Details on the Carnot Cycle

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1 Details on the Carnot Cycle he isothermal expansion (ab) and compression (cd): 0 ( is constant and U() is a function U isothermal of only for an Ideal Gas.) V b QH Wab nrh ln Va (ab : isothermal expansion) V d V c QC Wcd nrc ln nrc ln ( Vc Vd) Vc Vd (cd : isothermal compression)

2 Details on the Carnot Cycle he adiabatic expansion (bc) and compression (da): Qadiabatic 0 (by definition) From Section 19.8, we learned that V 1 const for adiabatic processes. V V 1 1 H b C c V V 1 1 H a C d (bc: adiabatic expansion) (da: adiabatic compression) Dividing these two equations gives, V V b a V V c d

3 Efficiency of the Carnot Cycle From definition, we have 1 e Q Q C H Using our results for Q C and Q H from the isothermal processes, nrc ln Vc Vd C ln Vc Vd e 1 1 nr ln V V ln V V From the adiabatic processes, we have H b a H b a V V V V b a c d ln ln V V c V V b d a 1 e QC 1 1 Q H C H ( must be in K) (Carnot Cycle only)

4 Efficiency of the Carnot Cycle e carnot 1 C H (Carnot Cycle) General Comments: Higher efficiency if either C is lower and/or H is higher. For any realistic thermal process, the cold reservoir is far above absolute zero and C > 0. hus, a realistic e is strictly less than 1! (No 100% efficient heat engine) Realistic heat engines must take in energy from the high reservoir for the work that it produces AND some heat energy must be released back to the lower reservoir. (Kelvin-Planck s Statement) skip

5 Internal Combustion Engine he Otto Cycle A fuel vapor can be compressed, then detonated to rebound the cylinder, doing useful work. animation

6 he Otto Cycle intake exhaust intake a b b, c c d exhaust For the two constant V processes: 2 and 4, we can calculate, QH U2 ncv( c b) 0 QC U4 ncv( a d) 0 r is the compression ratio (8 to 13)

7 he Otto Cycle Applying the definition of efficiency, 1 Q 1 ncv a d 1 e Q nc C a d H V c b c b Now, we can utilize the two adiabatic processes: 1 and 3, V V 1 1 a a b b and V V 1 1 c c d d a 1 1 b rv V 1 V rv c d 1 r a 1 b c r d 1

8 he Otto Cycle Substituting b and c into the efficiency equation, we have, e r r r a d a d a d c b d a a d e 1 1 r 1 Using a typical value for the compression ratio r = 8 and = 1.40 gives, e 0.56 (or 56%) Note: his is a theoretical value. Realistic gasoline engine typically has e ~ 35%.

9 he Diesel Cycle Key difference: No fuel in cylinder at the beginning of the compression stroke (process 1) Fuel is injected only moments before ignition in the power stroke No fuel until the end of the adiabatic compression can avoid pre-ignition Compression ratio r value can be higher (15 to 20) Higher temperature can be reached during the adiabatic compression Higher e and no need for spark plugs D steam engine

10 Entropy Recall from a Carnot Cycle, we have derived the following relationship: Q Q C H C H C 0 H H C Q Q Formally, we can rewrite this as, Q cycle 0 (We have absorbed the explicit sign back into the Q variable.) where Q represents the heat absorbed/released along the isotherm at temp.

11 Entropy Any reversible cycles can be approximated as a series of Carnot cycles!

12 Entropy his suggests that the following generalization to be true for any reversible processes, cycle dq r 0 where, dq r is the infinitesimal heat absorbed/released by the system at an infinitesimal reversible step at temp. cycle denotes the integration evaluated over one complete cycle.

13 Entropy We have seen this property previously, cycle du 0 Changes in the internal energy U over a closed cycle is zero! his is a consequence of the fact that U is a state variable and du for any processes depends on the initial and final states only. dqr hus the result 0 indicates that there is another state variable S cycle such that, ds dq and ds 0 cycle his new state variable S is called the entropy of the system.

14 Entropy: Disorder Recall that the 2 nd Law of hermodynamics is a statement on nature s preferential direction for systems to move toward the state of disorder. Let see how Entropy is a quantitative measure of disorder. From our previous derivation, the quantity dq / was from the isothermal branch of the infinitesimal Carnot cycle. Let look at an isothermal expansion of an ideal gas microscopically: Intuitively, as the gas expands into a bigger volume, the degree of randomness for the system increases since molecules now have more choices (spaces) for them to move around. One can associate the increase in randomness to the ratio: V dv or V V ( stays the same avg. KE stays the same)

15 Entropy: Disorder Since this is an isothermal process, we have the following relation from the 1 st Law: dq du 0 dw nr PdV dv V 1 dq dv nr V So, the newly introduced macroscopic variable S (entropy), ds dq S J / K is directly proportion to the degree of disorder of the system. ds is an infinitesimal entropy change for a reversible process at temperature. For any finite reversible process, the total entropy change S is, S f i dq

16 2 nd Law (Quantitative Form) W environment system Q S S S 0 tot sys env ( S 0 reversible; S 0 irreversible) tot he total entropy (disorder) of an isolated system in any processes can never decrease. Nature always tends toward the macrostate with the highest S (disorder) [most probable (later)] in any processes. tot

17 Entropy Changes for Different Processes 1. General Reversible Processes: f S ds i f i dq r Note: S is a state variable, S is the same for all processes (including irreversible ones) with the same initial and final states! P i f V NOE: in most applications, it is the change in entropy S which one typically needs to calculate and not S itself.

18 Entropy Changes for Different Processes 2. Reversible Cycles: dq r Scycle ds cycle cycle 0 3. Any Reversible Processes for an Idea Gas: 1 st Law gives, ( V, ) (, V) i i f f du dqr dw dqr du dw ncvd PdV nr dqr ncvd dv V (Note: hru the Ideal Gas Law, P is fixed for a given pair of & V.)

19 Entropy Changes for Different Processes Dividing on both sides and integrating, f f dqr ncv d S i dv nr V i We have, f Vf S ncv ln nrln i Vi

20 Entropy: Disorder (General Reversible Process) In our derivation of S for an ideal gas through a general reversible process, we have the following relation, ds dq nc d dv nr V r V d S hermal agitations dv S V Availability of space

21 Entropy Changes for Different Processes 4. Calorimetric Changes: dq mcd f dq S i f i mcd If c is constant within temperature range, f S mcln i If c is a function of, S m f i c d

22 Entropy Changes for Different Processes 5. During Phase Changes (or other isothermal Processes): dq S 1 dq ( stays constant during a phase change.) Q S ml

23 Entropy Changes for Different Processes 6. Irreversible Processes: Although for a given irreversible process, we cannot write ds = dq r /, S between a well defined initial state a and final state b can still be calculated using a surrogate reversible process connecting a and b. (S is a state variable!) Example 20.8: (adiabatic free expansion of an ideal gas) Since Q=W=0, U=0. For an ideal gas, this means that =0 also. Although Q=0, but S is not zero! Initial State a: (V,) Final State b: (2V,) V, 2 V,

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