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Home , Carnot cycle, Heat engine

PHY474

Engines and

Ulrich Z¨urcher∗ Physics Department, Cleveland State University, Cleveland, OH 44115 (Dated: October 7, 2008)

PACS numbers: 05.70

I. notation,

A heat engin is any device that absorbs heat and con- Qh = Qc + W. (2) verts parts of that into . The familiar internal If we use this equation to elimintae W , we have comustion engin used in automobiles does not actually absorb heat, but we can pretend that the thermal energy Q − Q Q e = h c = 1 − c . (3) comes from outside rather than inside and treat it as a Q Q heat . h h Unfortunately, only part of the energy absorbed as heat Thus the efficiency cannot be greater than 1, and can cann be converted to work by a . The reason equal 1 only if Qc = 0. is that the heat, as it flows in, brings also along To proceed, we must also invoke the second law, which which must somehow be disposed of before the cycle can tells us that the total entropy of the engine plus the sur- start over. To get rid of the entropy, every heat engine rounding can increase but not decrease. Since the state must dump some into its environment. The of the engine must be unchanged at the end of a cycle, work produced by the negine is the difference between the entropy it expels must be at least as much as the the heat absorbed and the waste heat expelled. We can entropy it absorbes. Now the entropy extracted from the say a great deal without knowing anything about how the hot reservoir is Qh/Th, while the entropy expelled to the engine actually works. cold reservoir is Qc/Tc. So the second law tells us The heat absorbed by the engine comes from a place Q Q Q T called hot reservoir, while the waste heat is dumped c ≥ h , or c ≥ c . (4) into the cold reservoir. The of these Tc Th Qh Th reservoirs, Th and Tc, are assumed fixed. We use the We conclude symbol Qh for the heat absorbed from the hot reservoir in some given period, and Q for the heat expelled to the c Tc cold reservoir. The net work done by the engine during e ≤ 1 − . (5) T this time will be W . All three of these symbols will rep- h resent positive quantities [opposite from the convention This is our desired result. in chapter 1!]. Problem 4.1. Rectangular cycle in PV -diagram with The benefit of operating a heat engine is the work pro- V2 = 3V1 and P2 = 2P1. The total work done by engine duced W . The cost of operation is the heat absorbed Qh. during cycle: We define the efficiency of an engine e as the benefit/cost ratio: W = (P2 − P1) · (V2 − V1) = P1 · 2V1 = 2P1V1. (6) benefit W e = = . (1) Note that heat is absorbed for part A and B of the cycle. cost Qh Part A: we have W = 0 [since V = const]. It follows We ask: for given values of T and T what is the max- 5 h c Q = ∆U = Nk(T − T ) imum possible efficiency? To answer this question, all 2 2 1 we need are the first and second law of , 5 5 = (P V − P V ) = (P − P ) V . (7) plus the assumption that the engine operates in cycles, 2 2 1 1 1 2 2 1 1 returning to its original state at the end of each cycle of operation. Part B: Work done W = −P2(V2 − V1). Thus The first law of thermodynamics tells us that the en- ergy is conserved. Since the state of the engine must be Q = U − W + P2(V2 − V1) (8) unchanged at the end of the cycle, the energy it absorbes 5 = Nk(T − T ) + P (V − V ) must be precisely equal to the energy it expels. In our 2 3 2 2 2 1 5 = (P V − P V ) + P (V − V ) 2 2 2 2 1 2 2 1 7 ∗ = P (V − V ). (9) Electronic address: [email protected] 2 2 2 1 2

So the total heat absorbed by the engine is and

Z V3 5 7 V3 Qh = (P2 − P1) V1 + + P2(V2 − V1) Qc = |W34| = P dV = NkTc ln . (16) 2 2 V4 V4 5 7 33 = P V + · 2P · 2V = P V . (10) 2 1 1 2 1 1 2 1 1 The efficiency follows

The efficiency is therefore Tc ln(V3/V4) e = 1 − . (17) Th ln(V2/V1) 2P V 4 e = 1 1 = ' 12%. (11) 33 For the adiabatic expansion we have 2 P1V1 33 f/2 f/2 In the above derivation, we have T2 = 2T1 and T3 = V3Tc = V2Th , (18) 3T2 = 6T1. We thus set Tc = T1 and Th = 6T1. This gives for the maximum efficiency, while for the adiabatic compression

f/2 f/2 T1 V4Tc = V3Th . (19) emax = 1 − = 83%. (12) 6T1 This gives V3/V4 = V2/V1 and the logarithm cancels. Problem 4.6 The problem with the ideal Carnot cy- II. cle is that it runs infinitely slowly and thus it provides infinitesimal amount of work. More realistic version ver- sion when the working is at Thw < Th as it We want the gas to absorb some heat Qh from the hot reservoir. In this process, the entropy of the reservoir de- absorbes heat from the hot reservoir, and at the temper- creases by Q /T , while the entropy of the gas increases ature Tcw > Tc as it expels heat to the cold reservoir. h h Rates of by Qh/Tgas. To avoid making any new entropy we would need to make Tgas = Th. This is not quite possible be- Q h = K(T − T ), (20) cause heat won’t flow between objects at the same tem- ∆t h hw perature. So we make Tgas just slightly less than Th, and Q c = K((T − T ). (21) keep the gas at this temperature [by letting it expand] as ∆t cw c it absorbs heat: isothermal expansion. Similarly, during the portion of the cycle when the We assume that no extra entropy is created except during gas is duming wast heat into the cold reservoir, we want heat transfers, i.e., Qh/Thw = Qc/Tcw, so that temperature to be only infinitesimally greater than Tc to T − T T avoid creating any new entropy. And as the heat leaves h hw = hw . (22) the gas, we need to compress it isothermally to keep it Tcw − Tc Tcw at this temperature. We assume that the time for the adiabatic processes can So we have an isothermal expansion at a temperature be neglected. The time required for the heat Q to flow just less than T , and an isothermal compression at tem- h h in is: perature just greater than Tc. But how do we get from one temperature to the other? We don’t want any heat Q ∆t = h . (23) to flow in or out when the gas is at intermediate tem- K(Th − Thw) peratures so these intermediate steps must be adiabatic: Carnot cycle. The total time for the cycle is then 2∆t. The Since we have follows

Q Q W Qh − Qc K(Th − Thw) h = c , (13) Power = = Th Tc 2∆t 2 Qh   K Tcw we have Q /Q = T /T , and the efficiency of the Carnot = 1 − (T − T ) . (24) c h c h 2 T h hw engine follows hw

T From eq (2.10) we have e = 1 − c . (14) Th T T cw = c . (25) Useful: direct computation. Thw 2Thw − Th We have It follows

Z V2   V2 K Tc Qh = |W12| = P dV = NkTh ln , (15) Power = 1 − (Th − Thw) . (26) V1 V1 2 2Thw − Th 3

We want to maximize the power and not the efficiency: We then have the inequality: dPower 1 T = 0 = (27) COP ≤ = c . (36) dThw Th/Tc − 1 Th − Tc 2 K 2Tc(Th − Thw) − (2Thw − Th) + Tc(2Thw − Th) . For our kitchen fridge: T = 298 K and T = 255 K. We 2 (2T − T )2 h c hw h get COP=5.9, i.e., for each joule of , the We get the quadratic equation: coolant can suck 5.9 J of heat from the inside of fridge. Heat dumped into kitchen is 6.9 J. 2 2 0 = −4Thw + 4ThThw + TcTh − Th . (28) Problem 4.14: . An electrical device The solutions are that a building by pumping heat in from the cold outside. It’s the same as a but its purpose Th 1p 1  p  is to warm the hot reservoir rather than to cool the cold Thw = ± TcTh = Th ± TcTh . (29) 2 2 2 reservoir [even though it does both]. We define Qh as < the heat pumped into cold reservoir, Qc heat taken from The minus sign is unphysical: consider Tc ∼ Th which (cold) outdoors, and W as the electrical energy used by would then give Thw ' 0 but Thw can’t be lower than Tc. pump. The benefit is the heating of inside and the cost For Tcw we find is the electric bill, Tc 1  p  T = T = T + T T . (30) Qh cw hw 2T − T 2 c c h COP = . (37) hw h W Now we can compute the efficiency: The first law gives, Qh = Qc + W so that Qc Tcw e = 1 − = 1 − Qh 1 Qh Thw COP = = > 1. (38) √ Q − Q 1 − Q /Q 1 (T + T T ) r T h− c c h = 1 − 2 c √ c h = 1 − c . (31) 1 The entropy expelled during a cycle must always be at 2 (Th + TcTh) Th least as great as the entropyabsorbed, so ◦ ◦ For Th = 600 C and Tc = 25 C, we have Qh Qc Tc Qh r298 ≥ , or ≥ . (39) e = 1 − = 42%. (32) Th Tc Th Qc 873 This gives for COP For the ideal (but unrealistic!) Carnot we get eideal = 1 − (298/873) = 66%. 1 T COP ≤ = h . (40) 1 − Tc/Th Th − Tc III. REFRIGERATORS For an electrical heater, all electrical energy is con- verted to heat Qh = W so that the COP =1. For a A refrigerator is a heat engine operated in reverse, heat pump operating between 25◦ C and 0◦ C, we have more or less: reverse arrows in flow diagram. The heat COP = 298/25 ' 12. While this is idealized and thus sucked out of the cold reservoir (the inside of the fridge) unrealistic, the heat pump is much more efficient than is Qc, while the electrical energy form wall outlet is W . direct conversion of electrical energy. Waste heat Qh damped into your kitchen. We define coefficient of performance (COP) as IV. REAL HEAT ENGINES benefit Q COP = = c . (33) cost W Note that the “heat reservoir” is produced internally We use the first law and find by burning . The result of burning is a gas at high temperature and , as if had absorbed heat from Q 1 COP = c = . (34) an external source. Qh − Qc Qh/Qc − 1 Otto engines [found in almost all gasoline-burning cars]; Note that there’s no obvious upper limit on this quan- see problem 4.18. Heat is added in step 2-3, while heat is tity yet; in particular, the first law allows the COP to be expelled in steps 4-1. No work is done during these steps greater than 1. Because entropy flows in opposite direc- since the is constant. Thus, tion [as compared to heat engine], the second law tells f f us Q = U − U = Nk(T − T ) = V (P − P )(41) h 3 2 2 3 2 2 2 3 2 Qh Qc Qh Th f f ≥ , or ≥ . (35) Qc = U4 − U1 = Nk(T4 − T1) = V1(P4 − P1)(42) Th Tc Qc Tc 2 2 4

The ratio Qc/Qh follows ∆Q1 and ∆Q2 by a reversible process. The change in entropy of the whole system is Q V (P − P ) c = 1 4 1 . (43) Qh V2(P3 − P2) ∆Q1 ∆Q2 ∆S1 + ∆S2 = + ≥ 0, (53) T1 T2 We have the equation for PV γ = const, or which we write as

γ γ γ γ P4V1 = P3V2 ,P1V1 = P2V2 . (44) ∆Q1T2 + ∆Q2T1 ≥ 0. (54)

Sove for P4 and P1: Since masses are same, we have  γ  γ  γ V2 V2 V2 ∆Q1 ∆T1 P4 − P1 = P3 − P2 = (P3 − P2) . = . (55) V1 V1 V1 ∆Q2 ∆T2 (45) The above ratio follows It follows  γ  γ−1 Qc V1 (P3 − P2) V2 V2 (∆T )T + (∆T )T = ∆(T T ) ≥ 0. (56) = = . (46) 1 2 2 1 1 2 Qh V2 (P3 − P2) V1 V1 This implies that the geometric mean of the tempera- : see problem 4.20. The heat input is (dur- tures cannot decrease during the process [though√ it may ing 2-3): increase]. The common final temperature is T1T2. The maximum energy that can be taken out is Qh = (U3 − U2) + P2(V3 − V2)   f T1 + T2 p = Nk(T − T ) + P (V − V ) Wmax = − T1T2 mc, (57) 2 3 2 2 3 2 2 f + 2 = P2(V3 − V2). (47) where c is the specific heat of water and m is the mass 2 of water. The expelled heat [“waste”] is during step 4-1:

f V. JOULE-THOMSON PROCESS, OR Qc = U4 − U1 = V1(P4 − P1). (48) 2 THROTTLING PROCESS The ratio is now Liquid pushed through a plug by a piston exerting Q (f/2)V (P − P ) 1 V (P − P ) c = 1 4 1 = 1 4 1 . (49) pressure Pi, while a second piston, exerting pressure Pf , Qh (f + 2/2)P2(V3 − V2) γ P2(V3 − V2) moves backwards making room. We have no heat flow during the process so that We have

γ γ γ γ Uf − Ui = Q + W = 0 + Wleft + Wright P4V1 = P2V3 ,P1V1 = P 2V2 . (50) = PiVi − Pf Vf (58) Solve for P4 and P1 and plug in: or, re-arranging the terms Q 1 V V γ V γ  c = 1 3 − 2 . (51) Qh γ V3 − V2 V1 V1 Uf + Pf Vf = Ui + PiVi. (59) This can be simplified to We defined the as H = U + PV , and thus con- clude that the enthalpy is constant for the throttling pro- Q V γ−1 1 (V /V )γ − 1 c = 2 · 3 2 . (52) cess: Qh V1 γ (V3/V2) − 1 Hf = Hi. (60) An unusual engine: Two thermally insulated contain- ers hold identical masses of water. The water is at For an H = U + PV = fNkT/2 + NkT = temperature T1 in one of them, but at temperature T2 (f/2+1)NkT which implies that the temperature is con- (T2 > T1) in the other. What is the maximum amount stant. For a , we have potential energy due to in- of work that this system can do if it used as a heat en- teractions between molecules, H = Upot +Ukin +PV : at- gine? Take the specific heat of water as constant over the tractive at long distances and repulsive at short distances. working range. Under normal circumstances attractive force dominates, Solution: We use ∆S ≥ ∆Q/T . Let ∆T1 and ∆T2 denote so that Upot < 0. Thus Upot becomes less negative as the small changes in temperatures of the bodies at tempera- pressure drops [gas is expanding], and the kinetic energy ture T1 and T2, respectively, as a result of receiving heat drops, i.e., liquid is cooling down.