Lecture 10: Heat Engines and Reversible Processes Last time we started discussing cyclic heat engines these are devices that convert heat energy into mechanical work We found that in general, heat engines take in energy at high temperature, due an amount of work W eng, and then exhaust energy at a lower temperature The 1st Law of Thermodynamics tells us the following about the internal energy of the engine: E int = Q W eng = 0 Minus sign because the engine does work on the outside world Zero because the process is cyclic
What about Q? the engine takes in energy, and emits energy Q c So, Q = Q c The 1st Law then says: Schematically: = Q E int h Q c W eng = 0 W eng = Q c Engine W eng Q c
Efficiency of the Heat Engine We define efficiency as the ratio of the energy supplied to the engine to the work done by it: e = W eng = Q c = 1 Q c For the steam locomotive, the efficiency is not 100 because Q c is not zero It turns out that that s true for any type of heat engine a heat engine that gets 50 efficiency is really good This experimental fact is the basis of the 2nd Law of Thermodynamics: No cyclic heat engine can convert heat to work at 100 efficiency
Heat Pumps and Refrigerators Doing mechanical work is not the only way we d like to manipulate heat imagine it s 95 o F outside, and we have a can of soda that s 85 o F we d prefer the soda to be even colder before we drink it in other words, we d like to transfer energy by heat from the soda to the warmer atmosphere But that s an example of a process that never happens if we wait, we d observe energy transfer in the other direction, and the soda would warm up to 95 o F This is the 2nd Law in action again. In fact, we can write the 2nd Law as: Heat will never spontaneously flow from a cold object to a warmer one
Even though the two ways of writing the 2nd Law sound very different, they are in fact equivalent as long as one is true, the other must be as well For example, imagine we had an imperfect heat engine but a perfect heat pump that could transfer heat to a higher temperature: Realistic heat engine: Perfect heat pump: Engine W eng Pump Q c Q c
We could then combine the two devices to get: = W eng W eng Q c Q c A perfect heat engine So if we were allowed to break the 2nd form of the 2nd Law, the first form would be broken as well
Refrigerators Of course, it is possible to cool down our can of soda by placing it in a refrigerator Due to the 2nd Law of thermodynamics, though, we must supply energy (in the form of work) to make the heat flow in the desired direction: Real refrigerator: W Q c
A good refrigerator will remove a lot of heat for a little amount of external work Q The coefficient of performance is defined as h a higher number means a better refrigerator W typically can achieve values of 5-6
Reversible Processes and the Carnot Engine We ve seen already that the 2nd Law of Thermodynamics rules out the construction of a perfect (100 efficient) heat engine Now we want to explore how close to perfection we can come To begin with, we ll define a reversible process this in one in which the system can retrace its steps on the PV diagram, and remain in equilibrium at all points: P V
No real processes are reversible whenever the pressure or volume changes, it takes time for the system to come back into equilibrium if any energy is lost due to friction or turbulence, external energy would have to be added to return the system to its initial state We can approximate a reversible process by considering a process that: proceeds slowly, so that the system is never far from being in equilibrium (and turbulent flow of the gas is supressed) uses well-lubricated moving parts to reduce friction By taking the ideal of an infinitely slow, infinitely lowfriction process, we can arrive at a reversible process at least in theory
The Carnot Cycle You might guess that the highest possible efficiency for a heat engine would be obtained when every part of the process is reversible this means that the heat engine forms a reversible cycle Such an arrangement is called a Carnot cycle, in honor of Sadi Carnot, who worked out the theory And you d be right This is Carnot s Theorem: For a given temperature difference, the most efficient heat engine possible is one that operates in a reversible cycle
Proof of Carnot s Theorem 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: W C =,C Q c,c = e C,C W V =,V Q c,v = e V,V
Let s say I set this up so that the Varnes engine provides the power for the Carnot refrigerator. Then: W C = W V e C,C = e V,V,C,V = e V e C > 1 Since we assumed at the start that e V > e C 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 In our idealized model of a heat engine, we do the following: 1. Absorb energy from high-temperature reservoir (isothermal at ) 2. Do work (i.e. gas expands) without any heat transfer (adiabatic) 3. Exhaust energy to low-temperature reservoir (isothermal at ) 4. 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: = nr B V A Work done: As heat is input: W 1 = nr A During adiabatic expansion: W 2 = 1 $ 1' ( P V P V C C B B ( )
as heat is exhausted: W 3 = nrt ln V C V D $ as gas is compressed W 4 = So the total work done is: 1 $ 1' ( P V P V A A D D ( ) W = W 1 +W 2 +W 3 +W 4 = nr A + 1 $ ' (1 P V ( P V C C B B ( ) + nr ln V C V D V D $ + 1 ' ( = nr A + nrt ln V $ C c + 1 $ ( ' (1 nrt ( nrt + nr h h V $ V $ $
This means the efficiency is: e = W = nr A + T ln V C c C V = 1+ D A nr B V A V D D V = 1' C A $ $
Using the properties of adiabatic and isothermal expansions, we have: P A V A = P B ; P C V C = P D V D P A V A = P D V D ; P B V B = P C V C P B V 1 A = P D V D P B V 1 B = P D V D V C V A 1 1 = V D 1 V C 1 V A = V D V C So the efficiency is: e = 1 ln V D $ V C ln V A $ ' ' = 1 ln V A $ ' ln V A $ ' = 1