#### Transcript Lecture Slides

ME 322: Instrumentation Lecture 20 March 6, 2015 Professor Miles Greiner myDAQ A/D converter, temperature uncertainty, First-order, centered numerical differentiation and random errors Announcements/Reminders • HW 7 due now – Did the computers and software in the ECC work the way they are supposed to? • HW 8 Due next Friday – Then Spring Break! • Please complete the Lab Preparation Problems and fully participate in each lab – For the final you will repeat one of the last three labs, including performing the measurements, and writing Excel, LabVIEW and PowerPoint, solo. A/D Converter Characteristics • Full-scale range VRL ≤ V ≤ VRU – FS = VRU - VRL – For myDAQ the user can chose between two ranges • ±10 V, ±2 V (FS = 4 or 20 V) • Number of Bits N – Resolves full-scale range into 2N sub-ranges – Smallest voltage change a conditioner can detect: • DV = FS/2N – For myDAQ, N = 16, 216 = 65,536 • ±10 V scale: DV = 0.000,310 V = 0.305 mV = 305 mV • ±2 V scale: DV = 0.000,061 V = 0.061 mV = 61 mV • Sampling Rate fS = 1/TS – For myDAQ, (fS)MAX = 200,000 Hz, TS = 5 msec Example • For a ±10 Volt, N = 2 bit A/D converter, what digitized voltages will it report for -∞<V<+∞ ? – M = 2N = __ sub-ranges • Break input range into __ steps • IOUT can be 0, 1, 2, 3 – Step size = 𝑉𝑅𝑈 −𝑉𝑅𝐿 𝑀 = 10𝑉 −(−10𝑉) 22 = 20𝑉 4 =5𝑉 – How do we interpret IOUT (VDigitized) and what is its uncertainty? IOUT 3 A/D Converter Transfer Function 2 1 0 -15 -10 -5 0 VIN [volts] 5 10 15 Input Resolution Error • The reported voltage is the center of the digitization sub-range in which the measured voltage is found to reside. – So the maximum error is half the sub-range size. • Inside the FS voltage range – 𝐼𝑅𝐸 = 1 𝐹𝑆 2 2𝑁 = 𝑉𝑅𝑈 −𝑉𝑅𝐿 2𝑁+1 • At edge or outside of FS range – 𝐼𝑅𝐸 → ∞ – To avoid this, estimate the range of voltage that must be measured before conducting an experiment, and choose appropriate A/D converter and/or signal conditioners. • The IRE is the uncertainty caused by the digitization process Summary of myDAQ Uncertainties Scale ±10 Volts ±2 Volts Absolute Absolute Accurcacy Accurcacy 23°C 18-28°C 22.8 mV 4.9 mV 38.9 mV 8.6 mV 0.1% FS 0.2% FS Measurd Shorted Voltage Error Input Resolution Error (IRE) 2.4 mV 0.9 mv 0.15 mV 0.03 mV 0.01 -0.02% FS 0.0008% FS • What are these? – AA: Maximum error of the voltage measurement reported by the manufacturer for all voltage levels • At different temperatures – MSVE: Maximum error measured at V = 0V for one device – IRE: Random error due to digitization process • Which best characterizes voltage uncertainty? Lab 7 Boiling Water Temperature in Reno • Water temperature uncertainty • Standard TC wire uncertainty – Larger of 2.2°C or 0.75% of measurement – Note: 0.0075 x 293°C = 2.2°C – For T < 293°C: wTC = 2.2°C; For T > 293°C: wTC = 0.0075*T • For ±10 Volts, measured shorted voltage uncertainty MSVU = 0.0024V – For TC signal conditioner SSC = 0.025 V/°C – wTsc = MSVU/SSC = 0.0024V/(0.025 V/°C) = 0.096°C • 𝑤𝑇 = 𝑊𝑇𝐶 2 + 𝑊𝑆𝐶 2 = 4.84 + .0092 =2.202°C ~ 2.2°C A/D Converters can be used to measure a long series of very rapidly changing voltage • Great for measuring a voltage signal – How voltage or measurand changes with time – Would be very difficult using a regular voltmeter • Allows determination of – Rates of Change and – Spectral (Frequency) Content • The voltage and time associated with each measurement has some error – It is associated with the centers of the voltage sub-range and sampling time. – Additional systematic and random errors as well • What can go wrong? Example Ti TB T(t) • A small thermocouple at initial temperature Ti is placed in boiling water at temperature TB • Its measured temperature versus time T(t) is shown • What caused the temperature to change? – What do you expect the time-dependent heattransfer rate to the thermocouple 𝑄 [joules/sec = watts] to look like qualitatively? – How can we determine it quantitatively? t [sec] 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0.012 0.013 0.014 0.015 0.016 0.017 0.018 0.019 T [oC] 20.599 20.387 20.646 20.316 20.905 20.528 20.716 20.858 20.693 20.905 20.669 20.811 20.811 20.716 20.246 20.646 20.387 20.387 20.693 20.222 1st Law of Thermodynamics • 𝑄−𝑊 = 𝑑𝑈 𝑑𝑡 = 𝑑 − 𝑑 𝑑𝑇 𝑚𝑐𝑇 = 𝑚𝑐 𝑑𝑡 𝑑𝑇 time-derivative (𝑡) 𝑑𝑡 • How to estimate a table of T versus t data? from a – ∆𝑡𝑆 is the sampling time step [sec] (TS) • First-order numerical differentiation – Centered differencing – 𝑑𝑉 𝑑𝑡 𝑡 𝑉 𝑡+∆𝑡𝐷 −𝑉 𝑡−∆𝑡𝐷 = lim ∆𝑡𝐷 →0 𝑡+∆𝑡𝐷 − 𝑡−∆𝑡𝐷 𝑉 𝑡+∆𝑡𝐷 −𝑉 𝑡−∆𝑡𝐷 = lim 2∆𝑡𝐷 ∆𝑡𝐷 →0 – ∆𝑡𝐷 is the differentiation time step [sec] • ∆𝑡𝐷 = 𝑚∆𝑡𝑆 , m = integer (1, 2, or ?) • Will we get the same result for different values of m? – What is the best value for m? (1, 10, 20, ?) U t [sec] 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0.012 0.013 0.014 0.015 0.016 0.017 0.018 0.019 𝑄 T [oC] 20.599 20.387 20.646 20.316 20.905 20.528 20.716 20.858 20.693 20.905 20.669 20.811 20.811 20.716 20.246 20.646 20.387 20.387 20.693 20.222 Sample Data • Lab 9 Transient Thermocouple Measurements – Download sample data – http://wolfweb.unr.edu/homepage/greiner/teaching/MECH322Instrumentation/L abs/Lab%2009%20TransientTCResponse/LabIndex.htm • Plot T vs t for t < 2 sec • Show how to evaluate and plot first-order centered derivatives with different differentiation time steps – Plot dT/dt vs t for m = 1, 10, 50 • Slow T vs t – for 0.95s < t < 1.05s and 25°C < T < 45°C – How do random errors affect “local” and “timeaveraged” slopes? Effect of Random Noise on Differentiation • Measured voltage has Real and Noise components – VM = VR+VN – 𝑑𝑉𝑀 𝑑𝑡 = 𝑉𝑅 −𝑉𝑁 𝑡+ − 𝑉𝑅 −𝑉𝑁 𝑡− 2∆𝑡𝐷 • ∆𝑉𝑅 = 𝑉𝑅𝑡+ − 𝑉𝑅𝑡− ≈ = ∆𝑉𝑅 +∆𝑉𝑁 2∆𝑡𝐷 𝑑𝑉𝑅 2∆𝑡𝐷 𝑑𝑡 • ∆𝑉𝑁 = 𝑉𝑁𝑡+ − 𝑉𝑁𝑡− ≈ 𝑤𝑉 ~ 𝐼𝑅𝐸 = • • • 𝑑𝑉𝑀 𝑑𝑡 𝑑𝑉𝑅 𝑑𝑡 𝑊𝑉 = + 2∆𝑡𝐷 𝑊𝑉 For small ∆𝑡𝐷 , 2∆𝑡𝐷 𝑑𝑉𝑅 𝑊𝑉 Want ≫ 𝑑𝑡 2∆𝑡𝐷 1 𝐹𝑆 2 2𝑁 RF, IRE, other random errors, does not increase with ∆𝑡𝐷 is large and random – Want ∆𝑡𝐷 to be large enough to avoid random error but small enough to capture real events – If wV is mostly IRE, then decreases as FS gets smaller and N increases Common Temperature Measurement Errors • Even for steady temperatures • Lead wires act like a fin, cooling a hot the surface compared to the case when the sensor is not there • The temperature of a sensor on a post will be between the fluid and duct surface temperature High Temperature (combustion) Gas Measurements Sensor h, TS, A, e Tgas QConv=Ah(Tgas– TS) TW TS QRad=Ase(TS4 -TW4) • Radiation heat transfer is important and can cause errors • Convection heat transfer to the sensor equals radiation heat transfer from the sensor – Q = Ah(Tgas – TS) = Ase(TS4 -TW4) • s = Stefan-Boltzmann constant = 5.67x10-8W/m2K4 • e = Sensor emissivity (surface property ≤ 1) • T[K] = T[C] + 273.15 • Measurement Error = Tgas – TS = (se/h)(TS4 -TW4) Problem 9.39 (p. 335) • Calculate the actual temperature of exhaust gas from a diesel engine in a pipe, if the measuring thermocouple reads 500°C and the exhaust pipe is 350°C. The emissivity of the thermocouple is 0.7 and the convection heat-transfer coefficient of the flow over the thermocouple is 200W/m2-C. • ID: Steady or Unsteady? • What if there is uncertainty in emissivity? Conduction through Support (Fin Configuration) TS T∞ h x L A, P, k T0 • Sensor temperature TS will be between those of the fluid T∞ and duct surface T0 – – • Fin Temperature Profile (from conduction heat transfer analysis): – – • • Support: cross sectional area A, parameter length P, conductivity k Convection heat transfer coefficient between gas and support h 𝑇(𝑥)−𝑇∞ 𝑇0 −𝑇∞ = cosh[𝑚𝐿 1−𝑥/𝐿 ] cosh 𝑚𝐿 cosh 𝑎 = 𝑒 𝑎 +𝑒 −𝑎 𝑚𝐿 = 2 ℎ𝑃 𝑘𝐴 𝐿 (dimensionless length) Sensor temperature at tip, 𝑇𝑆 = 𝑇(𝑥 = 𝐿) Dimensionless Tip Temperature Error from conduction 𝑇𝑆 −𝑇∞ 𝐸= – Decreases as 𝑚𝐿 = • • 𝑇0 −𝑇∞ = 1 – cosh 𝑚𝐿 , (want this to be small) L, h and P increase k and A decrease ℎ𝑃 𝑘𝐴 𝐿 decreases Example • A 1-cm-long, 1-mm-diameter stainless steel support (k = 20 W/mK) is mounted inside a pipe whose temperature is 200°C. The heat transfer coefficient between gas in the pipe and the support is 100 W/m2K, and a sensor at the end of the support reads 350°C. What is the gas temperature? Assume esensor =0 • Steady or unsteady • Radiation or Conduction errors Solution • Sensor temperature: • 𝑚𝐿 = 𝑇𝑆 −𝑇∞ 𝑇0 −𝑇∞ = 1 cosh 𝑚𝐿 ℎ𝑃 𝐿 𝑘𝐴 • What is given and what must be found? • What if esensor = 0.2? T TB Ti t=0 t Example A/D N= 2 ±10V Interpret: 𝐼𝑜𝑢𝑡 → 𝑉𝑜𝑢𝑡,𝐷 = 𝐼𝑂𝑢𝑡 + 1 2 𝑉𝑟𝑢 −𝑉𝑟𝑙 2𝑁 + 𝑉𝑐 Input Range (v) Iout Vout,D Max Error (V) -∞ to -5 -5 to 0 0 to 5 5 to ∞ 0 1 2 3 -7.5 -2.5 2.5 7.5 ∞ ± 2.5V ± 2.5 V ∞