H.2 Euler-Maclaur in Sum Formula
3.2 Carnot Cycle and Engines
In classical thermodynamics, the second law of thermodynamics is usually rationalized by considering processes involving the conversion of work to heat by engines that return to their original thermodynamic state after one cycle. Comparison is made to a hypothetical engine, known as a Carnot engine, which is imagined to execute a reversible cycle. The Carnot cyclepertains to an idealized engine in which the working substance is one mole6 of an ideal gas. There are four segments to the cycle, as depicted inFigure 3–1. All segments involve reversible processes, so the whole cycle is reversible. SegmentABis a reversible isothermal expansion in which an amount of heat|Q2| =Q2is extracted from a heat source at a high temperatureT2. SegmentBC is a reversible adiabatic expansion. SegmentCD
6We could also consider any fixed number of moles of an ideal gas.
36 THERMAL PHYSICS
p
V T2
T1 A
B
C D
FIGURE 3–1 The Carnot cycle in theV,pplane. The working substance is an ideal gas and the cycle consists of four reversible segments.ABis isothermal expansion at temperatureT2,BCis adiabatic expansion,CDis isothermal compression at temperatureT1, andDAis adiabatic compression. The figure is drawn forγ=5/3.
is a reversible isothermal compression in which an amount of heat|Q1| = −Q1is given up to a heat sink at temperature T1. Both the source and the sink are assumed to be heat reservoirs, so their temperatures do not change. Finally, segmentDAis a reversible adiabatic compression. In order for these segments to form a closed cycle, we can apply Eq. (2.27) to each of the adiabatic segments to obtain
T2VBγ−1=T1VCγ−1; T2VAγ−1=T1VDγ−1. (3.16) Division of one of these equations by the other and extraction of theγ−1 root gives
VA
VB =VD
VC. (3.17)
Combining Eq. (3.17) with the ideal gas law gives pA
pB =pD
pC, (3.18)
so the geometry of the cycle is completely known and simple to express.
On the adiabatic segmentBC,δQ=0 so we haveWBC= −UBC=CV(T2−T1). This exactly cancels the workCV(T1−T2)done by the gas on the other adiabatic segment. The work done by the gas on the isothermal expansion segmentABisRT2ln(VB/VA). Recalling thatUdepends only onTfor an ideal gas means thatUAB=0 for that segment, so
|Q2| =RT2ln(VB/VA). (3.19) Similarly, for the isothermal compression segmentCD, we obtain
|Q1| = −RT1ln(VD/VC)=RT1ln(VB/VA), (3.20) where Eq. (3.17) has been used in the last step. Dividing Eq. (3.19) by Eq. (3.20) we obtain
|Q1| T1 =|Q2|
T2
. (3.21)
For the entire cycle,U=0 so the total work done by the gas during the cycle isW= |Q2|−
|Q1|. The efficiency of the cycle is therefore η:= W
|Q2| =1−|Q1|
|Q2|=1−T1
T2
. (3.22)
This efficiency is always less than unity except for a heat sink at absolute zero, which is deemed to be impossible.
Let us examine the meaning of Eq. (3.21) in terms of the second law. Since the entropy is a function of state, we haveS=0 for a cycle. Applying Eq. (3.11) with the equality, for our reversible cycle, we obtain
0= Q2
T2 +Q1
T1 =|Q2| T2 −|Q1|
T1 (3.23)
in agreement with Eq. (3.21).
Beginning with the Carnot cycle, Fermi [1, chapter IV] proves a number of other things based on the Kelvin/Clausius postulates. These are used to rationalize the existence of the entropy and to formulate the second law. Here, we take the opposite approach by quoting the main results and demonstrating how they follow from the second law.
• Any reversible engine working between the same two temperatures T2 and T1 has the same efficiency as a Carnot engine.We follow the same procedure as we did in deriving Eq. (3.23) except that the amounts of heat are now|Q2|and|Q1|which might differ from those for a Carnot engine. Thus we obtain
0=Q2 T2 +Q1
T1 =|Q2| T2 −|Q1|
T1
. (3.24)
It follows that the ratio |Q1|/|Q2| = T1/T2 is the same as for a Carnot engine.
From Eq. (3.12) withU = 0, the amount of work done in the cycle is now W =
|Q2| − |Q1|, so
η:= W
|Q2| =1−|Q1|
|Q2|=1−T1
T2 =η. (3.25)
• Any irreversible engine working between the same two temperatures T2 and T1 has a smaller efficiency than a Carnot engine.This result follows by applying Eq. (3.11) with the inequality to obtain (superscriptifor irreversible)
0>Qi2 T2 +Qi1
T1 =|Qi2| T2 −|Qi1|
T1
, (3.26)
which leads to|Qi1|/|Qi2| > T1/T2. The amount of work done in the cycle is nowWi =
|Qi2| − |Qi1|, resulting in
ηi:= Wi
|Qi2| =1−|Qi1|
|Qi2|< η. (3.27)
38 THERMAL PHYSICS
• In a cycle of any reversible engine that receives heatδQ from a number of sources at temperature T,
δQ
T =0. (3.28)
This follows from Eq. (3.8) with the equality by recognizing thatS=0 for a cycle. In classical thermodynamics, Eq. (3.28) is deduced by arguing that any reversible cycle can be approximated to arbitrary accuracy by a very large number of small Carnot cycles.
It is actually Eq. (3.28) that was used to deduce that a state function, now known as the entropy, exists. By integrating from pointAto pointBalong some reversible path and the back again toAalong some other reversible path, we create a reversible cycle. Since the integral fromBtoAalong the return path is the negative of the integral fromAtoB along that path, it follows that
B
A
δQ T
reversible path I= B
A
δQ T
reversible path II. (3.29) Since the values of the integrals in Eq. (3.29) depend only on their end points, their integrand must be the differential of some function, namely dS=δQ/T, which is Eq. (3.6). In mathematics, 1/Twould be called an integrating factor forδQ.
• In a cycle of any irreversible engine that receives heatδQ from a number of sources at
temperature Ts,
δQ
Ts <0. (3.30)
This follows from Eq. (3.11) with the inequality by recognizing thatS=0 for a cycle.
Example Problem 3.1. Analyze a Carnot refrigerator in which heat|Q1| =Q1 is extracted (from the refrigerator) at a low temperature T1 and given to a Carnot engine running in reverse; then|Q2| = −Q2is extracted from that Carnot engine and given to a sink at higher temperatureT2.
Solution 3.1. The magnitudes|Q2|and|Q1|are still given, respectively, by Eqs. (3.19) and (3.20), so Eq. (3.21) still applies. But now an amount of workW= −W>0 must be doneonthe system, whereW=Q1+Q2= |Q1| − |Q2| = −W. Thus
|Q1| W = T1
T2−T1. (3.31)
We see that only a small amount of workWmust be provided to extract|Q1|from the refrigerator provided thatT1 is not too much lower thanT2. Since an amount of heat|Q2| = |Q1|(T2/T1) must be given up to the source, the cooling of a refrigerator can result in a large amount of heat given up to the surrounding room. Of course the process that takes place in an actual refrigerator is irreversible, so even a larger ratio of the removed heat to the workWis required than given by Eq. (3.31). Indeed, by using Eq. (3.26) for an irreversible engine, we obtain the inequality|Qi1|/W<T1/(T2−T1).
The considerations that led to Eq. (3.31) can also be applied to analyze a heat pump that adds an incremental amount of heat from an inexpensive source at temperatureT1to heat a room at temperatureT2. In that case, a more meaningful quantity is
|Q2| W = T2
T2−T1. (3.32)
Thus the heat pump will require only a small amount of work to provide|Q2|if the source temperatureT1is close toT2. For a real (irreversible) heat pump we would have|Qi2|/W <
T2/(T2−T1).