a cold body to a hot body. Lord Kelvin: An engine does not exist that operates in acycle and performs work by exchanging heat with only onereservoir!ClausiusKelvin

So any heat engine has got tolook like one (or more) of these devices:-

Heat Engine with OverallSpontaneous ProcessHeat withdrawn fromhot reservoir. Entropydecreases by qh/Th.If Tc Th, the entropy changecan be positive even if qc qh.The difference qh-qc can bewithdrawn as work.Heat added to cold sink.Entropy increases by qc/Tc.-

Efficiency of any EngineEfficiency, , of Engine: Work PerformedHeat Absorbed w / qh Heat (qc) dumped into the cold sink is wasted!(furnace exhaust up the chimney)-

ConcepTest #1Consider the engine on the left.What is the thermodynamicefficiency of this engine?A.14 %B.25 %C.33 %D.50 %E.What engine?-Now consider REVERSIBLE heat engines.

Cross Section of Double-Acting Watt Steam EngineThis is a REAL engine, with irreversible processes. We willbe concerned with REVERSIBLE engines (Carnot engines).

A generalized reversible heat engine

A generalized cyclic heat engineAll reversible engines operating between two heat bathshave the same efficiency.If not, then perpetual motion is ours! Proof is next.

Two reversible heat engines with different efficienciesLord Kelvin is NOT pleased!With Eff(B) Eff(A), it gives workw’ with nothing net to cold sink!

Carnot Engine (1825)French engineer Sadi CarnotThe engine consists of an ideal gas pistonoperating between two heat reservoirs.All steps are reversible!1. Isothermal Expansion2. Adiabatic Expansion3. Isothermal Compression4. Adiabatic CompressionHOTCOLD

Net work work done bysystem –work done onsystem

Efficiency of Carnot EngineAs before, we know the efficiency, , of a heat engine:Work Performed w Heat AbsorbedqhTrue for any engine, reversible or irreversible!However, for any reversible heat engine, we will find that Th TcThwhere Th and Tc are the temperatures of the hot and cold reservoirs.

ConcepTest #250 JThis diagram represents areversible cycle for an idealgas.What is the thermodynamicefficiency of the engine?A.80 %B.75 %C.60 %D.30 %E.10 %Now we look at the Carnot cycle steps in more detail.

Steps of Carnot CycleIsothermal expan A to B: Step 12 U qh w 0qh absorbed ( ) fromhot reservoir Thw done by system (–) V2 qh w nRTh ln V1 V2 S nR ln (positive) V1 q 0w done by system (–)w Cv(Tc-Th) S 0

Reversible Adiabatic Expansion Step 2q2 0 S2 đq2 2 T 0 U 0 w Cv TW2 Cv Th Tc We showed earlier that for anadiabatic expansion of ideal gas Tc Th CVnR Th V2 orV3 V2 Tc V3 CVnRNotice that I am trying to obtain V3 and V4in terms of V1 and V2, so that we can combine all stepsNow on to step 3, reversible isothermal compression.

Steps of Carnot Cycle U qh w 0qh absorbed ( ) fromhot reservoir Thw done by system (–) V2 qh w nRTh ln V1 V2 S nR ln (positive) V1 q 0w done by system (–)w Cv(Tc-Th) S 0qc discarded (–) into coldreservoir Tcw done on system ( ) U qh w 0 V4 qc w3 nRTc ln V3 V4 S nR ln (negative) V3

Reversible Isothermal Compression Step 3qc (q3) discarded (–) into cold reservoir Tcw done on system ( ) V4 w3 nRTc ln V3 V q3 nRTc ln 4 V3 V S nR ln 4 (negative) V3 Rewrite in terms of V1 and V2, using the adiabatic path expressions just developed. Th V1 Tq3 nRTc ln c Th V2 T c CvnR CvnR nRT ln V2 c V 1 Similarly, V2 w3 nRTc ln V1 V2 S nR ln V1 Now on to the final, adiabatic step!

Steps of Carnot Cycle U qh w 0q 0w done on system ( )w Cv(Th-Tc) S 0qh absorbed ( ) fromhot reservoir Thw done by system (–) V2 qh w nRTh ln V1 V2 S nR ln (positive) V1 q 0w done by system (–)w Cv(Tc-Th) S 0qc discarded (–) into coldreservoir Tcw done on system ( ) U qh w 0 V4 qc w nRTc ln V3 V4 S nR ln (negative) V3

Reversible Adiabatic Compression Step 4q 0 S 0w done on system ( )w Cv(Th-Tc)Finally, sum each component around the full cycle.

Summing up the Carnot CycleEntropy Stot đqđqđqđqđq TT 2T 3T 4T1 V V nR ln 2 0 nR ln 2 0 V1 V1 Stot đq 0TEfficiency WnetqinTh Tc Th V V nRTh ln 2 Cv Th Tc nRTc ln 2 Cv Th Tc V1 V1 V2 nRTh ln V1 The efficiency of an ideal reversible engine.It is not possible to do better!

Parameters for a possible Carnot 1.3118.95300D2.1511.44300

Summary of 4 Steps in Carnot Cycle Uqrevwrev SAB0RThlnVB/VA 1678 J–RThlnVB/VA– 1678 JRlnVB/VA4.19 J/KBCCv(Tc-Th)–1247 J0Cv(Tc-Th)–1247 J0CD0RTclnVD/VC–1259 J–RTclnVD/VC 1259 JRlnVD/VC–4.19 J/KDACv(Th-Tc) 1247 J0Cv(Th-Tc) 1247 J0Net0R(Th-Tc)lnVB/VA 419 J–R(Th-Tc)lnVB/VA– 419 J0Step

Entropy is a State FunctionThe treatment on the previous viewgraph showedthat S 0 for a full cycle of a Carnot (reversible)engine.This is one path on the PV diagram. If S was notzero on this closed path, we would know that Scould not be a state function. In order to provethat S is a state function, we must show that S 0for any reversible cycle.We do this next:

A generalized reversible, cyclic heat engineIf we could construct this arbitrary reversible, cyclic pathfrom a number of Carnot cycles, then we would have provedthat entropy is a state variable. We can do this!

A generalized reversible, cyclic heat engineCarnot cycleIn the limit of infinitesimal Carnot cycles, the heavy greenpath approaches arbitrarily close to the maroon path.So entropy is a state function; now look at engine efficiency.

Net work work done bysystem –work done onsystemIn example above, net work 1678 J 1247 J -1259 J – 1247 J 419 JWhat is the efficiency?

Efficiency of Carnot EngineThe efficiency is the ratio of the usable work outto the total heat input to the system.Waste heat is not usable in this context!So in this case the efficiency is 419 J/1678 J 0.2497Compare with (Th – Tc)/Th (400 -300)/400 0.25Now back to chemistry!

However, for anyreversible heat engine, we will find that where T h and T c are the temperatures of the hot and cold reservoirs. True for any engine, reversible or irreversible! ConcepTest #2 This diagram represents a reversible cycle for an ideal gas. What is the thermodynamic efficiency of the engine? A. 80 %