Presentation on theme: "VII. The second low of Thermodynamics"— Presentation transcript:
1 VII. The second low of Thermodynamics 1. Reversible and irreversible processesMechanicsThermodynamicsHeat can flow spontaneously from a hot object to a cold object;heat will not flow spontaneously from a cold object to a hot object.2. Heat enginesH.E. transforms heat partly into work or mechanical energyWorking substanceCycle process: ΔU = 0ΔU = Q - W = 0 W = Q = QH + QC = |QH| - |QC|THefficiency:|QH|Wengine3. Refrigerators|QC|coefficient ofperformance:TC
2 4. The second law of thermodynamics and heat engines It is impossible to build a heat engine with 100% efficiency5. The Carnot Cycle (reversible)Maximum efficiency:PabTHdcTLV6. The Carnot Cycle and the second law of thermodynamicsNo device is possible whose sole effect is to transform a given amount of heat completely into workNo device is possible whose sole effect is to transform heat from one system at temperature TL into a second system at higher temperature TH
3 Example 1: A nuclear power plant generate steam at 6300 C. The steam is sent through a series of turbines and exits at 1000 C.What is the maximum possible efficiency of the power station?TH = 630ºCTC = 100ºCεC - ?
4 Example 2: An ice-making machine operates in a Carnot cycle. It takes heat from water at 0.0C and rejects heat to a room at 24.0C.Suppose that 50.0 kg of water at 0.0C are converted to ice at 0.0C.How much energy must be supplied to the device?Heat of fusion of water is 334*103 J/kg.TH = 24.0CTC = 0.0Cm = 50.0 kgL = 334*103 J/kgW - ?
5 The thermal efficiency of the heat engine is ___ %. QuestionIn one cycle, a heat engine does 10 J of work and exhausts 20 J of waste heat.The thermal efficiency of the heat engine is ___ %.203350
6 The temperature of the hot reservoir is ___ K. QuestionIn a Carnot heat engine, the cold temperature reservoir is at 300 K (= 27 ˚C) and the thermal efficiency is 50%.4006008001000The temperature of the hot reservoir is ___ K.
7 QuestionA refrigerator has a coefficient of performance K = Work is done on the refrigerator at a rate Pwork = dW/dt = 100 Watts.50100200400The rate at which heat is removed from inside the refrigerator is d|QC|/dt = ___ Watts.
8 The internal-combustion engine A fuel vapor can be compressed, then detonated to rebound the cylinder, doing useful work.
9 r - compression ratio γ – CP/CV 6. Otto cycle Example: An automobile engine has a compression ratio of 10.0.The maximum theoretical thermal efficiency of the engine is ___ %. ( = 1.40 for air)