Lecture Slides

Report
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] = m/s2
–=


9.81 2

2 
 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
– =
2
 2
(want this)
• Forces on seismic mass, m
–
 = Fspring + Fdamper =  =
2 
 2

• z(t) = s(t) + yo + y(t) (location mass’s bottom surface)
• Fspring = -ky, Fdamper = -lv = -l(dy/dt)
• − −



=
2
 2
+0+
2
 2
=   +0+
2
 2
Dynamic Transfer Function
• −

−

2 
 2

2 
− 2

= ()

+

•
+  = −()
• 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:  =  + ℎ ;
• Particular Solution:  = −(m/k)a1 = constant
• Homogeneous Solution:
– my’’+ ly’ + ky = −  = 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
– 1,2 =
−± 2 −4
2
( = 2 )
•  = 1  1  + 2  2 
– 1 2 depend on initial conditions
• Accelerometer behavior (1 , 2 ) depends on damping l
• For no damping l = 0,
– =
−0± 02 −4
2
=±
=
−4
2

2 
= ±
=0


= ±
•  = −1 (imaginary number)
•  =

,

undamped natural angular frequency =2pfN
– Increases as stiffness k increases or mass m decreases
Un-damped Oscillations
•  = 1    + 2  −  =  t +Dcos 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 <  < 4
•
•

0<=
<1
2 
−± 2 −4
1,2 =
2
•
=
•
=
• =
−
±
2
−
±
2



=
−
−
2
±
 2
2
=
4−2
2
−
2
±   2 −
 2
2

 2 −
 2
2
– Damped natural angular frequency = 2pf
–  < 
•  = 1
 1
•  = 
−

2
+ 2
 −2 
=
−

2
( + )
sin( + ) - decaying sinusoid
Decaying Sinusoids
•  = 
–=
−

2
sin( + )
 2 −
 2
2
• As damping  increases, amplitude decays more
rapidly and the oscillatory frequency decreases
• Motion essentially stops after


2
> 5 (e-5 = 0.007)
– “Stop” time tstop ~ 10m/, decreases a  increases
Heavy Damping
• Critically Damped,  = 4;  =
– 1,2 =
−± 2 −4
2
=
−
2

2 
=1
<0
– Double, Real Roots,  = 
−

2
• Overdamped,  > 4;  =
+ 

2 
−

2
>1
– Distinct Real Roots, both negative,  =  1 +  2
• Both nearly eliminate oscillations
Response
a
a=a1
a=0
• Undamped  =
t

2 
=0
–  =  t +Dcos t ,  =
– oscillatory
  = 2
• Underdamped 0 <  < 1
−

2
–  =  sin(  + ),  =  2 −
– damped sinusoid (observe this in Lab 10)
 2
2
• Critically-damped  = 1, and Over-damped 1 < 
– not oscillatory
Response to Continuous “Shaking”
•   = , [before it was   = 0]
–  = shaking amplitude
–  = 2,  = shaking frequency
• Find response y(t) for all 
– For  → 0
• Expect () = −

()

=−



– For higher , expect lower amplitude and delayed response
• my’’+ ly’ + ky = -ma(t) = -m
– 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 = -m
• Expect: yP(t) = Bsin + Ccos (from experience)
• HW problem X2
– Plug this time-dependent function into the differential
equation of motion
– Collect the sin and cos terms,
– Find two equations by setting the coefficients of the sin
and cos terms to zero.
– Solve those equations for B and C to show that:
• =
• =
mA
(2
m−k
)2 +
l
2
l
l
2
(2 m−k)
A
(2 m−k)2 +
Solution
• yP(t) = Bsin+Ccos
–=
–=
mA
(2
m−k
)2 +
l
2
l
l
2
(2 m−k)
A
(2 m−k)2 +
•   =
A
l
(2 m−k)2 +
2
(2 m−k) sin +l
•   =  sin  + 
–  =
A
(2 m−k)2 +
l
2
;  =
– For no damping (l = 0), and  →
• For  → 0:   →
–  =
A

A
2

2 −
  =  : AP → ∞
−  =
A
−


=  
Compare to Quasi-Steady Solution


 
•


=
A
(2 m−k)2 +
l
A
k
2 
=
1
2
2 
 2
−1 +


– Insert Undamped Natural Frequency  =
•


1
=


2
2

2
 ; Damping ratio:  = 2

;

(Want this to be close to 1)
−1 + 2 

•  = 0.7 ℎ ℎ      with   = 1 +  ,   < 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 = −  = 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

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