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lecture04 - LECTURE 4 THERMODYNAMICS An Engineering...

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1 LECTURE 4 THERMODYNAMICS An Engineering Approach ENTROPY & EXERGY Dr. MinJun Kim Department of Mechanical Engineering & Mechanics Drexel University Spontaneous Change The direction of spontaneous change is that which 1. Leads to chaotic dispersal of the total energy 2. Moves from a state of low intrinsic probability towards one of greater probability. Work is needed to reverse a spontaneous process. We need a quantity – entropy – to describe energy dispersal, i.e. the probability of a state ± Spontaneous processes (irreversible) “generate” entropy. ± Reversible processes do not generate entropy – but they may transfer it from one part of the universe to another.
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2 Definition of Entropy Clausius Inequality The Clausius inequality states that The equality applies when there are no irreversibilities within the system as it executes the cycle and the inequality applies when internal irreversibilities are present. () b ( b )= - σ cycle - σ cycle = 0 - σ cycle < 0 - σ cycle > 0 No irreversibilities present within the system Irreversibilities present within the system Impossible
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3 Entropy T-s Diagram Entropy is an extensive property. Since entropy is a property, the change in entropy of a system in going from one state to another is the same for all processes, both internally reversible and irreversible, between these two states. s (T, p) s f (T) In the saturatation region the entropy may be calculated using the quality. In the absence of compressed liquid data, the value of the specific entropy can be
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4 T dS Equations TdS dQ T dQ dS pdV dW dW dU dQ REV INT REV INT REV INT REV INT REV INT = = = + = , , , , , ) ( ) / ( ) ( ) ( ) ( pdV dU TdS + = Vdp dH TdS = In the absence of overall system motion and the effect of gravity, an energy balance in differential form is At constant temperature and pressure, T dh ds / = T h h s s f g f g / ) ( = ± Entropy Change of an Ideal Gas TdS equations are used to evaluate the entropy change between two states of an ideal gas. = + = 2 1 1 2 1 1 2 2 2 1 1 2 1 1 2 2 ln ) ( ) , ( ) , ( ln ) ( ) , ( ) , ( T T p T T v p p R T dT T c p T s p T s v v R T dT T c v T s v T s When the specific heat is taken as constant, 1 2 1 2 1 1 2 2 1 2 1 2 1 1 2 2 ln ln ) , ( ) , ( ln ln ) , ( ) , ( p p R T T c p T s p T s v v R T T c v T s v T s p v = + = Entropy change of an incompressible substance, 1 2 1 2 ln T T c s s = ± incompressible, constant c
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5 Example (I) Water vapor at 400K and 0.1 MPa undergoes a process to 900K and 0.5 MPa. Determine the change in specific entropy, in kJ/kmol K, using (a) The superheated vapor table, (b) The ideal gas table for water vapor, (c) Integration with c p (T) from Table A-2. Entropy of Mixing Consider the mixing of two ideal gases:
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6 Entropy Change in Internally Reversible Process Entropy Change in Internally Reversible Process
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7 Example (II) Water initially a saturated liquid at 100 ° C is contained in a piston-cylinder assembely. The water undergoes a process to the corresponding saturated vapor state, during which the piston moves freely in the cylinder. If the change of state is brought about by heating the water as it undergoes an
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This note was uploaded on 02/28/2011 for the course MEM 310 taught by Professor Miller during the Winter '08 term at Drexel.

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lecture04 - LECTURE 4 THERMODYNAMICS An Engineering...

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