examples uncertainty measurement

Minia University

From the SelectedWorks of Dr. Adel A. Elbaset

Summer August 7, 2021

Examples Uncertainty measurementAdel A. Elbaset, Minia University

Available at: https://works.bepress.com/dr_adel72/105/

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إستشاري الاعمال الكهربية للمنشأت–إستشاري نظم الطاقة الجديدة والمتجدده –إستشاري نظم إدارة الجوده

Mobile : 01090343201 Email: [email protected]

Evaluation of measurement uncertainty

عادل عبدالباسط محمد . د. إعداد أ

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Example

UNCERTAINTY CALCULATION





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Prof. Dr. Adel A. ElbasetFaculty of Engineering, Minia University,

Mobile : 01090343201

[email protected]

[email protected]

مهندس استشارى

عادل عبدالباسط محمد/ الأستاذ الدكتور

-جددهإستشاري نظم الطاقة الجديدة والمت-إستشاري الاعمال الكهربية في المنشأت

إستشاري نظم إدارة الجوده

519229060سجل هندسي 1354/4استشارى رقم 7109/32قيد هـندسى





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Example No. 1






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Example

UNCERTAINTY CALCULATION FOR VOLTAGE

Product : MCB, 32 A, 240/ 415V, Single pole

Test : Short circuit test of MCB

Equipment used : Digitizer with Amplifier

Range used for calibration: 62.5 –1000 Volt

Accuracy : 0.16 % of Reading

Uncertainty of Digitizer with Amplifier from its calibration certificate : 0.281 %

Resolution : 0.0001 Volt





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Reading No. Voltage (Volts)

1 250.20

2 250.30

3 250.10

4 250.20

5 250.30

Type A Evaluation

Mean Rdg. (Volts) = 250.22 Volts (R1+R2+R3+R4+R5)/ 5

Standard deviation = 0.0836 Volts

Std. uncertainty Ur = 0.0374 Volts Standard Deviation/ sqrt(5)

Degree of freedom = V1 = 5-1 = 4

Std. uncertainty (% Ur) = 0.0149 % Ur *100/ Mean Reading





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Type B Evaluation

1. Uncertainty of Digitizer with Amplifier from its calibration certificate.

The distribution is normal and the coverage factor for approximately 95%

confidence level is 2

U1(%) = A1/ 2 = 0.281/ 2 = 0.141 %

U1 = 0.141 * 250.22 * 0.01 = 0.352 Volts

Estimate = 0.281* 250.22 * 0.01 = 0.703 Volts

Degree of freedom V2 = infinity





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2. Accuracy of Digitizer with Amplifier from its calibration certificate

A2 = 0.16 % of reading = 0.16 * 250.22 * 0.01 = 0.400 Volts

For rectangular distribution, the standard uncertainty

= U2 = A2/ sqrt(3) U2 = 0.231 Volts

% U2 = 0.0924 % U2 * 100/ Mean Reading





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3. Uncertainty due to resolution of Team Pro software = U3=

A3 = 0.0001/ 2 = 0.00005 Volts = Estimate

For rectangular distribution, the standard uncertainty = U3

=A3/ sqrt(3) U3 = 0.000028 Volts

% U3 = 0.0000115 % U3 * 100/ Mean Reading

Degree of freedom V4 = infinity





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Combined standard uncertainty (Uc)

𝑈𝑐 = 𝑈𝑟2 + 𝑈1

2 + 𝑈22 + 𝑈3

2

Uc = 0.423 Volts

% Uc = 0.169 % Uc * 100/ Mean Reading

Expanded Uncertainty at approximately 95% level of confidence, the

coverage factor k=2, Thus

U = k*Uc = 2 * 0.423 Volts

U = 0.85 Volts

% U = 0.34 % U * 100/ Mean Reading





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Source of

Uncertainty

Xi

Estimates

xi

Limits

%

Probability

Distribution

Type A or B

Factor

Standard

Uncertainty

(Volts)

Volts %

U1 0.281 Normal type B k=2 0.352 0.141

U2 0.16Rectangular

Type B Sqrt(3)0.231 0.0924

U30.00001

15

Rectangular

Type B Sqrt(3)

0.00002

80.0000115

Rep. Ur 250.22 - Normal Type A 0.0374 0.0149

Uc 0.423 0.169

Exp. Un. (Uexp) K=2 0.85 0.34

Uncertainty Budget

Result = (250.22 ±0.85) Volts Result = (250.22 ±0.852) Volts





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Example No. 2






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Problem Find the average, and average deviation for the 5 following data on the length of a pen, L.

Length (cm) |

12.2 0.02 0.0004

12.5 0.28 0.0784

11.9 0.32 0.1024

12.3 0.08 0.0064

12.2 0.02 0.0004

Sum 61.1 Sum 0.72 Sum 0.1880

Average

61.1/5 = 12.22

Average

0.14





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The pen has a length of

(12.22 + 0.14) cm or (12.2 + 0.1) cm [use average deviations]

Or (12.22 + 0.22) cm or (12.2 + 0.2) cm [use standard deviations].





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Example No. 3






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Problem: Find the average and the average deviation of the following measurements of a mass. This time there are N = 6 measurements, so for the standard deviation we divide by (N-1) = 5.

Mass (grams)

4.32 0.0217 0.000471

4.35 0.0083 0.000069

4.31 0.0317 0.001005

4.36 0.0183 0.000335

4.37 0.0283 0.000801

4.34 0.0017 0.000003

Sum 26.05 0.1100 0.002684

Average

4.3417

Average

0.022

The mass is

(4.342 + 0.022) g or

(4.34 + 0.02) g [using

average deviations] or

(4.342 + 0.023) g or

(4.34 + 0.02) g [using

standard deviations].





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Example No. 4






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Example #4: Resistance Measurement

A milliohm meter is used to measure the resistance of a current shunt

resistor. At the selected range of the meter for the measurement, the

calibration certificate states an uncertainty of ± 0.5 mΩ at 95 % of

confidence level. Effects of room temperature and humidity on the

measurement are found to be negligible.





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Degree of freedom, v = 9





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Type B evaluation:

The uncertainty of the calibration is ± 0.5 mΩ with 95 % of confidence

level (k = 1.96).





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Example No. 5






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Example #5: Temperature Measurement

A digital thermometer with a Type J thermocouple is used to measure the

temperature inside a temperature chamber. The temperature controller of

the chamber is set at 500oC.





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Digital thermometer specification:

Accuracy = ± 0.3oC.

Thermocouple specifications:

Temperature correction for the thermocouple at 500oC is 0.5 ± 1.0 oC

at 95% confidence level

Deviation due to immersion = ± 0.2 oC

Deviation due to drift = ± 0.3 oC





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The uncertainty of the thermocouple due to drift = ±0.3 OC. Assume

rectangular distribution, the standard uncertainty due to drift:





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A student measures the voltage shown.

What are the absolute, fractional and percentage uncertainties of his measurement? Find the precision.

▪ Absolute uncertainty = 0.001 V.

▪ Fractional uncertainty = 0.001/0.385 = 0.0026.

▪ Percentage uncertainty = 0.0026(100%) = 0.26%.

▪ Precision is 0.001 V.

EX:





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Thank you for your attention !

Any questions ?

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01090343201

examples uncertainty measurement

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