Example CalculaIon: SoluIon Cont Conversion of problem parameters to SI units Temperature: ... Example CalculaIon: SoluIon Cont Determine gas velocity at Lincoln, [r]
(1)Undergraduate Lectures on Material Balances for Chemical Engineers June 5-‐ 20, 2014 Chapter 5: Lecture 1: Two-phase Systems! Brian G Higgins Department of Chemical Engineering and Materials Science University of California, Davis (2) Measures of ConcentraIon In the analysis of gas phase system, it is o2en necessary to relate concentra5on to temperature T and pressure P This can be done through an equa,on of state (3) EquaIon of State For a N-‐component ideal gas mixture: number of moles of species A in volume V pA : partial pressure of species A R : gas constant (4) EquaIon of State Pressure p of the ideal gas mixture: (Dalton’s law) parIal pressures of species A in volume V Number of moles n in an ideal gas mixture: moles of species A in volume V Ideal gas law for mixture: total pressure total number of moles (5) Density of a Gas Mixture Density of a pure gas: Density of a ideal gas mixture: where (6) Example CalculaIon Flow ProperIes in a Gas Pipe Line (7) Example CalculaIon: SoluIon Lincoln, NE Glenpool, OK P0 , T P1 , T Control volume Z Species Balance: A ⇢Av A · n dA = Approximate natural gas as a single component, consis5ng of methane Z A ⇢CH4 v · n dA = For the specified control volume, our working equa5on is: Z A0 ⇢CH4 v · n dA + Z A1 ⇢CH4 v · n dA = (8) Example CalculaIon: SoluIon Lincoln, NE Glenpool, OK P0 , T P1 , T Control volume Thus the conserva5on of mass for the pipeline becomes (⇢CH4 )0 v0 A0 = (⇢CH4 )1 v1 A1 And the mass average velocity of the gas at Lincoln, NE is: A0 (⇢CH4 )0 v1 = v0 A1 (⇢CH4 )1 Note that the mass flow rate in the pipe line is a constant: ṁ0 = ṁ1 = (⇢CH4 )0 v0 A0 And likewise the molar flow rate is constant: (⇢CH4 )0 v0 A0 ṁ0 = Ṁ0 = Ṁ1 = M WCH4 M WCH4 (9) Example CalculaIon: SoluIon Cont Lincoln, NE Glenpool, OK P0 , T P1 , T Control volume Step 1: Express all variables in SI units psia = 6895 Pa MPa = 10 Pa Unit Conversions: 6895 Pa MPa (2100 psia) ⇥ ⇥ = 14.48 MPa psia 10 Pa F = 1.8 C + 32 K = C + 273.16 6895 Pa MPa (2900 psia) ⇥ ⇥ = 20.00 MPa psia 10 Pa (90 F 5C K 32 F) ⇥ ⇥ + 273.15 K = 305.4 K 9F C (10) Example CalculaIon: SoluIon Cont Conversion of problem parameters to SI units Temperature: Velocity: Pipe Diameter: Pipe Area: (45 F 5C K 32 F) ⇥ ⇥ + 273.15 K = 280.4 K 9F C 0.3048 m (50 ft/s) ⇥ = 15.24 m/s ft 0.0254 m D0 = D1 = (20 in) ⇥ = 0.508 m in ⇡D2 ⇡(0.508 m)2 A0 = A1 = = = 0.2027 m2 4 (11) Example CalculaIon: SoluIon Cont Calculate molar concentra5on of methane at : T0 , P0 m Pa 8.314 mol (305.4 K) v0 R T0 K C0 = = = = 1.2 ⇥ 10 n P0 19.98 ⇥ 10 Pa m3 /mol Calculate molar concentra5on of methane at : T1 , P1 C1 = v1 R T1 = = n P1 m3 Pa 8.314 mol K (280.4 K) 14.48 ⇥ 106 Pa = 1.53 ⇥ 10 m3 /mol Calculate gas densi5es at Glenpool, OK and Lincoln, NE (⇢CH4 )0 = M WCH4 16.043 g/mol = = 133.7 kg/m C0 1.2 ⇥ 10 m3 /mol M WCH4 16.043 g/mol (⇢CH4 )1 = = = 104.8 kg/m C1 1.53 ⇥ 10 m3 /mol (12) Example CalculaIon: SoluIon Cont Determine gas velocity at Lincoln, NE: A0 (⇢CH4 )0 v1 = v0 A1 (⇢CH4 )1 v1 = 133.7 kg/m 104.8 kg/m 3 (15.24 m/s) = 19.44 m/s Determine mass flow rate in pipe line: ṁ0 = ṁ1 = (⇢CH4 )0 v0 A0 ṁ0 = ṁ1 = (133.7 kg/m )(15.25 m/s)(0.2027 m2 ) = 413 kg/s Determine molar flow rate in pipe line: (⇢CH4 )0 v0 A0 ṁ0 = Ṁ0 = Ṁ1 = M WCH4 M WCH4 413 kg/s Ṁ0 = Ṁ1 = = 25.74 ⇥ 19 16.043 g/mol mol/s (13) Vapor Pressure of Pure Liquids p-‐V-‐T behavior of Methane: pvap (T ) VG VL criIcal pressure temperature isotherms saturaIon vapor pressure VL VG SaturaIon vapor pressure a funcIon of temperature only (14) Methods to Compute SaturaIon Vapor Pressure Antoine EquaIon: A, B, ✓ are constants for a given species Clausius-‐Clapeyron EquaIon: molar heat of vaporizaIon reference temperature (15) Example CalculaIon: Vapor Pressure of Methanol Determine the vapor pressure of methanol at 25C: Step 1: Use Antoine Equa5on Step2: Obtain Antoine parameters-‐ see Table A3 Appendix A Step 3: Compute using Antoine equa5on pM,vap = 102.10312 = 120.88 mmHg = 16.91 kPa (16) ApplicaIon of Equilibrium RelaIons Liquid-‐gas ExtracIon Process (gas scrubber) gas liquid liquid phase separa5on tray gas gas species A species B What is the composiIon in the liquid and gas phases? (17) Equilibrium RelaIonship for Binary Mixtures species A species B Equilibrium relaIon: pA = xA PA,vap parIal pressure of species A in gas phase Species B: saturaIon vapor pressure of species A mole fracIon of species A in liquid phase pB = xB PB,vap (18) Equilibrium RelaIonship for Binary Mixtures, cont species A species B What is the composiIon in gas phase? If gas phase is an ideal gas: pA V = nA RT parIal pressure of species A in gas phase moles of species A in gas phase Gas law for the mixture P V = n RT total pressure in gas phase total moles in gas phase nA pA yA = = n P (19) species A species B Working EquaIons for Binary Mixtures composiIon in gas phase: equilibrium relaIon (i): yA = pA /P pA = xA PA,vap Raoult’s Law: yA = xA Mole frac5on constraints for binary system: Equilibrium Rela5on (Raoult’s law): rela5ve vola5lity: ↵AB PA,vap = PB,vap ✓ PA,vap P ◆ (20) Henry’s Law (limiIng case of Raoult’s law) species A species B Equilibrium Rela5on (Raoult’s law): Species A in mixture is dilute ( x A << 1 ) such that + xA (↵AB Henry’s Law: 1) ⇡ yA = ↵AB xA (21)