14eq1

Name:	14eq1 - Three stage, two component distillation column
Source:	Cutlip, M. B. and Shacham, M, Problem Solving in Chemical
	Engineering with Numerical Methods (2nd Ed.), Prentice Hall Inc.,
Reference/s
Model:	14 implicit equations, indep. variables x11, x12, x13, x21, x22, x23, t1, t2, t3, tf, t0, V1, V2 and V3
	Average difficulty level
	Constraints: All the variables (except the temperatures: ti ) are nonnegative
	Discontinuities: Undefined for some negative temperature (ti) values
	Initial estimates: 1.(0.5, 0.4, 0.3, 0.3, 0.4, 0.5, 145, 190, 210, 200, 200, 1, 1, 1) ;
	2. (0.1, 0.1, 0.1,0.1, 0.1, 0.1,100, 100, 100, 100, 100, 5, 5, 5)
Solved by	Shacham, M., POLYMATH 5.1, build 19, April 18, 2001
Model Eqs.	Three stage, two component distillation column |POLVER05_1
EXCEL FILE	f(x11)=-((V1-L0)*k11+L1)*x11+V2*k12*x12 #
TEXT FILE	f(x12)=L1*x11-(V2*k12+L2)*x12+V3*k13*x13+z1*F #
POLYMATH FILE	f(x13)=L2*x12-(V3*k13+B)*x13 #
	f(x21)=-((V1-L0)*k21+L1)*x21+V2*k22*x22 #
	f(x22)=L1*x21-(V2*k22+L2)*x22+V3*k23*x23+z2*F #
	f(x23)=L2*x22-(V3*k23+B)*x23 #
	f(t1) = k11*x11+k21*x21-1 #
	f(t2) = k12*x12+k22*x22-1 #
	f(t3) = k13*x13+k23*x23-1 #
	f(tf) = k1f*z1+k2f*z2-1 #
	f(t0) = k10*k11*x11+k20*k21*x21-1 #
	f(V1) = -V1*hv1+V2*hv2-L1*hl1+L0*h0 #
	f(V2) = -V2*hv2+V3*hv3+hf+L1*hl1-L2*hl2 #
	f(V3) = -V3*hv3+Q+L2*hl2-L3*hl3 #
	L0=V1-D #
	L1=V2-D #
	L2=V3+F-D #
	L3=B #
	hl1 = t1*(29.6+0.04*t1)*x11+t1*(38.5+0.025*t1)*x21 #
	hv1=(8003+t1*(43.8-0.04*t1))*k11*x11+(12004+t1*(31.7+0.007*t1))*k21*x21 #
	hl2=t2*(29.6+0.04*t2)*x12+t2*(38.5+0.025*t2)*x22 #
	hv2=(8003+t2*(43.8-0.04*t2))*k12*x12+(12004+t2*(31.7+0.007*t2))*k22*x22 #
	hl3=t3*(29.6+0.04*t3)*x13+t3*(38.5+0.025*t3)*x23 #
	hv3=(8003+t3*(43.8-0.04*t3))*k13*x13+(12004+t3*(31.7+0.007*t3))*k23*x23 #
	hf = tf*(29.6+0.04*tf)*z1+tf*(38.5+0.025*tf)*z2 #
	h0 = t0*(29.6+0.04*t0)*k10*k11*x11+t0*(38.5+0.025*t0)*k20*k21*x21 #
	k11=10^(6.80776-935.77/((t1-32)*5/9+238.789))/P #
	k12 = 10^(6.80776-935.77/((t2-32)*5/9+238.789))/P #
	k13=10^(6.80776-935.77/((t3-32)*5/9+238.789))/P #
	k21 = 10^(6.85296-1064.84/((t1-32)*5/9+232.012))/P #
	k22=10^(6.85296-1064.84/((t2-32)*5/9+232.012))/P #
	k23=10^(6.85296-1064.84/((t3-32)*5/9+232.012))/P #
	k1f = 10^(6.80776-935.77/((tf-32)*5/9+238.789))/P #
	k2f = 10^(6.85296-1064.84/((tf-32)*5/9+232.012))/P #
	k10 = 10^(6.80776-935.77/((t0-32)*5/9+238.789))/P #
	k20 = 10^(6.85296-1064.84/((tf-32)*5/9+232.012))/P #
	F=1 #
	z1 = 0.40 #
	z2 = 1-z1 #
	B=0.75 #
	D=0.25 #
	P=760*120/14.7 #
	Q = 10000 #
	y11 = k11*x11 #
	y12 = k12*x12 #
	rec = y11*D/(z1*F) #
	x11(0)=0.5
	x12(0)=0.4
	x13(0)=0.3
	x21(0)=0.3
	x22(0)=0.4
	x23(0)=0.5
	t1(0)=145
	t2(0)=190
	t3(0)=210
	tf(0)=200
	t0(0)=200
	V1(0)=1
	V2(0)=1
	V3(0)=1
Variable/function values	Variable	Value	f(x)
	x11	0.5	0.087
	x12	0.4	0.037
	x13	0.3	-0.051
	x21	0.3	-0.031
	x22	0.4	0.258
	x23	0.5	-0.023
	t1	145	-0.507
	t2	190	-0.220
	t3	210	-0.126
	tf	200	-0.002
	t0	200	-0.305
	V1	1	6070.4
	V2	1	3212.9
	V3	1	742.2
	D	0.25
	L0	0.75
	L1	0.750
	B	0.75
	hl1	4398.9
	P	6204.1
	hl2	6114.2
	k21	0.2806
	hl3	6987.8
	k22	0.5376
	z1	0.4
	k23	0.6955
	k20	0.6128
	k11	0.817
	k12	1.411
	k13	1.753
	hv1	6927.7
	hv2	12331.3
	hv3	14716.8
	k1f	1.5759
	k2f	0.6128
	k10	1.5759
	F	1
	z2	0.6
	hf	8228
	L3	0.75
	L2	1.75
	h0	5288.1
	Q	10000
	y11	0.4083
	y12	0.5645
	rec	0.2552
Solution	Variable	Value	f(x)
	x11	0.57902594925514	1.11E-16
	x12	0.39569120229748	9.99E-16
	x13	0.27186578944464	-1.11E-15
	x21	0.42097405074479	0
	x22	0.60430879770252	3.33E-16
	x23	0.72813421055536	-2.22E-16
	t1	186.378525897475	6.66E-16
	t2	200.526868526979	0
	t3	211.486095297259	1.110E-15
	tf	200.167501447086	0
	t0	169.064015709595	0
	V1	1.08137588467567	-3.64E-12
	V2	1.06683987591090	1.82E-11
	V3	1.04952829661125	-2.18E-11
	D	0.25
	L0	0.83137588467567
	L1	0.81683987591090
	B	0.75
	hl1	7385.230868740510
	P	6204.081632653060
	hl2	8258.041508211440
	k21	0.51213933009160
	hl3	8931.054498916270
	k22	0.61698043668574
	z1	0.4
	k23	0.70841654598484
	k20	0.61413708445032
	k11	1.35469339962803
	k12	1.58495637623241
	k13	1.78094374630232
	hv1	15505.2876150983
	hv2	16469.3793109932
	hv3	17305.1918210042
	k1f	1.57879437332452
	k2f	0.61413708445032
	k10	1.10605654005438
	F	1
	z2	0.6
	hf	8235.9303882639
	L3	0.75
	L2	1.79952829661125
	h0	6290.0604895603
	Q	10000
	y11	0.78440263166929
	y12	0.62715329410047
	rec	0.49025164479331
Additional information	Some algorithms do not converge from initial guess 2