Lecture 3 Entropy and result of Euler’s homogenous theorem

The change of entropy over a Rankine cycle Clausius Duhem entropy equation W Turbine or Qin A Boiler Condenser -Qout B An open system Feed water pump B A Cycle process, a closed system Feed water tank An open system A=B A is the initial state B is the final state T is the boundary layer temperature between the system and the surroundings  = entropy generation [J/molK]

The entropy change S(B) – S(A) of a system The entropy change of an isolated system is always a positive one or zero. The entropy change of an open or a closed system may be a positive one, negative one or zero. Combined entropy change of the system and the surroundings is always a positive one or zero => we always generate entropy in real processe. Combined change of entropy is zero if nothing happens or the process is a reversible one (not possible in real life).

The change of entropy, s(T,p) Total differential of entropy where and => => If phase transition takes place it is included as follows

Example 1 Calculate the change of entropy for O2 from the initial sate A (p = 1bar, T = 300K) to the final state B (p = 3bar, T = 450K). O2 can be treated as an ideal gas. v = RT/p  = 1/T => Average cp (300..450K) = 29.983 J/(molK) => Volumetric thermal expansion coefficient

The third law of thermodynamics By Nernst, It is impossible to reach absolute zero of temperature with finite numbers of operations => it can never be reached. On the basis of the Nernst theorem can be “concluded” that the entropy of a pure, perfectly crystalline substance (element or compound) is zero at zero kelvin. The third law allows to calculate the absolute entropies for elements and compounds. Understanding the third law properly would require sufficient knowledge of the statistical thermodynamics.

Absolute entropy On the basis of the third law the absolute entropy can be defined for elements and compounds. For example, for N2, CO2 and Al, entropies at 0K and 1 bar (the standard state) are: s°[N2(s), T = 0 K] = s°[CO2(s), T = 0 K] = s°[Al(cr), T = 0 K] = 0 This definition is used when absolute entropies of elements and compounds are calculated

Calculation of the absolute entropy for N2 at 100K and 1bar = 27.17 + 6.43 + 23.38 + 11.42 + 11.41 + 72.13 + 7.87 = 159.81 J/(molK)

Absolute entropy of N2 in the JANAF table

Tabulated entropy values for H2O(g) The absolute entropy so or So [J/(molk)] is used. Pressure is 1 bar (standard state) NOTE: These values have not been taken from the Janaf table and therefore they may slighlty differ from Janaf table values

Entropy of formation Sfo For a compound the entropy of formation at 1bar and temperature T ( the standard state) is Sf° (T) = s°(compound) - isi° (elements, the most stable state) where s°is the absolute entropy at temperature T and i the stoichiometric coefficient Entropy of formation is not known for all substances. For example, contrary to the enthalpy of formation, the entropy of formation cannot be usually defined for solid fuels (e.g peat or wood).

Example 2 What is the entropy of formation for CH4 at 500K at the standard state? C + 2H2  CH4 Absolute entropies at 500K so(C) = 11.662 J/(molK) so(H2) 145.737 J/(molK) so (CH4) = 207.014 J/(molK) Sfo(CH4) = so(CH4) - so(C) - 2 so(H2) = 207.014 - 11.662 - 2145.737 = -96.122 J/(molK)

Entropy change in chemical reactions at the standard state For chemical reactions at the standard state, the entropy change is S° =  isi°(products) -  i si°(reactants) where si is the absolute entropy of the substance and i the stoichiometric coefficient

Example 3 Calculate the change of entropy for the following reaction at 500 K and 1 bar: CO + 1/2O2  CO2 So (500K) = so(CO2) – so(CO) – 1/2so(O2) = 234.901 – 198.883 -0.5198.883 = -74.329 J/(molK) CO2 500K, 1bar, 1 mol/s O2 500 K, 1 bar, 0.5 mol/s water 90oC, 2 bar, 1.66 kg/s Q CO 500 K, 1 bar, 1 mol/s water 45oC, 2 bar, 1.66 kg/s The entropy change (or entropy generation rate) over the ”boiler” is S = nCO2sCO2 + mwsw2 – nCOsCO – nO2sO2 – mwsw1 = -174.392 + 1.661192 – 1.66703 = 736 W/K => The total entopy change is a positive one as it has to be. Entropies for CO2 , CO and O2 have been taken from Janaf tables Entropies for water have been taken from steam tables

The dependence of the specific heat on the pressure => => v = molar volume [m3/mol] Derivate of the product dfg = f’g + g’f

Heat capacity's dependence on the pressure for an ideal gas, liquid and solid For an ideal gas v(T,p) = RT/p => => cp(T,p) = cp(T) Spefic heat capacities of liquids and solids only slightly depnd on the pressure. With good accuracy: cp,solid (T,p)  cp(T) and cp,liquid(T,p)  cp(T)

Heat capacity's dependence on the pressure for a real gas For real gases especially close to the critical point and presure may have a remarkable effect on cp For example, for water vapor close to the critical point (221bar and 647K) Vapor: cp (573.15 K, 1.0 bar) = 2.010 kJ/kgK Vapor: cp (573.15 K, 75 bar) = 4.686 kJ/kgK Vapor cp (573.15 K, 100 bar) = 5.692 kJ/kgK Vapor cp (573.15 K, 190 bar) = 5.350 kJ/kgK

Chemical system Chemical system T constant temperature [K] T, p p constant total pressure [Pa] ni the number of moles of each species (compound or element) in the system [mol] xi mole fractions of each species (compound or element) in the system [-] x1 + x2 +… xm = 1 T, p n1, n2,... nm (x1, x2....xm) Enthalpy of the system where - hi is molar enthalpy - Species in Finnish is “osalaji”

Euler’s theorem for homogenous functions in thermodynamics Using Euler’s homogenous theorem it is possible to show that state functions can be expressed as follows where where where where Homogenous theorem considers if si , hi ... in a mixture depends on other species.

Volume of mixture For solutions where Molar volumes of components in solutions may differ from molar volumes of pure components when they exist alone. For example, in ethanol-water solution molar volumes of ethanol and water are not the same as molar volumes of pure ethanol and water. Therefore homogenous theorem must be used in order to calculate total volume of a solution. For ideal gases where pi is the partial pressure of species in the gas mixture Molar volumes of species in an ideal gas mixture does not obey Euler’s homogenous theorem.