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ME 322: Instrumentation
Lecture 29
April 1, 2016
Professor Miles Greiner
Nuclear Safety, Accelerometers, Steady, Impulse,
Sinusoid responses
Announcements/Reminders
• Extra-Credit Opportunity
– 1%-of-grade extra-credit for active participation
– Open ended Lab 9.1
• proposals due now
• Possible elective course for ME students
– MSE 465/665: Fundamentals of Nuclear Power
Nuclear Power Plants
Used Nuclear Fuel
• Thin cladding confines highly radioactive used fuel
pellets and high pressure fission product gases
Transfer/Drying Operations
• Fuel is initially stored underwater while it’s radioactivity and heat
generation rates decrease
• After sufficient time a canister, which contains a support basket, is
placed in a transfer-cask, lowered into the pool, loaded with fuel,
covered, lifted out, and drained while helium is pumped in.
– Remaining moisture must be removed before the canister is filled with helium
gas, and sealed for storage or transport
• Canisters are transported in thick-walled packages designed to contain
the fuel, shield the surrounding and control criticality even under
severe accident conditions.
Transport Packages
• Regulatory Accident Conditions
– 30 ft drop onto unyielding surface
– 1 m drop onto a puncture bar
– 30 minute engulfment in an 800°C fire
• The package must continue to operate after these conditions
• What accelerations (shocks) can fuel cladding withstand?
– Measure what acceleration would it experience?
Acceleration Measurement
• Time-rate-of-change of velocity
–𝑎=
𝑑𝑉
𝑑𝑡
=
– [a] = m/s2
–𝑔=
𝑎
𝑚
9.81 2
𝑠
𝑑2 𝑥
𝑑𝑡 2
[dimensionless]
• Needed to quantify
– shock
– vibration amplitude and frequency
• Useful for
– Vibration analysis, resonance assessment
– Failure analysis
Measurement Devices
• Cantilever beam with end mass in a damping fluid
– Uses simple strain gages
– Low damping (continues to vibrate after acceleration, so add viscous fluid)
– Low stiffness and natural frequency (possible resonance effects)
• Piezoelectric accelerometer (used in Lab 10)
– Seismic mass increases/decreases compression of crystal,
• Compression causes charge [coulombs] to accumulate on its sides (piezoelectric effect)
• Changing charge can be measured using a charge amplifier
– High damping, stiffness and natural frequency
• But not effective for steady acceleration
Lab 10, Vibration of a Weighted
Cantilevered Beam
Weight
Accelerometer
Charge
Q=fn(y)
= fn(a)
Accelerometer Model
y = Reading
a
k [N/m]
l
[N/(m/s)]
y0
y
-m/k
a(t) = Measurand
• Un-deformed height y0 affected by sensor size (and gravity)
• Charge Q is affected by deformation y, which is affected by acceleration a
• If acceleration is constant then force on crystal is F = ma =,
– Spring (crystal) deformation: F = –ky
– So y = (-m/k)a, Static transfer function for constant a
• What is the dynamic response (transfer function) of y(t) to a(t)?
Moving Damped Mass/Spring System
z
s(t)
Inertial Frame
• We want to measure acceleration of an object, at sensor’s bottom surface
– 𝑎=
𝑑2𝑠
𝑑𝑡 2
(want this)
• Forces on seismic mass, m
–
𝐹 = Fspring + Fdamper = 𝑚𝑎 =
𝑑2 𝑧
𝑚 2
𝑑𝑡
• z(t) = s(t) + yo + y(t) (location mass’s bottom surface)
• Fspring = -ky, Fdamper = -lv = -l(dy/dt)
• −𝑘𝑦 −
𝑑𝑦
𝜆
𝑑𝑡
=𝑚
𝑑2𝑠
𝑑𝑡 2
+0+
𝑑2𝑦
𝑑𝑡 2
=𝑚 𝑎 𝑡 +0+
𝑑2𝑦
𝑑𝑡 2
Dynamic Transfer Function
• −𝑘𝑦
𝑑𝑦
−𝜆
𝑑𝑡
𝑑2 𝑦
𝑚 2
𝑑𝑡
𝑑2 𝑦
−𝑚 2
𝑑𝑡
= 𝑚𝑎(𝑡)
𝑑𝑦
+𝜆
𝑑𝑡
•
+ 𝑘𝑦 = −𝑚𝑎(𝑡)
• my’’+ ly’ + ky = -ma(t)
• Identify
Reading Measurand
– ? order
– Linear or non-linear?
– Homogeneous or Non-homogeneous?
• For steady or “quasi-steady” acceleration
–
–
y’’ ~ y’~ 0
y(t) = -(m/k)a(t)
• Algebraic (same as two slides ago)
• Non-dynamic, instantaneous response
• Want this, but how small must a(t) be to get this?
Response to Step change in Acceleration
a
a=a1
a=0
t
• Ideally: y(t) = -(m/k)a(t)= -(m/k)a1 = constant for t > 0
• What will the actual behavior be?
– my’’+ ly’ + ky = ma1
– General Solution: 𝑦 = 𝑦𝑃 + 𝑦ℎ ;
• Particular Solution: 𝑦𝑃 = −(m/k)a1 = constant
• Homogeneous Solution:
– my’’+ ly’ + ky = −𝑚𝑎 𝑡 = 0
• Solution
– Assume y = cebt , so y’ = cbebt , so y’’ = cb2ebt
• Know this from experience
• c, b unknown constants
– Plug into differential equation to find them
Characteristic Equation
• c(mb2 + lb + k)ebt = 0
=0
– 𝑏1,2 =
−𝜆± 𝜆2 −4𝑚𝑘
2𝑚
(𝜆𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = 2 𝑚𝑘)
• 𝑦 = 𝑐1 𝑒 𝑏1 𝑡 + 𝑐2 𝑒 𝑏2 𝑡
– 𝑐1 𝑐2 depend on initial conditions
• Accelerometer behavior (𝑏1 , 𝑏2 ) depends on damping l
• For no damping l = 0,
– 𝑏=
−0± 02 −4𝑚𝑘
2𝑚
=±
𝜁=
−4𝑚𝑘
2𝑚
𝜆
2 𝑚𝑘
= ±𝑖
=0
𝑘
𝑚
= ±𝑖𝜔𝑁
• 𝑖 = −1 (imaginary number)
• 𝜔𝑁 =
𝑘
,
𝑚
undamped natural angular frequency =2pfN
– Increases as stiffness k increases or mass m decreases
Un-damped Oscillations
• 𝑦 = 𝑐1 𝑒 𝑖𝜔𝑁 𝑡 + 𝑐2 𝑒 −𝑖𝜔𝑁 𝑡 = 𝐶𝑠𝑖𝑛𝜔𝑁 t +Dcos𝜔𝑁 t
• “Rings” forever (no damping)
– Not realistic, and would not be good for an accelerometer
• What happens if we add damping l > 0?
– more realistic
Underdamped, 0 < 𝜆 < 4𝑚𝑘
•
•
𝜆
0<𝜁=
<1
2 𝑚𝑘
−𝜆± 𝜆2 −4𝑚𝑘
𝑏1,2 =
2𝑚
•
=
•
=
• 𝜔=
−𝜆
±
2𝑚
−𝜆
±
2𝑚
𝑘
𝑚
𝑖
=
−
−𝜆
2𝑚
±𝑖
𝜆 2
2𝑚
=
4𝑚𝑘−𝜆2
2𝑚
−𝜆
2𝑚
± 𝑖 𝜔𝑁 2 −
𝜆 2
2𝑚
𝑖𝜔
𝜔𝑁 2 −
𝜆 2
2𝑚
– Damped natural angular frequency = 2pf
– 𝜔 < 𝜔𝑁
• 𝑦 = 𝑐1
𝑒 𝑏1𝑡
• 𝑦 = 𝐴𝑒
−𝜆
𝑡
2𝑚
+ 𝑐2
𝑒 −𝑏2 𝑡
=𝑒
−𝜆
𝑡
2𝑚
(𝐶𝑠𝑖𝑛𝜔𝑡 + 𝐷𝑐𝑜𝑠𝜔𝑡)
sin(𝜔𝑡 + 𝜙) - decaying sinusoid
Decaying Sinusoids
• 𝑦 = 𝐴𝑒
–𝜔=
−𝜆
𝑡
2𝑚
sin(𝜔𝑡 + 𝜙)
𝜔𝑁 2 −
𝜆 2
2𝑚
• As damping 𝜆 increases, amplitude decays more
rapidly and the oscillatory frequency decreases
• Motion essentially stops after
𝜆
𝑡
2𝑚
> 5 (e-5 = 0.007)
– “Stop” time tstop ~ 10m/𝜆, decreases a 𝜆 increases
Heavy Damping
• Critically Damped, 𝜆 = 4𝑚𝑘; 𝜁 =
– 𝑏1,2 =
−𝜆± 𝜆2 −4𝑚𝑘
2𝑚
=
−𝜆
2𝑚
𝜆
2 𝑚𝑘
=1
<0
– Double, Real Roots, 𝑦 = 𝐴𝑒
−𝜆
𝑡
2𝑚
• Overdamped, 𝜆 > 4𝑚𝑘; 𝜁 =
+ 𝐵𝑡𝑒
𝜆
2 𝑚𝑘
−𝜆
𝑡
2𝑚
>1
– Distinct Real Roots, both negative, 𝑦 = 𝐴𝑒 𝑏1𝑡 + 𝐵𝑒 𝑏2𝑡
• Both nearly eliminate oscillations
Response
a
a=a1
a=0
• Undamped 𝜁 =
t
𝜆
2 𝑚𝑘
=0
– 𝑦 = 𝐶𝑠𝑖𝑛𝜔𝑁 t +Dcos𝜔𝑁 t , 𝜔𝑁 =
– oscillatory
𝑘 𝑚 = 2𝜋𝑓𝑁
• Underdamped 0 < 𝜁 < 1
−𝜆
𝑡
2𝑚
– 𝑦 = 𝐴𝑒 sin(𝜔𝐷 𝑡 + 𝜙), 𝜔𝐷 = 𝜔𝑁 2 −
– damped sinusoid (observe this in Lab 10)
𝜆 2
2𝑚
• Critically-damped 𝜁 = 1, and Over-damped 1 < 𝜁
– not oscillatory
Response to Continuous “Shaking”
• 𝑎 𝑡 = 𝐴𝑠𝑖𝑛𝜔𝑡, [before it was 𝑎 𝑡 = 0]
– 𝐴 = shaking amplitude
– 𝜔 = 2𝜋𝑓, 𝑓 = shaking frequency
• Find response y(t) for all 𝜔
– For 𝜔 → 0
• Expect 𝑦(𝑡) = −
𝑚
𝑎(𝑡)
𝑘
=−
𝑚
𝐴𝑠𝑖𝑛𝜔𝑡
𝑘
– For higher 𝜔, expect lower amplitude and delayed response
• my’’+ ly’ + ky = -ma(t) = -m𝐴𝑠𝑖𝑛𝜔𝑡
– Homogeneous or Non-Homogeneous?
• y(t) = yh(t) + yP(t)
– yh(t) same as response to impulse
– yh(t) 0 after t > tstop
• How to find particular solution to whole equation?
Particular Solution
• myP’’+ lyP’ + kyP = -m𝐴𝑠𝑖𝑛𝜔𝑡
• Expect: yP(t) = Bsin𝜔𝑡 + Ccos𝜔𝑡 (from experience)
• HW problem X2
– Plug this time-dependent function into the differential
equation of motion
– Collect the sin𝜔𝑡 and cos𝜔𝑡 terms,
– Find two equations by setting the coefficients of the sin𝜔𝑡
and cos𝜔𝑡 terms to zero.
– Solve those equations for B and C to show that:
• 𝐶=
• 𝐵=
mA
(𝜔2
m−k
)2 +
l𝜔
2
l𝜔
l𝜔
2
(𝜔2 m−k)
𝑚A
(𝜔2 m−k)2 +
Solution
• yP(t) = Bsin𝜔𝑡+Ccos𝜔𝑡
–𝐶=
–𝐵=
mA
(𝜔2
m−k
)2 +
l𝜔
2
l𝜔
l𝜔
2
(𝜔2 m−k)
𝑚A
(𝜔2 m−k)2 +
• 𝑦𝑝 𝑡 =
𝑚A
l𝜔
(𝜔2 m−k)2 +
2
(𝜔2 m−k) sin𝜔𝑡 +l𝜔𝑐𝑜𝑠𝜔𝑡
• 𝑦𝑝 𝑡 = 𝐴𝑃 sin 𝜔𝑡 + 𝜙
– 𝐴𝑃 =
𝑚A
(𝜔2 m−k)2 +
l𝜔
2
; 𝑡𝑎𝑛𝜙 =
– For no damping (l = 0), and 𝜔 →
• For 𝜔 → 0: 𝑦𝑝 𝑡 →
– 𝐴𝑆 =
𝑚A
𝑘
𝑚A
𝑘2
𝜆𝜔
𝜔2 −𝑘
𝑘 𝑚 = 𝜔𝑁 : AP → ∞
−𝑘 𝑠𝑖𝑛𝜔𝑡 =
𝑚A
−
𝑠𝑖𝑛𝜔𝑡
𝑘
= 𝑦𝑆 𝑡
Compare to Quasi-Steady Solution
𝐴𝑃
𝐴𝑆
𝜔 𝜔𝑁
•
𝐴𝑃
𝐴𝑆
=
𝑚A
(𝜔2 m−k)2 +
l𝜔
A
k
2 𝑚
=
1
2
𝜔2 𝑚
𝜆𝜔 2
−1 +
𝑘
𝑘
– Insert Undamped Natural Frequency 𝜔𝑁 =
•
𝐴𝑃
𝐴𝑆
1
=
𝜔
𝜔𝑁
2
2
𝜔
2
𝑘 𝑚; Damping ratio: 𝜁 = 2
𝜆
;
𝑚𝑘
(Want this to be close to 1)
−1 + 2𝜁 𝜔
𝑁
• 𝜁 = 0.7 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝜔 𝜔𝑁 with 𝐴𝑃 𝐴𝑆 = 1 + 𝜀 , 𝜔 𝜔𝑁 < 0.3
End 2016
Piezo-electric Accelerometer
F
b
a
Quartz produce change Q [coulombs] on surface when compressed.
d ≡ Piezoelectric constant
Charge Amplifiers
•Measure charge while drawing very little current. (which dissipates charge)
•Sensor needs to keep producing charge. (Can’t be used for a steady acceleration.)
Frequency > fmin
•Only accurate for fmin < f < fN
For lab 20 < Hz < 2000
Sensitivity at 100 Hz
Rapidly Changing a(t)
a
v
0
t
0
• Step change in v(t)
• Huge acceleration a at t = 0, but a(t) = 0 afterward
– Ideally, after t = 0, y(t) = -(m/k)a(t)= 0
• my’’+ ly’ + ky = −𝑚𝑎 𝑡 = 0
– Homogeneous
• Solution
– Assume y = cebt , so y’ = cbebt , so y’’ = cb2ebt
• Know this from experience
• c, b unknown constants
– Plug into differential equation to find them
t