1

Example 1

A liquid-liquid extraction process conducted in the Electrochemical Materials Laboratory involved the extraction of nickel from the aqueous phase into an organic phase. A typical set of experimental data from the laboratory is given below.

NI AQUEOUS PHASE, a ( g /l ) 2 2.5 3

NI ORGANIC PHASE, g ( g/ l ) 8.57 10 12

ASSUMING g IS THE AMOUNT OF NI IN THE ORGANIC PHASE AND a IS THE AMOUNT OF NI IN THE AQUEOUS PHASE, THE QUADRATIC INTERPOLANT THAT ESTIMATES g IS GIVEN BY g= x1 a2+x2 a+x3 , 2≤a≤3

THE SOLUTION FOR THE UNKNOWNS x1 , x2 , AND x3 IS GIVEN BY

[ 4 2 16 .25 2.5 1

9 3 1 ][ x1

x2

x3]=[8 .57

1012 ]

FIND THE VALUES OF x1 , x2 , AND x3 USING THE GAUSS-SEIDEL METHOD. ESTIMATE THE AMOUNT OF NICKEL IN THE ORGANIC PHASE WHEN 2.3 G/L IS IN THE AQUEOUS PHASE USING QUADRATIC INTERPOLATION. USE

[ x1

x2

x3]=[111 ]

AS THE INITIAL GUESS AND CONDUCT TWO ITERATIONS.

Solution:-REWRITING THE EQUATIONS GIVES

x1=8 . 57−2 x2− x3

4

x2=10−6 . 25 x1−x3

2. 5

x3=12−9 x1−3x2

1ITERATION #1GIVEN THE INITIAL GUESS OF THE SOLUTION VECTOR AS

2

[ x1

x2

x3]=[111 ]

WE GET

x1=8 .57−2×1−1

4 =1. 3925

x2=10−6 .25×1 .3925−1

2 .5 =0 . 11875

x3=12−9×1. 3925−3×0 . 11875

1 =−0 .88875

THE ABSOLUTE RELATIVE APPROXIMATE ERROR FOR EACH x i THEN IS

|∈a|1=|1 . 3925−11 .3925

|×100

=28 . 187 %

|∈a|2=|0. 11875−10 . 11875

|×100

=742.11%

|∈a|3=|−0 .88875−1−0. 88875

|×100

=212.52 %AT THE END OF THE FIRST ITERATION, THE ESTIMATE OF THE SOLUTION VECTOR IS

[ x1

x2

x3]=[ 1. 3925

0 . 11875−0. 88875]

AND THE MAXIMUM ABSOLUTE RELATIVE APPROXIMATE ERROR IS 742 .11% .ITERATION #2THE ESTIMATE OF THE SOLUTION VECTOR AT THE END OF ITERATION #1 IS

[ x1

x2

x3]=[ 1. 3925

0 . 11875−0. 88875]

NOW WE GET

x1=8 . 57−2×0 .11875−(−0. 88875 )

4 =2 .3053

x2=10−6 .25×2 . 3053−(−0 . 88875 )

2 .5

3

=−1 . 4078

x3=12−9×2.3053−3×(−1 . 4078 )

1 =−4 .5245

THE ABSOLUTE RELATIVE APPROXIMATE ERROR FOR EACH x i THEN IS

|∈a|1=|2 . 3053−1 .39252. 3053

|×100

=39 .596 %

|∈a|2=|−1.4078−0 .11875−1.4078

|×100

=108 .44 %

|∈a|3=|−4 . 5245−(−0 .88875 )

−4 . 5245|×100

=80 .357 %AT THE END OF THE SECOND ITERATION, THE ESTIMATE OF THE SOLUTION VECTOR IS

[ x1

x2

x3]=[ 2. 3053

−1.4078−4 .5245 ]

AND THE MAXIMUM ABSOLUTE RELATIVE APPROXIMATE ERROR IS 108 . 44 % .CONDUCTING MORE ITERATIONS GIVES THE FOLLOWING VALUES FOR THE SOLUTION VECTOR AND THE CORRESPONDING ABSOLUTE RELATIVE APPROXIMATE ERRORS.

ITERATION x1 |∈a|1 % x2 |∈a|2 % x3 |∈a|3 %123456

1.39252.30533.97757.058412.75223.291

28.186739.596042.04143.64944.64945.249

0.11875–1.4078–4.1340–9.0877–18.175–34.930

742.1053108.435365.94654.51049.99947.967

–0.88875–4.5245–11.396–24.262–48.243–92.827

212.5280.35760.29653.03249.70848.030

AFTER SIX ITERATIONS, THE ABSOLUTE RELATIVE APPROXIMATE ERRORS ARE NOT DECREASING MUCH. IN FACT, CONDUCTING MORE ITERATIONS REVEALS THAT THE ABSOLUTE RELATIVE APPROXIMATE ERROR

CONVERGES TO A VALUE OF 46 . 070 % FOR ALL THREE VALUES WITH THE SOLUTION VECTOR DIVERGING FROM THE EXACT SOLUTION DRASTICALLY.

ITERATION x1 |∈a|1% x2 |∈a|2 % x3 |∈a|3 %32 2 .1428×108 46 . 0703 −3 .3920×108 46 . 0703 −9 . 1095×108 46 . 0703

THE EXACT SOLUTION VECTOR IS

4

[ x1

x2

x3]=[ 1 . 14

−2.278 . 55 ]

TO CORRECT THIS, THE COEFFICIENT MATRIX NEEDS TO BE MORE DIAGONALLY DOMINANT. TO ACHIEVE A MORE DIAGONALLY DOMINANT COEFFICIENT MATRIX, REARRANGE THE SYSTEM OF EQUATIONS BY EXCHANGING EQUATIONS ONE AND THREE.

[ 9 3 16 .25 2. 5 1

4 2 1 ] [ x1

x2

x3]=[12

108. 57]

ITERATION #1GIVEN THE INITIAL GUESS OF THE SOLUTION VECTOR AS

[ x1

x2

x3]=[111 ]

WE GET

x1=12−3×1−1

9 =0 .88889

x2=10−6 . 25×0. 88889−1

2. 5 =1 .3778

x3=8 .57−4×0 . 88889−2×1.3778

1 =2 .2589

THE ABSOLUTE RELATIVE APPROXIMATE ERROR FOR EACH x i THEN IS

|∈a|1=|0. 88889−10 . 88889

|×100

=12 .5 %

|∈a|2=|1 .3778−11 .3778

|×100

=27 .419 %

|∈a|3=|2. 2589−12 . 2589

|×100

=55 .730 %AT THE END OF THE FIRST ITERATION, THE ESTIMATE OF THE SOLUTION VECTOR IS

[ x1

x2

x3]=[0. 88889

1.37782 .2589 ]

and the maximum absolute relative approximate error is 55 .730 % .

5

ITERATION #2THE ESTIMATE OF THE SOLUTION VECTOR AT THE END OF ITERATION #1 IS

[ x1

x2

x3]=[0. 88889

1.37782 .2589 ]

NOW WE GET

x1=12−3×1 . 3778−1×2. 2589

9 =0 .62309

x2=10−6 .25×0. 62309−1×2. 2589

2.5 =1 .5387

x3=8 .57−4×0 . 62309−2×1. 5387

1 =3 .0002

THE ABSOLUTE RELATIVE APPROXIMATE ERROR FOR EACH x i THEN IS

|∈a|1=|0.62309−0 .888890 .62309

|×100

=42. 659 %

|∈a|2=|1 . 5387−1 .37781. 5387

|×100

=10. 460 %

|∈a|3=|3.0002−2. 25893 . 0002

|×100

=24 .709 %AT THE END OF THE SECOND ITERATION, THE ESTIMATE OF THE SOLUTION IS

[ x1

x2

x3]=[0.62309

1. 53873 . 0002 ]

AND THE MAXIMUM ABSOLUTE RELATIVE APPROXIMATE ERROR IS 42 . 659 % .CONDUCTING MORE ITERATIONS GIVES THE FOLLOWING VALUES FOR THE SOLUTION VECTOR AND THE CORRESPONDING ABSOLUTE RELATIVE APPROXIMATE ERRORS.

ITERATION x1 |∈a|1 % x2 |∈a|2 % x3 |∈a|3 %12345

0.888890.623090.487070.421780.39494

12.542.65927.92615.4796.7960

1.37781.53871.58221.56271.5096

27.41910.4562.75061.25373.5131

2.25893.00023.45723.75763.9710

55.73024.70913.2207.99285.3747

6

6 0.38890 1.5521 1.4393 4.8828 4.1357 3.9826

AFTER SIX ITERATIONS, THE ABSOLUTE RELATIVE APPROXIMATE ERRORS SEEM TO BE DECREASING. CONDUCTING MORE ITERATIONS ALLOWS THE ABSOLUTE RELATIVE APPROXIMATE ERROR DECREASE TO AN ACCEPTABLE LEVEL.

ITERATION x1 |∈a|1 % x2 |∈a|2 % x3 |∈a|3 %

199200

1.13351.1337

0.0144120.014056

–2.2389–2.2397

0.0348710.034005

8.51398.5148

0.0106660.010403

THIS IS CLOSE TO THE EXACT SOLUTION VECTOR OF

[ x1

x2

x3]=[ 1 . 14

−2.278 . 55 ]

THE POLYNOMIAL THAT PASSES THROUGH THE THREE DATA POINTS IS THEN

g (a )=x1 (a )2+x2 (a )+ x3

=1 .1337 (a )2+(−2 .2397 ) ( a )+8 .5148WHERE g IS THE AMOUNT OF NICKEL IN THE ORGANIC PHASE AND a IS THE AMOUNT OF NICKEL IN THE AQUEOUS PHASE.

WHEN 2 .3 g/ l IS IN THE AQUEOUS PHASE, USING QUADRATIC INTERPOLATION, THE ESTIMATED AMOUNT OF NICKEL IN THE ORGANIC PHASE IS

g (2. 3 )=1. 1337 (2. 3 )2+ (−2.2397 )×(2.3 )+8 . 5148 =9.3608 g/ l

7

Example 2

A trunnion has to be cooled before it is shrink fitted into a steel hub.

Figure 1 Trunnion to be slid through the hub after contracting

.

The equation that gives the temperature T f to which the trunnion has to be cooled to obtain the desired contraction is given byf (T f )=−0. 50598× 10−10 T f

3+0 . 38292× 10−7 T f2+0 . 74363× 10−4 T f+0 .88318× 10−2=0Use

the bisection method of finding roots of equations to find the temperature T f to which the trunnion has to be cooled. Conduct three iterations to estimate the root of the above equation. Find the absolute relative approximate error at the end of each iteration and the number of significant digits at least correct at the end of each iteration.

Solution:-FROM THE DESIGNER’S RECORDS FOR THE PREVIOUS BRIDGE, THE TEMPERATURE TO WHICH THE TRUNNION

WAS COOLED WAS −108 ° F . HENCE ASSUMING THE TEMPERATURE TO BE BETWEEN −100 ° F AND −150 ° F , WE HAVE

T f , ℓ=−150 ° F , T f , u=−100 ° F

CHECK IF THE FUNCTION CHANGES SIGN BETWEEN T f , ℓ AND T f , u .

f (T f , ℓ )= f (−150 )

=−0 . 50598× 10−10(−150)3+0 .38292× 10−7 (−150)2

+0 . 74363× 10−4 (−150 )+0 .88318× 10−2

=−1. 2903× 10−3

8

f (T f , u)=f (−100 )

=−0 .50598× 10−10(−100)3+0. 38292× 10−7 (−100)2

+0 .74363× 10−4 (−100)+0 . 88318× 10−2

=1 .8290× 10−3

HENCE

f (T f , ℓ ) f (T f , u)= f (−150 ) f (−100 )=(−1. 2903× 10−3 ) (1. 8290× 10−3)<0

SO THERE IS AT LEAST ONE ROOT BETWEEN T f , ℓ AND T f , u THAT IS BETWEEN −150 AND −100 .ITERATION 1THE ESTIMATE OF THE ROOT IS

T f ,m=T f , ℓ+T f ,u

2

=

−150+(−100 )2

=−125f (T f , m)=f (−125 )

=−0.50598× 10−10(−125 )3+0 . 38292× 10−7 (−125 )2

+0 .74363× 10−4 (−125)+0 .88318× 10−2

=2 . 3356×10−4

f (T f , ℓ ) f (T f , m )=f (−150 ) f (−125 )=(−1 .2903× 10−3) (2 .3356× 10−4)<0

HENCE THE ROOT IS BRACKETED BETWEEN T f , ℓ AND T f , m , THAT IS, BETWEEN −150 AND −125 . SO, THE LOWER AND UPPER LIMITS OF THE NEW BRACKET ARE

T f , ℓ=−150 , T f , u=−125

AT THIS POINT, THE ABSOLUTE RELATIVE APPROXIMATE ERROR |∈a| CANNOT BE CALCULATED, AS WE DO NOT

HAVE A PREVIOUS APPROXIMATION.ITERATION 2THE ESTIMATE OF THE ROOT IS

T f ,m=T f , ℓ+T f ,u

2

=

−150+(−125 )2

=−137.5f (T f , m)= f (−137 . 5 )

=−0 .50598× 10−10(−137 .5 )3+0 .38292× 10−7(−137 .5 )2

+0 .74363× 10−4 (−137 .5 )+0 .88318× 10−2 = −5 .3762 × 10−4

f (T f ,m) f (T f , u)=f (−137 .5 ) f (−125 )= (−5 .3762× 10−4 ) ( 2.3356× 10−4 )<0

9

HENCE, THE ROOT IS BRACKETED BETWEEN T f , m AND T f , u , THAT IS, BETWEEN −125 AND −137 .5 . SO THE LOWER AND UPPER LIMITS OF THE NEW BRACKET ARE

T f , ℓ=−137. 5 , T f ,u=−125

THE ABSOLUTE RELATIVE APPROXIMATE ERROR |∈a| AT THE END OF ITERATION 2 IS

|∈a|=|T f , m

new−T f , mold

T f , mnew |×100

=|

−137 . 5−(−125)−137 . 5

|×100

=9. 0909 %NONE OF THE SIGNIFICANT DIGITS ARE AT LEAST CORRECT IN THE ESTIMATED ROOT OF

T f , m=−137 . 5AS THE ABSOLUTE RELATIVE APPROXIMATE ERROR IS GREATER THAT 5 % .ITERATION 3THE ESTIMATE OF THE ROOT IS

T f ,m=T f , ℓ+T f ,u

2

=

−137 . 5+(−125 )2

=−131. 25f (T f , m)=f (−131 .25 )

=−0. 50598× 10−10(−131 .25 )3+0 . 38292× 10−7 (−131. 25)2

+0 . 74363× 10−4 (−131 .25 )+0 . 88318× 10−2 = −1 .54303 × 10−4

f (T f , ℓ ) f (T f ,m )=f (−125 ) f (−131 .25 )=(2 .3356× 10−4 ) (−1 .5430× 10−4 )<0

HENCE, THE ROOT IS BRACKETED BETWEEN T f , ℓ AND T f , m , THAT IS, BETWEEN −125 AND −131 .25 . SO THE LOWER AND UPPER LIMITS OF THE NEW BRACKET ARE

T f , ℓ=−131.25 , T f ,u=−125

THE ABSOLUTE RELATIVE APPROXIMATE ERROR |∈a| AT THE ENDS OF ITERATION 3 IS

|∈a|=|T f , m

new−T f , mold

T f , mnew |×100

=|

−131 . 25−(−137 . 5)−131 .25

|×100

=4 .7619 %THE NUMBER OF SIGNIFICANT DIGITS AT LEAST CORRECT IS 1.SEVEN MORE ITERATIONS WERE CONDUCTED AND THESE ITERATIONS ARE SHOWN IN THE TABLE 1 BELOW.

TABLE 1 ROOT OF f ( x )=0 AS FUNCTION OF NUMBER OF ITERATIONS FOR BISECTION METHOD.

10

ITERATION T f , ℓ T f , u T f , m|∈a|% f (T f , m)

12345678910

−150−150−137.5−131.25−131.25−129.69−128.91−128.91−128.91−128.81

−100−125−125−125−128.13−123.13−123.13−128.52−128.71−128.71

−125−137.5−131.25−128.13−129.69−128.91−128.52−128.71−128.81−128.76

---------9.09094.76192.43901.20480.606060.303950.151750.0758150.037922

2 .3356×10−4

−5 .3762×10−4

−1 .5430×10−4

3 . 9065×10−5

−5 .7760×10−5

−9 . 3826×10−6

1 .4838×10−5

2 .7228×10−6

−3 .3305×10−6

−3 . 0396×10−7

AT THE END OF THE10 th ITERATION,

|∈a|=0 .037922 %HENCE, THE NUMBER OF SIGNIFICANT DIGITS AT LEAST CORRECT IS GIVEN BY THE LARGEST VALUE OF m FOR WHICH

|∈a|≤0 . 5×102−m

0 . 037922≤0 . 5×102−m

0 .075844≤102−m

log (0 . 075844 )≤2−mm≤2−log (0 .075844 )=3 .1201

SO

m=3THE NUMBER OF SIGNIFICANT DIGITS AT LEAST CORRECT IN THE ESTIMATED ROOT OF −128 .76 IS 3.

11

Example 3

A SOLID STEEL SHAFT AT ROOM TEMPERATURE OF 27o C IS NEEDED TO BE CONTRACTED SO THAT IT CAN BE SHRUNK-FIT INTO A HOLLOW HUB. IT IS PLACED IN A REFRIGERATED CHAMBER THAT IS MAINTAINED AT −33o C . THE RATE OF CHANGE OF TEMPERATURE OF THE SOLID SHAFT θ IS GIVEN BY dθdt

=−5 . 33×10−6¿ (−3 . 69×10−6 θ4+2.33×10−5 θ3+1.35×10−3 θ2 ¿ )¿¿

¿

θ (0 )=27 °CUSING EULER’S METHOD, FIND THE TEMPERATURE OF THE STEEL SHAFT AFTER 86400 SECONDS. TAKE A STEP SIZE OF h=43200 SECONDS.

Solution:-

dθdt

=−5 . 33×10−6¿ (−3 . 69×10−6 θ4+2. 33×10−5 θ3+1.35×10−3 θ2 ¿ )¿¿

¿

f ( t ,θ )=−5 . 33×10−6¿ (−3 .69×10−6θ4+2. 33×10−5 θ3+1.35×10−3 θ2 ¿)¿¿

¿THE EULER’S METHOD REDUCES TO

θi+1=θi+ f ( ti , θi )h

FOR i=0 , t 0=0 , θ0=27θ1=θ0+f ( t0 ,θ0) h

=27+f (0 , 27 ) 43200

=27+¿¿

=27+(−0 . 0020893 ) 43200 =−63.258 °C

θ1 IS THE APPROXIMATE TEMPERATURE AT

t=t 1= t0+h=0+43200=43200 sθ (43200 )≈θ1=−63 .258 °C

FOR i=1 , t 1=43200 , θ1=−63 . 258θ2=θ1+f (t 1 , θ1) h

=−63.258+ f (43200 ,−63 .258 ) 43200

12

=−63.258+¿¿ =−63. 258+(−0. 0092607 ) 43200

=−463 .32 °Cθ2 IS THE APPROXIMATE TEMPERATURE AT

t=t 2= t1+h=43200+43200=86400 sθ (86400 )≈θ2=−463 . 32° C

FIGURE 1 COMPARES THE EXACT SOLUTION WITH THE NUMERICAL SOLUTION FROM EULER’S METHOD

FOR THE STEP SIZE OF h=43200 .

FIGURE 1 COMPARING EXACT AND EULER’S METHOD.

THE PROBLEM WAS SOLVED AGAIN USING SMALLER STEP SIZES. THE RESULTS ARE GIVEN BELOW IN TABLE 1.

TABLE 1 TEMPERATURE AT 86400 SECONDS AS A FUNCTION OF STEP SIZE, h .

STEP

SIZE, h θ (86400 ) Et |∈t|%

864004320021600108005400

153.52463.3229.54227.79526.958

127.42437.223.44211.69620.85870

488.211675.213.1896.49883.2902

FIGURE 2 SHOWS HOW THE TEMPERATURE VARIES AS A FUNCTION OF TIME FOR DIFFERENT STEP SIZES.

13

FIGURE 2 COMPARISON OF EULER’S METHOD WITH EXACT SOLUTION FOR DIFFERENT STEP SIZES.

WHILE THE VALUES OF THE CALCULATED TEMPERATURE AT t=86400 s AS A FUNCTION OF STEP SIZE ARE plotted in Figure 3.

FIGURE 3 EFFECT OF STEP SIZE IN EULER’S METHOD.

THE SOLUTION TO THIS NONLINEAR EQUATION AT t=86400 s IS

θ(86400 )=−26 .099 °C

14

Example 4

A solid steel shaft at room temperature of 27oC is needed to be contracted so that it can be shrunk-fit into a hollow hub. It is placed in a refrigerated chamber that is maintained at −33o C . The rate of change of temperature of the solid shaft θ is given bydθdt

=−5 . 33×10−6¿ (−3 . 69×10−6 θ4+2.33×10−5 θ3+1.35×10−3 θ2 ¿ )¿¿

¿

θ (0 )=27 °CUsing Euler’s method, find the temperature of the steel shaft after 86400 seconds. Take a step size of h=43200 seconds.

Solution:-

dθdt

=−5 . 33×10−6¿ (−3 . 69×10−6 θ4+2. 33×10−5 θ3+1.35×10−3 θ2 ¿ )¿¿

¿

f ( t ,θ )=−5 . 33×10−6¿ (−3 .69×10−6θ4+2. 33×10−5 θ3+1.35×10−3 θ2 ¿ )¿¿

¿THE EULER’S METHOD REDUCES TO

θi+1=θi+ f ( ti , θi )h

FOR i=0 , t 0=0 , θ0=27θ1=θ0+f ( t0 ,θ0) h

=27+f (0 , 27 ) 43200

=27+¿¿

=27+(−0 . 0020893 ) 43200 =−63.258 °C

θ1 IS THE APPROXIMATE TEMPERATURE AT

t=t 1= t0+h=0+43200=43200 sθ (43200 )≈θ1=−63 .258 °C

FOR i=1 , t 1=43200 , θ1=−63 . 258θ2=θ1+f (t 1 , θ1) h

=−63.258+ f (43200 ,−63 .258 ) 43200

15

=−63.258+¿¿ =−63. 258+(−0. 0092607 ) 43200

=−463 .32 °Cθ2 IS THE APPROXIMATE TEMPERATURE AT

t=t 2= t1+h=43200+43200=86400 sθ (86400 )≈θ2=−463 . 32° C

FIGURE 1 COMPARES THE EXACT SOLUTION WITH THE NUMERICAL SOLUTION FROM EULER’S METHOD

FOR THE STEP SIZE OF h=43200 .

Figure 1 Comparing exact and Euler’s method.

THE PROBLEM WAS SOLVED AGAIN USING SMALLER STEP SIZES. THE RESULTS ARE GIVEN BELOW IN TABLE 1.

TABLE 1 TEMPERATURE AT 86400 SECONDS AS A FUNCTION OF STEP SIZE, h .

STEP

SIZE, h θ (86400 ) Et |∈t|%

864004320021600108005400

153.52463.3229.54227.79526.958

127.42437.223.44211.69620.85870

488.211675.213.1896.49883.2902

FIGURE 2 SHOWS HOW THE TEMPERATURE VARIES AS A FUNCTION OF TIME FOR DIFFERENT STEP SIZES.

16

FIGURE 2 COMPARISON OF EULER’S METHOD WITH EXACT SOLUTION FOR DIFFERENT STEP SIZES.

WHILE THE VALUES OF THE CALCULATED TEMPERATURE AT t=86400 s AS A FUNCTION OF STEP SIZE ARE PLOTTED IN FIGURE 3.

FIGURE 3 EFFECT OF STEP SIZE IN EULER’S METHOD.

THE SOLUTION TO THIS NONLINEAR EQUATION AT t=86400 s IS

θ(86400 )=−26 .099 °C