UNCERTAINTY ANALYSIS

Some form of analysis must be performed on all experimental data. The analysis may be a simple verbal appraisal of the results, or it may take the form of a complex theoretical analysis of the errors involved in the experiment and matching of the data with fundamental physical principles.

A method of estimating uncertainty in experimental results has been presented by Kline and McClintock [1]. The method is based on careful specifications of the uncertainties in the various primary experimental measurements. For example, a certain pressure reading might be expressed as

P =100 kN/m2 +/- 1kN/m2

When the plus or minus notation is used to designate the uncertainty, the person making this designation is stating the degree of accuracy with which he or she believes the measurement has been made.

Suppose a set of measurements is made and the uncertainty in each measurement may be expressed with the same odds. These measurements are then used to calculate some desired results of the experiments. We wish to estimate the uncertainty in the calculated result on the basis of the uncertainties in the primary measurements. The result R is a given function of the independent variables x1, x2, x3, , xn. Thus,

R = R(x1, x2, x3, ,xn) (1)

Let wR be the uncertainty in the result and w1, w2, , wn be the uncertainties in the independent variables. If the uncertainties in the independent variables are all given with the same odds, then the uncertainty in the result having these odds is given as

(2)

Example 1. Calculation of electric power

The electric power is calculated as

P = EI

where E and I are measured as

E = 100 V +/- 2 V

I = 10 A +/- 0.2 A

The nominal value of the power is 100 x 10 = 1000 W. The uncertainty in this value is calculated by applying Eq. (2). The various terms are

wE=2 V

wI=0.2 A

Thus, the uncertainty in the electric power is

WP=[(10)2(2)2+(100)2(0.2)2]1/2 = 28.28 W or 2.83 percent.

Example 2. Uncertainty of a copper wire.

The resistance of a certain size of a copper wire is given as

R = Ro[1+a(T-20)]

Where Ro = 6 W +/- 0.3 percent is the resistance at 20oC, a=0.004oC-1 +/- 1 percent is the temperature coefficient of resistance, and the temperature of the wire is T = 30 +/- 1oC. Calculate the resistance of the wire and its uncertainty.

The nominal resistance is

R = 6[1+0.004(30-20)] = 6.24W

The uncertainty in this value is calculated by applying Eq. (2). The various terms are

wRo = 6x0.003 = 0.018 W

wa = 0.004x0.01 = 4x10-5oC-1

wT = 1oC

Thus the uncertainty in the resistance is

WR = [(1.04)2(0.018)2+(60)2(4x10-5)2+(0.024)2(1)2]1/2 = 0.0305 W or 0.49 percent.

References

  1. Kline, S. J., and F. A. McClintock: "Describing Uncertainties in Single-Sample Experiments", Mech. Eng., p. 3, January 1953

2. Holman, J. P.: "Experimental Methods for Engineers", McGraw-Hill, Sixth Edition, 1994