Transcript Lecture Slides

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
𝑛
𝑖=1 𝑋𝑖
≈𝜇=
1
𝑁
𝑁
𝑖=1 𝑋𝑖
• Sample Standard Deviation
–
𝑠=
1
𝑛−1
𝑛
𝑖=1(𝑋𝑖
− 𝑋)2 ≈ 𝜎 =
1
𝑁
𝑁
𝑖=1(𝑋𝑖
− 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 𝑥1 ≤ 𝑥 ≤ 𝑥2
• Let 𝑧𝑖 =
𝑥𝑖 −𝜇
𝜎
≈
𝑥𝑖 −𝑋
𝑠
(# of SDs from mean)
• 𝑃 𝑧1 < 𝑧𝑖 < 𝑧2 = 𝐼 𝑧2 − 𝐼 𝑧1
– 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?
Standard
Reading, hS
[in WC]
Transmitter
Output, IT
[mA]
0
0.5328
1.0597
1.5617
2.0863
2.5295
1.9637
1.5483
0.9211
0.5216
0
0.5619
0.9595
1.4562
1.9927
2.6214
2.1092
1.6423
1.0696
0.5315
0
4
6.88
9.72
12.48
15.34
17.83
14.66
12.35
9.03
6.83
4.01
7.09
9.18
11.92
14.84
18.3
15.43
12.89
9.86
6.88
4.02
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
– 𝑠𝑦,𝑥 =
𝑛 (𝑦 −𝑎𝑥 −𝑏)2
𝑖
𝑖=1 𝑖
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 ± 𝑠𝑥,𝑦 (68%) (Units and significant figures!)
or 𝑥 = 𝑥 ± 2𝑠𝑥,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 𝑠𝑥,𝑦 , where
• 𝐼 = 0.75 − 0.5 = 0.25, 𝑠𝑜 𝑧 = 0.675
Stand. Dev. of Best-Fit Slope and Intercept
• 𝑠𝑎 = 𝑠𝑦,𝑥
𝑛
𝐷𝑒𝑛𝑜
(68%)
• 𝑠𝑏 = 𝑠𝑦,𝑥
( 𝑥𝑖 )2
𝐷𝑒𝑛𝑜
(68%)
– where Deno = 𝑛 𝑥𝑖2 −
𝑥𝑖 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 𝑥𝑖 find the best estimate for its value
𝑥𝑖 , and its uncertainty 𝑤𝑥𝑖 = 𝑤𝑖 with a certaintylevel (probability) of pi
– 𝑥𝑖 = 𝑥𝑖 ± 𝑤𝑖 𝑝𝑖 𝑖 = 1,2, … 𝑛
– Note: pi increases with wi
• The best estimate for the results is:
– 𝑅 = 𝑓𝑛(𝑥1 , 𝑥2 , 𝑥3 ,…, 𝑥𝑛 )
• The confidence interval for the result is
– 𝑅 = 𝑅 ± 𝑤𝑅 (𝑝𝑅 ) units
• Find 𝑤𝑅 𝑎𝑛𝑑 𝑝𝑅
𝑥
Statistical Analysis Shows
• 𝑤𝑅,𝐿𝑖𝑘𝑒𝑙𝑦 =
𝑛
𝑖=1
𝑤𝑅𝑖
2
=
𝑛
𝑖=1
𝛿𝑅
𝛿𝑥𝑖 𝑥
𝑖
2
𝑤𝑖
• 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 𝑛𝑖=1 𝑥𝑖 𝑒𝑖 where a and ei are constants
• The likely fractional uncertainty in the result is
–
𝑊𝑅,𝐿𝑖𝑘𝑒𝑙𝑦 2
𝑅
=
𝑛
𝑖=1
𝑊𝑖 2
𝑒𝑖
𝑥𝑖
(𝑝𝑅 )
– 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
–
𝑊𝑅,𝑀𝑎𝑥
𝑅
=
𝑛
𝑖=1
– Don’t use this unless told to.
𝑊𝑖
𝑒𝑖
𝑥𝑖
(100%)
Instruments
U-Tube Manometer
∆𝑃 = 𝑃1 − 𝑃2 = 𝜌𝑚 − 𝜌𝑠 𝑔ℎ
𝐼𝑓 𝜌𝑠 << 𝜌𝑚
𝑇ℎ𝑒𝑛: ∆𝑃 = 𝜌𝑚 𝑔ℎ
DP = 0
Measurand
Fluid
Air (1 ATM, 27°C)
Water (30°C)
Hg (27°C)
• Power product?
Reading
𝝆 𝒌𝒈 𝒎𝟑
1.774
995.7
13,565
Inclined-Well Manometer
𝐴
∆𝑃 = 𝑃1 − 𝑃2 = 𝜌𝑚 − 𝜌𝑠 𝑔 𝑠𝑖𝑛𝜃 + 𝑡 R
𝐴𝑤
If 𝑠𝑖𝑛𝜃 ≫
𝐴𝑡
𝐴𝑤
and 𝜌𝑚 ≫ 𝜌𝑠
𝑇ℎ𝑒𝑛: ∆𝑃 = 𝜌𝑚 𝑔𝑅𝑠𝑖𝑛𝜃
1
𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛: 𝑅 =
∆𝑃
𝜌𝑚 𝑔𝑠𝑖𝑛𝜃
Strain Gages
𝑑𝑅𝑖
= 𝑆𝜀 + 𝑆𝑇 ∆𝑇
𝑅
• Electrical resistance changes by small amounts when
– They are strained (desired sensitivity)
• Strain Gage Factor: S = 1 + 2υ + Cstrain
– 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
-
+
R3
+
-
• When R1 ≈ R3 ≈ R2 ≈ R4, then 𝑉0 ≈ 0
• Small changes in Ri cause small changes in 𝑉0
–
𝑉0
𝑉𝑆
=
1
4
𝑑𝑅1
𝑅1
−
𝑑𝑅2
𝑅2
• If gages are in all 4 legs
– with
–
𝑉0
𝑉𝑠
𝑑𝑅𝑖
= 𝑆𝜀𝑖
𝑅𝑖
1
= 𝑆 𝜀1
4
+
𝑑𝑅3
𝑅3
−
𝑑𝑅4
𝑅4
+ 𝑆𝑇 ∆𝑇𝑖 (S and ST same)
− 𝜀2 + 𝜀3 − 𝜀4 + 𝑆𝑇 ∆𝑇1 − ∆𝑇2 + ∆𝑇3 − ∆𝑇4
Quarter Bridge
+
-
• Only one leg (R3) has a strain gauge
–
𝑑𝑅3
𝑅3
= 𝑆3 𝜀3 + 𝑆𝑇3 ∆𝑇3
• Other legs are fixed resistors
–
•
𝑉0
𝑉𝑆
𝑑𝑅1
𝑅1
=
=
1
4
𝑑𝑅2
𝑅2
=
𝑑𝑅4
𝑅4
=0
𝑆𝜀3 + 𝑆𝑇 ∆𝑇3
Undesired Sensitivity
-
R3
+
Half Bridge
+
-
R3
• Use gages for R2 (-) and R3 (+)
– Place R3 on deform specimen; ε3, ΔT3
– Place R2 on identical but un-deformed; ε2=0, ΔT2 =ΔT3
–
𝑉0
𝑉𝑆
=
1
4
𝑆 𝜀3 − 𝜀2 + 𝑆𝑇 ∆𝑇3 − ∆𝑇2
=
1
𝑆𝜀3
4
Automatic temperature
compensation
+
Beam in Bending: Half Bridge
ε3
ε2 = -ε3
ε2 = -ε3
•
𝑉0
𝑉𝑆
=
1
4
𝑆 𝜀3 − 𝜀2 + 𝑆𝑇 ∆𝑇3 − ∆𝑇2
=
1
𝑆(2𝜀3 )
4
𝑉0
1
= 𝑆𝜀3
𝑉𝑆
2
– Twice the output amplitude as quarter bring, with temperature
compensation
Beam in Bending: Full Bridge
3
2
•
•
𝑉0
𝑉𝑠
𝑉0
𝑉𝑠
=
=
1
+
-
-
R3
4
+
1
4
𝑆 𝜀1 − 𝜀2 + 𝜀3 − 𝜀4 + 𝑆𝑇 ∆𝑇1 − ∆𝑇2 + ∆𝑇3 − ∆𝑇4
1
4
𝑆 4𝜀3 + 𝑆𝑇 0
= e3
= -e3
= -e3
= DT3
• V0 is 4 times larger than quarter bridge
– And has temperature compensation.
= DT3
= DT3
Tension Configuration (HW)
2
+
3
4
1
ε1 = ε3
ε4 = ε2 = -υ ε3
• What would happen if all four were parallel?
-
R3
+
Beam Surface Strain
• Bending: 𝜀3 =
𝜎3
𝐸
=
1 𝑀𝑦
𝐸 𝐼
y
Neutral
Axis
=
𝑇
𝑚𝑔𝐿( )
2
𝑇3 𝑊
𝐸
12
=
6𝐿𝑔
𝑚
2
𝐸𝑇 𝑊
F = mg
W
σ
L
• Tension:𝜀 =
𝜎
𝐸
=
𝐹/𝐴
𝐸
=
𝐹
𝐴𝐸
F
• Could be used for force-measuring devices
T
Fluid Speed V (Pressure Method)
PS
PT > PS
PT > PS PS
V
• Pitot Tube Transfer function: ∆𝑃 =
• To use: 𝑉 = 𝐶
2∆𝑃
𝜌
1
𝜌𝑉 2
2
(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
–𝜌=
𝑃
𝑅𝑇
=
𝑃 𝑀𝑀
𝑅𝑈 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
Nozzle
Venturi Tube
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 2
𝐴2
2
1 − b4
• Less uncertainty for larger 𝑄 than for small ones
To use
• Invert the transfer function:
– 𝑄 = 𝐶𝐴2
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)
– 𝑅𝑒𝐷 =
𝑉1 𝐷𝜌
𝜇
=
𝑚
𝜋
𝜌 𝐷2
4
𝐷𝜌
𝜇
=
4𝑚
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
106 𝛽
𝑅𝑒𝐷
0.5
(see restrictions in Text)
• Orifice: page 349, Eqn. 10.13
– C = 0.5959 +
0.0312b2.1 -
91.71𝛽 2.5
8
0.184b +
0.75
𝑅𝑒𝐷
(0.3 < b < 0.7)
Correlation Coefficient
Student T
If N >30 use student t