Entropy and the Second Law of Thermodynamics
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1 Entropy and the Second Law of hermodynamics Reading Problems 6-, 6-2, 6-7, 6-8, 6-6-8, 6-87, 7-7-0, 7-2, , 7-46, 7-6, 7-89, 7-, 7-22, 7-24, 7-30, 7-55, 7-58 Why do we need another law in thermodynamics? nswer: While the st law allowed us to determine the quantity of energy transfer in a process it does not provide any information about the direction of energy transfer nor the quality of the energy transferred in the process. In addition, we can not determine from the st law alone whether the process is possible or not. he second law will provide answers to these unanswered questions. process will not occur unless it satisfies both the first and the second laws of thermodynamics.. Direction: Consider an isolated system where Q = W =0 From the st law we know (U + KE) = U 2 E E 2 same both ways 2 2 possible gas : cold system propeller & gas rotating impossible gas 2: warm system propeller & gas stationary he physical interpretation of this is:
2 State : Most of the energy is in a highly organized form residing in the macroscopic KE of the propeller and the rotating gas. State 2: ll of the energy is now in a disorganized form residing in the microscopic E, i.e. U of the propeller and the gas. he process 2 has resulted in a higher state of molecular chaos. ENROPY is the thermodynamic property that describes the degree of molecular disorder in matter. Hence, S 2 >S. Entropy can be considered a quantitative index that describes the quality of energy. he process 2 is impossible because it would require disorganized KE to produce macroscopically organized KE. hat is S 2 <S which is impossible for an isolated system. thermodynamic process is only possible if it satisfies both the st and 2nd laws simultaneously. 2. Quality of Energy: heat engine produces reversible work as it transfers heat from a high temperature reservoir at H to a low temperature reservoir at L. If we fix the low temperature reservoir at L = 300 K, we can determine the relationship between the efficiency of the heat engine, η = Q L Q H = L H as the temperature of the high temperature reservoir changes. In effect we are determining the quality of the energy transferred at high temperature versus that transferred at low temperature. H Q H reversible heat engine W rev Q L L 2
3 L (K) H (K) η Since the purpose of the heat engine is to convert heat energy to work energy, we can clearly see that more of the high temperature thermal energy can be converted to work. herefore the higher the temperature, the higher the quality of the energy. Second Law of hermodynamics he second law of thermodynamics states: he entropy of an isolated system can never decrease. When an isolated system reaches equilibrium, its entropy attains the maximum value possible under the constraints of the system Definition P S = S gen = S 2 S 0 2nd law the 2nd law dictates why processes occur in a specific direction i.e., S gen cannot be ve he second law states, for an isolated system: (ΔS) system +(ΔS) surr. 0 where Δ final initial Gibb s Equation From a st law energy balance when KE and PE are neglected Energy Input = Energy Output + Increase in Energy Storage 3
4 δq amount = δw + du differential () We know that the differential form of entropy is ds = δq (2) δw = PdV (3) Combining Eqs., 2 and 3 ds = du + PdV ds = du + Pdv }{{ } per unit mass 2nd Law nalysis for a Closed System (Control Mass) MER ER ER dw dq CM We can first perform a st law energy balance on the system shown above. du = δq + δw () For a simple compressible system δw = PdV (2) From Gibb s equation we know ER ds = du + PdV (3) 4
5 Combining (), (2) and (3) we get ER ds = δq herefore (ds) CM = storage δq ER entropy flow + dp S production Integrating gives (S 2 S ) CM = Q 2 ER + S gen 0 where Q 2 ER - the entropy associated with heat transfer across a finite temperature difference, i.e. >0 5
6 2nd Law nalysis for Open Systems (Control Volume) S= -s m FR ER S= -dq ER S=0 m dq MER dw S CV CV ER S= dq dq ER m FR S= s m isolated Sgen 0 δs gen =(ΔS) sys +(ΔS) sur δs gen =ΔS }{{ CV + } system or as a rate equation ( ) s m 2 + s m 2 δq 2 + δq 2 ER ER surroundings Ṡ gen = ( ) ds dt CV + sṁ + Q ER OU sṁ + Q ER IN his can be thought of as generation = accumulation + OU IN 6
7 Reversible Process Example: Slow adiabatic compression of a gas process 2 is said to be reversible if the reverse process 2 restores the system to its original state without leaving any change in either the system or its surroundings. idealization where S 2 = S S gen =0 2 > increased microscopic disorder V 2 <V reduced uncertainty about the whereabouts of molecules Reversible S gen=0 + diabatic P rocess Q=0 Isentropic P rocess S =S 2 7
8 Calculation of Δs in Processes process path d=ds 2 ds s the area under the curve on a s diagram is the heat transfer for internally reversible processes q int,rev = 2 ds and q int,rev,isothermal = Δs abulated Calculation of Δs for Pure Substances Calculation of the Properties of Wet Vapor: Use ables -4 and -5 to find s f,s g and/or s fg for the following s =( x)s f + xs g s = s f + xs fg Calculation of the Properties of Superheated Vapor: Given two properties or the state, such as temperature and pressure, use able -6. Calculation of the Properties of a Compressed Liquid: Use able -7. In the absence of compressed liquid data for a property s,p s f@ 8
9 Calculation of Δs for Incompressible Materials - for an incompressible substance, dv =0,andC p = C v = C ds = du = C d s 2 s = 2 C( ) d Δs = C avg ln 2 where C avg =[C( )+C( 2 )]/2 Calculation of Δs for Ideal Gases For an ideal gas with constant C p and C v Ideal Gas Equation Pv = R du = C v d u 2 u = C v ( 2 ) dh = C p d h 2 h = C p ( 2 ) here are 3 forms of a change in entropy as a function of & v, & P,andP & v. s 2 s = C v ln 2 + R ln v 2 v = C p ln 2 R ln P 2 P = C p ln v 2 v + C v ln P 2 P y setting the above equations to zero (isentropic, i.e. Δs =0) we can obtain the isentropic equations 2 = ( ) (k )/k ( ) (k ) P2 v = P v 2 where k = C p /C v which can be found tabulated in able -2 for various gases. 9
10 he Carnot Cycle an ideal theoretical cycle that is the most efficient conceivable based on a fully reversible heat engine - it does not include any of the irreversibilities associated with friction, viscous flow, etc. in practice the thermal efficiency of real world heat engines are about half that of the ideal, Carnot cycle H L Q Q s s s H L P = P P = P 4 Q W in out Process State Points Description Pump 2 isentropic compression from L H to return vapor to a liquid state Heat Supply 2 3 heat is supplied at constant temperature and pressure Work Output 3 4 the vapor expands isentropically from the high pressure and temperature to the low pressure Condenser 4 the vapor which is wet at 4 has to be cooled to state point he cycle is totally reversible. he reversed Carnot cycle is called the Carnot refrigeration cycle. 0
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