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Introduction to Chemical Engineering Calculations
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños
Lecture 2.
Regression and Nonlinear Axes
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
2
Regression and Nonlinear Axes2
What is regression analysis?
A technique for modeling and analyzing the relationship between 2 or more variables. Usually, 1 variable is designated as the independent variable.
Purpose: Establish a mathematical relation between the variables.
t
a0
n=1
C(t) = e
2 nπx n x n yu(x,y) = f(x)sin dx sin sinha sinh (nπb/a) a a a
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
3
Regression and Nonlinear Axes2
Linear Regression
An equation of a line is used to establish the relationship between 2 variables x and y.
y = mx + b
where
y,x = dependent and independent variables, respectively.
m, b = slope and intercept, respectively.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
4
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
Consider a rotameter calibration data:
Rotameter Reading(independent variable)
Flow Rate (mL/min)(dependent variable)
10 20.0
30 52.1
50 84.6
70 118.3
90 151.0
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
5
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
6
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
Calculating the slope and intercept using the method of least squares:
xy x y2
xx x
xx y xy x2
xx x
S - S SΔy m = = Δx S - S
S S - S S b =
S - S
slope :
intercept :
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
7
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
Calculating the S-values:
1
1
2 2
1
1
1 (average of x values) =
1 (average of y values) =
1 (average of x values) =
1 (average of xy values) =
nx i
in
y iin
xx iin
xy i ii
S xn
S yn
S xn
S x yn
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
8
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
For the previous data set:
2 2 2 2 2
1 = 10 30 50 70 90 = 505
1 = 20.0 52.1 84.6 118.3 151.0 85.25
1 = 10 30 50 70 90 = 33005
10 20 30 52.1 50 84.61 = 5572.85 70 118.3 90 151.0
x
y
xx
xy
S
S
S
S
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE
9
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
Calculating the slope and intercept:
2
2
5572.8 - 50 85.2 = 1.641
3300 - 50
3300 85.2 - 5572.8 503.15
3300 - 50
m
b
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE10
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
Therefore, the mathematical relation for flow rate in terms of the rotameter reading is:
y = 1.641x + 3.15
At any given value of x, a corresponding value for y can easily be calculated using this equation.
For x = 25, y = 1.641(25) + 3.15 = 44.175
for x = 250, y = 1.641(250) + 3.15 = 413.40
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE11
Regression and Nonlinear Axes2
Two-Point Linear Interpolation
Given the following set of data:
x 1.0 2.0 3.0 4.0
y 0.3 0.7 1.2 1.8
Supposed we want to know two values of y when:
1. x = 2.5 (x is between the range of data points: interpolation)
2. x = 5.0 (x is outside the range of data points: extrapolation)
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE12
Regression and Nonlinear Axes2
Two-Point Linear Interpolation
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE13
Regression and Nonlinear Axes2
Two-Point Linear Interpolation
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE14
Regression and Nonlinear Axes2
Two-Point Linear Interpolation
If points 1, 2, and A lie on the same line, then
Slope (1-A) = Slope (1-2)
If slope is defined as y/x, then,
1 2 1
1 2 1
y - y y - y=x - x x - x
Rearranging the equation,
11 2 1
2 1
x - xy = y + y - yx - x
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE15
Regression and Nonlinear Axes2
Two-Point Linear Interpolation
Two-point interpolation can be used when extracting values from a tabular data.
Consider the following tabular data:
x 1 2 3y 1 4 8
Find the value of y when x = 1.3.
1.3 - 1y = 1 + 4 - 1 = 1.92- 1
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE16
Regression and Nonlinear Axes2
Fitting Nonlinear Data
For simple non-linear equations, linearization can be done by manipulating or plotting the data so that it can be expressed analogously as a linear equation.
Original equation: y = mx2 + bLinearized Form: y = mw + b (with w = x2)
Original equation: sin y = m(x2 – 4) + bLinearized Form: u = aw + b
(with w = x2 – 4 and u = sin y)
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE17
Regression and Nonlinear Axes2
Fitting Nonlinear Data
A mass flow rate M(g/s) is measured as a function of temperature T(0C).
T(0C) 10.0 20.0 40.0 80.0
M 14.76 20.14 27.73 38.47
There is reason to believe that m varies linearly with the square root of T:
M = aT1/2 + b
Determine the values and units of a and b.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE18
Regression and Nonlinear Axes2
Fitting Nonlinear Data
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE19
Regression and Nonlinear Axes2
Fitting Nonlinear Data
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE20
Regression and Nonlinear Axes2
Fitting Nonlinear Data
Linearizing the data set:
T 10.0 20.0 40.0 80.0
T1/2 3.162 4.472 6.325 8.944
M 14.76 20.14 27.73 38.47
The values of a (slope) and b (intercept) can be determined using the method of least squares as discussed previously.
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE21
Regression and Nonlinear Axes2
Fitting Nonlinear Data
Calculate for SX, SY, SXX, and SXY:
X (T1/2) Y (M) X2 XY
3.162 14.76 10.00 46.68
4.472 20.14 20.00 90.07
6.325 27.73 40.00 175.38
8.944 38.47 80.00 344.09
SX = 5.726 SY = 25.275 SXX = 37.50 SXY = 164.05
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE22
Regression and Nonlinear Axes2
Fitting a Straight Line to a Set of Data Points
Calculate the slope (a) and intercept (b):
2
2
(164.05) - (5.726)(25.275) a = 4.100(37.50) - 5.726
(37.50)(25.275) - (164.05)(5.726) b = 1.798(37.50) - 5.726
slope :
intercept :
Apply dimensional consistency to get the units of a and b:
a = 4.100 g/(s-0C1/2) and b = 1.798 g/s
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE23
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE24
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE25
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE26
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE27
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE28
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE29
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE30
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
1. If y versus x data appear linear on a semilog plot, then ln(y) versus x would be linear on a rectangular plot, and the data can therefore be correlated by an exponential function:
y = aebx
2. If y versus x data appear linear on a log plot, then ln(y) versus ln(x) would be linear on a rectangular plot, and the data can therefore be correlated by a power function:
y = axb
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE31
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
A plot of F versus t yields a straight line that passes through the points (t1 = 15, F1 = 0.298) and (t2 = 30, F2 = 0.0527) on a log plot. Determine the mathematical equation that relates F and t.
If linear on a logarithmic plot, then,
F = atb
Linearizing the equation,
ln (F) = ln (a) + b ln(t)
Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE32
Regression and Nonlinear Axes2
Nonlinear Axes: The Logarithmic Coordinates
Determine a and b:
2 1
2 1
Δ ln(F) ln(F /F ) ln(0.0527/.298)b = = = = -2.50Δ ln(t) ln(t /t ) ln(30/15)
To determine a:
ln (a) = ln (F1) – b ln(t1) or ln (a) = ln (F2) – b ln(t2)
Using the first data point,
ln (a) = ln (0.298) – b ln(15) = 5.559a = exp (5.559) = 260
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