Application of type B and Monte Carlo methods for UA: examples for discharge and volume measurements in a circular sewer pipe Jean-Luc BERTRAND-KRAJEWSKI Université de Lyon, INSA Lyon, LGCIE (Laboratory of Civil and Environmental Engineering), 34 avenu des Arts, F-69621 Villeurbanne cedex, France email: [email protected]

1. INTRODUCTION In this example, the uncertainty in the discharge and volume measured in a circular sewer pipe is calculated by means of two methods: i) the GUM type B method by application of the law of propagation of uncertainties LPU (ISO, 2009a) and ii) the Monte Carlo method (ISO, 2008, 2009b). For pedagogical reasons, most algebra and calculations are presented in detail.

2. DISCHARGE MEASUREMENT Let consider a circular concrete sewer pipe with radius R = 0.5. It is assumed i) that the pipe is circular and not affected by any deformation, and ii) that there are no deposits on the invert. The discharge Q (m3/s) is then given by 2⎤ ⎡ h⎞ ⎛ h⎞ h⎞ ⎛ ⎛ Q(R, h, U ) = S (h)U = R 2 ⎢Arccos ⎜1 − ⎟ − ⎜1 − ⎟ 1 − ⎜1 − ⎟ ⎥ U ⎢ R⎠ ⎝ R⎠ R⎠ ⎥ ⎝ ⎝ ⎣ ⎦

eq. 1

where h (m) is the water level and U (m/s) the mean flow velocity. Calculations have been made with the following values: R = 0.5 m, u(R) = 0.001 m h = 0.7 m, u(h) = 0.005 m U = 0.8 m/s, u(U) = 0.05 m/s. Appendix 5.1 explains how the standard uncertainties u(R), u(h) and u(U) have been estimated. The resulting discharge Q = 0.4697 m3/s. Note here that all results in the paper will be given with 4 or more digits only for illustration and comparison purposes. Under real conditions of application, one, two or three digits would be sufficient: the additional ones appear in italic characters in numerical values. However, it is of course recommended to keep the maximum number of digits in all intermediate calculations.

2.1 TYPE B ESTIMATION All measured variables R, h and U are measured independently with different instruments and are not correlated. Consequently, the law of propagation of uncertainty (LPU) can be written 2

2

⎛ ∂Q ⎞ 2 ⎛ ∂Q ⎞ 2 ⎛ ∂Q ⎞ u (Q ) 2 = u ( R ) 2 ⎜ ⎟ ⎟ + u ( h) ⎜ ⎟ + u (U ) ⎜ ∂ R ∂ h ⎝ ⎠ ⎝ ⎠ ⎝ ∂U ⎠

2

eq. 2

The partial derivatives and their numerical values are equal to h⎞ ⎛ ⎛ ∂Q ⎞ 2 ⎜ ⎟ = 2UR Arccos ⎜1 − ⎟ − 2U 2hR − h = 0.852 638 427 096 m2/s R ∂ R ⎝ ⎠ ⎝ ⎠

eq. 3

⎛ ∂Q ⎞ 2 ⎜ ⎟ = 2U 2hR − h = 0.733 212 111 192 m2/s ∂ h ⎝ ⎠

eq. 4

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JLBK – Feb. 2011

h⎞ ⎛ ∂Q ⎞ ⎛ 2 2 ⎜ ⎟ = R Arccos ⎜1 − ⎟ − ( R − h) 2hR − h = 0.587 229 807 114 m2 ∂ U R ⎝ ⎠ ⎝ ⎠

eq. 5

In case algebra is considered too difficult, the above exact analytical expressions can be replaced by numerical estimations of partial derivatives by applying a second order approximation operator (several digits are given only for comparison between exact and approximated values): ⎛ ∂Q ⎞ Q( R + δ R , h, U ) − Q( R − δ R , h, U ) 0.469792 - 0.469775 ≈ ≈ 0.852 638 427 287 m2/s ⎜ ⎟= 2δ R 2.10 − 5 ⎝ ∂R ⎠

eq. 6

with δ R = 10 −5 m ⎛ ∂Q ⎞ Q( R, h + δ h , U ) − Q( R, h − δ h , U ) 0.469791 − 0.469776 ≈ ≈ 0.733 212 111 117 m2/s ⎜ ⎟= 2δ h 2.10 − 5 ⎝ ∂h ⎠

eq. 7

with δ h = 10 −5 m ⎛ ∂Q ⎞ Q( R, h, U + δU ) − Q( R, h, U − δU ) 0.469789 - 0.469777 ≈ ≈ 0.587 229 807 111 m2 ⎜ ⎟= 2δU 2.10 − 5 ⎝ ∂U ⎠

eq. 8

with δU = 10 −5 m/s

It is important to note that δx values should be chosen in such a way that δx