CL-201Chapter 2 Introduction to Engg Calculations [Compatibility Mode]

8/10/2019 CL-201Chapter 2 Introduction to Engg Calculations [Compatibility Mode]

1/20

Lectures

Introduction to Engineering


2/20


3/20


4/20


5/20

Range, Sample mean, sample variance, Sample

standard deviation

Let N measured values of a property or

variables are: x1, x2, x3, ..xN, then Range:

N11

minmax XXR

j

jN

NN 121 .........

1

2

1

21

)(1

1

......1

N

jj

NX

XXN

Ns

Sam le standard deviation2

)(1

N

X XXs1 j


6/20


7/20

Calibration of Equipment

Concentr Electrical140

ation C conductivity

K100

120

. .

2.0 0.7

3.0 1.2

80

ncentration

Slope

4.0 1.8y = 0.5x - 0.25

R = 0.992

40

Co

Best fit

0

20

0 10 20 30 40 50 60 70 80

Electrical Conductivity


8/20

Calibration of Rotameter

Flowrate Rotameter 160

V (Lit/m) Reading (R)

10 20 120

140

.

50 84.6

70 118 60

80

100

V,

Lit/m

V = 1.641R + 3.15

20

40

0

0 20 40 60 80 100

RRotameter

To measure

qu owra e


9/20

finding the best fitting straight line through a

set of points

baXbaXfY ),.........,,(

iii ebaXY )( ei

nn

02)(

2

n

aXbYe

ii

i

i

11

nnn202

)( 2

naXbY

e

1 ia

i

ii

i

i

i

i yxxax

111

Solve and Find: a and b

1 ib


10/20

22

iiii

xxna

22

iiiii yxxxyb

ii


11/20

xx

xy

SSSS

SS

r

2

2 Sum squaresCorrelation

Coefficient,

n

i

ixx xxSS1

2)(

SS

saSE )(

n

i

iyy yySS1

2)(x 21

2

)()( yyxxSSi

n

ixy

xxSSn

SS

SS

S xx

xy

yy

n

xx1

in 1


12/20

Fit a least square line to the following data

and show that sum of error is zero

y 2 5 3 8 7

Solutionx y xy x = . x+ . y-y =e

1 2 2 1 2.4 -0.4

2 5 10 4 3.7 1.3

3 3 9 9 5 -2

4 8 32 16 6.3 1.7

5 7 35 25 7.6 -0.6

nn 5 b + 15 a = 25

x=15 y=25 xy=88 x^2=15 ei=0

Solving

i

i

i

i yxanb11

15 b + 55 a = 88

b =1.1, a =1.3

iii

ii

ii

yxxaxb11

2

1 Then

Y = 1.1+1.3 X


13/20

-

Non linear Linear equation : Make

a e on near o near orm as:

y = ax2+b Y = aX + b y as Y, x2 as X

2=a/x+b Y = aX + b 2 asY, 1/x as X

1/y=a(x+3)+b Y = aX + b 1/y as Y, (x+3) as X

siny = a (x2-4) Y = aX + b siny as Y, (x2-4) as X, b as 0

y=1+x(mx2+n)1/2 Y = aX + b (y-1)2/x2 as Y, x2 as X; m as a

and n as b

= b

sides i.e.

Ln(y )= bln(x) +Ln(a)

im lies Y = aX + b

, ,

b and b as a

Y = aX + b


14/20

uD

u

P2

y = ax + b

y = Ln (2uP

),x=ln(

uD),b=Ln(),a=

y x x2 xy (xi

xbar)^2

(yi

ybar)^2

(xixbar)(yi

ybar)

Predicted

y

ei^2

4.636

8.418

70.871

39.025

0.313

0.019

1.810

4.631

2.478E

05

4.656 8.519 72.577 39.669 0.210 0.014 1.547 4.656 4.056E08

4.689 8.662 75.033 40.613 0.100 0.007 1.132 4.693 1.653E05

4.816 9.170 84.080 44.159 0.037 0.002 0.600 4.822 3.299E05

4.893

9.441

89.141

46.196

0.215

0.014

1.706

4.891

4.269E

06

4.948 9.655 93.220 47.770 0.459 0.031 2.522 4.945 6.575E06

= = = = = = = = =

28.637 53.866 484.921 257.432 1.333 0.086 0.339 28.637 8.519E05

. . . .

a 0.254 SSyy 0.086 sigma^2x 0.222

alfa 0.083 SSxy 0.339 sigma^2y 0.014

^ ^ . . .


15/20

t 10 15 20 25 30 35 40

F 0.8222 0.2984 0.1453 0.0832 0.0527 0.0359 0.0257

Take logarithm of the model

log F = a log(t)+log(k)onsider log F as Y, log t as x, log k as b

ThenY = aX + b

n

i

i

n

i

i yxanb11

n

iii

n

ii

n

ii yxxaxb

11

2

1


16/20

Y=log(F) X=log(t) X^2 XY

0.08503 1 0.08503 0.08503

0.52525 1.176091 0.61775 0.61775

0.8376 1.30103 1.08974 1.08974

1.07988 1.39794 1.5096 1.5096

. . . .

1.4452 1.544068 2.23148 2.23148

1.59018 1.60206 2.54756 2.54756

=

6.84096 =

9.498311 =

9.96867 =

9.96867

n

i

i

n

i

i yxanb

11

7b+9.4983a = -6.8409 Solve

a = -2.5

n

i

ii

n

i

i

n

i

i yxxaxb11

2

1

9.4983b-9.8696a= -9.9686b = 2.415

So, k = antilog (b) = 260

So, The fitted equation will be F = 260t-2.5


17/20

. , . , . , . , . , . ,

2.45


18/20

Problem No. 2.36: Find the values of k and c

of equation PVk = c when fitted with exp dataas:

P(mmHg) 760 1140 1520 2280 3040 3800

V(cm3) 48.3 37.4 31.3 24.1 20 17.4

Also find goodness of fit


19/20

PVk = C Ln(P) + kln(V) = ln(C) y = ax+b y = ln(P), x = - ln(V)

a = k, b = ln(c)

P V x=ln(V) y=ln(P) x'^2 x'y' (x'ix'bar) 2 (y'iy'bar)^2

(x'x'bar)(y'

y'bar) Predictedy' Ei^2

760.000 48.300 3.877 6.633 15.034 25.720 0.300 0.749 0.474 6.637 1.306E05

1140.000 37.400 3.622 7.039 13.116 25.492 0.085 0.212 0.134 7.039 2.196E07

1520.000 31.300 3.444 7.326 11.859 25.230 0.013 0.030 0.020 7.319 5.085E05

2280.000 24.100

3.182 7.732 10.126

24.605 0.022 0.054 0.034 7.731 1.946E06

3040.000 20.000 2.996 8.020 8.974 24.025 0.111 0.271 0.174 8.024 1.816E05

3800.000 17.400 2.856 8.243 8.159 23.545 0.224 0.553 0.352 8.243 3.313E08

Sum 12540.000 178.500 19.977 44.993 67.270 148.616 0.755 1.869 1.188 44.993 8.426E05

x'bar 3.330 S 0.005 a 1.573

y'bar 7.499 SE(a) 0.005 b 12.736

SSxx 0.755 SE(b) 0.018

SSyy 1.869 r^2 0.9999549

SSxy 1.188

So, a = k = 1.573, c = exp(b) = 3.40 105


20/20

Problem 2.45:

Fit D = D0exp(-E/RT) where R = 1.987

T 347 374.2 396.2 420.7 447.7 471.2

T D x=1/RT y=ln(D) x'^2 x'y' (x'ix'bar) 2 (y'iy'bar)^2

(x'x'bar)(y'

y'bar) Predictedy' Ei^2

. . . . . .

347.000 1.340E06 1.450E03 13.523 2.104E06 0.020 4.327E08 2.127 3.034E04 13.580 3.246E03

374.200 2.500E06 1.345E03 12.899 1.809E06 0.017 1.053E08 0.697 8.563E05 12.812 7.628E03

396.200 4.550E06 1.270E03 12.300 1.614E06 0.016 7.793E10 0.056 6.583E06 12.268 1.056E03

. . . . . . . . . . .

447.700 1.407E05 1.124E03 11.171 1.264E06 0.013 1.397E08 0.798 1.056E04 11.204 1.027E03

471.200 1.999E05 1.068E03 10.820 1.141E06 0.012 3.037E08 1.548 2.168E04 10.795 6.321E04

Sum 2457.000 5.097E

05

7.454E

03

72.387 9.361E

06 0.091 1.010E

07 5.378 7.360E

04

72.387 1.672E

02

x'bar 0.001 S 0.065 a 7284.269 E=a 7284.269

y'bar 12.065 SE(a) 203.391 b 3.015 D0=exp(b) 0.05

SSxx 1.010E07 SE(b) 0.254

SSyy 5.378 r^2 0.9968912

SSxy 0.001

CL-201Chapter 2 Introduction to Engg Calculations [Compatibility Mode]

Documents

Transcript of CL-201Chapter 2 Introduction to Engg Calculations [Compatibility Mode]