Transcript Lecture Slides

ME 322: Instrumentation
Lecture 25
March 16, 2016
Professor Miles Greiner
Thermocouple response to sinusoidally varying temperature, radiation error, 2
examples
Ran out of time during last example
Undergraduate Research Awards
• An opportunity for undergraduates to apply for
funding to support research, scholarship or
creative activity.
• Proposals Deadline April 11, 2016.
• The first application writing workshop will be this
Friday, March 18 at 3:00 PM in 104 MIKC.
• http://environment.unr.edu/undergraduateresearch
/opportunities/nvura.html
• I’m interested in working with students,
especially ones who are doing well in this class
– NASA Satellite Thermal Management
– Used Nuclear Fuel Packaging Safety
Announcements/Reminders
• This week in lab: Discretely Sampled Signals
– How is the lab going?
– Next Week: Spring Break
– After break: Transient Temperature Measurements
• HW 9 is due Monday after break
• Midterm II, Wednesday, March 30, 2016
– Review Monday after break
TC Response to Sinusoidally-Varying
Environment Temperature
tD
TENV
TTC
T
• For example, a TC in an engine cylinder or exhaust
• “Eventually” the TC will have
– The same average temperature and unsteady frequency
as the environment temperature. However
• Its unsteady amplitude may be less than the environment
temperature’s.
• The TC temperature peak may be delayed by time tD
Heat Transfer from Environment to TC
Environment Temp
TE(t)
Q = hA(TE–T)
T
D=2r
Heat Transfer to TC
• Environment Temperature: TE = M + Asin(wt)
– w= 2pf = 2p/T (Assume M, A, and w are known)
• 𝑄−𝑊 =
𝑑𝑈
𝑑𝑡
=
𝑑𝑇
𝜌𝑐𝑉
𝑑𝑡
= ℎ𝐴 𝑇𝐸 − 𝑇
– Divide by hA, and
– Let the TC time constant be 𝜏 =
•
𝑑𝑇
𝜏
𝑑𝑡
+ 𝑇 = 𝑇𝐸 𝑡
𝜌𝑐𝑉
ℎ𝐴
=
𝜌𝑐𝐷
6ℎ
(for sphere)
Identify this equation
– 1st order, linear differential equation, non-homogeneous
– For steady fluid temperature 𝑇𝐹 we got, 𝜏
𝑑𝑇
𝑑𝑡
+ 𝑇 = 𝑇𝐹
Solution
𝑑𝑇
𝜏
𝑑𝑡
•
+ 𝑇 = 𝑇𝐸 𝑡 = 𝑀 + 𝐴𝑠𝑖𝑛(𝜔𝑡)
• Solution has two parts
– Homogeneous and Particular (non-homogeneous)
– T = T H + TP
• Homogeneous solution
–
𝑑𝑇𝐻
𝜏
𝑑𝑡
+ 𝑇𝐻 = 0
−𝑡
– Solution: 𝑇𝐻 = 𝐴𝑒 𝜏
– Decays with time, so not important after initial transient
• Particular Solution (to whole equation)
– Assume 𝑇𝑃 = 𝐶 + 𝐷𝑠𝑖𝑛 𝜔𝑡 + 𝐸𝑐𝑜𝑠(𝜔𝑡) (but 𝐶, 𝐷, 𝐸 =?)
𝑑𝑇𝑃
𝑑𝑡
•
= 𝜔𝐷𝑐𝑜𝑠 𝜔𝑡 − 𝜔𝐸𝑠𝑖𝑛(𝜔𝑡)
• Plug into non-homogeneous differential equation
– Find constants C, D and E in terms of M, A and 𝜔
Particular Solution
𝑑𝑇𝑃
𝜏
𝑑𝑡
•
+ 𝑇𝑃 = 𝑀 + 𝐴𝑠𝑖𝑛(𝜔𝑡)
• Plug in assumed solution form: 𝑇𝑃 = 𝐶 + 𝐷𝑠𝑖𝑛
𝜔𝑡 + 𝐸𝑐𝑜𝑠(𝜔𝑡)
𝜏[𝜔𝐷𝑐𝑜𝑠 𝜔𝑡 − 𝜔𝐸𝑠𝑖𝑛(𝜔𝑡)] + [𝐶 + 𝐷𝑠𝑖𝑛 𝜔𝑡 + 𝐸𝑐𝑜𝑠(𝜔𝑡)] =
𝑀 + 𝐴𝑠𝑖𝑛(𝜔𝑡)
• Collect terms:
𝑐𝑜𝑠 𝜔𝑡 𝜏𝜔𝐷 + 𝐸 +𝑠𝑖𝑛 𝜔𝑡)(−𝜏𝜔𝐸 + 𝐷 − 𝐴 + 𝐶 − 𝑀 = 0
=0
=0
• Find C, D and E in terms of A, M and 𝜔
–
–
–
–
–
C=M
E = −𝜏𝜔𝐷
−𝜏𝜔 2 𝐷 + 𝐷 = 𝐴
𝐴
D =
𝜏𝜔 2 +1
−𝜏𝜔𝐴
E=
𝜏𝜔 2 +1
=0
For all
times
Result
• 𝑇𝑃 = 𝐶 +
𝐷𝑠𝑖𝑛 𝜔𝑡
• 𝑇𝑃 = 𝑀 +
𝐴
𝑠𝑖𝑛
2
𝜏𝜔 +1
• 𝑇𝑃 = 𝑀 +
𝐴
[𝑠𝑖𝑛
2
𝜏𝜔 +1
• 𝑇𝑃 = 𝑀 +
𝐴
𝜏𝜔
2 +1
+
𝜔𝑡 +
𝐸𝑐𝑜𝑠(𝜔𝑡)
−𝜏𝜔𝐴
𝑐𝑜𝑠(𝜔𝑡)
2
𝜏𝜔 +1
𝜔𝑡 − (𝜏𝜔)𝑐𝑜𝑠 𝜔𝑡 ]
𝑠𝑖𝑛 𝜔𝑡 − 𝜙
– where tan(𝜙) = 𝜏𝜔
• Same frequency and mean as environment
temperature
– TE = M + Asin(wt)
– But delayed and attenuated
Compare to Environment Temperature
tD
T
𝜏
•
𝑇𝐸 𝑡 = 𝑀 + 𝐴𝑠𝑖𝑛(𝜔𝑡)
•
𝑇𝑃 = 𝑇𝑇𝐶 = 𝑀 +
•
𝑠𝑖𝑛 𝜔𝑡 − 𝜙 ; tan(𝜙) = 𝜏𝜔
Same mean value and frequency
–
𝑇𝑇𝐶 = 𝑀 + 𝐴 𝑇𝐶 𝑠𝑖𝑛 𝜔𝑡 − 𝜙 ;
If 1 >> 𝜏𝜔 = 𝜏2𝜋𝑓 =
𝐴 𝑇𝐶 =
𝐴
𝜏𝜔 2 +1
2𝜋𝜏
𝑇
𝑇 ≫ 2𝜋𝜏
Then 𝑇𝑇𝐶 = 𝑀 + 𝐴𝑠𝑖𝑛 𝜔𝑡
–
•
𝜏𝜔 2 +1
–
•
•
𝐴
𝜏𝜔 = 2𝜋 𝑇
Minimal attenuation and phase lag
Otherwise 𝐴 𝑇𝐶 =
–
𝐴
𝜏𝜔 2 +1
<𝐴
𝜏
if 𝜏𝜔 < 0.1 (𝜏 < T/20p) , then 𝐴 𝑇𝐶 /A > 0.995 (1/2% attenuation)
•
Delay time: when 0 = 𝜔𝑡𝐷 − 𝜙
•
𝑡𝐷 = 𝜙
𝜔
=𝑇
arctan(𝜏𝜔) 𝑡𝐷
;
𝑇
2𝜋
=
arctan(𝜏𝜔)
2𝜋
=
arctan(𝜏𝜔)
360
≤ 0.25
𝜏𝜔 = 2𝜋 𝑇
Example
• A car engine runs at f = 1000 rpm. A type J
thermocouple with D = 0.1 mm is placed in one of
its cylinders. How high must the convection
coefficient be so that the amplitude of the
thermocouple temperature variations is 90% as large
as the environment temperature variations? If the
combustion gases may be assumed to have the
properties of air at 600°C, what is the required
Nusselt number?
• ID: Steady or Unsteady?
Material Properties
Common Temperature Measurement Errors
• Even for steady temperatures
• Lead wires act like a fin, cooling 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
• TC temperature changes until convection heat transfer
to sensor equals radiation heat transfer from 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 200 W/m2-C.
• ID: Steady or Unsteady?
• What if there is uncertainty in emissivity?