Lecture 08 s

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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

21

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.

-5

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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.