Heat Engines and the Second Law of Thermodynamics

Size: px

Start display at page:

Download "Heat Engines and the Second Law of Thermodynamics"

Pamela Preston
6 years ago
Views:

1 Heat Engines and the Second Law of hermodynamics here are three equivalent forms of the second law of thermodynamics; will state all three, discuss: I. (Kelvin-Planck) It is impossible to construct an engine, operating in a cycle, the only effect of which is to remove heat from a reservoir and transform it entirely into work. (See HRW page 49.) II. (Clausius) It is impossible to construct an engine, operating in a cycle, the only effect of which is to remove heat from a cold reservoir and transfer it to a warmer one. (See HRW, page 495.) III. One can define a state function, the entropy S, such that d d S, S = For any process whatever, S 0 ; in particular S = 0 reversible process S > 0 irreversible process he entropy is thus a quantitative measure of irreversiblilty. Remember our two examples: adiabatic free expansion, and dropping a hot object into a bucket of water. Let s examine these notions by considering the nature of a heat engine how does one turn heat into work; and what is all this talk about cycles. Steam Engine alve alve 2 to atmosphere piston and cylinder water burner

2 his engine, like all heat engines, works in a cycle:. valve open, valve 2 closed, piston at top of cylinder: Steam at high pressure from the high temperature boiler enters the cylinder. 2. Since the pressure of the steam is higher that atmospheric pressure, the steam expands, pushes the piston down, and DOES POSIIE WORK. 3. close valve, open valve 2: o continue to do work, the piston must return to its original position. Hence the external world (i.e., us) pushes the piston up, does a (smaller) negative work, rejects the used steam, now at a lower pressure, to the atmosphere. Exercise: What might this process look like on a p diagram? Consider a more abstract version of this cycle:. Heat is taken from some system, or reservoir, at a high temperature. 2. Net work is done on the external world. 3. Left-over heat is returned to a second, colder reservoir one at a lower temperature. Definition: A reservoir is some system which can absorb or reject any amount of heat without having its temperature change significantly. For most processes, Lake Superior is a pretty good heat reservoir. o show the nature of this cycle more clearly, consider an abstract heat engine a copper cylinder and piston, with water inside. We can move this system back and forth from a hot plate to a block of ice. water 2 Hot plate h Ice c Now consider our cycle:. Heat is conducted from the hot reservoir into the cylinder. he water is heated to steam at some high temperature and pressure (say, 50 K). he steam, under pressure, expands, pushes up the piston, and 2

3 2. does work. 3. Eventually, the piston reaches the top of the cylinder. Now, we slide the cylinder over to the block of ice. A heat 2 moves from the cylinder to the ice. he used steam in the piston condenses, the piston falls to its original position. We can now slide the cylinder back and run the cycle again. Note the following: A. he engine necessarily operates in a cycle that is, it must return to the same point on the p diagram if it is to continue doing useful work. B. Since the engine returns to its starting point, we know that over one cycle, E int = 0. Consequently, from the first law of thermodynamics (that is, from the conservation of energy), E = 0 = W int 2 Be sure that you understand the signs. Consequently, W = 2. Notice that one implication of this equation is that heat cannot be transformed entirely into work some heat must be rejected to the cold reservoir! We can define the efficiency of such an engine: W 2 2 eff = = =. We often represent an engine with the following diagram: h 2 W c 3

4 Here we have the beginning of the Kelvin-Planck statement of the second law. he rest will follow shortly, when we answer the question: What is the best we can do? We can never make the efficiency =. But how large can we make it? One way to find out is through a careful analysis of a Carnot cycle. We ll postpone that method for a while. Instead, consider the third formulation of the second law, involving the entropy. Recall the definition: d ds = (units are J/K; is always in K). herefore 2 d S2 S = S =. Consider first a SPECIAL CASE: = constant. in this case, S =. > 0 heat is absorbed Recall our sign convention: < 0 heat is given off Example: What is the entropy change as kg of liquid water turns into steam at 00 C (or K)? hus, one can say that J/kg kg S = = = J/K K latent heat S = or S = Lm In fact, many tables list not the latent heat of steam as a function of temperature, but instead the entropy changes at the boiling point. (Remember that the boiling point is not a fixed temperature; it varies as a function of external pressure, as we have seen.) Continue the example. Now suppose that we do the process the other way kg of steam at K condenses into water. For this case, J/kg kg S = = = J/K K What happened? I claimed that S 0 always. Long pause to reflect! 4

5 Here a problem arose because I did not consider the entire process. Clearly a heat J is rejected from the steam as it condensed. WHA HAPPENED O HA HEA?? Answer: It must have been absorbed by a reservoir at some LOWER temperature. Hence we did not calculate the complete entropy change. Suppose the heat was absorbed by a reservoir at 99 C (372.6 K). hen apparently J J Stotal = + = =.9 J/K K In other words, we must consider the entire process. What we have done, then, in cooling steam, is to transform some heat from a hot reservoir h to a cooler one c, so that h c S = + = > 0. h c h c Be sure you understand the signs. In this process, of course, no work at all is done. Note also that S 0 as h c. hus, a process is which heat is transferred across a vanishingly small temperature difference is reversible. Consider our abstract heat engine again: water 2 Hot plate h Ice c Consider only the engine: After a complete cycle, Eengine = 0 and Sengine = 0 because both E and S are state functions, and the engine returns to its original state. herefore the total entropy change (physicists are sometimes a bit grandiose, and speak of the total entropy change of the universe) is S = Shot plate + Sice 2 = + > 0 h Note, among other things, that in order for S > 0, we must have 2 0 that is, some of the heat must be rejected to the cold reservoir (Kelvin-Planck statement). c 5

6 Consider the implications for efficiency: We found earlier that W eff = 2 =. But from the above, we also have or h c 2 or c h 2 c or, finally, h 2 c And therefore 2 eff = c h he equality corresponds to the very best we can do zero total entropy change! Any real engine will have a smaller efficiency that is, will transform a smaller fraction of heat into work. Here, we have a connection between heat engines, and our notions of reversible and irreversible processes an ideal heat engine is reversible, and is also the most efficient engine possible. In the above equation, the equality holds. A Carnot engine is an example of such an ideal engine. A careful treatment of the Carnot engine shows that both of the following statements are true: total entropy change is zero; and all heat transfers are at constant temperature that is, all heat transfers are across vanishingly small temperature differences! hus, the Carnot engine is important because it is the best we can do! Here is a sketch of a Carnot cycle: a h d H b L c 6

7 See Fig. 2-8, page 489 of HRW for a more complete version of this figure. Extra Credit problem: Here is the extra-credit problem I would like you to do: Consider a Carnot cycle. he notation is the same as in Fig (a) Show that the work done along the isotherm ab is d W pd nr nr b b ab = = H H a = a ln (b) Show that the work done along the adiabat bc is c c K Wbc = pd = d Kc Kb b = γ b γ Simplify this result and show that P b b P c c Wbc = = nr( H L ) γ γ γ ( ) (c) Repeat the calculation for the second adiabat, and show that the total work along the two adiabats is zero one is the negative of the other. (d) Similarly, show that the total work done along the two isotherms is b d b c Wnet = nr H ln + C ln = nr H ln C ln a c a d hen use Eq , in the form = γ γ H b L c and b a to show = γ γ H a L d net ( ) W = nr (e) Finally, recall the definition of efficiency, W ε=. H H L ln b Also, remember that along an isotherm, = W, because the change in internal energy is zero. As a result, show that the efficiency ε of the Carnot cycle is given by a 7

8 H L L ε= =. H H 8

Physics 202 Homework 5

Physics 202 Homework 5 Apr 29, 2013 1. A nuclear-fueled electric power plant utilizes a so-called boiling water reac- 5.8 C tor. In this type of reactor, nuclear energy causes water under pressure to boil