#### Transcript Lecture Slides

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