Lect- 8-9 Uncertainity Analysis

8/17/2019 Lect- 8-9 Uncertainity Analysis

1/47

Error

The difference between the measured value and truevalue is called “Error”

Hence, it is important to know how close the measured

value is to the true value.

True value Measured value Errors= ±

Systematic Errors:

These errors have some

identifiable source and may becorrected by calibration. Likezero drift, sensitivity drift, etc.

Random Errors:

These errors can not be sourced out and

hence can not be corrected.Due to temperature change, humidity,wind, vibrations, electromagnetic field,etc.

Since, this error arises from a multiplesources, it is impossible to quantify.


2/47

Uncertainty Analysis

Since the total error includes random error, which

are uncertain, errors are usually referred to as

uncertainties.

For any experimental study, uncertainties analysismust be performed and reported along with the

measurements.


3/47

Estimation of Random Errors

Since random errors are not deterministic, a

probabilistic approach is used.

Probabilistic approach involve a probabilitydistribution

Gaussian or Normal distribution is commonly used

The function is constructed by taking several samples

of a certain measurement and constructing a

frequency distribution function

Counting the number of occurrences of a certain

value in an interval.


4/47

Propagation of Uncertainties

Consider the measurement of an effective resistance of

a circuit comprising three resistors in series

So, R1, R2 and R3, have to be measured.

There would be uncertainties in the measurement of each of

these resistances, which will affect the Reff .


5/47

Kline & McClintock’ Method, 1953.

Consider a variable N (dependent) which is calculated from

various measurements such as u1, u2, u3,u4…….un and

governed by the function

The total uncertainty (in N) ∆n would include uncertaintiesof ui ; ie ∆u1, ∆u2,∆u3,∆u4, ∆un

Therefore;

1 1 2 2( , ,......, )

n n N N f u u u u u u+ ∆ = + ∆ + ∆ + ∆

1 2 3( , , ,....., )n N f u u u u=


6/47

Expand f(u1, u2, u3…un) by Taylor’s series;

Ignoring second and higher derivatives;

1 1 2 2

1 2 1 2

1 2

2 31 1

( , ,....., )

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

( ) ( )

n n

n

n

n

f u u u u u u

f f f u u u u u

u u

f u O u O uu

+ ∆ + ∆ + ∆ =

∂ ∂+ ∆ + ∆ +

∂ ∂

∂+ ∆ + ∆ + ∆∂

1 2

1 2

.... nn

f f f N u u u

u u u

∂ ∂ ∂∆ = ∆ + ∆ + + ∆

∂ ∂ ∂


7/47

Overall error is given by

Sometimes the estimated error may be wrong . Some time

systematic errors may have equal magnitude in OPPOSITE

sign; they may tend to cancel out each other. So, a

reasonable estimate would be the Residual Sum of Squareserror or root-sum-square (RSS) given by;

22 2

1 2

1 2

... nn

f f f N u u u

u u u

∂ ∂ ∂∆ = ∆ + ∆ + + ∆

∂ ∂ ∂

RMS: Measure of the magnitude of a varying quantity


8/47

In statistics, the residual sum of squares (RSS) is the sum of

squares of residuals. It is also known as the sum of squared

residuals (SSR) or the sum of squared errors of prediction (SSE). It

is a measure of the discrepancy between the data and an estimationmodel. A small RSS indicates a tight fit of the model to the data.

In general, total sum of squares = explained sum of squares +

residual sum of squares


9/47

Example problem

Calculate the uncertainty in head loss hl expressed as2

2

l

flV h

gd

=

Given uncertainties in l,v and d are 2%, 4% and 1%. Ignore

the uncertainties in f and g

2 2 2

l ll

h h f h l v d

l v d

∂ ∂ ∂ ∆ = ∆ + ∆ + ∆ ∂ ∂ ∂

Solution

2

;2

l lh h fv

wherel gd l

∂= =

∂

2

2 2

l lh h fvl

v gd v

∂= =

∂

2

22

l lh h fv l

d gd d

−∂= = −

∂


10/47

1/ 22 2 2

1

4l l

l v d h h

l v d

∆ ∆ ∆ ∆ = + +

[ ] [ ] [ ]

1/ 22 2 21

0.02 0.04 0.014l lh h

∆ = + +

[ ] [ ] [ ]

1/ 2

2 2 210.02 0.04 0.01 0.034

l

l

hh

∆

= + + =

Total Error = 3 %

2

;

2

l lh h fv

l gd l

∂= =

∂

2

2 2

l lh h fvl

v gd v

∂= =

∂

2

22

l lh h fv l

d gd d

−∂= = −

∂


11/47

Points to be noted:

In many mechanical problems, we assume the value

of g is 9.8.

However, in reality; there is an uncertainty in the

value of g, which has to be accounted where the

situation arises.

So, expect for universal constants, all other

parameters used for estimating a quantity to be

considered while calculating the total uncertainties.


12/47

Find uncertainties is the following calculated quantities

Flow rate in fully developed laminar region

4

128

pd Q

l

π ∆=

Grashof Number; Gr

2 3

2

( )s

r

g T T lG

ρ β

µ

∞−

=

Given uncertainties in ∆∆∆∆p, d, µµµµ, l, ρρρρ, ββββ and T are 4%, 2%,3%, 2%, 4%, 5% and ±2 °°°°C


13/47

Airflow rate of 17m3/h through a pipe of 60 mm ID at 20oC is measuredusing a square edged orifice (ββββ = 0.4). A pressure drop observed is

157.85 N/m2 with ±±±± 0.4%. If the area of orifice is maintained within 0.2 %,estimate the design stage uncertainty in the flow rate. Assumeaccuracies of Cd and ρρρρ are ±±±± 0.5%. Estimate the total error in themeasurement for Cd = 0.63 and P = 0.97 bar abs and R= 287 J/kg K.

For an orifice, Q = CdA(2∆p/ ρ)1/2Solution

1/ 2

2 2 2 2( ) ( ) ( ) ( )

2 2

d C Q p A

d

uu uuu

Q C A p

ρ

ρ

∆

= ± + + +

∆

; ; ;2 2d d

Q Q Q Q Q Q Q Q

C C A A p p ρ ρ

∂ ∂ ∂ ∂= = = = −

∂ ∂ ∂ ∆ ∆ ∂

2 22 2

( )d d

Q Q Q Q

Q C A PC A P ρ ρ

∂ ∂ ∂ ∂ ∆ = ∆ + ∆ + ∆ ∆ + ∆ ∂ ∂ ∂∆ ∂


14/47

0.005 ; 0.002

0.004 ; 0.005

d

d

C A

C A

p

p

ρ

ρ

∆ ∆= =

∆∆ ∆= =

∆

1/ 2

2 2 2 2( ) ( ) ( ) ( )2 2

d C Q p A

d

uu uuu

Q C A p

ρ

ρ

∆

= ± + + + ∆

Total uncertainty

1/ 22 2 2 2(0.005) (0.002) (0.5 0.005) (0.5 0.004)

Qu

X X Q

= ± + + +

0.63%Q

u

Q

= ±

3

95000

287 2931.13 /

P

RT

X

kg m

ρ

ρ

=

⇒=

=

Total Error

2 32 2 157.850.63 (0.06) 0.03 / 4 1.13d

PQ C A m hr

π

ρ

∆ ×= = × =


15/47

Properties of the Normal or

Gaussian Function The area under the Gaussian Curve is 1, hence the

maximum probability is 1. i.e., the probability of any real

value to lie between -∞∞∞∞ to + ∞∞∞∞ is 1.

The area lies under [-σσσσ, +σσσσ] is 68%: the change that ameasurement will be out of the given limits is 1 in 3 times

The area lies under [-2σσσσ, +2σσσσ] is 95.4% ; in every 20measurements, one may fall out of the given limits

The area lies under [-3σσσσ, +3σσσσ] is 99.7% ; onemeasurement in every 300 measurements may fall outside

of the limits


16/47

Gaussian function narrows with decreasing σσσσ


17/47

Procedure for expressing experimental uncertainty

Step 1 : Take sufficient large number of samples x

Step 2 : Obtain mean xm, and standard deviation σσσσStep 3 : Express measurement as x = xm ±±±± n σσσσ


18/47

Graphical Presentation of Data

1. Graph usually serves to communicate knowledge from the

author to the readers .

2. Graphs plays a vital role in testing the theoretical calculations

against real experiments results.

3. There are some situation, where it is very difficult to conduct

experiments through out the range of parameters. Hence,

graphs are highly useful to interpolate / extrapolate the

output in the missed regions ( where the exact details of out

put is not known).


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General rules for plotting graphs

1. The graphs should be plotted in such a way that the reader shouldunderstand the conveyed information with out any difficulties.

2. The axes should have clear labels i.e. name of quantities, units, andtheir symbol if necessary

3. Axes should be clearly numbered and should have clear tick markswith significant numerical divisions (Sub division also should beclearly mentioned)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 2 4 6 8 10 12

Number of cycles

Absorption temperature = 27°C

Absorption pressure = 80 bar

MmNi4.6Al0.4

ma=0.4 kg

H y d r o g e n s t o r a g e c a p a c i t y ( w t %

)

Fig.1 Activation characteristics of MmNi4.6Al0.4


20/47

4. Use scientific notation to avoid placing of too many digits on thegraph.

5. When plotting on log coordinates, use real logarithmic axes. Logscales should have tick marks at powers of 10 .

6. Choice of the scales in axes should be based on the relativeimportance of the variations shown in the results and the also based

on the availability of data.

Fig.2 Effect of delivery pressure on compressor volumetric efficiency

y = -0.1827x2 + 11.864x - 173.56

R2 = 1

0

5

10

15

20

25

0 5 10 15 20 25 30 35 40 45

Delivery pressure ( bar)

C o m p r e s s o r w o r k ( k

J )

Compressor work

MmNi4.6Al0.4

Tc = 20°C, Ps = 10 bar

ma = 0.4 kg, Th = 85°C

y = -0.1827x2 + 11.864x - 173.56

R2 = 1

8

10

12

14

16

18

20

28 30 32 34 36 38 40 42


C o m p r e s s o r w o r k ( k

J )

Compressor work

MmNi4.6Al0.4

Tc = 20°C, Ps = 10 bar

ma = 0.4 kg, Th = 85°C


21/47

Fig. Stages of compression at supply pressure 15 bar

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 50 100 150 200 250 300 350 400 450 500

Time (s)

H y d r o g e n c o m p r e s s e d ( g )

0.001

0.01

0.1

1

10

100

75°C

95°C

85°C

75°C

95°C

85°C

Hydrogen compressed

Hyderogen compression rate

MmNi4.6Al0.4

Tc = 20°C, Ps=5 bar

ma = 0.4kg

H y d r

o g e n c o m p r e s s i o n r a t e L

o g ( g / m i n . )

Fig. Stages of compression at supply pressure 15 bar

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 50 100 150 200 250 300 350 400 450 500

Time (s)

H y d r o g

e n c o m p r e s s e d ( g )

0

1

2

3

4

5

6

7

8

75°C

95°C

85°C

75°C

95°C

85°C

Hydrogen compressed

Hyderogen compression rate

MmNi4.6Al0.4Tc = 20°C, Ps=5 bar

ma = 0.4kg

H y d r o g e n c o

m p r e s s i o n r a t e ( g / m i n . )

Use of Logarithmic scale

same data without Logarithmic

scale


22/47

7. Use specific symbols for the experimental data points.

Analytical results are expressed by using straight lines.8. Indicate the maximum uncertainty of the estimated quantity.

9. When the two different dependent variables are compared

with a common parameter, a secondary Y axis should be used.10.When several curves are plotted on a single graph, use

different pattern of line such as solid line, dotted line, dashedline etc. In case of experimental results use different legends

for differentiating the curves.

11.Other independent parameters should be specified in thegraphs. Lettering on the graphs should be clearly visible.

12. Title of the graph should be placed at the bottom and the titlesshould be properly numbered. Choice of the scales in axesshould be based on the relative importance of the variationsshown in the results.


23/47

Fig.20. Effects of hot fluid temperature and supply

pressure on compressor efficiency

0

1

2

3

4

5

6

7

8

9

70 75 80 85 90 95 100Hot fluid / Heat source temperatures (°C)

C o m p r e s s o r e

f f i c i e n c y ( % )

P = 5 Bar

P = 10 BarP = 15 Bar

MmNi4.6Al0.4

Tc = 20°Cma = 0.4 kg

Experimental valuesNumerical simulation

s

s

s

Representation of experimental and numerical simulation


24/47

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 1.2

Concentration (H/M)

=

=

=

=

E q u i l i b

r i u m p r e s s u r e ( b a r )

10°C

15°C

20°C

25°C

MmNi4.6Fe0.4

MmNi4.6Al0.4

Fig.4. PCT characteristics of the two mischmetal based alloys


25/47

0

5

10

15

20

25

30

35

40

45

0 200 400 600 800 1000 1200 1400 1600

Time (s)

H y d r o g e n s t o r a g e p r e s s u

r e ( b a r )

0

2

4

6

8

10

12

14

16

C o m p r e s s i o n r a t e ( g / m i n )

MmNi4.6Al0.4

U = 1000 W/m2K

Tc = 20°C

Ps = 10 bar

Th = 85°C

Vs = 3.8 litre

Fig. Cyclic performance of hydrogen compressor (Storage volume 3.8 l)

Hydrogen storage pressure

Compression rate

Use of secondary Y axis


26/47

Choosing X –Y Coordinates

Linear - Linear

Liner – Logarithmic ( Semi log )

Logarithmic – Logarithmic (Full log)

Polar coordinates ( Quantities varies with an angle)


27/47


28/47

Selection of axis

Semi log Scale


29/47

2 1 residual

total

SS R

SS

= −

( )2

1

n

residual i iiSS E P

== −∑

( ) 2

1totalSS n σ = − ×

( )2

12

n

ii E E

n

σ =

−

= ∑

σ = standard deviation

E i= Experimental value

Pi= Predicted value

n = total numberof data

Correlation coefficient = R and

Coefficient of determination = R2


30/47

METHOD OF LEAST SQUARESMethod of obtaining the correlation between the quantity to be

measured (y) and the variable by graphical analysis.

y ax b= + ( ) 2

1

n

i iiS y ax b

== − + ∑

For the best accuracy of fit, S should tend to zero

Goodness of the fit is revealed by correlation coefficient, r.

2

,

21

y x

y

rσ σσ σ

σ σσ σ

= −

[ ]1/2

2

1, ;

2

n

i ici y x

y y

nσ σσ σ = − =

− ∑ [ ]

1/22

1

1

n

i mi y

y y

nσ σσ σ = − =

− ∑

yi are actual values

yic are computed from correlation equation


31/47

Vertical least squares fitting proceeds by finding the sum of the squares of thevertical deviations of a set of data points . The error E is defined as

( , ) f a b a bx= +

2 2

1

( , ) [ ( )]n

i

i

E a b y a bx=

= − +∑


32/47

2

1

( )2 [ ( )] 0

n

i i

i

E y a bx

a =

∂= − − + =

∂ ∑

2

1

( )2 [ ( )] 0

n

i i i

i

E y a bx x

b =

∂= − − + =

∂ ∑

1 1

n n

i i

i i

na b x y= =

+ =∑ ∑

2

1 1 1

n n n

i i i i

i i i

a x b x x y− = =

+ =∑ ∑ ∑

11

2

1

1 1

nn

iiii

n nn

i i ii

i i

n x ya

b x x x y

==

=

= =

=

∑ ∑

∑ ∑ ∑


33/47

1

11

2

1

1 1

n

niiii

n nn

i i ii

i i

n x ya

b x x x y

−

==

== =

=

∑ ∑∑ ∑ ∑

( )

2

1 1 1 1

22

1 1 11 1

1n n n n

i i i i ii i i i

n n nn n

i i i ii i ii ii i

y x x x ya

b n x y x yn x x

= = = =

= = == =

− = −−

∑ ∑ ∑ ∑∑ ∑ ∑∑ ∑

The 2x2 matrix is of the form


34/47

Observation

1. For a perfect fit σσσσy,x = 0; here r = 1. There is no variationbetween the estimated values and the values obtained from

the correlation.

2. r = 0 indicates a poor fit or the values are widely scattered

around a straight line. Hence, the value of r should be closer

to 1 for a good fit.

3. It is necessary to observe the behavior of the fit and also

their range of scatter. If the data scatter but still appear to

follow a liner relationship, then a fit is also said to be poor

one.


35/47

1 exp n

Q A

RT ε σ ε σ ε σ ε σ

= −

1ln ln ln Q A n

RT ε σ ε σ ε σ ε σ = + −

1ln ln lniQ

E A n RT ε σ ε σ ε σ ε σ = − + +

The minimization of the error is carried out by finding the

partial derivatives of w.r.t ln(A1 ), n and Q/R andsetting them equal to zero, where N is the number of

observations. Taking the partial derivatives, the following

equations are obtained

2

1

N

ii =∑

---------(1)

Example


36/47

( )11 1 1 1ln ln ln ln n n n n

i i i i

Q A n

RT σ ε σ ε σ ε σ ε

= = = =+ − =∑ ∑ ∑ ∑

( ) ( ) ( )( )2

11 1 1 1

lnln ln ln ln ln ln

n n n n

i i i i

Q A n

RT

σ σσ σ σ σ σ ε σ σ σ ε σ σ σ ε σ σ σ ε = = = =

+ − = ∑ ∑ ∑ ∑

( )1

21 1 1 1

lnln ln n n n ni i i i

A n Q

T T RT T

ε εε ε σ σσ σ = = = =

− − + = −∑ ∑ ∑ ∑

In the matrix form, this can be written in the form

( )1 1

1

12

1 1 1 1

121 1 1

1ln

lnln

lnln (ln ) ln ln

ln1 ln 1

N N

N i i

i

N N N N

i i i i

N N N N

ii i i

N T A

nT

Q

R T T T T

σ σσ σ ε εε ε

σ σσ σ σ σ σ ε σ σ σ ε σ σ σ ε σ σ σ ε

ε εε ε σ σσ σ

= =

=

= = = =

=− = =

− − = − −

∑ ∑ ∑

∑ ∑ ∑ ∑

∑∑ ∑ ∑


37/47

the matrix form,

( )1 1

11

2

1 1 1 1

121 1 1

1

ln lnln

lnln (ln ) ln ln

ln1 ln 1

N N

N i i mi

N N N N

mi i i i

N m N N N

ii i i

N T A

nT

Q

R T T T T

σ σσ σ ε εε ε

σ σσ σ σ σ σ ε σ σ σ ε σ σ σ ε σ σ σ ε

ε εε ε σ σσ σ

= =

=

= = = =

=− = =

− − = − −

∑ ∑ ∑∑ ∑ ∑ ∑

∑∑ ∑ ∑

This is of the form [A][x] = [b]

where the elements of [A] and [b] are obtained from the experimental

data. The three unknown parameters , lnA1 , n and Q1 /R of equationare determined by solving

[ ] [ ] [ ]1

x A b−

=

1 exp n Q A

RT ε σ ε σ ε σ ε σ

= −


38/47

y = 12.804x - 51.611

R2 = 1

0

50

100

150

200

250

0 5 10 15 20 25Milliampere (mA)

Applied pressure vs milliampere

Linear (Applied pressure vs milliampere)

Trendline

A

p p l i e d P r e s s u

r e ( b a r )

Range 0-200 bar

Fig. Calibration of pressure transducer

20


39/47

y = -0.0457x2 + 2.2782x - 9.35

R2 = 1

6

8

10

12

14

16

18

15 20 25 30 35 40 45


C o m p r e s s o r w o r k ( k J )

Compressor work

MmNi4.6Al0.4

Tc = 20°C, Ps = 10 bar

ma = 0.4 kg, Th = 85°C

y = 41.718e-0.0362x

R2 = 0.7532

6

8

10

12

14

16

18

20

22

15 20 25 30 35 40 45


C o m p r e s s

o r w o r k ( k J )

Compressor work

MmNi4.6Al0.4

Tc = 20°C, Ps = 10 bar

ma = 0.4 kg, Th = 85°C

Different

Patterns of fit


40/47

Different pattern

of fits

Reference:

J.P.Holman


41/47

Different pattern

of fits

Reference:J.P.Holman


42/47

General Considerations in Data Analysis

Elimination of inconsistent data: Analysis the data

carefully. Try to eliminate the inconsistent data or the error

data. Suppose the measurement itself consists of many

inconsistent data, the entire experiments may be repeated.

Estimate the Uncertainty : Detailed uncertainty analysis

should be performed before doing the data analysis.

Anticipate the results from the theory : Before trying toobtain the correlations of the experimental data, investigators

should carefully review the theoretical background of the

estimated quantity. This may be useful for determining the

graphical formats, etc. e.g. First order instrument

Correlate the data: Critical review of the results.


43/47

Kline & McClintock’ Method, 1953.

Consider a variable N,

The total uncertainty (in N), ∆N was defined by root sum squares

1 2 3( , , ,....., )n N f u u u u=

22 2

1 2

1 2

...n

n

f f f N u u u

u u u

∂ ∂ ∂∆ = ∆ + ∆ + + ∆

∂ ∂ ∂


44/47

a. Uncertainties for product functions

In cases where the result function takes the form of the product

of respective primary variables raised to exponent expressed as

1 2 3

1 2 3

1 2 3 1

1 2 3

( .......... )

( )..........−

=

∂=

∂

a a a an

n

a a a ai an

i i n

i

N f u u u u

N u u u a u u

u

The partial differentiation results

Dividing by N,1 i

i i

N

N u u

∂=

∂Inserting in the equation for ∆ N

1/22

i i

i

a u N

N x

∆∆ = ∑


45/47

b. Uncertainties for addition function

When the result function has an additive form, N will

be expressed as

1 1 2 2 . . . . . . . . . .= + + + =

∂=

∂

∑n n i i

i

i

N a u a u a u a u

N a

u

( ) ( )

1/22

1/22 2.

∂ ∆ = ∆ = ∆ ∂

∑ ∑i i ii

N N u a uu


46/47

Practical use of uncertainty analysis

1. For revealing the confidence level in the results.

2. For selection of measurement method

3. For instrument selection


47/47

1. The resistance of a copper wire is given as R = R0[1+α(T-20)].Where Ro = 6Ω ± 0.3% is the resistance at 20°C.

α = 0.004°C ± 1%, T = 30 ± 1°C.

Calculate the uncurtaining in the resistance of the wire.

2. A resistor has an nominal stated value of 10 Ω ± 1%. A voltage isimpressed on the resistor and the power dissipation is to be

calculated in two different ways. (i) Only voltage is measured (ii)

both Voltage and current will be measured. Calculate the uncertainty

in power for each case when the measured values of E and I are

E = 100V ± 1% (for both case and I= 10A ±1%.

Sample Problems

Documents

Lect- 8-9 Uncertainity Analysis