lecture 08 s
TRANSCRIPT
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1824 N. L. S. Carnot (1796~1832)q1
W
q2
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Carnot cycle
(Carnot cycle) (cyclic loop)
(isothermal) (adiabatic)
T1 T2=- wcyc / q1
rev irrev
1
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T
T=
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CH03 Fig3.13
Carnot Heat Engine:On one cycle the
engine receives heat|q1| from the high T1reservoir, rejects heat
to the low temperaturesink, and does work |w|on its surroundings.
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CH03 Fig3.14
Plot of P versus V forthe working fluid in a
Carnot engine. Heat|q1| is absorbed inthe isothermal
expansion at T1, andheat |q2| is evolved inthe isothermalcompression at T1.
The other two stepsare adiabatics.
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Carnot cycle
Step 1, isothermal reversible expansion at the Th.
Step 2, reversible adiabatic expansion in which thetemperature falls from Th to Tc .
Step 3, isothermal reversible compression at Tc,
Step 4, adiabatic reversible compression, whichrestores the system to its initial state at the Th.
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Carnot cycle
Step 1, isothermal expansion at the Th:
w1 = - nRThln(VB/VA); U1 = q1+w1 = 0
Step 2, adiabatic expansion from Th to Tc :
q2 = 0; U2 = w2 = n Cv (Tc Th);
Step 3, isothermal reversible compression at Tc,w3 = - nRTcln(VD/VC); U3 = q3+w3 = 0
Step 4, adiabatic compression, from Tc to Th :
q4 = 0; U4 = w4 = n Cv (Th Tc);
0 = Ucyc = wcyc + qcyc; qcyc = - wcyc
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Carnot cycle
qcyc = - wcyc , that is q1 + q3 = -(w1 + w2 + w3 + w4)
= nRThln(VB/VA) - nCv(Tc -Th) + nRTcln(VD/VC) - nCv(Th Tc)
= nRThln(VB/VA) nRTcln(VC/VD)For two adiabatic steps 2 and 4:
(VC/VB) = (Th/Tc)Cv/R; (VD/VA) = (Th/Tc)Cv/R
So, VC/VB = VD/VA or VC/VD = VB/VA
- wcyc = q1 + q3 = nR (Th-Tc) ln(VB/VA)
q1 = - w1 = nRTh ln(VB/VA)
= - wcyc / q1 = (Th-Tc) /Th = 1 (Tc /Th)
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A general cycle can be dividedinto small Carnot cycles. Thematch is exact in the limit ofinfinitesimally small cycles.Paths cancel in the interior ofthe collection, and only theperimeter, an increasingly good
approximation to the true cycleas the number of cyclesincreases, survives. Because theentropy change around every
individual cycle is zero, theintegral of the entropy aroundthe perimeter is zero too.0==
perimeter
rev
all
rev
T
q
T
q
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Thermodynamic Temperature
For a reversible engine working between a hotsource Th and a cold sink at temperature T, then
T = (1 - ) Th
This expression enabled Kelvin to define thethermodynamic temperature scale in terms of theefficiency of a heat engine, a purely mechanical
basis.
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Suppose an energy qh (for example,20 kJ) is supplied to the engine and qcis lost from the engine (for example, qc= -15 kJ) and discarded into the cold
reservoir. The work done by the engineis equal to w=-(qh + qc); for example,-[ 20 kJ + (-15 kJ) ]= -5 kJ . The
efficiency is the work done divided bythe heat supplied from the hot source.
= - w / qh
All reversible engines have the sameefficiency regardless of theirconstruction.
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When energy leaves a coldreservoir as heat, the entropy ofthe reservoir decreases. Whenthe same quantity of energyenters a hotter reservoir, theentropy increases by a smaller
amount. Hence, overall there isan decrease in entropy and theprocess is spontaneous. Relative
changes in entropy are indicatedby the sizes of the arrows.
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The flow of energy asheat from a coldsource to a hot sink is
not spontaneous. Asshown here, theentropy increase of
the hot sink is smallerthan the entropyDECREASE of the cold
source, so there is anet decrease inentropy.
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The processbecomes feasible ifwork is provided to
add to the energystream. Then theincrease in entropy
of the hot sink canbe made to cancelthe entropy
decrease of theCOLD source.
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Carnot cycle
rev irrev
B A Th Tc Th
qh Tc qcqc : wB=qh-qc wA=qh-qcA B
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Suppose two reversibleengines are coupled
together and runbetween the same tworeservoirs.
C l
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Carnot cycle
A > B wA > wBB A wB
w = wA wB
ThTc qc-qc w=wA - wB
Kelvin wA wB I R
A B Th Tc A, B > A. B
B < A. A = B .
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The demonstration of theequivalence of theefficiencies of all reversible
engines working betweenthe same thermalreservoirs is based on the
flow of energy representedin this diagram. If engine Ais more efficient thanengine B, not all the workthat A produces is neededfor B to work.
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The net result is the coldsink unchanged, hotsource lost some energy
|qh qh|, and work hasbeen produced (w). Thenet effect of theprocesses is the
conversion of heat intowork without there beinga need for a cold sink:
this is contrary to theKelvin statement ofSecond Law.
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The process is spontaneousif overall the entropy of theglobal, isolated system (the
system plus itssurroundings) increases, sothat of the surroundingsmust increase in order for
the process to bespontaneous, which meansthat energy must pass from
the system to thesurroundings as heat.Therefore, less work than
U can be obtained.
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In this process, theentropy of the systemincrease; hence we can
afford to lose someentropy of thesurroundings. That is,some of their energy may
be lost as heat to thesystem. This energy canbe returned to them aswork. Hence the workdone can exceed U.