#### 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
– 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
• 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.
• 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
• 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
∞
```