The%202nd%20Law

The%202nd%20Law - The second law of thermodynamics...

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The second law of thermodynamics Lingchong You
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From 1 st law to 2 nd law E = q + w – If I can somehow reduce the energy of a system by E, I can then extract work. I can also do this continuously through a cyclic process (for each cycle E = 0) – For one thing, the work I extract will never exceed what’s reduced in the energy of the system (First law: I can’t win ), even if I can reduce heat loss (q) to zero. – Well, even the clause in red can’t be realized: the reduced energy can’t be 100% converted into useful work in a cyclic process – there is always gonna be some heat loss (Second law: I can’t get even).
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T hot T cold engine w q total q cold Consider an engine trying to extract work by operating between a hot reservoir and a cold reservoir. Two important points: 1) In order to do any work through a cycle, we must have T hot > T cold. 2) w < q total; But, what’s the maximum work that can be extracted?
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The Carnot Cycle : an idealized heat reversible engine cycle using ideal gas I State 1 State 2 State 4 State 3 II III IV T cold T hot I. 1 Æ 2: isothermal expansion II. 2 Æ 3: adiabatic expansion III. 3 Æ 4: isothermal compression IV. 4 Æ 1: adiabatic compression 4 1 4 1 i i i i ww qq = = = = In each cycle: Extracted work Consumed energy
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Step I, isothermal expansion : 22 11 hot 1 1 hot 2 ln 0 VV dw PdV nRT V w PdV dV nRT = −⇒ = −= = < ∫∫ For ideal gas, E = 0 along an isotherm. Thus, according to 1 st law: 2 1 1 1 1 1 hot 1 0l n 0 V Eqw q wn R T V = ∆=+ = > The system acquires heat and converts it to an equal amount of work during the first step. Step II, adiabatic expansion : 2 2 2 2 2 cold hot 0; () V q wE qE n C TT = =∆ = Thus, the system does work without acquiring any heat, resulting in a decrease in its temperature. Step III, isothermal compression : 4 3 3 3c o l d 4 33 ln 0 0 V V V w PdV nRT V qw = > =− < The system will release heat while receiving equal amount of work from the surroundings (cold reservoir). Step IV, adiabatic compression : 4 4 4 4 4 hot cold V q n C T T = = The system recovers its temperature while acquiring work from surroundings
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4 3 1 hot cold hot cold hot cold 1 24 3 1 hot cold l n ()l n () ln ln 0 (consistent with 1st law) VV i i V V ww n R T n C T T n R T n C T T V V nRT nRT qw E = == + + + =+ =− ∆= Total extracted work: 1. Overall, a Carnot cycle clearly obeys the 1 st law. The surroundings (hot reservoir and cold reservoir) supplies the system with heat (q), which is converted into equal amount of work (w). 2. However, there’s something permanent that has happened. Consider the amount of heat extracted from the hot reservoir (from step 1 ): 2 1 hot 1 ln V qn R T V = Total heat extracted from hot reservoir Not all of this is converted to w. Instead, there’s a fraction that goes into the cold reservoir: 4 3c o l d 3 ln V R T V = Total heat “wasted” on the cold reservoir
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42 31 ln ln VV =− We shall see: Why? (see next slide) Thus, we can estimate the efficiency of this process: 3c o l d 1 1 hot maximum work efficiency =1 1 1 heat extracted qT w qq T = <
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The%202nd%20Law - The second law of thermodynamics...

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