applied numerical analysis.pptx
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7/29/2019 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