Cubic root using excel
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how to find roots in the cubic equation using ms excel, examples and
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Cubic root using excel
- 2. Peng-Robinson • Given: R, Tc, Pc, a Ω 𝑎 = 0.457235 Ω 𝑏 = 0.077796 1. Calculate ac and b using equation 𝑎 = 𝑎 𝑐α(𝑇) = Ω 𝑎(𝑅T 𝐶)2 P 𝑐 α(T) 𝑏 = Ω 𝑏 𝑅T 𝐶 P 𝑐 α(T)= 1 + 𝑚 1 − 𝑇𝑟0.5 2 m= 0.37464 + 1.54226 ∗ ꙍ − 0.26992 ∗ ꙍ2 For heavier compound(ꙍ>0.49) m=0.3796 + 1.485 ∗ ꙍ − 0.1644 ∗ ꙍ2 + 0.01667 ∗ ꙍ3 2. Calculate A and B from equation 𝐴 = 𝑎𝑃 (𝑅𝑇)2 Z= 𝑝𝑣 𝑅𝑇 𝐵 = 𝑏𝑃 𝑅𝑇 2 2 2 RT a P v b v bv b
- 3. 3. Writing equation in terms of Z factor, A and B from PR equation 𝑃 = 𝑅𝑇 𝑣−𝑏 − 𝑎 𝑣 𝑣+𝑏 +𝑏 𝑣−𝑏 𝑃 v − b v2 + 2bv − b2 = RT v2 + 2bv − b2 − av + ab 𝑃 𝑣3 + 𝑏𝑣2 − 3𝑏2 𝑣 + 𝑏3 = 𝑅𝑇𝑣2 + 2 𝑅𝑇 𝑏𝑣 − 𝑅𝑇𝑏2 − 𝑎𝑣 + 𝑎𝑏 multiplying both side of equation by( 𝑃2 𝑅𝑇 3) and then rearranging the equation: - 𝑍3 + 𝐵 − 1 𝑍2 + 𝐴 − 3𝐵2 − 2𝐵 𝑍 + 𝐵3 + 𝐵2 − 𝐴𝐵 = 0 We can have three possibilities • Three distinct real values of Z, from which we will take the extreme 2 values. • One real and two complex values, from which we will consider the real one.(weather compressibility is for liquid or gas will be based on density of fluid) • Three equal values of Z, which means single phase.
- 4. Finding 1st root 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 q = 3𝑏−𝑎2 3 r = 2𝑎3−9𝑎𝑏+27𝑐 27 d = 𝑞3 27 + 𝑟2 4 case 1: d=0 𝑥1 = 2 − 𝑟 2 1 3 − 𝑎 3 case 2: d>0 𝑥1 = − 𝑟 2 + 𝑑 1 2 1 3 + − 𝑟 2 − 𝑑 1 2 1 3 − 𝑎 3 case 3: d<0 𝑥1 = 2 − 𝑞 3 1 2 𝑐𝑜𝑠 Ɵ 3 − 𝑎 3 & cos(Ɵ) = 3𝑟 2𝑞 − 3 𝑞 1 2 Equation box
- 5. Let α be first root of the equation: - Then above equation can be reduced to (x-α)(𝑎1 𝑥2 + 𝑏1 𝑥 + 𝑐1) Expanding the equation: – 𝑎1 𝑥3 + (𝑏1−α𝑎1)𝑥2 + (𝑐1−α𝑏1)𝑥 − α𝑐1 Comparing above equation with the previous one: - 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 We have 𝑎1 = 1 𝑏1 =a+α 𝑐1 = − 𝑐 α New quadratic equation obtained is: - 𝑥2 + 𝑎 + α 𝑥 − 𝑐/α 𝑥 = (− 𝑏 + − 𝑏2−4𝑎𝑐 1 2 2𝑎 ) if 𝑏2 − 4𝑎𝑐 > 0 the roots are real and unequal if 𝑏2 − 4𝑎𝑐 = 0 the roots are real and equal if 𝑏2 − 4𝑎𝑐 < 0 the roots are imaginary
- 6. Example case 1 A B C D 1 3 3 1 Enter coefficient here Ax3+Bx2+Cx+D=0 Comparing d for different cases a b c q r d Cos(Ɵ) Ɵ α(root) 3 3 1 0 0 0 #DIV/0! #DIV/0! -1.000000 𝑎1 𝑏1 𝑐1 𝑏2 − 4𝑎𝑐 Root 1 Root 2 1 2 1 0 -1 -1 Comparing d for different casesSolving Cubic equation D=0 Solving quadratic equation Case 1
- 7. Example case 2 A B C D 125 -100 25 -2 Enter coefficient here Ax3+Bx2+Cx+D=0 a b c q r d Cos(Ɵ) Ɵ α (root) -0.8 0.2 -0.016 -0.013333 -0.000592 5.558E-22 1 #NUM! 0.4 𝑎1 𝑏1 𝑐1 𝑏2 − 4𝑎𝑐 Root 1 Root 2 1 -0.6 0.08 0 0.2 0.2 Comparing d for different casesSolving Cubic equation D>0 Solving quadratic equation Case 2
- 8. Example case 2 A B C D 1 -0.25 -1.23 -2.35 Enter coefficient here Ax3+Bx2+Cx+D=0 Comparing d for different cases a b c q r d Cos(Ɵ) Ɵ α (root) -0.25 -1.23 -2.35 -1.25083 -2.45366 1.4326593 4.56 #NUM! 1.736997 𝑎1 𝑏1 𝑐1 𝑏2 − 4𝑎𝑐 Root 1 Root 2 1 1.48699 1.352909 -3.20047 FALSE FALSE Comparing d for different cases D>0 Solving quadratic equation Solving Cubic equation Case 2
- 9. Example case 3 A B C D 1 -1 0.32 -0.032 Enter coefficient here Ax3+Bx2+Cx+D=0 Comparing d for different cases a b c q R d Cos(Ɵ) Ɵ α (root) -1 0.32 -0.032 -0.013333 0.0005925 -1.28E-21 -1.00 3.1415925 0.4000000 𝑎1 𝑏1 𝑐1 𝑏2 − 4𝑎𝑐 Root 1 Root 2 1 -0.5999 0.079999 0.039999 0.399999 0.2 Comparing d for different casesSolving Cubic equation D<0 Solving quadratic equation Case 3
- 10. Saturation pressure calculation The fugacity of a component in the vapor phase and the liquid phase is basically a measure of the potential for transfer of component between the phases. For example, a higher fugacity of a component in the vapor phase compared to liquid phase indicates that liquid accepts the component from vapour phase. The equality of component fugacities in the vapor phase and liquid means zero net transfer of component between the two phases. Therefore, a zero net transfer for all the components or the equality of component fugacities in the two phases implies thermodynamic equilibrium of a hydrocarbon system(saturation pressure), which is mathematically expressed as : 𝑓𝑙= 𝑓𝑣 Fugacity – • A thermodynamic property of a real gas which if substituted for the pressure or partial pressure in the equations for an ideal gas gives equations applicable to the real gas. 𝑓 = 𝑝 ∗ 𝑒𝑥𝑝 𝑍 − 1 − 𝑙𝑛 𝑍 − 𝐵 + 𝐴 80.5 ∗ 𝐵 𝑙𝑛 𝑍 + 𝐵 1 − 20.5 𝑍 + 𝐵 1 + 20.5
- 11. Start Input data Tc,Pc,Ꙍ Specify T Calculate a=(ac*α) Solve EOS to get Zv , ZL(refer eq on slide no 3) Calculate fv , fL(refer eq on slide no 10) 𝐴𝐵𝑆 1 − 𝑓𝑙 𝑓𝑣 2 = 0 Record Psaturation Stop Yes No Adjust P 1 root or 3 equal roots Then single phase exist Minimum root is ZV Middle root is neglected Maximum root is ZL 3 roots 1 roots initial guess of P Calculate A,B Calculate b,m,ac Calculate α(T)
- 12. The molar gibbs energy and fugacity of pure component Starting from equality of molar gibbs energies in coexisting phases 𝐺 L(T,P)= 𝐺 v(T,P)………….1 Since : d 𝐺 = - 𝑆 dT+𝑉dP So that 𝜕𝐺 𝜕𝑇 𝑃 = −𝑆..................2 And 𝜕𝐺 𝜕𝑃 𝑇 = 𝑉………………….3 We presume that we have equation of state from which we can compute V as a function of T and P, only equation 3 will be considered further. Integration of equation 3 between any two pressure P1 and P2(at constant temperature) yields 𝐺(T1,P2)- 𝐺(T1,P1)=∫ 𝑃1 𝑃2 (𝑉𝑑𝑃)……4 If the fluid consideration were an ideal gas, then 𝑉 𝐼𝐺 =RT/P, so that 𝐺 𝐼𝐺 (T1,P2)-GIG(T1,P1)=∫ 𝑃1 𝑃2 𝑅𝑇 𝑃 𝑑𝑃……5 Subtracting eq 5 from 4 gives [𝐺(T1,P2)- 𝐺 IG(T1,P2)]-[𝐺(T1,P1)- 𝐺 IG(T1,P1)]=∫ 𝑃1 𝑃2 𝑉 − 𝑅𝑇 𝑃 𝑑𝑃……..6
- 13. Now setting P1=0 equal to zero , (2) recognizing that at P=0 all fluids are ideal gases so that 𝐺(T1,P=0)= 𝐺 IG(T1,P=0), and equation 4 omitting all subscript yields 𝐺(T,P)- 𝐺 IG(T,P)= ∫0 𝑃 𝑉 − 𝑅𝑇 𝑃 𝑑𝑃……….7 Definition of Fugacity 𝑓 = 𝑃 exp 𝐺 T,P −𝐺 IG T,P RT = 𝑃𝑒𝑥𝑝 1 𝑅𝑇 ∫ 0 𝑃 𝑉 − 𝑅𝑇 𝑃 𝑑𝑃 …..8 Definition of fugacity coefficient 𝜙 = 𝑓 𝑃 exp 𝐺 T,P −𝐺 IG T,P RT = 𝑒𝑥𝑝 1 𝑅𝑇 ∫ 0 𝑃 𝑉 − 𝑅𝑇 𝑃 𝑑𝑃 …..9 In fact, all these equations of states are in form in which pressure is explicit function of volume and temperature. Therefore, it is useful to have equation relating the fugacity to an integral over volume (rather than pressure). We obtain such an equation by starting with equation 9 and Equation 𝑑𝑃 = 1 𝑉 𝑑 𝑃𝑉 − 𝑃 𝑉 𝑑𝑉 at constant temperature in form 𝑑𝑃 = 1 𝑉 𝑑 𝑃𝑉 − 𝑃 𝑉 𝑑𝑉 = 𝑃 𝑍 𝑑𝑍 − 𝑃 𝑉 𝑑𝑉……10 To change the variable of integration we obtain: ln 𝑓 𝑇,𝑃 𝑃 = ln 𝜙 = 1 𝑅𝑇 ∫ 𝑉=∞ 𝑉 𝑅𝑇 𝑉 − 𝑃 𝑑𝑣 − ln 𝑍 + 𝑍 − 1 …….11