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Page 1: APPLIED NUMERICAL ANALYSIS.pptx

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APPLIED NUMERICAL ANALYSIS

BATCH 10

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PROBLEM 1

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SOLUTION

(a) Finding out the entropy corresponding to a specific

volume of 0.108 using Linear Interpolation

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SOLUTION

(b) Using Quadratic Interpolation

x0 = 0.10377 

x1 = 0.11144 

x2 = 0.1254

f(x0) = 6.4147

f(x1) = 6.5453f(x2) = 6.7664

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SOLUTION

b1 = 17.027394b0 = 6.4147

b2 = -54.98245573

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SOLUTION

Substituting b0, b1, b2, x0, x1 and x2,

f 2(x) = 6.4147 + 17.0273794(x-0.10377)

 – 54.98245573(x-0.10377)(x-0.11144)

where x = 0.108

f 2(0.108) = 6.487525876 kJ / kg Kwhich is the entropy

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SOLUTION

(c) Finding out the specific volume corresponding toan entropy of 6.6 kJ / kg K using InverseInterpolation

Substituting f 2(x) = 6.6 in the formula derived usingQuadratic Interpolation

f 2(x) = 6.4147 + 17.0273794(x-0.10377)

 – 54.98245573(x-0.10377)(x-0.11144)

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SOLUTION

We get, x = 0.1147707903 which is the specificvolume

Substituting f 1(x) = 6.6 in the formula derived forlinear interpolation,

We get, x = 0.1162393607 which is the specificvolume

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PROBLEM 2

During a survey, you are required to compute the area of the

field shown in figure. Use Simpson’s rules to determine the area.

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SOLUTION

DISTANCE (NORTH TO SOUTH) ORDINATE (WEST TO EAST)

0 0

200 600

400 880

600 950

800 1100

1000 1600

1200 1900

1400 21001600 2300

1800 2600

2000 2950

2200 3150

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SOLUTION

DISTANCE (NORTH TO SOUTH) ORDINATE (WEST TO EAST)

2400 3200

2600 3200

2800 3150

3000 3000

3200 2800

3400 2700

3600 2600

3800 25254000 2525

4200 2430

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SOLUTION

Use Simpson’s 1/3 rd rule for the first 18 segments and Simpson’s 3/8the rule for the remaining 3 segments.

Area1 = (3600 – 0) [0 +4(600+950+1600+2100+2600+3150+3200+3000+2700) +

2(880+1100+1900+2300+2950+3200+3150+2800+2600)]/4 =7917333.333 sq. ft

Area2 = (4200 – 3600)[2600+3(2525+2525)+2430]/8

= 1513500 sq. ft

Total = 9430833 sq. ft