Carnot theorem and its corollary

Most energy extracted from the fuel in power plants is dumped to the environment as waste heat, here using a large cooling tower.

Heat engine extracts work from heat reservoirs. gasoline & diesel engines fossil-fueled & nuclear power plants jet engines Perfect heat engine: coverts heat to work directly. Heat dumped 2 nd law of thermodynamics ( Kelvin-Planck version ): There is no perfect heat engine. No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

 Carnot cycle Idealized thermodynamic cycle consisting of four reversible processes (working fluid can be any substance): The four steps for a Carnot Heat Engine are:  Reversible isothermal expansion (1-2, T H = constant)  Reversible adiabatic expansion (2-3, Q = 0, T H  T L )  Reversible isothermal compression (3-4, T L =constant)  Reversible adiabatic compression (4-1, Q=0, T L  T H ) Carnot cylce.ppt Modified 10/9/02

Work done by the gas =  PdV, i.e. area under the process curve Work done on gas =  PdV, area under the process curve subtract Net work dV>0 from  PdV>0 Since dV<0  PdV<0 T L = const.

 “Carnot theorem states that no heat engine working in a cycle between two constant temperature reservoirs can be more efficient than a reversible engine working between the same reservoirs.”  In other words it means that all the engines operating between a given constant temperature source and a given constant temperature sink, none, has a higher efficiency than a reversible engine.

Q-Wa Q-Wb Q-Wa Q-Wb Heat source at T1 Heat Engine A Heat sink at T2 Heat source at T1 Heat Engine B Heat Engine A Heat pump B Heat sink at T2

Wa-Wb Combined system of A&b Heat sink at T2

 Suppose there are two engines Ea and Eb operating between the given source at temperature T1 and the given sink at temperature T2.  Let EA be any irreversible heat engine and EB be any reversible heat engine. We have to prove that efficiency of heat engine EB is more than that of heat engine EA.  Suppose both the heat engines receive same quantity of heat Q from the source at temperature T1. Let W A and WB be the work output from the engines and their corresponding heat rejections be (Q – WA) and (Q – WB) respectively.

Assume that the efficiency of the irreversible engine be more than the reversible engine i.e. ηA > ηB. Hence,  WA / Q > WB / Q  I.e. WA > WB  Hence it is concluded that reversible engine working between same temperature limits is more efficient than irreversible engine thereby proving Carnot’s theorem

(1) “All reversible heat engines operating between the same two heat reservoirs must have the same efficiency.”  Proof: Consider two reversible engines R1&R2 are working between the same reservoir at temp T1&T2,and receive same amount of heat Q from high temp reservoir fig (1). Let assume that R1 is more efficient than R2,and lower efficient R2is reversed to operate as a heat pump as shown fig (2).  n th(R1) > n th(R2)  W R1> W R

Q- W R1 Q-W R2 Q- W R1 Q-W R2 Heat source at T1 Heat sink at T2 Heat source at T1 Heat Engine R2 Heat Engine R1 Heat pump R2 Heat sink at T2 Heat Engine R1 W R1 W R2 W R1 –W R2 W R2

 W R1 -W R2 Combined system of R1&R2 Heat sink at T2

 The heat engine R1, absorbs heat Q from source,and heat pump R2 delivers same amount of heat Q to heat source.  So we can eliminate the high temp source and consider combined system as shown fig.(3),  It absorbs heat (Q-W R2 )-(Q-W R1 ) =W R1 -W R2 from heat sink and produce equivalent amount of work WR1-WR2.  This combined system results in a PMM2, and it violates second low of thermodynamic  Therefore,all All reversible heat engines operating between the same two heat reservoirs must have the same efficiency.”

(2) “ The efficiency of any reversible heat engine operating between two thermal reservoirs does not depend on nature of working fluid and depends only on the temperature of the reservoirs. “  Proof- consider, heat engine E and heat pump R, working between the same thermal reservoirs as shown in Fig(a). The efficiency is same because both are reversible engines and work on the Carnot cycle.  Efficiency depends only upon the temperature of the reservoirs. so, Work is produced by engine E equal to work is required to heat pump R.

 Now assumed that, efficiency of engine E be increased by changing nature of working substance. It is as shown in Fig (b),means that the engine E produces more work and rejects less heat to sink.  (Q-W E )< (Q-W R )  W E > W R

W E =W R W net= W E -W R W R Q- W E Q-W R Q- W E Q-W R Heat source at T1 Heat sink at T2 Heat source at T1 Heat Engine R Heat Engine E Heat pump R Heat sink at T2 Heat Engine E

Combined system of E & R Heat sink at T2  W E -W R

 However engine E receives Q amount of heat from source and pump R delivers same amount of heat to source. Therefore, we can eliminate high temperature source and combined system as shown in Fig c, receives W E -W R amount of heat from sink and produces same amount of work.  This, violates second law of the thermodynamics. Therefore it is concluded that efficiency does not depend on any properties of working fluid other than temperature of reservoirs.

A Carnot engine extracts 240 J from its high T reservoir during each cycle, & rejects 100 J to the environment at 15  C. How much work does the engine do in each cycle? What’s its efficiency? What’s the T of the hot reservoir? work done efficiency 