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8/17/2019 Lect- 8-9 Uncertainity Analysis
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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.
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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.
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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.
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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 .
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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=
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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
∂ ∂ ∂∆ = ∆ + ∆ + + ∆
∂ ∂ ∂
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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
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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
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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
−∂= = −
∂
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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
−∂= = −
∂
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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.
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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
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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
∂ ∂ ∂ ∂ ∆ = ∆ + ∆ + ∆ ∆ + ∆ ∂ ∂ ∂∆ ∂
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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
∆ ×= = × =
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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
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Gaussian function narrows with decreasing σσσσ
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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
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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
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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
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
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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
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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.
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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
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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
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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
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Choosing X –Y Coordinates
Linear - Linear
Liner – Logarithmic ( Semi log )
Logarithmic – Logarithmic (Full log)
Polar coordinates ( Quantities varies with an angle)
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Selection of axis
Semi log Scale
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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
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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
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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=
= − +∑
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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
==
=
= =
=
∑ ∑
∑ ∑ ∑
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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
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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.
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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
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( )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
= =
=
= = = =
=− = =
− − = − −
∑ ∑ ∑
∑ ∑ ∑ ∑
∑∑ ∑ ∑
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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
= −
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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
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y = -0.0457x2 + 2.2782x - 9.35
R2 = 1
6
8
10
12
14
16
18
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 = 41.718e-0.0362x
R2 = 0.7532
6
8
10
12
14
16
18
20
22
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
Different
Patterns of fit
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Different pattern
of fits
Reference:
J.P.Holman
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Different pattern
of fits
Reference:J.P.Holman
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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.
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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
∂ ∂ ∂∆ = ∆ + ∆ + + ∆
∂ ∂ ∂
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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
∆∆ = ∑
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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
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Practical use of uncertainty analysis
1. For revealing the confidence level in the results.
2. For selection of measurement method
3. For instrument selection
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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