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203/4c18:1 Chapter 18: The Second Law of Thermodynamics Directions of a thermodynamic process Reversible processes: Thermodynamic processes which can be.

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Presentation on theme: "203/4c18:1 Chapter 18: The Second Law of Thermodynamics Directions of a thermodynamic process Reversible processes: Thermodynamic processes which can be."— Presentation transcript:

1 203/4c18:1 Chapter 18: The Second Law of Thermodynamics Directions of a thermodynamic process Reversible processes: Thermodynamic processes which can be reversed with a small change in conditions. Requires near equilibrium conditions (quasi-equilibrium) that take place slowly (quasi-statically). Irreversible processes generally non-equilibrium processes have a preferred direction (towards increasing “disorder, increasing entropy).

2 203/4c18:2 Heat Engines working substance: material which undergoes heat addition and extraction, performs mechanical work, may undergo phase transitions. (examples: steam, air, “ideal gas”) cyclic processes: thermodynamic processes are repeated on working substance Low temperature Reservoir High temperature Reservoir W QHQH QCQC

3 203/4c18:3 Goal: Analyze several cyclic processes, and by example study the process of analyzing the application of thermodynamics. Low temperature Reservoir High temperature Reservoir W QHQH QCQC

4 203/4c18:4 Otto Cycle Approximation of gasoline engine. a-b adiabatic compression *V  > rV b-c constant volume heat addition *Q H c-d adiabatic expansion d-a constant volume heat release a-e, e-a exhaust, intake QHQH QCQC W e b c d a V rV p

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8 203/4c18:8 Idealized Diesel Cycle Approximation of diesel engine. a-b adiabatic compression *rV  > V b-c constant pressure heat addition *Q H c-d adiabatic expansion d-a constant volume heat release a-e, e-a exhaust, intake QHQH QCQC W c e b d a V rV p

9 203/4c18:9 Problem 38: Consider a Diesel cycle that starts with 1.20L of air at 300K and a pressure of 1.00E5 Pa. If T c is 1200K, derive an expression for the efficiency in terms of the compression ratio r.

10 203/4c18:10 Refrigerators Heat Engines run backwards Low temperature Reservoir High temperature Reservoir W QCQC QHQH

11 203/4c18:11 Low temperature Reservoir High temperature Reservoir W QCQC QHQH

12 203/4c18:12 The Second Law of Thermodynamics “It is impossible for any system to undergo a cyclic process in which it absorbs heat from a reservoir and converts the heat completely into mechanical work.” Low temperature Reservoir High temperature Reservoir W QHQH

13 203/4c18:13 Low temperature Reservoir High temperature Reservoir QCQC The Second Law of Thermodynamics, equivalent form for refrigeration: “It is impossible for any process to have as its sole result the transfer of heat from cooler body to a warmer one.” QCQC

14 203/4c18:14 Low temperature Reservoir High temperature Reservoir W QHQH QCQC QHQH Low temperature Reservoir High temperature Reservoir W QCQC QHQH W Q QCQC

15 203/4c18:15 Entropy S A quantitative measure of disorder. A new thermodynamic variable. (  S cyclic = 0) For reversible processes (T may or may not be constant) Adiabatic Process => dS = 0

16 203/4c18:16 Melt ice at 0°C:  S = ? (L F = 3.34x10 5 J/Kg) Heat water from 0°C to 100°C :  S = ? (c = 4190 J/Kg-°C) Boil water at 100°C:  S = ? (L V = 2.256x10 6 J/Kg)

17 203/4c18:17 T-S diagrams: similar to p-V diagrams dQ = T dS T S Isothermal Adiabatic constant volume constant pressure

18 203/4c18:18 dQ = T dS  Q = “area under curve” on a T- S diagram QQ T S

19 203/4c18:19 Cyclic Processes Q net = “enclosed area” on a T- S diagram = W net (first law, cyclic process) Q net = W net T S

20 203/4c18:20 T- S diagrams and efficiency Q net = “enclosed area” on a T- S diagram = W net Q net = W net T S QHQH QCQC

21 203/4c18:21 QHQH QCQC Consider heat engines operating between two extremes of temperature, T C and T H. THTH TCTC Increase efficiency by Increasing Q H. Decreasing Q C.

22 203/4c18:22 TCTC The most efficient engine cycle operating between two specified temperatures: Carnot Cycle a-b: Isothermal Compression at T C. |Q C | = T C  S b-c: Adiabatic Compression to T H. c-d: Isothermal Expansion at T H. |Q H | = T H  S d-a: Adiabatic Expansion to T C. THTH SS a b c d T S Best (highest coefficient of performance) refrigerator operating between two specified temperatures is a Carnot Cycle

23 203/4c18:23 Example 18-2: A Carnot engine takes 2000 J of heat from a reservoir at 500 K. does some work, and discards the remainder to a heat reservoir at 300 K. How much work does it do, how much heat is discarded, and what is the thermal efficiency?

24 203/4c18:24 Example 18-3:.200 moles of an ideal gas (  = 1.40) undergoes a Carnot cycle operating between 400 K and 300 K. The initial pressure (just before the isothermal expansion) is 10.0E5 Pa, and during the isothermal expansion, the volume doubles.

25 203/4c18:25 Best (highest coefficient of performance) refrigerator operating between two specified temperatures is a Carnot Cycle (reversed Carnot Engine)


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