skf2133-chapter2n
Transcript of skf2133-chapter2n
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Numerical Methods for Numerical Methods for Chemical EngineersChemical EngineersChapter 2: Approximations and ErrorsChapter 2: Approximations and Errors
Saharudin Saharudin HaronHaron
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Dimensional HomogeneityValid equation must be dimensionally homogenousBoth sides of equation must have same dimensions
V(m/s) = Vo(m/s) + g(m/s2)t(s)
Example : Consider this equation D(ft) = 3t(s) + 4What is the dimension and unit for constants 3 and 4?
Constant 3 length/time Constant 4 length
ft/sft
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Scientific NotationLarge and very small number often encountered in process calculationsConvenient way to represent such number is to use scientific notationExpressed between 0.1 and 10 and a power of 10
123 000 000 = 1.23 x 108 (or 0.123 x 109)0.000028 = 2.8 x 10-5 (or 0.28 x 10-4)
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Significant FigureThe significant of a number are the digits from the first nonzero on the left to either
the last digit (zero or nonzero) on the right if there is decimal pointthe last nonzero digit of the number if there is no decimal point
2300 or 2.3 x 103 has two significant figures2300. or 2.300 x 103 has four significant figures2300.0 or 2.3000 x 103 has five significant figures23040 or 2.304 x 104 has four significant figures0.035 or 3.5 x 10-2 has two significant figures0.03500 or 3.500 x 10-2 has four significant figures
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Rule of thumb on significant figureMultiplication and division
The answer should be equal to the lowest significant figure of any multiplicands or divisors
(3) (4) (7) (3)(3.57)(4.286) = 15.30102 15.3
(2) (4) (3) (9) (2) (2)(5.2 x 10-4)(0.1635 x 107)/(2.67) = 318.426966 3.2 x102 = 320
rounding the last digit 5 to even number1.35 => 1.41.25 => 1.2
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Calculate the following and give your answer according to the rule of significant figure:
a) (3.4)(2.75)b) (1.76)(0.00120)/(78.2)c) (5.400 x 103)/27d) (9.83 x 107)(0.2 x 10-5)
Significant Figure(assignment in class)
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Accuracy and PrecisionAccuracy - How closely a computed or measured value agrees with the true valuePrecision - How closely individual computed or measured values agree with each other
the number of significant figures representing a quantitythe spread in repeated computations or measurements of a particular value
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Accuracy and PrecisionIncreasing accuracy
Incr
easi
ng P
rec i
sion
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
ErrorsNumerical error arise from the use of approximation to represent exact mathematical operations and quantities
Truncation Error - approximation of mathematical proceduresRound-off error - approximation of exact numbers
The relation between exact, or true and the approximateTrue value = Approximation + errorEt = true value - approximation
Short comings - no magnitude how big the error isTo rectify the problem - Fractional Relative Error = Error/True valuePercent relative error is given by ε t =
true errortrue value
100%
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
ErrorsExample: The measured lengths of a bridge and a pen are 9999 cm and 9 cm respectively. If the true values are 10,000 cm and 10 cm respectively, compute, (a) true error and (b) true percent relative error for each case.Solution:a) The error for measuring the bridge is:
Et = 10,000 – 9999 = 1 cmand for the pen is:Et = 10 – 9 = 1 cm
b) The percent Relative Error for the bridge is:εt = 1/10,000 * 100 = 0.01 % and for the pen is:εt = 1/10 * 100% = 10 %
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Approximation of errorIn actual situation the true value is rarely availableTypically the case when we investigate the theoretical behavior of technique of simple system
The challenge of numerical method is to find the approximation errorSolving this by using iterative approach by using the previous value as basis
The iteration is stop until |εa| < εs (the prespecified tolerance)Tolerance to at least n significant figures is given as εs = (0.5 X 10 2-n)%
εa =approximate error
approximation100%
εa =present approximation − previous approximation
present approximation100%
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Error EstimatesExponential function can be estimated by
(Maclaurin Series)
Find the estimate of e 0.5 with εs to 3 significant figures. The true value for this function is 1.648721271.
Results from iteratione 0.5 = 1 + 0.5 = 1.5
ex = 1+ x +
x2
2!+
x 3
3!+L+
x n
n!
ε t =1.648721271 −1.5
1.648721271100% = 9.02 εa =
1.5 −11.5
100% = 33.3%
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Error using iterative method
Terms Result et ,% ea ,%
1 1 39.32 1.5 9.02 33.33 1.625 1.44 7.694 1.645833333 0.175 1.275 1.648437500 0.0172 0.1586 1.648697917 0.00142 0.0158
After six terms are included, the approximate error falls below ea = 0.05 %
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Evaluate e-8.3 using two approches :
and
and calculate the errors. Compare the answers with the true value of 2.485168 x 10-4 after 6 iteration.
...-4!
3!
-2
1432 xxxxe x ++−=−
...4!
3!2
1
11432
+++++==−
xxxxee x
x
Error using iterative method(assignment in class)
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Round-off ErrorsRelated to the manner in which numbers are stored in computerDue to the problems of word, string of binary digits and bits
In every day life we are using the base-10 system0,1, 2, 3, 4, 5, 6, 7, 8, 9
For larger quantity combination of this digit is used.86409 = (8 X 104) + (6 X 103) + (4 X 102) + (0 X 101) + (9 X 100)
The computers are using the system of binary unit or base-211 in computer (base -2) is equivalent to 3 in the base-10 system
Decimal -173 on a 16 bit computer using signed method
1 0 0 0 0 0 0 0 1 0 1 0 1 1 0 1
e,722,,7 π
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Range of IntegersProblem Statement
Determine the range of integers in base-10 that can be presented on a 16 - bit computer
Solution
Of the 16 bits, the first bit holds the sign. The remaining 15 bits can hold binary numbers from 0 to 11111111111111. Converted to a decimal integer which equals to 32767. Thus the ranging of 16-bit computer will be :
-32,767 to 32767
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Truncation ErrorOccur because an exact mathematical procedure is estimatedExample
first order equation in falling parachute problem using finite different method
It occur due to the difference equation only approximates the true value of the derivative
dvdt
≈∆v∆t
=v ti+1( )− v t( )[ ]
ti+1 − ti( )
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Taylor Series ExpansionUse to predict a function value at one point in terms of the function value and its derivatives at another point.
zero order estimationNormally use to estimate parameter or constant
first order estimationUse to predict linear equation
second order
To capture some of the curvature that the function might exhibit
f (xi+1) ≈ f (xi )
f (xi+1) ≈ f (xi ) + ′ f (xi )(xi+1 − xi )
f (xi+1) ≈ f (xi ) + ′ f (xi )(xi+1 − xi ) +
′ ′ f 2
(xi)( xi+1 − xi)2
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Higher Order Taylor Series Expansion
( )
1)1(
321
)!1()(
!)(
!3)(
!2)()()(
++
+
++
++′′′
+′′
+′+=
nn
nin
iiiii
hn
f
hn
xfhxfhxfhxfxfxf
ξ
L
Where h = x i+1 - xi
This function can evaluate any function with nth-order approximation
The last term in the equation mean the truncation error with the order of
)( 1+= nn h0R
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Remainder of Taylor Expansion
If we truncated the Taylor Series expansion after the zero term
The remainder or error of this zero order prediction will be
Ro ≈ ′ f (xi)h +
′ ′ f (xi )2!
h2 +′ ′ ′ f (xi )3!
h3 +L
f (xi+1) ≈ f (xi )
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Zero Order Prediction & Remainder
Zero order prediction
Exact prediction
f(x)
x i+1
Ro
f(xi )
xi x
h
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Taylor Series Approximation of PolynomialUse zero- through fourth-order Taylor series expansions to approximate the function:
f(x) = -0.1x4 – 0.15x3 – 0.5x2 – 0.25x + 1.2
from xi = 0 with h = 1.
That is, predict the function’s value at xi+1 = 1. (f(1) = 0.2 - true value)
Taylor series with n = 0 (zero-order);
f (xi+1) ≅ f (xi)
Thus, f(1) ≈ 1.2
Et = 0.2 – 1.2 = - 1.0
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Taylor series with n = 1 (first-order)
f (xi+1) ≅ f (xi) + f’(xi) (xi+1 – xi)
The first derivative must be determined & evaluated at x = 0;
f ’ (0) = -0.4(0.0)3 – 0.45(0.0)2 – 1.0(0.0) – 0.25
= -0.25
Therefore, the first-order approximation:
f (xi+1) ≈ 1.2 – 0.25h
f (1) ≈ 0.95
This results in a reduction of the Et to;
Et = 0.2 – 0.95 = -0.75
Taylor Series Approximation of Polynomial
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Taylor Series Approximation of PolynomialTaylor series with n = 2 (second-order)
f (xi+1) ≅ f (xi) + f’(xi ) ( xi+1 – xi ) + f’’ ( xi+1 – xi )2
The second derivative must be determined & evaluated at x = 0;
f’’ = -1.2(0.0)2 – 0.9(0.0) – 1.0 = -1.0
Therefore, the second-order approximation:
f (xi+1) ≈ 1.2 – 0.25h – 0.5 h2
f (1) ≈ 0.45
The Et is reduced further to = 0.2 – 0.45 = -0.25
2!)i(x
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Taylor Series Approximation of PolynomialAdditional terms would improve the approximation even more. Inclusion of 3rd and 4th derivatives results in exactly the same equation we started with:
f (x) = 1.2 – 0.25h – 0.5h2 – 0.15h3 – 0.1h4
where the remainder term is;
R4 = f (5) (ξ )/5! h5 = 0
R4 = 0 because the 5th derivative of a 4th-order polynomial is zero.
Consequently, the Taylor series expansion to the fourth derivative yields an exact estimate at xi+1 = 1:
f (1) = 1.2 - 0.25(1) - 0.5(1)2 - 0.15(1)3 - 0.1(1)4 = 0.2
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Approximation of f(x)
0.5
1.0
f(x)
xi = 0 xi+1 = 1
zero order
first order
second order
True
0x
h
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Taylor Series Approximation of Polynomial(assignment in class)
Use zero- through third-order Taylor series expansions to predict f(2) for
f(x) = 25x3 – 6x2 + 7x - 88
from x = 1.
Compute the true percent relative error εt for each approximation.
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Chemical Engineering Numerical Method© copyright PSE- FKKKSA, UTM
Trade-off between Round-off and Truncation error
Round-off error
Truncation error
Total errorPoints of diminishingreturn
log
erro
r
log step size