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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] = 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