Heat Engines A heat engine is a device that absorbs energy as heat and does work on its surroundings while going through a cycle (so that it can do it again). [Typical types of work are turning a motor, lifting a material (e.g. pumping oil or water from the ground), turning wheels, etc.] Consider a fluid going through a cycle. (The fluid = the system = the engine.) The work on the system W = - P dV. Since we want the system to do work, we want W < 0. Therefore, an engine cycle is a clockwise loop on a PV diagram Let |Weng| = work done by the system in one cycle: |Weng| -W = + P dV = area enclosed by cycle. .
Since it is a cycle, Eint = Eint(f) – Eint(i) = 0. From 1st Law: total Q + W = 0 for the cycle. Q = - W = |Weng| During some parts of the cycle, heat may flow into the system, e.g. from a hot reservoir, (Qh) and at other parts it may exhaust heat, e.g. into a cold reservoir, (Qc): Q = |Qh| - |Qc| 1st Law: |Qh| - |Qc| = |Weng| i,f The engine efficiency e |Weng| / |Qh| = 1 - |Qc| / |Qh| What we want What we pay for What we waste
Kelvin-Planck Statement of the 2nd Law of Thermodynamics: The engine efficiency e |Weng| / |Qh| = 1 - |Qc| / |Qh| What we want What we pay for What we waste As industrial revolution began in 18th century, there were many attempts to create a “perfect engine”, with e = 1, i.e. |Qc| = 0. [“Perpetual motion machine of the second kind”]. By the end of the century, experience had shown this was impossible: Kelvin-Planck Statement of the 2nd Law of Thermodynamics: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work.
Kelvin-Planck Statement of the 2nd Law of Thermodynamics: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work. Real Heat Engine ( **The system (the engine) goes through a cycle: Eint = 0 (in arrows) = (out arrows). Objects in the surroundings do NOT go through a cycle: some reservoirs lose energy (via heat), some gain energy (via heat), and something has work done on it by the engine.
For an engine Eint (cycle) = 0 |Weng| = Q: 1st Law: Eint = Q + W For an engine Eint (cycle) = 0 |Weng| = Q: i.e. you cannot do better than break even. 2nd Law: |Weng| = |Qh| - |Qc| and |Qc| 0: i.e. you cannot even break even.
Express e in terms of P0 and V0. Problem: Calculate the efficiency of this quasi-static cycle for n moles of a high temperature diatomic gas (CP = 9/2 R, CV = 7/2 R, = 9/7). Express e in terms of P0 and V0. e = |Weng| / |Qh| = 1 - |Qc| / |Qh| ab: adiabatic: Q = 0, so only QC step is bc and the only heat in step is ca: |Qh| = |Qca| = nCV (Ta – Tc) |Qc| = |Qbc| = nCP(Tb-Tc) Adiabatic: PV = constant e = 1 - |Qc| / |Qh| = 1 – (CP/CV) (Tb-Tc)/(Ta-Tc) = 1 - (Tb/Tc-1) / (Ta/Tc -1) To find the temperature ratios, use the adiabatic piece, P0V0 = Pb(4V0) Pb = P0 (1/4) = 0.168 P0 (using = 9/7) Ta/Tc = P0/Pb = 1/0.168 = 5.94 Tb/Tc = 4 e = 1 – (9/7) (4-1) / (5.94 -1) = 0.22
Problem: Calculate the efficiency of this cycle for n moles of a diatomic ideal gas near room temperature (CP = 7/2 R, CV = 5/2 R, = 7/5). [Express e in terms of P0, V0, a, b.] Two heat in steps: |Qh| = |Q41 + Q12| Two heat out steps: |Qc| = |Q23 + Q34| |Qh| = n[CV (T1-T4) + CP (T2-T1)] = nR [5/2 (bP0V0 – P0V0)/nR + 7/2 (abP0V0- bP0V0)/nR] = P0V0 [7/2 (a-1)b + 5/2(b-1)] |Qc| = n[CV(T2-T3) + CP(T3-T4)] = P0V0 [5/2 (b-1)a + 7/2(a-1)] V0 aV0 P0 bP0 e = 1 - |Qc| / |Qh| e = 1 – [5/2 (b-1)a + 7/2 (a-1)] /[7/2 (a-1)b + 5/2(b-1)] e = [ab – b – a +1] / [7/2 (a-1)b + 5/2(b-1)] e = [(a-1) (b-1)] / [7/2 (a-1)b + 5/2(b-1)] |Weng|/P0V0 [so could have skipped calculating Qc.]
Gasoline Engine The material at start of cycle (gasoline + air) is different from the material at end (burnt fuel +air), so not really “thermodynamic” cycle,, can model as a cycle by assuming material at intake is “same” as exhausted material (not great approx!). Cycle steps are NOT really quasi-static, but we’ll analyze as if they are. (Efficiency we calculate will therefore be larger than actual efficiency.)
“Otto Cycle” Intake of air/gasoline mixture: V2 V1 (O A) Compression: adiabatic (fast) compression from V1 V2 (A B) Ignition: Spark plug fires and gas/air mixture explodes very quickly (so V constant). Qh is the chemical energy released by the combustion. (BC).
d) Power stroke: Hot gas pushes piston down rapidly ( adiabatically) (C D) This is the step in which work is done, which gets transmitted to the wheels. e) “Heat exhaust”: Exhaust valve opens so hot material (carrying energy) is ejected. We’ll model as if only energy (heat) leaves, not material so this is the Qc step. (D A). f) Exhaust stroke: remaining spent material is ejected as cylinder closes (V1 V2) (A O) [Note we are assuming that steps (f) and (a) “cancel”, allowing system to be analyzed as a thermodynamic cycle.]
If one assumes that the air/gas mixture is an ideal gas (also poor approximation) with CP/CV = , then (example 22.5 on p. 667): e = 1 – (V2/V1)-1 . e.g. compression ratio V1/V2 = 8 and = 1.4, e = 0.56. In practice, e < 20%: non-ideal gas, friction, heat escapes through cylinder walls, not quasi-static, …
Refrigerators and Heat Pumps Refrigerators, air conditioners, and heat pumps are devices which absorb heat from cold reservoirs (Qc) and exhaust it into hot reservoirs (Qh). For a refrigerator (or air conditioner), the goal is to remove the heat from the cold reservoir (inside of refrigerator or house) that may have leaked in due to imperfect insulation, radiation, etc. For a heat pump, the goal is return heat to the hot reservoir (inside of house) that may have leaked out due to imperfect insulation, radiation, etc. It would be great if these goals, i.e. having heat flow from the cold reservoir to the hot reservoir, could be accomplished without doing any work, but experience has shown that this is impossible: heat spontaneously flows from hot to cold, but never from cold to hot. [This is another statement of the second law.]
Clausius Statement of Second Law of Thermodynamics: It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work. [The Kelvin-Planck Statement and the Clausius Statement are mathematically equivalent!] Real Heat Pump or Refrigerator or refrigerator condenser evaporator
The directions of the arrows for Qh, Qc, and W are the opposite as the directions in an engine, so in a PV diagram, the cycle for a refrigerator or heat pump is counter clockwise (e.g. positive work is done on the system (the refrigerant).
Refrigerator (air conditioner): Coefficient of Performance = |Qc|/|W| Heat Pump: Coefficient of Performance = |Qh|/|W| COP(heat pump) = 1/e(engine) COP(refrig) = (|Qh|-|W|)/|W| = |Qh|/|W| -1 = 1/e(engine) -1 What you want What you pay for
COP(heat pump) = 1 /e(engine) = 1/ 0.22 = 4.5 Problem: Calculate the COP of this cycle, used as either a refrigerator or heat pump, for n moles of a high temperature diatomic gas (CP = 9/2 R, CV = 7/2 R, = 9/7). e(engine) = 0.22 Adiabatic: PV = constant COP(heat pump) = 1 /e(engine) = 1/ 0.22 = 4.5 [Note an ideal electric furnace has |Qh|=|W|, so COP = 1] COP(refrig)= 1/e(engine) -1 = 3.5
Practical refrigerators and heat pumps use a material with a low, pressure dependent, boiling point and high latent heat of vaporization. It evaporates next to the cold reservoir, where its pressure is low, absorbing the latent heat and condenses at the hot reservoir, where its pressure is high, exhausting the latent heat The traditional refrigerant was “Freon” (chlorofluorcarbon: CCl2F2). However, if this escapes into atmosphere, it cause breakdown of the ozone layer. Freon is being replaced with hydrofluorocarbon (F4C2H2). These do not work well if the cold reservoir is at temperatures near 0oC, so heat pumps are generally augmented with electric furnaces for days near/below freezing.