Oneeq19b

Name:	Oneeq19b - Compressibility factor from the RK equation - one solution
Source:	Cutlip, M. B. and Shacham, M, Problem Solving in Chemical
	Engineering with Numerical Methods (2nd Ed), Prentice Hall Inc., 1999
Reference/s
Model:	1 implicit equation, indep. variable name: z
	Average difficulty level
	Constraints: z>0
	Discontinuities: none in the range of interest
	Initial range: zmin=-0.5, zmax=1.2
Solved by	Shacham, M., POLYMATH 5.1, build 19, April 1, 2001
Model Eqs.	Compressibility factor from the RK equation - one solution |POLVER05_1
EXCEL FILE	f(z)=z^3-z^2-Q*z-r #
TEXT FILE	P = 200 #
POLYMATH FILE	Pc=33.5 #
	T= 631 #
	Tc=126.2 #
	Pr=P/Pc #
	Tr=T/Tc #
	Asqr=0.4278*Pr/(Tr^2.5) #
	B=0.0867*Pr/Tr #
	Q=B^2+B-Asqr #
	r=Asqr*B #
	p=(-3*Q-1)/3 #
	q=(-27*r-9*Q-2)/27 #
	R=(p/3)^3+(q/2)^2 #
	V=if(R>0)then((-q/2+sqrt(R))^(1/3))else(0) #
	WW=if(R>0)then(-q/2-sqrt(R))else(0) #
	psi1 = if(R<0)then(arccos(sqrt((q^2/4)/(-p^3/27))))else(0) #
	W = if(R>0)then(sign(WW)*(abs(WW))^(1/3))else(0) #
	z1 = if(R<0)then(2*sqrt(-p/3)*cos((psi1/3)+2*3.1416*0/3)+1/3)else(0) #
	z2 = if(R<0)then(2*sqrt(-p/3)*cos((psi1/3)+2*3.1416*1/3)+1/3)else(0) #
	z3 = if(R<0)then(2*sqrt(-p/3)*cos((psi1/3)+2*3.1416*2/3)+1/3)else(0) #
	z0=if(R>0)then(V+W+1/3)else(0) #
	z(min)=-.5, z(max)=1.2
Variable/function values	Variable	Value	f(x)
	z	0.35	-1.0835E-01
	P	200
	Pc	33.5
	T	631
	Tc	126.2
	Pr	5.970149254
	Tr	5
	Asqr	0.045687875
	B	0.103522388
	Q	0.068551398
	r	0.004729718
	p	-0.401884731
	q	-0.101654258
	R	0.000179362
	V	0.400457283
	WW	0.037434511
	psi1	0
	W	0.334521515
	z1	0
	z2	0
	z3	0
	z0	1.068312132
Root	Variable	Value	f(x)
	z	1.06831213384200	2.89E-16
	P	200
	Pc	33.5
	T	631
	Tc	126.2
	Pr	5.970149254
	Tr	5
	Asqr	0.04568787490270
	B	0.10352238805970
	Q	0.06855139798660
	r	0.00472971791530
	p	-0.40188473132000
	q	-0.10165425798490
	R	0.00017936221260
	V	0.40045728293280
	WW	0.03743451115230
	psi1	0
	W	0.33452151531760
	z1	0
	z2	0
	z3	0
	z0	1.06831213158400
Additional Information	The plot can be interpreted as there is a double root in the
	vicinity of z =0. The analytical solution shows however that
	there is only a single root. The absolute function value
	is minimal at z=-0.03267 where f(z) = -0.0035
Function plot