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• Gabriel Vallespin

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NONLINEAR PROBLEM

The Redlich-Kwong equation of state is given by p

RT a  v  b v(v  b) T

where R = the universal gas constant [= 0.518 kJ/(kg K)], T = absolute temperature (K), p = absolute pressure (kPa), and v = the volume of a kg of gas (m3/kg). The parameters a and b are calculated by

a  0.427

R 2TC 2.5 pC

b  0.0866R

TC pC

where pc = 4600 kPa and Tc = 191 K. As a chemical engineer, you are asked to determine the amount of methane fuel that can be held in a 3-m3 tank at a temperature of −40 ◦C with a pressure of 65,000 kPa.

Use a root-locating method of your choice to calculate v and then determine the mass of methane contained in the tank.

GIVEN:

T  40C  273  233K R  0.518kJ / (kgK) p  65000kPa

pC  4600kPa TC  191K vL  0.0026 vU  0.003

REQUIREMENTS: A. the specific volume v and the mass using the exact solution of finding the roots B. the specific volume v and the mass using the bisection method with 3 iterations with the absolute relative approximate error in percent C. the specific volume v and the mass using the newton-raphson method with 3 iterations with the absolute relative approximate error in percent D. the specific volume v and the mass using the secant method with 3 iterations with the absolute relative approximate error in percent E. the specific volume v and the mass using the regula-falsi method with 3 iterations with the absolute relative approximate error in percent

SOLUTION: f(v) 

RT a  p v  b v(v  b) T

R 2TC 2.5 (0.518)2 (191)2.5 Where a  0.427  0.427  12.5577832 pC 4600 And b  0.0866R

TC 191  0.0866(0.518)  0.001862615 pC 4600

f(v) 

(0.518)(233) 12.5577832   65000 v  0.001862615 v(v  0.001862615) 233

f(v) 

120.694 0.822687731   65000 v  0.001862615 v(v  0.001862615)

A. To determine the specific volume v and the mass using the exact solution of finding the roots, With the aid of MATLAB application, the roots are: 0.002807378m3/kg -0.000475273m3/kg -0.000475273m3/kg Since the true value ranges from 0 to 3m3, Therefore, the true value for the specific volume is 0.002807378m3/kg. The mass can be calculated using this formula: m

V where V = 3m3 and v = 0.002807378m3/kg v

m

3 0.002807378

m  1068.612776762kg

B. To determine the specific volume v and the mass using the bisection method with 3 iterations with the absolute relative approximate error in percent, For the 1st iteration, To check,

f(vU ) * f(v L )  0 f(0.003) * f(0.0026)  0

1528.082313056 * 27774.156684279  0 424391400.311505500  0 Then the root is bounded and the approximated specific volume is, vM 

vU  v L 0.003  0.0026  2 2

vM  0.0028m3 / kg

Therefore the approximated mass is, mM 

V 3  v M 0.0028

mM  1071.428571429kg

For the 2nd iteration, To check, f(vM ) * f(v L )  0 f(0.0028) * f(0.0026)  0

740.554048453 * 27774.156684279  0 20568264.174918566  0 Then the root is not bounded and the approximated specific volume is,

vM1 

vU  v M 0.003  0.0028  2 2

vM1  0.0029m3 / kg

Therefore the approximated mass is, mM1 

V v M1



3 0.0029

mM1  1034.482758621kg

For the absolute relative approximate error in percent in specific volume,

|A |

| present  past | | 0.0029  0.0028 | * 100%  * 100% present 0.0029

|A | 3.448275862%

For the absolute relative approximate error in percent in mass,

|A |

| present  past | * 100% present

|A |

| 1034.482758621  1071.428571429 | * 100% 1034.482758621

|A | 3.571428571%

For the 3rd iteration, To check, f(vM1 ) * f(vM )  0 f(0.0029) * f(0.0028)  0

8820.592728173 * 740.554048453  0 6087793.225534353  0 Then the root is bounded and the approximated specific volume is,

v M1  v M 0.0029  0.0028  2 2

v M 2 

vM2  0.00285m3 / kg

Therefore the approximated mass is, mM2 

V v M2



3 0.00285

mM2  1052.631578947kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.00285  0.0029 * 100  * 100 present 0.00285

A  1.754385965% For the absolute relative approximate error in percent in mass,

A 

present  past 1052.631578947  1034.482758621 * 100  * 100 present 1052.631578947

A  1.724137931%

C. To determine the specific volume v and the mass using the newton-raphson method with 3 iterations with the absolute relative approximate error in percent,

f ' v  

120.694 12.5577832(2v  0.001862615)  2 (v  0.001862615) 233(v  v  0.001862615 )2

For the 1st iteration, The approximated specific volume is, vI  0.003m3 / kg

vI1  vI 

f  vI  15280.082313056  0.003  f '  vI  62901414.370263815

vI1  0.002757079m3 / kg

Therefore the approximated mass is, mI 

V vI

mI 

3 0.003

mI  1000.000000000kg mI1 

V 3  vI1 0.002757079

mI1  1088.108102815kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.002757079  0.003 * 100  * 100 present 0.002757079

A  8.810810281% For the absolute relative approximate error in percent in mass,

A 

present  past 1088.108102815  1000.000000000 * 100  * 100 present 1088.108102815

A  8.097366667% For the 2nd iteration, The approximated specific volume is, vI1  0.002757079m3 / kg

vI2  vI1 

f  vI1  5343.353111776  0.002757079  f '  vI1  113446040.096620220

vI2  0.002804179m3 / kg

Therefore the approximated mass is, mI2 

V 3  vI2 0.002804179

mI2  1069.831697331kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.002804179  0.002757079 * 100  * 100 present 0.002804179

A  1.679636000% For the absolute relative approximate error in percent in mass,

A 

present  past 1069.831697331  1088.108102815 * 100  * 100 present 1069.831697331

A  1.708343988% For the 3rd iteration, The approximated specific volume is, vI2  0.002804179m3 / kg

v I 3  v I  2 

f  v I 2  319.343532363  0.002804179  f '  v I 2  100250902.714098590

vI3  0.002807364m3 / kg

Therefore the approximated mass is, mI3 

V 3  vI3 0.002807364

mI3  1068.617937198kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.002807364  0.002804179 * 100  * 100 present 0.002807364

A  0.113451622% For the absolute relative approximate error in percent in mass,

A 

present  past 1068.617937198  1069.831697331 * 100  * 100 present 1068.617937198

A  0.113582235%

D. To determine the specific volume v and the mass using the secant method with 3 iterations with the absolute relative approximate error in percent, vI  0.003m3 / kg vI1  0.0026m3 / kg

For the 1st iteration, The approximated specific volume is, vI1  vI 

f  vI  [vI  vI1 ] f  vI   f(vI1 )

f  vI   f  0.003   1528.082313056

f  vI1   f  0.0026   27774.156684279 vI1  0.003 

f  0.003  [0.003  0.0026] f  0.003   f(0.0026)

vI1  0.003 

1528.082313056[0.003  0.0026] 1528.082313056  27774.156684279

vI1  0.002858039m3 / kg

The approximated mass is, mI1 

V 3  vI1 0.002858039

mI1  1049.670856945kg

For the 2nd iteration, The approximated specific volume is, vI2  vI1 

f  vI1  [vI1  vI ] f  vI1   f(vI )

f  vI1   f  0.002858039   4727.975225643 f  vI   f  0.0026   27774.156684279 vI2  0.002858039 

f  0.002858039  [0.002858039  0.0026] f  0.002858039   f(0.0026)

vI2  0.002858039 

1528.082313056[0.002858039  0.0026] 1528.082313056  27774.156684279

vI2  0.002820503m3 / kg

The approximated mass is, mI2 

V 3  vI2 0.002820503

mI2  1063.640089963kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.002820503  0.002858039 * 100  * 100 present 0.002820503

A  1.330828636% For the absolute relative approximate error in percent in mass,

A 

present  past 1063.640089963  1049.670856945 * 100  * 100 present 1063.640089963

A  1.313342093% For the 3rd iteration, The approximated specific volume is, v I 3  v I 2 

f  vI2  [vI2  vI1 ] f  vI2   f ( vI1 )

f  vI2   f  0.002820503   1283.432429450 f  vI1   f  0.002858039   4727.975225643 vI3  0.002820503 

f  0.002820503 [0.002820503  0.002858039] f(0.002820503)  f(0.002858039)

vI3  0.002820503 

1283.432429450[0.002820503  0.002858039] 1283.432429450  (4727.975225643)

vI3  0.002806517m3 / kg

The approximated mass is, mI3 

V 3  vI3 0.002806517

mI3  1068.940561746kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.002806517  0.002820503 * 100  * 100 present 0.002806517

A  0.498335374% For the absolute relative approximate error in percent in mass,

A 

present  past 1068.940561746  1063.640089963 * 100  * 100 present 1068.940561746

A  0.495862162%

E. To determine the specific volume v and the mass using the regula-falsi method with 3 iterations with the absolute relative approximate error in percent, For the 1st iteration, To check,

f(vU ) * f(v L )  0 f(0.003) * f(0.0026)  0

1528.082313056 * 27774.156684279  0 424391400.311505500  0 Then the root is bounded and the approximated specific volume is, vR  vU 

f  v U  v L  v U  f  vL   f  vU 

vR  0.003 

( 1528.082313056)(0.0026  0.003) 27774.156684279  (1528.082313056)

vR  0.002858039m3 / kg

Therefore the approximated mass is,

mR 

V 3  v R 0.002858039

mR  1049.670856945kg

For the 2nd iteration, To check,

f(vR ) * f(v L )  0

f  0.002858039 * f (0.0026)  0

4727.975225643 * 27774.156684279  0

424391400.311505500  0 Then the root is bounded and the approximated specific volume is, v R 1  v R 

f  v R  v L  v R  f  vL   f  vR 

vR 1  0.002858039 

( 4727.975225643)(0.0026  0.002858039) 27774.156684279  (4727.975225643)

vR1  0.002820503m3 / kg

Therefore the approximated mass is, mR 1 

V v R 1



3 0.002820503

mR1  1063.640089963kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.002820503  0.002858039 * 100  * 100 present 0.002820503

A  1.330826452%

For the absolute relative approximate error in percent in mass,

A 

present  past 1063.640089963  1049.670856945 * 100  * 100 present 1063.640089963

A  1.313342093% For the 3rd iteration, To check, f(vR 1 ) * f(vL )  0

f  0.002820503 * f (0.0026)  0

1528.082313056 * 27774.156684279  0

-42441197.589292817  0 Then the root is bounded and the approximated specific volume is, v R 2  vR 1 

f  v R 1  v L  vR 1  f  v L   f  v R 1 

vR 2  0.002820503 

( 1283.432429450)(0.0026  0.002820503) 27774.156684279  (1283.432429450)

vR2  0.002810764m3 / kg

Therefore the approximated mass is, mR 2 

V vR 2



3 0.002810764

mR2  1067.325582912kg

For the absolute relative approximate error in percent in specific volume,

A 

present  past 0.002810764  0.002820503 * 100  * 100 present 0.002810764

A  0.346489424% For the absolute relative approximate error in percent in mass,

A 

present  past 1067.325582912  1063.640089963 * 100  * 100 present 1067.325582912

A  0.345301659%

SUMMARY: NONLINEAR APPROXIMATION METHOD

BISECTION METHOD

NEWTONRAPHSON METHOD

SECANT METHOD

REGULA-FALSI METHOD

REQUIREMENT

ITERATION

SPECIFIC VOLUME (m3/kg)

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

MASS (kg) SPECIFIC VOLUME (m3/kg) MASS (kg) SPECIFIC VOLUME (m3/kg) MASS (kg) SPECIFIC VOLUME (m3/kg) MASS (kg)

APPROXIMATE VALUE 0.0028 0.0029 0.00285 1071.428571429 1034.482758621 1052.631578947 0.002757079 0.002804179 0.002807364 1088.108102815 1069.831697331 1068.617937198 0.002858039 0.002820503 0.002806517 1049.670856945 1063.640089963 1068.940561746 0.002858039 0.002820503 0.002810764 1049.670856945 1063.640089963 1067.325582912

|∈A| (%) --3.448275862 1.754385965 --3.571428571 1.724137931 8.810810281 1.679639000 0.113451622 8.097366667 1.708343988 0.113582235 --1.330828636 0.498335374 --1.313342093 0.495862162 --1.330826452 0.346489424 --1.313342093 0.345301659