Teneq2a

Name: Teneq2a - Combustion of Propane in Air - R = 10

Source: Mentjes, K. and Morgan, A. P., ACM. Trans. Math. Soft. 16(2) 143(1990)

Reference/s Shacham, M., N. Brauner and M. B. Cutlip, “ A Web-based Library for

Testing Performance of Numerical Software for Solving NLEs"

Presented at Escape-11 conference, Kolding, Denmark, May 27-30, 2001

Model: 10 implicit equations, indep. variables n1 to n10

Lower difficulty level

Constraints: ni are nonnegative for I=1,2,…10

Discontinuities: Undefined for negative values of n1,n2,n3,n4

Initial estimates: 1.(1.5, 2, 35, 0.5, 0.05, 0.005, 0.04, 0.003, 0.02, 5) ; 2.(2, 2, 10, 1, 1, 2, 0, 0, 0, 0);

3. (1, 1, 10, 1, 1, 1, 0, 0, 0, 0)

Solved by Shacham, M., POLYMATH 5.1, build 19, April 13, 2001

Model Eqs. Combustion of Propane in Air - R = 10 |POLVER05_3

EXCEL FILE
f(n1) = n1+n4-3 #Mol of Carbon Dioxide - Carbon Balance

TEXT FILE
f(n2) = 2*n1+n2+n4+n7+n8+n9+2*n10-R #Mol of Water - Oxygen Balance

POLYMATH FILE
f(n3) = 2*n2+2*n5+n6+n7-8 #Mol of Nitrogen - Hydrogen Balance

f(n4) = 2*n3+n9-4*R #Mol of Carbon Monoxide - Nitrogen Balance

f(n5) = K5*n2*n4-n1*n5 #Mol of Hydrogen - Equilibrium Expression

f(n6) = K6*sqrt(n2*n4)-sqrt(n1)*n6*sqrt(p/nt) #Hydrogen atom - Equilibrium Expression

f(n7) = K7*sqrt(n1*n2)-sqrt(n4)*n7*sqrt(p/nt) #Hydroxyl Radical - Equilibrium Expression

f(n8) = K8*n1-n4*n8*(p/nt) #Oxygen Atom - Equilibrium Expression

f(n9) = K9*n1*sqrt(n3)-n4*n9*sqrt(p/nt) #Mol Nitric Oxide - Equilibrium Expression

f(n10) = K10*n1^2-n4^2*n10*(p/nt) #Mol of Oxygen - Equilibrium Expression

nt = n1+n2+n3+n4+n5+n6+n7+n8+n9+n10 #Total Number of Moles of Combustion Products

K5 = 0.193 #Equilibrium Constant at 2200 K

K6 = 2.597e-3 #Equilibrium Constant at 2200 K

K7 = 3.448e-3 #Equilibrium Constant at 2200 K

K8 = 1.799e-5 #Equilibrium Constant at 2200 K

K9 = 2.155e-4 #Equilibrium Constant at 2200 K

K10 = 3.846e-5 #Equilibrium Constant at 2200 K

R = 10 #Air to Fuel Ratio

p = 40 #Pressure (atm.)

n1(0)=1.5

n2(0)=2

n3(0)=35

n4(0)=0.5

n5(0)=0.05

n6(0)=0.005

n7(0)=0.04

n8(0)=0.003

n9(0)=0.02

n10(0)=5

Variable/function values Variable Value f(x)

n1 1.5 -1.000

n2 2 5.563

n3 35 -3.855

n4 0.5 30.020

n5 0.05 0.118

n6 0.005 -0.0032

n7 0.04 -0.0210

n8 0.003 -0.0013

n9 0.02 -0.0076

n10 5 -1.133

nt 44.118

K5 0.193

K6 0.002597

K7 0.003448

K8 0.00001799

K9 0.0002155

K10 0.00003846

R 10

p 40

Solution Variable Value f(x)

n1 2.91572542389522 0

n2 3.96094281080888 -1.78E-15

n3 19.9862916465515 0

n4 8.42745761047765E-02 0

n5 2.20956017698935E-02 0

n6 7.22766590884200E-04 -4.337E-19

n7 3.32004082515745E-02 -3.469E-18

n8 4.21099693391800E-04 -6.776E-21

n9 2.74167068969179E-02 8.674E-19

n10 3.11467752270064E-02 5.42E-20

nt 27.0622378157901

Additional information Only constrained algorithm that keeps the values of n1,n2,n3 and n4

positive throughout the iterations converges from initial guess 3.

Name:	Teneq2a - Combustion of Propane in Air - R = 10
Source:	Mentjes, K. and Morgan, A. P., ACM. Trans. Math. Soft. 16(2) 143(1990)

Reference/s	Shacham, M., N. Brauner and M. B. Cutlip, “ A Web-based Library for
	Testing Performance of Numerical Software for Solving NLEs"
	Presented at Escape-11 conference, Kolding, Denmark, May 27-30, 2001

Model:	10 implicit equations, indep. variables n1 to n10
	Lower difficulty level
	Constraints: ni are nonnegative for I=1,2,…10
	Discontinuities: Undefined for negative values of n1,n2,n3,n4
	Initial estimates: 1.(1.5, 2, 35, 0.5, 0.05, 0.005, 0.04, 0.003, 0.02, 5) ; 2.(2, 2, 10, 1, 1, 2, 0, 0, 0, 0);
	3. (1, 1, 10, 1, 1, 1, 0, 0, 0, 0)

Solved by	Shacham, M., POLYMATH 5.1, build 19, April 13, 2001

Model Eqs.	Combustion of Propane in Air - R = 10 \|POLVER05_3
EXCEL FILE	f(n1) = n1+n4-3 #Mol of Carbon Dioxide - Carbon Balance
TEXT FILE	f(n2) = 2n1+n2+n4+n7+n8+n9+2n10-R #Mol of Water - Oxygen Balance
POLYMATH FILE	f(n3) = 2n2+2n5+n6+n7-8 #Mol of Nitrogen - Hydrogen Balance
	f(n4) = 2n3+n9-4R #Mol of Carbon Monoxide - Nitrogen Balance
	f(n5) = K5n2n4-n1*n5 #Mol of Hydrogen - Equilibrium Expression
	f(n6) = K6sqrt(n2n4)-sqrt(n1)n6sqrt(p/nt) #Hydrogen atom - Equilibrium Expression
	f(n7) = K7sqrt(n1n2)-sqrt(n4)n7sqrt(p/nt) #Hydroxyl Radical - Equilibrium Expression
	f(n8) = K8n1-n4n8*(p/nt) #Oxygen Atom - Equilibrium Expression
	f(n9) = K9n1sqrt(n3)-n4n9sqrt(p/nt) #Mol Nitric Oxide - Equilibrium Expression
	f(n10) = K10n1^2-n4^2n10*(p/nt) #Mol of Oxygen - Equilibrium Expression
	nt = n1+n2+n3+n4+n5+n6+n7+n8+n9+n10 #Total Number of Moles of Combustion Products
	K5 = 0.193 #Equilibrium Constant at 2200 K
	K6 = 2.597e-3 #Equilibrium Constant at 2200 K
	K7 = 3.448e-3 #Equilibrium Constant at 2200 K
	K8 = 1.799e-5 #Equilibrium Constant at 2200 K
	K9 = 2.155e-4 #Equilibrium Constant at 2200 K
	K10 = 3.846e-5 #Equilibrium Constant at 2200 K
	R = 10 #Air to Fuel Ratio
	p = 40 #Pressure (atm.)
	n1(0)=1.5
	n2(0)=2
	n3(0)=35
	n4(0)=0.5
	n5(0)=0.05
	n6(0)=0.005
	n7(0)=0.04
	n8(0)=0.003
	n9(0)=0.02
	n10(0)=5

Variable/function values	Variable	Value	f(x)
	n1	1.5	-1.000
	n2	2	5.563
	n3	35	-3.855
	n4	0.5	30.020
	n5	0.05	0.118
	n6	0.005	-0.0032
	n7	0.04	-0.0210
	n8	0.003	-0.0013
	n9	0.02	-0.0076
	n10	5	-1.133
	nt	44.118
	K5	0.193
	K6	0.002597
	K7	0.003448
	K8	0.00001799
	K9	0.0002155
	K10	0.00003846
	R	10
	p	40

Solution	Variable	Value	f(x)
	n1	2.91572542389522	0
	n2	3.96094281080888	-1.78E-15
	n3	19.9862916465515	0
	n4	8.42745761047765E-02	0
	n5	2.20956017698935E-02	0
	n6	7.22766590884200E-04	-4.337E-19
	n7	3.32004082515745E-02	-3.469E-18
	n8	4.21099693391800E-04	-6.776E-21
	n9	2.74167068969179E-02	8.674E-19
	n10	3.11467752270064E-02	5.42E-20
	nt	27.0622378157901

Additional information	Only constrained algorithm that keeps the values of n1,n2,n3 and n4
	positive throughout the iterations converges from initial guess 3.