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Mo u   quantified
 

Mo u quantified

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    Mo u   quantified Mo u quantified Presentation Transcript

    • Premise • Measured value = repeatable component + Quantifying measurement random component uncertainty • Random component – includes influence of all factors affecting N L Ricker measurement precision – causes uncertainty in measured value – leads to uncertainty in calculated property based on the measured value. NIST Convention Quantifying standard uncertainty• NIST = National Institute of Standards and Technology • NIST allows two approaches• ChemE 436 follows NIST convention – Type A -- statistical evaluation – details: http://physics.nist.gov/cuu/Uncertainty/ • Based on at least two true replicates• Report each measured value as • xx.xx = xi (mean of replicates) xx.xx ± ui • ui = si (standard deviation of mean) where • Also report νi (degrees of freedom) xx.xx = best estimate of true value – Type B -- “other” (appropriate number of sig. figs.!) • Best estimate of the above -- see later examples ui is standard uncertainty (a number, same sig. figs.) i represents the variable being measured 1
    • Equations for Type A evaluation Example ni Excel function:Sample 1Mean xi = ∑ xi, k Your evaluation of the activated carbon n i k =1 =AVERAGE( ) adsorption system consists of 4 replicates:Sample ni 1StandardDeviation si = n − 1 k =1 ( ) ∑ xi, k − x i 2 =STDEV( ) k Pin PoutMean 1 1030 211 siStandard si = 2 1220 252Deviation ni 3 985 197Degrees of ν i = ni − 1Freedom 4 1120 237 ni = number of replicates for variable i NOTE: each measured value has 3 significant figures xi,k = value of kth replicate Example -- sample standard deviation Example -- mean value ni 1 1 ni si = ( ∑ xi, k − xi 2) (Excel: STDEV) xi = ∑ xi, k (Excel’s AVERAGE function) n − 1 k =1 n i k =1 1 1 s Pin = [ (1030 − 1090 ) 2 + (1220 - 1090 )2 x Pin = [1030 + 1220 + 985 + 1120 ] 4 −1 4 + (985 - 1090 )2 + (1120 - 1090 )2 ] = 1088.75 = 103.9531 = 1090 (rounded to 3 significant figures = ROUND(1088.75, -1) ) = 100 (rounded -- one’s digit is not significant) Similarly Similarly x Pout = 224 s Pout = 25 (rounded -- one’s digit is significant) 2
    • Example -- mean standard deviation si Example -- reporting results si = ni 103 . 9531 Pin Pout s Pin = 4 = 51.97656 Result 1090 ± 50 224 ± 12 = 50 (rounded -- one’s digit is not significant) D.O.F. 3 3Similarly s Pout = 12 (rounded -- one’s digit is significant) Example Type B evaluation Type B evaluation • You are using a thermometer • Judgment based on: • The manufacturer claims accuracy = ±1 oC. – data obtained in a similar experiment – Assume: as measured accuracy rating, not a – known “typical” instrument performance standard uncertainty. – manufacturers specifications – Assume the manufacturer has been – calibration report conservative, so larger errors are very unlikely. – uncertainties assigned to reference data – Lacking other information, assume all errors in taken from handbooks this range are equally probable. – etc. 3
    • Uniform (rectangular) probability Moments of a probability Probability 1 Example with distribution function x1 = 1 f(x) 0.5 x2 = 3 • Zeroth moment -- area under f(x): 0 ∞ µ 0 = ∫ f ( x )dx 0 1 2 3 4 −∞ x Random variableFormal definition: (measurement) For a rectangular probability distribution:f (x) = 0 − ∞ < x ≤ x1 1 x1 and x2 represent limits of 1f (x) = ∞ x 2 − x1 x1 ≤ x ≤ x 2 measurement uncertainty (on both µ 0 = ∫− ∞ f ( x )dx = (x 2 − x1 ) = 1 sides of the measured value) x 2 − x1f (x) = 0 x2 ≤ x < ∞ First moment (“mean”) Second moment (“variance”) ∞ ∫ (x − µ ) f (x )dx 2 ∞ ∫ xf ( x )dx σ 2 = −∞ µ= −∞ µ0 µ0 For a rectangular distribution: For a rectangular distribution: σ2 = (x2 − x1 )2 (variance) x +x 12 µ= 1 2 x 2 − x1 2 σ = (standard deviation) 2 3 4
    • Using assumed rectangular Assuming a triangular probability distribution in Type B evaluation distribution x 2, i − x1, i f(x) ui ≈ σ i = Concept: smaller errors more probable 2 3 2 Example: accuracy = ±1 oC implies x 2 − x1 x 2 − x1 σ = 2 6 x 2, i − x1, i 2 ui ≈ σ i = = = 0.58 oC 2 3 2 3 0 x x1 x x2 Assuming a Normal distribution Comparison 1  1  x − µ 2  N (µ , σ ) ⇔ f ( x ) = exp −    σ 2π   x 2, i − x1, i  2 σ   Rectangular ui ≈ σ i = = 2 = 0.6 (rounded) 1 2 3 2 3 Example: N(µ =1.5, σ = 0.5) 0.8 Note: xi ,2 − xi,1 2 0.6 Triangular ui ≈ σ i = = = 0.4 (rounded) x>µ+3σ 2 6 2 6f(x) 0.4 x<µ−3σ µ 0.2 µ−σ xi, 2 − xi,1 2 µ+σ very unlikely! Normal ui ≈ σ i = = = 0.3 (rounded) 0 6 6 0 1 2 3 x Result depends on assumptions. (No “right” answer.) State and justify your assumptions. Thus, assume x2 − x1 = 6 σ Rectangular is the most conservative (largest uncertainty). 5