Chemical Exergy

1 Lecture in Chemical Exergy By: Dr. Giuma M. Fellah Department of Mechanical and Industrial Engineering Faculty of Engineering-Al_Fateh University Tripoli-Libya E-mail: gfellah2008@yahoo.com Modeling the Environment The environment is considered to be in stable equilibrium. Associated with it a unique temperature, a unique pressure, and a unique chemical potentials of the substances that making it up. Those values do not change as a result of the processes under consideration. All substances under consideration are formable from the substances that making up the environment. It is advantageous to select as environmental components, the most common ones, as they occur naturally, since then their concentrations are known with a high degree of accuracy. By assigning zero exergies to the most common components, others have positive exergies which reflect their economic values. Standard Environment and Standard Chemical Exergy There is no one theoretically correct environment. That means, simply, a specified environment which is good for one place and one application may not be good for another place and another application. An extensive study may be required to determine the environment for an application in a certain location, then a series of calculations is performed to calculate the exergy values. It is helpful to select a standard environment-that is, an environment for all locations and all applications. Exergy relative to this environment can be calculated and then tabulated for the future use. This exergy is called the Standard Chemical Exergy. It is claimed the deviation of the standard environment from the actual one, produces a significant errors, only in a few cases. 2 Szarugt, J. suggested a standard environment by assuming a standard temperature and standard pressure and a number of substances, one for each chemical element. The concentrations of the selected substances are chosen as the average values in the natural environment. The reference substances are selected from three obvious regions: gaseous components of the atmosphere; ionic and non ionic substances from the oceans; solid substances from the outer layer of the earth to a few miles depth. The chemical energy of a mixture can be calculated such that: Ψchemical Ni Ψchemical = ∑ N i  µ i,0 − µ i,00    i is the chemical exergy. is the number of moles, i =1,2,3,…….. µ i,0 is the chemical potential of the element “i” in the restricted dead state. µ i,00 is the chemical potential of the element “i” in the environmental dead state. For a mixture of ideal gas at T0 and P0 , we may write: µ i , 00 ∫ µ i ,0 dµ i = RT0 ∫ d ln( Pi ) , Pi , 00 then Pi , 0 P  µ i ,0 − µ i ,00 = RT0 ln i ,0  ,  Pi ,00  R is the universal gas constant. Pi,0 and Pi,00 are the partial pressures of the substance “i” at the restricted dead state and at the environmental dead state respectively y P  µ i ,0 − µ i ,00 = RT0 ln i ,0 0   y i ,00 P0  where yi,0 and yi,00 are the mole fractions of the substance “i” at the restricted dead state and at the environmental dead state respectively The chemical exergy can be written as: y  Ψchemical = RT0 ∑ N i ln i ,0   y i,00  i 3 For standard environment the values of yi,00 are specified, thus the standard chemical exergy can be calculated for the given mixture. per unit mixture mole we may write y  ψ chemical = RT0 ∑ y i ,0 ln i ,0   y i ,00  i For pure substance (not a mixture), the value of yi,0 is one, and the formula becomes: Given R = 8.3145 Ψchemical = − RT0 ln( y i ,00 ) kJ , T0 = 298 K, and different values for the mole kmol. K fractions at different relative humidity, the chemical exergy is calculated for selected elements Substance Φ = 60% ψ chemical Φ = 80% ψ chemical Φ =100% ψ chemical N2 0.7662 659.84695 0.7613 675.74335 0.7564 691.74239 O2 0.2055 3920.5208 0.2042 3936.2448 0.2029 3952.0691 CO2 0.0003 20098.599 0.0003 20098.599 0.0003 20098.599 H2O 0.0188 9846.2115 0.0250 9140.0141 0.0313 8583.1654 Others 0.0092 0.0092 0.0091 Chemical exergy of Material which is not a reference substance (the concentration of the material in the environment is unknown), can be calculated by selecting a reference substance for it, then chemical exergy can be calculated with respect to this reference substance. For example the concentration of the carbon in the environment is unknown. The carbon dioxide is selected as the reference substance for the carbon, hence C + O 2 ⇒ CO 2 Then the chemical exergy of the carbon can be calculated, such that: ψ c, chemical  ( y O2 ,00 ) ν O 2   = ∆G + RT0 ln ν CO 2   ( y CO2 )  ∆G = ν O 2 g O 2 − ν CO 2 g CO 2 , the change in the standard Gibbs function. 4 g O 2 = 0.0 g CO2 = −394380 kJ kmol then ∆G = 394380 kJ kmol Using the foregoing table then we may have Φ = 60% Φ = 80% Φ = 100% 410558.08 410542.35 410526.53 kJ/kmol kJ/kmol kJ/kmol ψ c, chemical = ψ c , chemical = ψ c, chemical = Szargut reported the chemical exergy of carbon is 410260 kJ/kmol. Fuel hydrocarbons Example: Find the standard chemical exergy of n-Nonane and compare with that given by Szargut page 306. C 9 H 20 + 14O 2 ⇒ 9CO 2 + 10H 2 O ∆G = 14g O 2 − 9g CO 2 − 10g H 2 O ∆G = 5835320 ψ C 9 H 20 , chemical kJ kmol   ( y O 2 ,00 ) ν O 2   = ∆G + RT0 ln ν H 2O  ν CO 2  (y (y H 2O )  CO 2 ,00 )  For relative humidity equal to 60%, the result is ψ C 9 H 20 , chemical = 6059.7822 kJ / mol Szargut reported for the same substance ψ C 9 H 20 , chemical = 6064.9 kJ / mol The slight difference is due to slight difference in the environmental data. For water vapor, the change in chemical potential can be found as µ v , 00 ∫ µ g ,0 dµ v = RT0 ∫ d ln( Pv ) Pv , 00 Pg , 0 5 The lower limit of the integration is selected as the state of the saturated vapor at the standard temperature. Note also the chemical potential of pure substance is equal to the Gibbs function, i.e µ g ,0 = g g ,0 . Where g g ,0 = h g ,0 − T0 sg ,0 µ v ,00 = g g ,0 + RT0 ln( y v ,00 P0 Pg ,0 ) Pg,0 is the saturation pressure corresponding to T0. The chemical exergy for water vapor becomes ψ v , chemical = µ g ,0 − µ v ,00 ψ v , chemical = RT0 ln( Pg ,0 y v ,00 P0 ) ψ v , physical = ( h T, P − h g ,0 ) − T0 (sT, P − sg ,0 ) Note also since: g g ,0 = g f ,0 then h g ,0 − T0 sg ,0 = h f ,0 − T0 s f ,0 when we have a compressed liquid then: h T, P = h f , T + v f ( P − Psaturated , T ) Example calculate the chemical and total exergy for water (T0 = 298 K, P0 = 0.1 MPa): 1- superheated steam at 14 MPa and 793 K. 2-saturated vapor at 373 K. 3-saturated liquid at 373 K. 4-copmressed liquid at 373 K and 0.5 MPa solution 1- from steam tables we get: h T, P = 3377.8 sT, P = 6.4610 h g ,0 sg,0 kJ kg kJ = 2547.2 kg Pg,0 = 0.003169 MPa kJ kg. K kJ = 8.5580 kg. K 6 for relative humidity equal to 60 % ψ v , chemical = 8.3145 * 298 * ln( . ψ v , chemical = 7187 0.003169 kJ ) = 1293.73 0.0188 * 01 . kmol kJ kg . ψ v , physical = (3377.8 − 2547.2) − 298 * (6.4610 − 8.5580) = 1455506 ψ v , total = 1527.38 ψ v , chemical ψ v , total kJ kg kJ kg * 100 = 4.94% 2-from steam tables we get . h g ,373K = 26761 kJ kg sg,373K = 7.3549 kJ kg. K the chemical exergy is the same as in part (1) ψ v , physical = (26761 . − 2547.2) − 298 * ( 7.3549 − 8.5580) = 487.238 ψ total = 559.29 ψ v , chemical ψ v , total kJ kg * 100 = 12.85% 3-from steam tables we get: h f ,298 K = 104.89 h f ,373K kJ kg kJ = 419.04 kg s f,298K = 0.3674 s f,373K kJ kg. K kJ . = 13069 kg. K the chemical exergy is the same as before. ψ f , physical = (419.04 − 104.89) − 298 * (13069 . − 0.3674) = 34.18 ψ f , total = 106.049 kJ kg kJ kg kJ kg 7 ψ f , chemical ψ f , total * 100 = 67.77% 4-from steam tables we get m3 v f =.0010435 kg h 373K ,0.5MPa = 419.04 + 0.0010435 * (0.5 − 010133 . ) * 103 = 419.46 . − 0.3674) = 34.60 ψ f , physical = (419.46 − 104.89) − 298 * (13069 ψ f , total = 106.47 kJ kg kJ kg kJ kg The result is almost the same as in part (3). We conclude from the foregoing example that chemical exergy may has a substantial contribution to total exergy as the given state approaches the restricted dead state-that is, as the reduction in the contribution of the physical exergy to total one.