Heat Engines A heat engine – any device that is capable of converting thermal energy (heating) into mechanical energy (work). We will consider an important class of such devices whose operation is cyclic. Heating – the transfer of energy to a system by thermal contact with a reservoir. Work – the transfer of energy to a system by a change in the external parameters (V, EM fields, etc.). The main question we want to address: what are the limitations imposed by thermodynamic on the performance of heat engines? Perpetual Motion Machines are Impossible Perpetual Motion Machines of the first type violation of the First – these designs seek to create the energy Law (energy required for their operation out of nothing. conservation) Perpetual Motion Machines of the second type – these designs extract the energy violation of the Second required for their operation in a manner Law that decreases the entropy of an isolated system. hot reservoir Word of caution: for non-cyclic processes, 100% of heat can be TH transformed into work without violating the Second Law. impossible cyclic heat Example: an ideal gas expands heat engine isothermally being in thermal contact work with a hot reservoir. Since U = const at T = const, all heat has been transformed into work. Despite the fact that successful isolated system perpetual motion devices are physically impossible in terms of the current understanding of the laws of physics, the pursuit of perpetual motion remains popular. Essential parts of a heat engine (any continuously operating reversible device generating work from “heat”) An engine can get rid of all the entropy received from the hot reservoir by transferring only part hot reservoir, Th of the input thermal energy to the cold reservoir. heat Q heat absorbed h work W work produced entropy Qc heat expelled heat Thus, the only way to get rid of the accumulating entropy is to absorb more internal energy in cold reservoir, Tc the heating process than the amount converted to work, and to “Working substance” – the system that “flush” the entropy with the flow absorbs heat, expels waste energy, and of the waste heat out of the does work (e.g., water vapor in the steam system. engine) An essential ingredient: a temperature difference between hot and cold reservoirs. W Efficiency of an engine: e ⌘ Qh What is the maximum possible efficiency for given Th and Tc? Qh = Qc + W Qc The First Law tells e =1 us than the efficiency − Qh cannot exceed 1 The total entropy (of engine + environment) cannot decrease. Since the state of the engine must be unchanged at the end of the cycle, expelled entropy ≧ absorbed entropy. hot reservoir, Th ∆Sh ∆Sc Q ≥ ΔS = h Q h T h h Qc Qh heat Tc ≥ Th work Tc W e 1 entropy − Th heat QC ΔSC = QC TC cold reservoir, TC • Any difference TH –TC ≠ 0 can be exploited to generate mechanical energy. • The greater the TH –TC difference, the more efficient the engine. • Energy waste is inevitable. 0 0 Example: In a typical nuclear power plant, TH = 300 C (~570K), TC = 40 C (~310K), and the maximum efficiency emax=0.45. If the plant generates 1000 MW of “work”, its waste heat production is at a rate ⎛⎞1 δδδδQQWWCH=−=⎜⎟−≈1 1220 MW ⎝⎠e i.e., more fuel is needed to get rid of the entropy then to generate useful power. Perfect Engines (no extra S generated) The Carnot cycle is a theoretical thermodynamic cycle proposed by Carnot in 1823. It is the most efficient cycle for converting a given amount of thermal energy into work. ∆Sh = ∆Sc P Sadi Carnot (1796-1832) 1 1 – 2 isothermal expansion (in contact with T ) absorbs h 2 – 3 isentropic expansion to T heat C 3 – 4 isothermal compression (in contact with TC) 2 4 – 1 isentropic compression to Th (isentropic ≡ adiabatic+quasistatic) Th 4 3 TC TC e =1− rejects heat max TH V S On the S -T diagram, the work done is 3 2 the area of the loop: ∫dU = 0 = ∫TdS − ∫ PdV entropy 4 1 Qh = Th(Sh Sc) contained in gas − T T T Qc = Tc(Sh Sc) c h − W =(T T )(S S ) h − c h − c T e =1 c − Th Carnot's theorem: No engine operating between two heat reservoirs can be more efficient than a Carnot engine operating between those same reservoirs. The entropy of the reservoir decreases by Qh/Th; the entropy of the agent increases by Qh/Tgas. To avoid making new entropy, Tgas needs to be as close to Th as possible. By the same token, , Tgas needs to be as close to Tc as possible. A generalized thermodynamic cycle taking place between a hot reservoir at temperature Th and a cold reservoir at temperature Tc. By the second law of thermodynamics, the cycle cannot extend outside the temperature band from Tc to Th. The area in red Qc is the amount of energy exchanged between the system and the cold reservoir. The area in white W is the amount of work energy exchanged by the system with its surroundings. The amount of heat exchanged with the hot reservoir is the sum of the two. If the system is behaving as an engine, the process moves clockwise around the loop, and moves counter- clockwise if it is behaving as a refrigerator. The efficiency of the cycle is the ratio of the white area (work) divided by the sum of the white and red areas (heat absorbed from the hot reservoir). A Carnot cycle taking place between a hot reservoir at temperature Th and a cold reservoir at temperature Tc. Refrigerators The purpose of a refrigerator is to hot reservoir, TH make thermal energy flow from cold to hot. The coefficient of performance for Q a fridge: S H heat dumped Δ H = QH T QC QC 1 H COP ≡ = = W QH − QC QH / QC −1 heat T COP ≤ COP = C work W max TH −TC entropy heat work supplied Q T H ≥ H QC TC QC ΔSC = Qc heat sucked off TC COP is the largest when TH and TC are close to each other! cold reservoir, TC For a typical kitchen fridge TH ~300K, TC~ 250K ⇒ COP ~ 6 (for each J of el. energy, the coolant can suck as much as 6 J of heat from the inside of the freezer). P 1 rejects heat We can create a refrigerator by running a Carnot engine 2 backwards: the gas extracts heat from the cold reservoir and TH deposit it in the cold reservoir. 4 3 TC absorbs heat V .