2300_les13 - V. The Second Law of Thermodynamics A....

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Unformatted text preview: V. The Second Law of Thermodynamics A. Introduction 1. The first law: energy is conserved 2. The second law: certain processes do occur and certain processes don’t a. example of a mechanical process (assume adiabatic, pistoncylinder device) If the pins are removed, our Gas at Vacuum experience tells us P pressure P which way the piston will move. b. example of a thermal process Insulating If the barrier is barrier removed, our experience tells us TL which way heat will flow. Low T reservoir TH High T reservoir Lesson 13, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics c. example of a less obvious process It was easy to predict what would happen in the previous two examples, based on our experience. Consider a less obvious example: Compressed air Cold air Magic Vortex Tube Hot air How can we use our experience to know whether this process is possible? Is there a way to generalize our experience with simple processes so that we can make decisions about more complex processes? Lesson 13, Geof Silcox, Chemical Engineering, University of Utah 1 V. The Second Law of Thermodynamics B. Heat Engines 1. Definition of heat engine: a cyclic device that converts heat to work a. First Law: energy balance over engine High T source, TH Qin Heat engine Qout Low T sink, TL Wout ∆E = Q − W = 0 Qin − Qout − Wout = 0 b. Examples include the steam power plant, automotive and diesel engines 2. Second Law: is it possible or not? Can Qout be zero? What is the best we can do? Lesson 13, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics B. Heat Engines 3. Thermal-energy reservoirs: hypothetical closed systems with the following properties a. Interact with surroundings only via heat transfer. b. Their temperature remains constant and uniform. c. Examples: a furnace, the atmosphere, a lake, the ocean, twophase systems. High T reservoir Qin Heat engine Qout Low T reservoir Wout Lesson 13, Geof Silcox, Chemical Engineering, University of Utah 2 V. The Second Law of Thermodynamics B. Heat Engines 4. A simple steam power cycle (continuous cyclic process) Simple, ideal vapor power cycle (Rankine cycle, see Ch. 10). Vapor Q in Shaded area is net work out. 3 4 Liquid Pump Vaporliquid mix. Q out 2 log(P) Turbine W T ,out 3 Boiler Liquid W P,in 2 1 4 1 Condenser Saturated liquid line log(v) Saturated vapor line Lesson 13, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics 5. Energy balance over system (the steam power cycle) dE = Qin − Qout + WP ,in − WT ,out = 0 dt Qin − Qout − Wnet ,out = 0 or Qin − Qout − Wnet ,out = 0 6. Performance of device Performance = desired result WT ,out − WP ,in = required input Qin (2-41) (6-3) Lesson 13, Geof Silcox, Chemical Engineering, University of Utah 3 V. The Second Law of Thermodynamics 7. The performance of a heat engine is called the thermal efficiency, ηth ηth = Wnet ,out Wnet ,out = Qin Qin (6-4) ηth = Wnet ,out Qin − Qout Q = = 1 − out Qin Qin Qin (6-5) The performance or efficiency must be determined experimentally. The typical automobile engine has a thermal efficiency of about 25%. This means that 25% of the chemical energy in the gasoline can be converted to mechanical work. Lesson 13, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics 8. Example of a modern, coal-fired steam power plant (1000 MWe) Qin (From combustion chamber) Wnet,out = 1000 MWe Boiler Turbine WT,out Qout = 1275 MW WP,in Pump Condenser (To stream, ocean lake or cooling towers) Lesson 13, Geof Silcox, Chemical Engineering, University of Utah 4 V. The Second Law of Thermodynamics 8. Example of a modern, coal-fired steam power plant (1000 MWe) Qin − Qout − Wnet ,out = 0 ∴ Qin = Qout + Wnet ,out = 1275 + 1000 = 2275 MW ηth = Wnet ,out Qin = 1000 = 0.44 2275 ηcombust = Qin 2275 = = 0.91 HHV • mcoal 2500 = 1000 = 0.40 (2-43) 2500 ηgen ≅ 1 ηoverall = ηthηcombustηgen = Wnet ,out HHV • mcoal Where HHV is the higher heating value of the coal (MJ/kg, see p. 79) and mcoal is its mass flow rate (kg/s). Note that ηgen ≅ 1. Lesson 13, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics 9. The Kelvin-Plank Statement of the Second Law of Thermodynamics ηth = Q Wout Qin − Qout = = 1 − out Qin Qin Qin Source at T Qin Heat engine Wout,net What is the best we can do? Can Qout be zero? Kelvin-Plank Statement (Section 6-3): It is impossible for any process that operates on a cycle to receive heat from a single reservoir and perform a net amount of work on the surroundings. This means that Wout,net ≤ 0. Lesson 13, Geof Silcox, Chemical Engineering, University of Utah 5 V. The Second Law of Thermodynamics 10. A clarifying note on power requirements and efficiencies If we refer to a 0.25-hp fan, driven by a small electric motor with an efficiency of 54%, does that mean that the fan draws 0.25-hp of electric power or does it draw 0.25/0.54 hp? It turns out that the latter is correct. The motor will draw about 0.5-hp of electric power. Lesson 13, Geof Silcox, Chemical Engineering, University of Utah 6 ...
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