how to find roots in the cubic equation using ms excel, examples and

1. Finding
Cubic equation solution

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