Outline. Property diagrams involving entropy. Heat transfer for internally reversible process

Transcription

1 Outline roperty diagrams involving entropy What is entropy? T-ds relations Entropy change of substances ure substances (near wet dome) Solids and liquids Ideal gases roperty diagrams involving entropy Recall on a -v diagram the area under the curve represents the boundary work for a quasiequilibrium process w = dv Recall equality part of Clasius inequality or definition of entropy δq ds = δq = TdS Q = TdS T integrate int rev Heat transfer for internally reversible process On a T-s diagram the area under the curve represents the heat transfer for an internally reversible process

2 T-S diagram of a Carnot cycle Two reversible isothermal (T=constant) and two reversible adiabatic (s=constant) processes Recall area under curve on T-s diagram represents heat transfer for internally reversible process For cycle, net heat transfer equals net work Level of molecular disorder (entropy) of substance increases as it melts or evaporates Entropy is related to total number of possible microscopic states for a given macroscopic state How certain are we about molecular position? What is entropy? Disorder (entropy) increases during heat transfer Heat is form of disorganized energy During heat transfer net entropy increases Increase in entropy of cold body more than offsets decrease in entropy of hot body There is no entropy transfer associated with work

3 Third law of thermodynamics and absolute entropy Third law of thermodynamics Entropy of a pure crystalline substance at absolute zero temperature is zero (no uncertainty about molecular state) Entropy determined relative to this state is called absolute entropy Use of entropy (disorganization, uncertainty) is not limited to thermodynamics e.g. studying for exam, library, textbooks, army, friction at work, angry words, etc. Derivation of T-ds relations Consider closed, stationary system and differential form of first law δq δw = du Consider quasiequilibrium boundary work and recall W dv δ = Substitute work expression into first law δ Q = du + dv T-ds relations (cont.) Consider an internally reversible process TdS = δq Substitute heat expression into first law TdS = du + dv Divide through by system mass Tds = du + dv 3

4 T-ds relations (cont.) Recall definition of enthalpy dh = d( u + v) = du + dv + vd = Tds + vd Tds = dh vd In summary, we have two T-ds relations: Tds = du + dv Tds = dh vd Implications of T-ds relations Entropy is a property T-ds relations express change in entropy in terms of changes in other properties roperties are process path independent Hence, T-ds relations relate entropy changes to changes in other properties for any process between any two states T-ds relations very useful for evaluating change in entropy when property relations (v, T,, u, and h) are available This is true for incompressible substances (solids and liquids) and ideal gases Entropy change and isentropic relations pure substances ure substances (near wet dome) Any process: s = s s ( ) kj / kg K Isentropic proces: s = 0 s = s 4

5 Entropy change and isentropic process incompressible substances Incompressible substances (solids/liquids) du CdT v = constant dv = 0 ds = = T T dt T Liquids/solids: s = s s = C( T ) Cav ln kj / ( kg K ) T T T Isentropic proces: s = 0 s = s = C ln = 0 T = T av T Temperature of incompressible substance remains constant during isentropic process Isentropic process of an incompressible substance is also isothermal (model for solids/liquids) Entropy change and isentropic process ideal gas Tds = du + dv = CvdT + RT ideal gas v dt dv dt v ds = Cv + R s = Cv( T ) R ln T v + T v Tds = dh vd = CdT RT ideal gas dt d dt ds = C R s = C( T ) R ln T T d dv Entropy change and isentropic process ideal gas (cont.) Recall we derived two entropy change formulas for ideal gas dt v s = Cv( T ) + R ln T v dt s = C( T ) R ln T Further evaluations depends upon whether or not you assume constant or variable specific heats 5

6 Entropy change ideal gas Constant specific heats dt v T v s = Cv( T ) + R ln C ln + R ln T v T v Variable specific heats v, av dt T s = C( T ) R ln C ln R ln T T T T T p T T = 0 T = 0, av C dt dt dt O = C C = s ( T ) s ( T ) = s T T T s s s = s s R s T o o o ln ( ( ) in A-/3) O o o Entropy change and isentropic process ideal gas; constant specific heat T s = Cv, av ln + R ln T isentropic v T R v v ln = ln = ln T Cv v v T T v = v s= const k v R Cv T T s = C, av ln R ln = T T s= const ( k ) / k Validity of isentropic relations 6

7 Entropy change and isentropic process ideal gas; variable specific heat s s = s s R ln 0 = s s R ln o o o o isentropic o o o o o r = ln = exp = = o R exp( s / R) r s s R = r s= const r v v r s= const r s s exp( s / R) v = r = v (A-7 for air); ( v T / ) r Use of isentropic relations variable specific heats Isentropic relation formula summary Constant specific heats k ( k ) / k k T v T v = ; = ; = T v s const T s const v = = s= const Variable specific heats v v = ; = v v r r s r s r 7

8 Example A kg block initially at 350C is quenched in an insulated tank that contains 00 kg of water at C. Assuming the water that vaporizes during the process condenses back in the tank, determine the entropy change during this process. Solution Model tank as closed system Model iron and water (initial and final states) as incompressible substances Find final equilibrium temperature using st law Q W = U U = 0 = U + U [ mc T T ] [ mc T T ] 0 = ( ) + ( ) iron ()(0.45)( T 350 C) + (00)(4.8)( T ) = 0 T = 6.C iron water water Solution Second law for entropy change T 99. Siron = mcave ln = ()(0.45) ln = 3.96 kj / K T 63 T 99. Swater = mcave ln = (00)(4.8) ln = 5.9 kj / K T 95 S = S + S = = S total iron water gen Improve accuracy using specific heats at average temperature 8

9 Example Oxygen gas is compressed in a pistoncylinder device from an initial state of 0.8m3/kg and 5C to a final state of 0.m3/kg and 87C. Determine the entropy change of the oxygen during this process. Assume constant specific heats. Solution Assume ideal gas; use constant specific heat at average temperature Tave = = 49K Cv, ave = 0.69 kj / kg K T v s = Cv, ave ln + R ln T v 560K 0. = (0.69) ln + (0.598) ln 98K 0.8 = 0.05 kj / kg K Example Air is compressed steadily by a 5-kW compressor from 00ka and 7C to 600ka and 67 C at a rate of.6kg/min. During this process, some heat transfer takes place between the compressor and the surrounding medium at 7C. Determine the rate of entropy change of the air during this process. 9

10 Solution Use variable specific heats T = 90K o s =.688 ( A ) = 00ka T = 440K s = 600ka o =.0887 ( A ) o o Sɺ sys = mɺ s s R ln = (.6 / 60) (0.87) ln(600 /00) = kw / K ( ) Summary T-s diagram important! Area under curve represents heat transfer only for an internally reversible process Entropy is related to disorder T-ds relations Entropy change... ure substance near wet dome use tables Incompressible substances (solids/liquids) Ideal gases Summary Key formulas (where did they come from?) ure substances s = s s Solids/liquids dt T s = s s= Ideal gases C( T ) Cav ln T T Constant specific heats T v T s = Cv, av ln + R ln or s = C, av ln R ln T v T Variable specific heats o o o s = s s R ln ( s ( T ) in A-7) 0

11 Summary Isentropic relations s = s s = 0 Constant specific heats k ( k ) / k k T v T v = ; = ; = T v s const T s const v = = s= const Variable specific heats v v = ; = v v r r s r s r