ME 322: Instrumentation Lecture 13 February 17, 2016 Professor Miles Greiner Midterm I Review

Announcements/Reminders HW due now Labs This week: Lab 6 Elastic Modulus Measurement Next Week: Monday only: Lab 6 No lab on the other days In two weeks: Lab 6 Wind Tunnel Flow Rate and Speed Only 4 wind tunnels (currently constructing one more) Sign-up for 1.5 hour slots with your partner this week in lab Section 2 students will take exam next-door (OB 101)

This week

Midterm 1 Friday Open book, plus bookmarks, plus one page of notes If you have an e-book, you must turn off internet 4 problems, some have parts Each part like HW or Lab calculations Remember significant figures, uncertainty, units, confidence level Be able to use your calculator to find Sample average and sample standard deviation Linear regression YFIT = aX + b Review Session Marissa Tsugawa, Thurs., Feb. 18, starting at 7:30pm (location?) Handout: last year’s midterm problems These problems will not be on the exam Neither Marissa nor I will not provide answers or solutions for this See me after class today regarding special needs

Multiple Measurements of a Quantity Quad Area [m2] Do not always give the same results. Affected by Measurand and Uncontrolled (random) and Calibration (systematic) errors. Patterns are observed if enough measurements are acquired Bell-shaped probability distribution function The sample may exhibit a center (mean) and spread (standard deviation) Statistical Analysis can be applied to any “randomly varying” processes

Statistics Find properties of an entire population of size N (which can be ∞) using a smaller sample of size n < N. Sample Mean 𝑋 = 1 𝑛 𝑖=1 𝑛 𝑋 𝑖 ≈𝜇= 1 𝑁 𝑖=1 𝑁 𝑋 𝑖 Sample Standard Deviation 𝑠= 1 𝑛−1 𝑖=1 𝑛 ( 𝑋 𝑖 − 𝑋 ) 2 ≈𝜎= 1 𝑁 𝑖=1 𝑁 ( 𝑋 𝑖 −𝜇) 2 How can we use these statistics? The mean characterizes the best estimate of the measurand The standard deviation characterizes the measurement imprecision (repeatability) 𝑋= 𝑋 ±𝑠 units 68% Confidence Interval Confidence Level

Example Problem Find the probability that the next sample will be within the range 𝑥 1 ≤ 𝑥 ≤ 𝑥 2 Let 𝑧 𝑖 = 𝑥 𝑖 −𝜇 𝜎 ≈ 𝑥 𝑖 − 𝑋 𝑠 (# of SDs from mean) 𝑃 𝑧 1 < 𝑧 𝑖 < 𝑧 2 =𝐼 𝑧 2 −𝐼 𝑧 1 I(z) on Page 146 *Bookmark* Useful facts: I(0) = 0, I(∞) = 0.5, I(-z) = -I(z)

“Typical” Problems Find the probability the next value is within a certain amount of the mean (“symmetric”) Find the probability the next value is below (or above) a certain value (one sided) If one more value is acquired, what is the likelihood it is above the mean? How much must be added to a measurement so the sum will have a specified-likelihood to be above the mean?

Instrument Calibration Experimental determination of instrument transfer function Record instrument reading y for a range of measurands x (determined by a standard) Use least-squares method to fit line yF = ax + b (or some other function) to the data. Hint: Use calculator to find a and b unless told (remember Units) Determine the standard error of the estimate of the Reading for a given Measurand 𝑠 𝑦,𝑥 = 𝑖=1 𝑛 ( 𝑦 𝑖 −𝑎 𝑥 𝑖 −𝑏) 2 𝑛−2 Hint: Lean to calculate this efficiently (use table format) For partial credit, you may want to write this formula even if you use a calculator to find the quantity.

To use the calibration Make a measurement and record instrument reading, 𝑦 Invert the transfer function to find the best estimate of the measurand 𝑥 =( 𝑦 −𝑏)/𝑎 Determine standard error of the estimate of the Measurand for a given Reading sx,y = sy,x/a (Units!) Confidence interval 𝑥= 𝑥 ± 𝑠 𝑥,𝑦 (68%) (Units and significant figures!) or 𝑥= 𝑥 ± 2𝑠 𝑥,𝑦 95% … Calibration Removes calibration (bias, systematic) error Quantifies imprecision (random error) but does not remove it

What is the likelihood 𝑥 will be above the true value of the measurand How much do you need to add to 𝑥 to be 75% sure it is above the true value? 𝑥= 𝑥 +𝑧 𝑠 𝑥,𝑦 , where 𝐼=0.75−0.5=0.25, 𝑠𝑜 𝑧=0.675

Stand. Dev. of Best-Fit Slope and Intercept 𝑠 𝑎 = 𝑠 𝑦,𝑥 𝑛 𝐷𝑒𝑛𝑜 (68%) 𝑠 𝑏 = 𝑠 𝑦,𝑥 ( 𝑥 𝑖 ) 2 𝐷𝑒𝑛𝑜 (68%) where Deno=𝑛 𝑥 𝑖 2 − 𝑥 𝑖 2 Not in the textbook (write it in) Hint: Learn to calculate this efficiently (use table format) wa = ?sa (95%)

Propagation of Uncertainty Consider a calculation based on uncertain inputs R = fn(x1, x2, x3, …, xn) For each input 𝑥 𝑖 find the best estimate for its value 𝑥 𝑖 , and its uncertainty 𝑤 𝑥 𝑖 = 𝑤 𝑖 with a certainty-level (probability) of pi 𝑥 𝑖 = 𝑥 𝑖 ± 𝑤 𝑖 𝑝 𝑖 𝑖=1,2,…𝑛 Note: pi increases with wi The best estimate for the results is: 𝑅 =𝑓𝑛( 𝑥 1 , 𝑥 2 , 𝑥 3 ,…, 𝑥 𝑛 ) The confidence interval for the result is 𝑅= 𝑅 ± 𝑤 𝑅 ( 𝑝 𝑅 ) units Find 𝑤 𝑅 𝑎𝑛𝑑 𝑝 𝑅 𝑥

Statistical Analysis Shows 𝑤 𝑅,𝐿𝑖𝑘𝑒𝑙𝑦 = 𝑖=1 𝑛 𝑤 𝑅 𝑖 2 = 𝑖=1 𝑛 𝛿𝑅 𝛿 𝑥 𝑖 𝑥 𝑖 𝑤 𝑖 2 In this general expression Confidence-level for all the wi’s, pi (i = 1, 2,…, n) must be the same Confidence level of wR,Likely, pR = pi is the same at the wi’s

General Power Product Uncertainty If 𝑅=𝑎 𝑖=1 𝑛 𝑥 𝑖 𝑒 𝑖 where a and ei are constants The likely fractional uncertainty in the result is 𝑊 𝑅,𝐿𝑖𝑘𝑒𝑙𝑦 𝑅 2 = 𝑖=1 𝑛 𝑒 𝑖 𝑊 𝑖 𝑥 𝑖 2 ( 𝑝 𝑅 ) Square of fractional error in the result is the sum of the squares of fractional errors in the inputs, multiplied by their exponent. If not a power product, use general formula (previous slide) The maximum fractional uncertainty in the result is 𝑊 𝑅,𝑀𝑎𝑥 𝑅 = 𝑖=1 𝑛 𝑒 𝑖 𝑊 𝑖 𝑥 𝑖 (100%) Don’t use this unless told to.

Instruments

U-Tube Manometer ∆𝑃= 𝑃 1 − 𝑃 2 = 𝜌 𝑚 − 𝜌 𝑠 𝑔ℎ 𝐼𝑓 𝜌 𝑠 << 𝜌 𝑚 ∆𝑃= 𝑃 1 − 𝑃 2 = 𝜌 𝑚 − 𝜌 𝑠 𝑔ℎ 𝐼𝑓 𝜌 𝑠 << 𝜌 𝑚 𝑇ℎ𝑒𝑛: ∆𝑃= 𝜌 𝑚 𝑔ℎ DP = 0 Measurand Reading Fluid Air (1 ATM, 27°C) Water (30°C) Hg (27°C) 𝝆 𝒌𝒈 𝒎 𝟑 1.774 995.7 13,565 Power product?

Inclined-Well Manometer If 𝑠𝑖𝑛𝜃≫ 𝐴 𝑡 𝐴 𝑤 and 𝜌 𝑚 ≫ 𝜌 𝑠 𝑇ℎ𝑒𝑛: ∆𝑃= 𝜌 𝑚 𝑔𝑅𝑠𝑖𝑛𝜃 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛: 𝑅= 1 𝜌 𝑚 𝑔𝑠𝑖𝑛𝜃 ∆𝑃

Strain Gages 𝑑 𝑅 𝑖 𝑅 =𝑆𝜀+ 𝑆 𝑇 ∆𝑇 𝑑 𝑅 𝑖 𝑅 =𝑆𝜀+ 𝑆 𝑇 ∆𝑇 Electrical resistance changes by small amounts when They are strained (desired sensitivity) Strain Gage Factor: S=1+2υ+ C strain Their temperature changes (undesired sensitivity) Solution: Subject “identical” gages to the same environment so they experience the same temperature change and the same temperature-associated resistance change. Incorporate gages into a Wheatstone bridge circuit that cancels-out the temperature effect

Wheatstone Bridge Output Voltage, VO - + - + When R1 ≈ R3 ≈ R2 ≈ R4, then 𝑉 0 ≈0 Small changes in Ri cause small changes in 𝑉 0 𝑉 0 𝑉 𝑆 = 1 4 𝑑 𝑅 1 𝑅 1 − 𝑑 𝑅 2 𝑅 2 + 𝑑 𝑅 3 𝑅 3 − 𝑑 𝑅 4 𝑅 4 If gages are in all 4 legs with 𝑑 𝑅 𝑖 𝑅 𝑖 =𝑆 𝜀 𝑖 + 𝑆 𝑇 ∆ 𝑇 𝑖 (S and ST same) 𝑉 0 𝑉 𝑠 = 1 4 𝑆 𝜀 1 − 𝜀 2 + 𝜀 3 − 𝜀 4 + 𝑆 𝑇 ∆ 𝑇 1 −∆ 𝑇 2 +∆ 𝑇 3 −∆ 𝑇 4

Quarter Bridge Only one leg (R3) has a strain gauge + - - + Only one leg (R3) has a strain gauge 𝑑 𝑅 3 𝑅 3 = 𝑆 3 𝜀 3 + 𝑆 𝑇3 ∆ 𝑇 3 Other legs are fixed resistors 𝑑 𝑅 1 𝑅 1 = 𝑑 𝑅 2 𝑅 2 = 𝑑 𝑅 4 𝑅 4 =0 𝑉 0 𝑉 𝑆 = 1 4 𝑆 𝜀 3 + 𝑆 𝑇 ∆ 𝑇 3 Undesired Sensitivity

Half Bridge + - - + Use gages for R2 (-) and R3 (+) R3 Place R3 on deform specimen; ε3, ΔT3 Place R2 on identical but un-deformed; ε2=0, ΔT2 =ΔT3 𝑉 0 𝑉 𝑆 = 1 4 𝑆 𝜀 3 − 𝜀 2 + 𝑆 𝑇 ∆ 𝑇 3 −∆ 𝑇 2 = 1 4 𝑆 𝜀 3 Automatic temperature compensation

Beam in Bending: Half Bridge ε3 ε2 = -ε3 ε2 = -ε3 𝑉 0 𝑉 𝑆 = 1 4 𝑆 𝜀 3 − 𝜀 2 + 𝑆 𝑇 ∆ 𝑇 3 −∆ 𝑇 2 = 1 4 𝑆 (2𝜀 3 ) 𝑉 0 𝑉 𝑆 = 1 2 𝑆 𝜀 3 Twice the output amplitude as quarter bring, with temperature compensation

Beam in Bending: Full Bridge 3 1 + - - + 2 4 𝑉 0 𝑉 𝑠 = 1 4 𝑆 𝜀 1 − 𝜀 2 + 𝜀 3 − 𝜀 4 + 𝑆 𝑇 ∆ 𝑇 1 −∆ 𝑇 2 +∆ 𝑇 3 −∆ 𝑇 4 𝑉 0 𝑉 𝑠 = 1 4 𝑆 4 𝜀 3 + 𝑆 𝑇 0 V0 is 4 times larger than quarter bridge And has temperature compensation. = e3 = -e3 = -e3 = DT3 = DT3 = DT3

Tension Configuration (HW) 2 3 + - - + 4 1 ε1 = ε3 ε4 = ε2 = -υ ε3 What would happen if all four were parallel?

Beam Surface Strain Bending: 𝜀 3 = 𝜎 3 𝐸 = 1 𝐸 𝑀𝑦 𝐼 = 𝑚𝑔𝐿( 𝑇 2 ) 𝐸 𝑇 3 𝑊 12 = 6𝐿𝑔 𝐸 𝑇 2 𝑊 𝑚 y F = mg Neutral Axis W L T σ Tension:𝜀= 𝜎 𝐸 = 𝐹/𝐴 𝐸 = 𝐹 𝐴𝐸 Could be used for force-measuring devices F

Fluid Speed V (Pressure Method) PT > PS PT > PS PS PS V Pitot Tube Transfer function: ∆𝑃= 1 2 𝜌 𝑉 2 To use: 𝑉=𝐶 2∆𝑃 𝜌 (Power product?) C accounts for viscous effects, which are small Assume C = 1 unless told otherwise Less uncertainty for larger 𝑉 than for small ones

How to Find Density Ideal Gases Liquids 𝜌= 𝑃 𝑅𝑇 = 𝑃 𝑀𝑀 𝑅 𝑈 𝑇 ; 𝑉=𝐶 2∆𝑃 𝜌 =𝐶 2 ∆𝑃 𝑅 𝑇 𝑃 P = PS = Static Pressure R = Gas Constant = RU/MM Ru = Universal Gas Constant = 8.314 kJ/kmol K MM = Molar Mass of the flowing Gas For air: R = 0.2870 kP*m3/kg*K T = Absolute Temperature = T[°C] + 273.15 Can plug this into speed formula Liquids 𝜌=𝑓𝑛 𝑇 ≠𝑓𝑛(𝑃) Tables

Water Properties (Appendix B of Text) Be careful with header and units

Volume Flow Rate, Q Variable-Area Meters Venturi Tube Nozzle Orifice Plate Measure pressure drop Δ𝑃 at specified locations Diameter in pipe D, at throat d Diameter Ratio: b = d/D < 1 Ideal (inviscid) transfer function: Δ𝑃= 𝜌 𝑄 𝐴 2 2 2 1−b4 Less uncertainty for larger 𝑄 than for small ones

To use Invert the transfer function: C = Discharge Coefficient 𝑄=𝐶 𝐴 2 2Δ𝑃 𝜌 1−β4 = 𝐶(pd2/4) 1−β4 2Δ𝑃 𝜌 C = Discharge Coefficient C = fn(ReD, b = d/D, exact geometry and port locations) 𝑅𝑒 𝐷 = 𝑉 1 𝐷𝜌 𝜇 = 𝑚 𝜌 𝜋 4 𝐷 2 𝐷𝜌 𝜇 = 4 𝑚 𝜋𝐷𝜇 = 4𝜌𝑄 𝜋𝐷𝜇 Need to know Q to find Q, so iterate Assume C ~ 1, find Q, then Re, then C and check…

Discharge Coefficient Data from Text Nozzle: page 344, Eqn. 10.10 C = 0.9975 – 0.00653 10 6 𝛽 𝑅𝑒 𝐷 0.5 (see restrictions in Text) Orifice: page 349, Eqn. 10.13 C = 0.5959 + 0.0312b2.1 - 0.184b8+ 91.71 𝛽 2.5 𝑅𝑒 𝐷 0.75 (0.3 < b < 0.7)

Correlation Coefficient Student T If N >30 use student t