MECH 261/262 Measurement Lab (& Statistics)

MECH 261/262 Measurement Lab (& Statistics) January 6, 2010 Final Exams and Midterm Tests in Reverse Chronological Order until 2004 (FD1)/2612Exams/ExaMT781.tex 1 Introduction On the following pages you will find all questions and answers, as they were to appear, on completed work sheets handed in by the student at the end of that final exam or midterm test. Worked out solutions will be added eventually but it has taken me six years to compile this lot in some semblance of order so don t expect miracles. Furthermore some questions are (very) bad and some answers are wrong. Unfortunately that s life but at least no attempt is made here to cover up such mistakes and assurances are given that most of these are caught before they accrue to the detriment of a student s final grade. In a nutshell, any small consensus of students answers to a particular question, that differ from mine, triggers a re-solution by me and independently by a senior TA. Credit is thus given for correct answers. Furthermore allowance is made for bad questions by awarding full marks if the student made a reasonable attempt or, better still, identified the question as seriously flawed. An example solution is found on the page immediately after Q.1,2,3 on the 04-02-19 midterm. Notice how Q.3 is solved and due to ambiguity of that question, 3 out of the possible 6 multiple choices were deemed to be correct, acceptable answers. (FD1)/Exams/ExaMT781.tex 1

1. A tire normally at 2At has a slow valve leak. Pressure is measured as 1.9At. 3 minutes later it s at 1.8At. To slowly get to a garage takes 1 hour. Tire is destroyed if it gets to 0.6At. Can you chance it? Leak behaves as a 1st order system. Find how long it takes to go from 1.9At to 0.6At. about 64min. 2. A solid round steel bar is used to make a torque transducer. What diameter of bar is needed to sustain a shear stress of 207MPa when a torque of 100Nm is applied? = 13.5mm 100Nm =? 3. A farmer boiling sap to make maple syrup knows he can siphon it from the evaporator pan when the boiling temperature, raised by the dissolved sugar, reaches 125C. A thermocouple with one red and one white lead is used. His millivoltmeter is near a thermometer that reads 82 F. What meter reading should be expected when the syrup is ready? 5.210mV (FD3)MLXA043 Final Exam 10-04-29 (circle one) MECH 261 262

4. Compare the expected output voltage of RTD bridges using a 2V power supply, 2 lead wires (each length). All other resistors are 100 including the RTD itself (when measured at 0C). The RTD is at 390C. What is Vo with...... a two lead bridge... a three lead bridge 0.422V 0.417V Weight 0.5kg Piston... a four lead bridge 0.4125V 13.030 kg 5. Find mass of weight necessary to exert pressure Oil of 1MPa on oil in dead weight pressure tester. 145.05 psi 13 mm What is that in psi (pounds/sq. in)? 144.7 (hard way) 6. An impact tube measures air stream velocity and is connected to a manometer inclined at 20 and filled with fluid with SG=0.8. The manometer reads 17mm, the thermometer reads 70 F and the barometer reads 29in.Hg. Express the air stream velocity in ft./s, mph, m/s or kmh. 17mm 29.02 /s or 19.79mph 8.845m/s or 31.844kmh v 20 Final Exam 10-04-29 (FD3)MLXB043 (circle one) MECH 261 262

1. Rare events like sighting a whale are often modeled with a Poisson distribution. If average weekly number of sightings at Tadousac is 2.6/week what is the mean and standard deviation of this distribution? 2.6 and its square root 1.612 What s the probability of sighting <2? 0.267... >5? 0.049 2. A sample of 50 airline passengers on the Montreal-Toronto run, taken at random at both PET and Pearson at various times of day, week and year, reveals an average mass of 68kg with sample standard deviation of 4.3kg. At a confidence level of 90% what is the error tolerance of average passenger mass based on this sample mean? I.e., how much "off" the actual population mean mass might this figure of 68kg be? +or-1kg 3. Since the experiment above involves 50 separate trips to either the Montreal or Toronto airports, which is costly, it is suggested to weigh only 20 passangers. The accountants claim 10 are enough however an error in the mean estimate of more than +or-2kg at this confidence level is excessive. State the error in the mean in both cases. n=20 n=10 +or-1.66kg +or-2.49kg 4. What is the probability of drawing a 4-of-a-kind poker hand on a 5-card deal? 0.024% or 1 in 4165 Final Exam 10-04-29 key (FD3)STXA043 MECH 262

2010 Winter Measurement (& Statistics) Final Answers and Solutions (FD1)FXW10Sol043.tex April 29, 2010 1 Problem 1. A first order system goes from an initial state of 1.9At to 1.8At in 3 minutes. How long does it take to go down to 0.6At? First, find the time constant C t. Set t = C t and note that for the 3 minute interval X = 1.9 1.8 1.9 = 0.1 1.9. e T 1 e T With C t in hand solve for X = (1.9 0.6)/1.9. t X = e( C ) ( t 1 X = 0 1 0.1 ) e 3/Ct 1 = 0 C e ( t C ) t = 55.486min. t 1.9 e ( t C t ) 1 e ( t C t ) 1.3 1.9 = 0 t = 63.958min. So one might make it to the repair shop before the tire goes critically flat. 2 Problem 2. The tensile and compressive gauges on a toque cell are at 45 to the round steel shaft axis. They sustain the same stress magnitude in tension and compression as the maximum shear stress, i.e., on the surface of the bar at r o. Γr o J 3 Problem 3. τ = 0 100r πr 4 /2 207000000 = 0 200 r3 = 207000000π r = 13.5mm The red-white lead insulation identifies the couple as Fe-con type J and the 82 F temperature must be converted to Celsius. (82 32) 5 9 = 250 9 = 27.777C Couple output at 82 F= 27.77C with reference to 0C is calculated with linear interpolation between 20C and 40C. 1.019 + 250 9 20 40 20 (2.058 1.019) = 1.4230555... Similarly, couple output at 125C, with reference to 0C, is interpolated between 120C and 140C. 6.359 + The difference between these, hence the required reading, is 4 Problem 4. 125 120 (7.457 6.359) = 6.6335 140 120 6.6335 1.4230555... = 5.210444... mv This problem involves substituting V s = 2V, R 2 = R 3 = 100Ω, R RT D = 246.05Ω (the tabulated resistance at 390C), respectively, into Eqs, (9.11), (9.12) and (9,13). The equations below were derived from scratch circuit analysis. They are identical to the three above taken from the text. Equations from the text give correct numerical 1

values but they turn out negative which means the meter polarity shown in the three illustrations in the text produce negative voltage on the terminals shown as positive. First a two-wire solution. Next for three wires. And finally for four wires. V o = V o = V o = V s(r RT D R 2 ) 2(R RT D + R 2 ) = 2(246.05 100) 2(246.05 + 100) = 0.422049V V s (R RT D R 2 ) 2(246.05 100) = 2(R RT D + 2R L + R 2 ) 2(246.05 + 4 + 100) = 0.417226V V s (R RT D R 2 ) 2(246.05 100) = 2(R RT D + 4R L + R 3 ) 2(246.05 + 8 + 100) = 0.412512V Notice that inclusion of the lead wire resistances successively lowers the sensor resistance R RT D by not counting the lead wire resistance as part of the sensor. 5 Problem 5. Pressure is force f divided by area a = πr 2 of piston. Force in this case is weight (w + 0.5)g of mass m of 0.5kg piston plus the applied weight. Required pressure p is 1000000Pa. p = f a = mg a = (w + 0.5)9.81 0.0065 2 π w = 106 (0.0065 2 )π 9.81 0.5 = 13.030 3 kg ( 127.827 3 N) As regards converting 1MPa to psi, the conversion table handed out in class gives 1psi=6.894 10 3 Pa so 10 6 Pa is just the reciprocal 1 0.006894 = 145.05 4psi On the other hand, using common equivalents between pounds (lbs) and kg (2.2lbs=1kg) and inches and mm (25.4mm=1in) gives 13.53 2.2 π(6.5/25.4) 2 = 144.68psi 6 Problem 6. Inclined manometer at 20deg. reads 17mm of fluid, SG=0.8. Impact tube is measuring air velocity at T = 70 F and 29in-Hg. Find velocity. Taking R the air gas constant as 53.3lbs(force)-ft/lbs(mass) R and water density as 62.3lbs/ft 3 at T = 70 F write the energy equation. ρ f is air density, ρ m is manometer fluid density, v is air flow velocity, g is conversion from kg force units to N and h is vertical height of fluid in manometer. Don t confuse R, meaning Rankin or absolute Fahrenheit degrees with R a gas constant, in this case for air. rho f v 2 2 2 ρ m gh = pv2 2RT ρ mgh = 29 29.22 (14.696)144v 2 2(53.3)(70 + 460) Solving produces v = 29.02ft/s=19.79mph=8.845m/s=31.844kmh. End of Measurement Part 0.8(62.3)32.1617 sin 20 12(25.4) = 0 2

Beginning of Statistics Part 7 Question 1. The mean of a Poisson distribution is the average. The standard deviation is just the square-root µ = 2.6, σ = 2.6 = 1.612 The probability of seeing exactly x whales per week is The probability of seeing less than two is P (x) = λx e λ λ! The probability of seeing more than five is P (x < 2) = P (0) + P (1) = λ0 e 2.6 0! + λ1 e 2.6 1! = 0.26738 P (x > 5) = 1 (P (0) + P (1) + P (2) + P (3) + P (4) + P (5) ( 2.6 0 e 2.6 = 1 + 2.61 e 2.6 + 2.62 e 2.6 + 2.63 e 2.6 + 2.64 e 2.6 + 2.65 e 2.6 ) 0! 1! 2! 3! 4! 5! 8 Question 2. = 0.049037 A sample of 50 airline passengers with mean mass of µ = 68kg and standard deviation σ = 4.3kg can be found with Eqs. (6.46) to have a mean confidence interval µ = x ± z α/2 σ n at a confidence level of 90% by using Table 6.3 backwards. Looking for the integral value of exactly 0.5 we find, via linear interpolation, that it corresponds to the row z = 1.6, halfway between the columns under 0.04 and 0.05 indicating z = 1.645. With Eq. (6.46) this gives µ = 68 ± 1.645 4.3 50 = 68 ± 1kg 9 Question 3. This time we must us Student s t distribution because n < 30. This time we use Table 6.6 that is indexed directly on v = n 1 and headed with the tail fractions α/2. The pertinent equation is Eq. (6.51). µ = x ± t α/2 S n For n = 20, v = 19 and under α/2 = 0.05 for a 90% confidence interval we read t = 1.729 so we have µ = 68 ± 1.729 4.3 20 ±1.66kg which is less than ±2kg. On the other hand. with n = 10, v = 9 under α/2 = 0.05 one finds t = 1.833. Now µ = 68 ± 1.833 4.3 10 which is clearly greater than the prescribed tolerance on the mean. ±2.49kg 3

10 Question 4. There are only 13 Fours-of-a-Kind in an ordinary 52 card deck that offers 52C 5 = 2598960 possible 5-card combinations. However the fifth card may be any of the remaining 48 having removed the 4 that make up the ones of same denomination. Therefore of the 2598960 possible hands 13 48 are Fours-of-a-Kind. The probability of being dealt one, without intermediate removal of cards like when dealing to more than one receiver, is 13 48 2598960 0.024% or 1 chance among 4165 hands, on average. 4

R 2 iv 1. A four active arm strain gauge bridge with 300 SR4 gauges is shown. Gauge factor = 2. This is the circuit of the force transducer below. The block of steel has cross + section area of 2x2cm and subject to compressive force of 60kN. R 1 a) Label the spaces i,ii,iii,iv properly with R 1,R 2,R 3,R 4. R i b) What is the compressive stress and strain on the transducer block? M 150MPa 724.6 R 2 R 4 - -60kN iv MLT03a - + 5V iii i iii R 3 ii R 1 ii R 4 +60kN R 3 c) The 4 resistance values are R 299.565 1 R 300.130 2 R 300.130 3 R 299.565 4 d) If R what is i 8 8 the reading of M? 4.711mV e) What is the sensitivity of this force transducer? 2. The RMS value of the 1kHz wave form is 2V. What is its peak-to-peak voltage? 8.94V V 2 0 t -8 V=kt 2 0.5ms 1ms Midterm Test 10-03-01, 40min. 78.5 V/kN (circle one) MECH 261 262

1. A 300 strain gauge bridge with an output of 5mV is connected to an instrumentation amplifier with a 10k input impedance and a gain of 20dB. It has an output impedance of 10 and is connected to a millivoltmeter with a 2k input impedance. What is the meter reading? 48.302mV 5mV 300 + 10k 20dB 10 2k?mV 2. The number $AD7649 is mapped into a 24-bit register shown below. "$" means base 16 or "hexadecimal". Show its binary image. Interpret this as a 2 s complement number + and write its equivalent decimal magnitude. -5409207 0 1 0 1 0 1 1 0 1 0 1 1 1 0 1 1 0 0 1 0 0 1 0 0 1 3. Design a low-pass 1st order Butterworth filter with R 1=7.2k R 2 =22k f c =40Hz. C= 180.9nF What is the DC gain? 3.055/+9.701dB And at 120Hz? 1.019/+0.159dB R 2 MLT03c R 1 C Midterm Test 10-03-03, 40min. (circle one) MECH 261 262

1. There is a known 5% probability that oversize pistons required by an automotive engine rebuilder are either too large or too small to fit the rebored cylinder blocks. 55 pistons are bought to fill an order for 10 rebuilt 5-cylinder VW motors. What is the probability that enough good ones are in the lot of 55? 94.4% 2. Given the points A(3,3,-4), B(7,8,-5), C(9,4,0), D(10,7,-2) fin d a) The centroid of this four point system. {4:-29:22:-11} c) Write the characteristic equation. 2 3 207736832-1151680 +1936 - =0 b) Its "inertia" matrix numerical elements I 508-184 -204 xx -I xy -I xz -164 -I I 696 88 xy yy -I yz -204 88 -I xz -I yz I 732 zz d) Given the eigenvalues 967.766, 343.713, 624.521 write a pair of equations whose solutions are the direction numbers of the best fitting plane s normal vector. -459.766e -184e -304e =0 1 2 3-184e -271.766e +88e =0 1 2 3 Use the top 2 rows of the inertia matrix. STT03e Midterm Test 10-03-05, 40min. MECH 262

MECH 261/262 Measurement Lab & Statistics Midterm Test Solutions 2010 1 Question 1 Monday March 2, 2010 1 a) i = R 1 or R 4, ii = R 2 or R 3, iii = R 4 or R 1, iv = R 3 or R 2 The compressive gauges ( ) are oriented in the direction of the applied, compressive force. The tensile, Poisson gauges ( ) are oriented in the direction normal to the applied force. 1 b) 1 c) σ = F A = 60000N 0.02 2 m 2 = 150MPa, ɛ = σ E = 150 106 Pa 207 10 9 Pa = 724.6µɛ R 1 = R 4 = R 0 1 + F G ɛ = 300 6 = 299.565Ω 1 + 2 724.6 10 R 2 = R 3 = R 0 (1 + F G µ ɛ) = 300 (1 + 2 0.3 724.6 10 6 ) = 300.130Ω Note that µ is Poisson s ratio, ɛ is strain while µɛ has been used to mean micro-strain. 1 d) Calculate the voltage difference between the junction of resistors R 1 and R 2 and that between resistors R 3 and R 4. V s is the power supply voltage of 5V. M is the meter reading. ( R2 M = V s R ) ( ) 4 300.130 299.565 = 5 = 4.711mV R 1 + R 2 R 3 + R 4 300.130 + 299.565 1 e) Sensitivity S is defined as output/input or M/F where F is the load of 60kN. S = M F = 4.711mV 60kN = 78.5µV/kN 2 Question 2 Monday Take a symmetrical half-cycle to integrate. Call the limits of integration 0 to 1. It does not matter. It does not have to be seconds or radians or whatever. Note that v(t) = kt 2 so v(t) 2 = k 2 t 4. The RMS voltage is given as V RMS = 2V. 1 k V RMS = (k 2 t 4 )dt = 2 t 5 ] 1 0 = k = 2V, k = 2 5 5 5 0 Notice that peak-to-peak goes as much negative as positive. So V 1 2 PtP = k = 2 5 4.472V. The signal is bi-polar so V PtP 8.944V. 1

3 Question 1 Wednesday Input V i to the amplifier is given in terms of the given output of the source V s = 5mV, its output impedance R s = 300Ω and the input impedance of the amplifier R i = 10kΩ. ( ) Ri V i = V s R s + R i Output V o of the amplifier depends on the gain G and V i. Note that G db = +20 is given. V o = GV i, G = 10 G db/20 = 10 Finally the meter reading V L is determined by V o, the meter input impedance R L = 2kΩ and the amplifier output impedance R o = 10Ω. Putting it all together ( ) ( ) ( ) ( ) Ri RL 10 2 V L = GV s = 10 5 = 48.302mV R s + R i R L + R o 10.3 2.01 4 Question 2 Wednesday The hexadecimal number translates to straight binary as $AD7649 [1010][1101][0111][0110][0100][1001] Since the leading bit is [1] this is a negative, 2 s complement number. Reversing all the bits and adding [1] produces the positive equivalent of the negative number [0101][0010][1000][1001][1011][0111] Converting back to base-16 gives This translates to $5289B7 5 16 5 + 2 16 4 + 8 16 3 + 9 16 2 + 11 16 1 + 7 16 0 = 5242880 + 131072 + 32786 + 2304 + 176 + 7 = ( )5409207 5 Question 3 Wednesday To get the required capacitance C, rearrange Eq. (3.26) in Wheeler & Ganji. C = 1 2πf c R 2 = 1 2π 40 22000 = 180.9nF 2

The DC gain G o is just the ratio G 2 : G 1. It is then expressed in db. G o = R 2 R 1 = 22000 7200 = 3.055., G o db = 20 log 10 G o = +9.702dB To get the gain at 120Hz, use Eq. (3.27) to get the magnitude ratio G : G o. G = 1 1 = 2πfCR 2 2π 120 22000 = 1 3 G o So G 120Hz = 3.055. /3 1.0185 or +0.1594dB. This ends the measurement lab part of the test 3

6 Question 1 Friday This is a binomial distribution issue like many quality control problems. The gamble concerns to odds of there being 6 bad eggs among a lot of 55 given a defect expectation/mean probability of 5% or 1 in 20. Count up the exactlys from 1 to 5 and subtract from unity. P 0 = 55 C 0 0.05 0 0.95 55 = 0.0595 P 1 = 55 C 1 0.05 1 0.95 54 = 0.1723 P 2 = 55 C 2 0.05 2 0.95 53 = 0.2449 P 3 = 55 C 3 0.05 3 0.95 52 = 0.2277 P 4 = 55 C 4 0.05 4 0.95 51 = 0.1558 P 5 = 55 C 1 0.05 5 0.95 50 = 0.0836 P 50 = 0.0595 + 0.1723 + 0.2449 + 0.2277 + 0.1558 + 0.0836 = 0.944 7 Question 2 Friday To get the centroid one just adds up the 4 point coordinates and appends the count of 4 as a homogenizing coordinate on the left. ( 29 A(3, 3, 4), B(7, 8, 5), C(9, 4, 0), D(10, 7, 2) G{4 : 29 : 22 : 11} 4, 22 4, 11 ) 4 The last 3 coordinates are subtracted from 4 times the original, 4 given coordinates. That way one avoids fractions/decimals. A ( 17, 10, 5), B ( 1, 10, 9), C (7, 6, 11), D (11, 6, 3) Next, compute the inertia matrix elements. I xx = ( 10) 2 + ( 5) 2 + 10 2 + ( 9) 2 + ( 6) 2 + 11 2 + 6 2 + 3 2 = 508 I yy = ( 17) 2 + ( 5) 2 + ( 1) 2 + ( 9) 2 + 7 2 + 11 2 + 11 2 + 3 2 = 696 I zz = ( 17) 2 + ( 10) 2 + ( 1) 2 + 10 2 + 7 2 ( 6) 2 + 11 2 + 6 2 = 732 I xy = ( 17)( 10) + ( 1)10 + 7( 6) + 11(6) = 184 I xz = ( 17)( 5) + ( 1)( 9) + 7(11) + 11(3) = 204 I yz = ( 10)( 5) + 10( 9) + ( 6)11 + 6(3) = 88 Then populate the inertia matrix, subtract the diagonal λ-matrix and take the determinant to get the characteristic equation. 508 λ 184 204 [I] = 184 696 λ 88 204 88 732 λ = 207736832 1151680λ + 1936λ2 λ 3 = 0 This produces three eigenvalues. Since the first is of greatest magnitude it belongs to the eigenvector (normal) to the best fitting plane, the one that minimizes the sum of least squares distances with respect to the 4 given points. λ 1 = 967.77, λ 2 = 343.71, λ 3 = 624.52 4

The test, as distributed, contained a misprinted first eigenvalue of λ 1 = 767.77. Those who used it, to otherwise correctly calculate the pair of equations required to compute that principal eigenvector, suffer no penalty. Using the first two rows of [I] for the eigenvector e = [e 1 e 2 e 3 ] equations produces 459.77e 1 184e 2 204e 3 = 0 184e 1 271.766e 2 +88e 3 = 0 Solving homogeneously and normalizing on e 1 gives the eigenvector e[1 1.0888 1.2717]. With this as plane normal the centroid G can be used to obtain the best fit plane equation, again normalizing on the constant term. 1 0.0525x + 0.0572y + 0.0668z = 0 (MECH261/262)MLSCTSol10.tex 5

1. Design a lowpass Butterworth filter with a gain of 0dB, corner frequncy 30Hz and input resistor R =10k. C 1 R = 10k C= 530.5nF Hint:- The unit of capacitance is "Farad" (F). 2 Resistors are big; capacitors are small. R 1 R 2 2. This is a "1st order" filter. What is the gain in db if the input signal frequency is 60Hz? If it is desired to reduce (attenuate) a 60Hz (noise) signal down to 1% G db = about -7db of its original amplitude or less, what is the smallest even number n n= 6,7,8 (order of the filter) that can do this? 741 3. A diaphragm type pressure transducer with a 2cm chmber diameter, 1mm diaphragm thickness sustains a diaphagm material tensile stress of 200MPa. If the pressure difference being measured is 3.92MPa what is the dome height? h= 0.491mm Material is assumed 2 to be steel, E=200GPa. (FD1)MLXA84N p Final Exam 08-04-14 (circle one) MECH 261 262

4. A Copper-Constantan thermocouple reads 7.934mV when the reference junction is 0.6 0.2 measured as 23.7C with an accurate thermometer. What is the temperature of the measuring junction if the reading is positive? What if it is negaitive? (FD1)MLXB84N -320.24C Why might you consider the negative R R@25C (130 ) 0.25 0 20 40 60 80 100 120 C 192.18C 1 0 Cu Final Exam 08-04-14 (circle one) MECH 261 262 ±7.934mV reading result to be unreliable? It s off scale. Con. Cu The table is in 50C increments. 23.7C 5. Calibrate the thermistor using the three given points and equation 1/T=A+BlnR+C(lnR) 3 3 6. Here s a "pissing" tank of water. A= 0.5343453399 Where does the stream land? 1 V B= -0.1980715346 (7.5,2) Find s= 917mm V@25C (2.1V) C= 0.004073451 2 for reference only H=1m Hint: A three-point calibration. H 2 O h=0.3m 0 s

7. Statistics have been gathered in the Montr al M tro to show your probability of experiencing service interruption during any given week is 5%. What s the chance you ll suffer no delays during a 3-week period? 0.857375 What s the probability you ll have exactly one frustrating wait during 4 consecutive weeks? 0.171475 What s the chance of getting caught at least once in 5 weeks? 0.2262190625 8. On average, 84 people visit the bar at Thompson House during the 4-hour period of our weekly visit. About every every 20 minutes a colleague (or I go) goes to order a pitcher or two of beer. Estimate the probability that the delegate will find a line of 5 or more before him; we never ask ladies to buy. 0.8270083949 Don t count our table among the 84. 9. A sample of 10 from a batch of electrical motors is tested and found to have a mean efficiency of 91% with S=æ0.8%. What is 95% confidence interval for the mean efficiency for the entire batch? 0.57 If it s desired to get the confidence interval up to 98%, how many motors should be tested? 16 This last one s straight from the book. If you re lucky, you ve already done and checked it and just have to copy down the answer. Final Exam 08-04-14 key (FD1)STXA84N MECH 262

2008 Winter Measurement Final Answers and Solutions (FD1)FXW08Sol84n.tex April 16, 2008 1 Problem 1. A first order Butterworth filter with DC gain (f = 0Hz) of G = 1 (0dB) requires R 2 = R 1 = 10kΩ because any capacitor C in parallel with feedback resistor R 2 has X C. Then Eq. (3.26), p.65, Wheeler & Ganji (W+G) specifies the corner frequency f c as 1 f c = 2πCR 2 Solving, C = 0.531 10 6 530nF 2 Problem 2. First and higher order, n, low-pass Butterworth filter attenuation is given by Eq. (3.20), p.51, W+G. At n = 1 the 60Hz noise is down to 45% of its input level. f c = 30, f = 60. G 1 60Hz = 1 1 + ( f f c ) 2n = 1 5 0.4472 ( 6.989dB) With n = 6 we just miss the 1% noise specification while n = 8 seems a bit overkill. Do n =odd filters exist? Anyway, n = 6, 7, 8 all seem to be acceptable answers. 3 Problem 3. G 6 60Hz = 1 1 + G 8 60Hz = 1 1 + ( f f c ) 2n = ( f f c ) 2n = 1 4097 0.0156 ( 36.12dB) 1 65537 0.0039 ( 48.16dB) The pressure transducer diaphragm is of thickness t = 1mm, radius ρ = 100mm and sustains a material tensile stress σ = 200MPa when a p = 3.92MPa pressure is applied to the instrument. The Young s modulus specified as E = 200GPa is a red herring, i.e., is not a required specification. The governing design equation, derived in the web-based notes can be solved for h, the height of the spherical bubble. p σ = 4 th ρ 2 + h 2 h2 4σt p h + ρ2 = 0 h = 0.4912 and 203.6mm Note the two solutions that account for the small and big parts of the bubble partitioned by the undeformed, flat diaphragm. 4 Problem 4. Given a Copper-Constantan (Type T, blue-red) thermocouple whose reference junction is at 23.7C and whose output is, in the first instance, +7.934mV and, in the second -7.934mV, the temperature of the measuring junction 1

is estimated by interpolating output voltages in Table 9.2, p.277, W+G in column under T. First, the output, with respect to 0C at 23.7C is obtained. V 23.7 = V 20 + T actual 20 (V 40 V 20 ) = 0.789 + 40 20 23.7 20 (1.611 0.789) = 0.94107mV 40 20 Notice that in the top diagram in Fig. 1 since the output is positive and the reversed polarity, the reference junction Cu +7.934mV Con 23.7C Cu -7.934mV Con 23.7C Figure 1: Thermocouple Polarity output opposes that of the measuring junction so the output, had the reference junction been at 0C, is really 0.94107 + 7.934 = 8.875mV and this falls between the 180C and 200C entries in the table. Interpolation produces 180 + 8.875 8.235 (200 180) = 192.18C 9.286 8.235 In the second case the output of the measuring junction is negative, the same polarity as the reference junction, so the net output of the measuring junction contribution is only 7.934 ( 0.941) = 6.993mV Since the table stops at -250C we must use the interval between -250C and -200C to make the temperature estimate by backward extrapolation. 6.993 ( 6.181) 250 + 50 = 320.24C 5.603 ( 6.181) There are two reasons why the result is unreliable. Firstly, it is inadvisable to use a thermocouple outside its prescribed, tabulated range. Secondly the result is meaningless because absolute zero is about -273.3C. 5 Problem 5. Fig. 9.24(b) on p.287, W+G has been reproduced, more or less, and three points are located to obtain the coefficients A, B, C in Eq. (9.15), p.285, W+G that relates thermistor temperature T to its resistance R. 1 T = A + B ln R + C(ln R)3 Thermistor resistance is expressed as a ratio of its resistance R = 130Ω at 25C plotted against T expressed in units (C=Celsius). The three linear equations obtained are written below at T = 7.5, 25, 60C and R = 260, 130, 32.5Ω, respectively. A + B ln 260 + C(ln 260) 3 1 7.5 = 0 A + B ln 130 + C(ln 130) 3 1 25 = 0 A + B ln 32.5 + C(ln 32.5) 3 1 60 = 0 2

Simultaneous solution provides the following unique result. 6 Problem 6. A = 0.534345, B = 0.198072, C = 0.00407345 The water has a static head of 1 0.3 = 0.7m from the liquid surface in the tank to the horizontally emerging stream. First, this initial velocity v is calculated. 1 2 v2 x = H h v x = 2g(H h) = 2 9.81 0.7 = ±3.7059m/s Then the time to fall vertically through h = 0.3m is calculated. Finally, the distance s traveled horizontally at v x is t = 2h/g = 2 0.3/9.81 = ±0.24731s s = v x t = 3.7059 0.24731 = 0.9165m Notice the ± results for v x and t are inevitable when one uses the energy equation. Since the two kinetic energy degrees of freedom are conveniently orthogonal one may ignore the sign. 7 Problem 7. This is a dice-throwing or binomial problem with the mean probability of delay P d = 0.05. No delays in three consecutive weeks, n = 3, is given by the exact, single term contribution of P 3 d0 = n C r P r d (1 P d ) n r = 3 C 0 0.05 0 (1 0.05) 3 = 0.857 or about an 86% probability of experiencing three consecutive weeks with no interruption of your trip. To experience one bad trip in four weeks, n = 4, is still a single term calculation. P 4 d1 = 4 C 1 0.05 4 (1 0.05) (4 1) = 0.171 So exactly one delay in four consecutive weeks is a 17% probability. To estimate the probability of at least one delay is still a one term problem. It is the complement of zero delays in five weeks, n = 5. P 1 d5 = 1 5 C 0 0.05 0 1 0.05 (5 0) = 0.226 At least one delay in five consecutive weeks is almost a 23% probability. 8 Problem 8. This is a Poisson s distribution problem based on a total of 84 arrivals in 4 hours and the period of interest is 20 minutes so the arrival rate per period is i = 7 events per period and we wish to assess the probability that there will be a line-up of at least 5 people in any given 20 minute period. The probability, like in the binomial case of in Problem 7, of exactly i in line in any period is P i = e λ λ i i! so the complement of up-to-four-in-line is represented by the following summation. ( ) 7 1 e 7 0 0! + 71 1! + 72 2! + 73 3! + 74 = 0.827 4!... an almost 83% probability. 3

9 Problem 9. This is a Student s t test to establish a confidence interval of 95% on a sample of n = 10, v = n 1 = 9 motors that exhibit a mean efficiency x = 91% with a sample standard deviation S = ±0.8%. Be careful, this efficiency is the measured quantity. Do not confuse it with the confidence interval. The tolerance on the mean at this level of confidence is obtained with Table 6.6, p.144, W+G under the column α/2 = 0.025 in the row v = 9. The value at the intersection, i.e., t α/2 = 2.262. Therefore the tolerance on the efficiency is T r = ± t α/2s 2.262 0.8 = = ±0.57% n 10 To determine the sample size necessary to get the confidence interval 98% one examines the column downward under α/2 = 0.010 and looks for the first value 2.262. This occurs in the row v = 15, n = 16 at t α/2 = 2.602. 4

1. A four active arm strain gauge bridge with 300 SR4 gauges that have gauge factor of 2 + R 2 R 1 is loaded to a nominal maximum 1000. What is the resistance of each gauge? R 299.4 R 300.6 R 300.6 R 299.4 1 2 3 4 R 4 M R i - - + 5V R 3 2. What is the sensitivity of this system if... R i 8 10 V/ R =10k 9.709 V/ R =2k 8.696 V/ i 3. A 0 to 10V ADC returns a reading of :- 9 8 7 6 5 4 3 2 1 0 If this instrument is perfect 1 0 0 1 1 0 1 1 1 0 what is the highest 6.085V and lowest 6.075V possible voltage this could represent? 9 8 7 6 5 4 3 2 1 0 Convert this to a Gray coded integer. 1 1 0 1 0 1 1 0 0 1 Now to a 2 s complement number. 9 8 7 6 5 4 3 2 1 0 1 1 1 0 1 1 0 0 1 0 0 1 0 Finally to a 6-digit 10 s complement integer. i (FD1)MLT83c 9 9 9 3 7 8 Midterm Test 08-03-03, 40min. (circle one) MECH 261 262

4. A certain amplifier is specified as having a GBP of 25MHz. If it is incorporated into a system meant to provide a gain of +50dB, what is the high frequency corner frequncy? about 79kHz 5. A modern fever thermometer must "settle" to within 0.05C of its true reading in 1s after a normal patient, with a temperature of 37C, sticks it under the tongue. Assuming an ambient (room) temperature of 20C (in all cases) how quickly must such a thermometer reach 39.5C when suddenly placed in an under-the-tongue-simulator calibration bath at 40C? 0.632855s V s =100Vdc + (FD1)MLT83e R 1 R 2 R i M 6. Choose the resistors R 1 and R 2 to make a 20:1 voltage divider, i.e., reduce the 100V supply to 5V across R 2 when the meter is not there, i.e., R i. You can use resistors of any value but all are rated at 0.25W, maximum. 8 R 36.1k 1 R 1.9k 2 If R =20k what will the meter read? 4.586V (Forget about the small increase in i power due to decreased overall Bonus: The new Jeep Cherokee uses resistance.) 16.5L/100 km. What s that in miles per Midterm Test 08-03-05, 40min. U.S. gallon? 14.26 (circle one) MECH 261 262

7. I ve found that in Europe my VISA card fails to read properly about 10% of the time. So I ve taken to carrying two cards and they re equally (un)reliable. If I make 100 transactions during my trip and ask the merchant to try the second card whenever the first fails what is the most likely number of two-card failures I can expect? 1 is most likely What is the probability of this happening? 0.3697 What is the probability that I complete my trip without experiencing any two-card failures? 0.366 8. Here are three different types of thermocouple output in mv at five temperatures. 20C 60 100 140 180 0.789mV 2.467 4.277 6.204 8.235 1.192 3.683 6.317 9.078 11.949 1.019 3.115 5.268 7.457 9.667 type "T" type "E" type "J" Find the linear least squares best fit coefficients a and b for each using a simple minimization of 2 y (not a normal distance squared minimization) a= 0.04657 a= 0.06727 a= b= -0.2628 b= -0.2835 b= 0.054095-0.104298 9. Which type is most sensitive? E Least? T Calculate the standard deviation of the five measurements for each thermocouple from their respective best fit line. T 0.05503 Midterm Test 08-03-07, 40min. What desireable property does E 0.05932 type J have? It s the most J 0.01787 nearly linear. (FD1)STT83g MECH 262

1 Question 1 MECH 261/262 Measurement Lab & Statistics Midterm Test Solutions 2008 March 26, 2008 If gauge factor is 2 and strain on all 4 gauges is 0.001 and their original resistance is 300Ω the up/down change will be ±(2 0.001 300) = ±0.6Ω. I.e., R 1 = R 4 = 299.4Ω, R 2 = R 3 = 300.6Ω. 2 Question 2 If the meter input impedance R i is infinite and V s = 5V the sensitivity is given by ( R2 + R ) 4 Vs R 1 + R 2 R 3 + R 4 1000µɛ = 10µV µɛ If R i = 10kΩ and 2kΩ, respectively, the following seven simultaneous equations, based on Kirchhoff s node (current) and branch (voltage) laws, are solved simultaneously for V 1 and V 2. i 1 i 2 i i = 0, i 4 i 3 i i = 0, 5 V 1 299.4i 1 = 0, V 1 300.6i 2 = 0 This results in and V 1 V 2 R i i i = 0, 5 V 2 300.6i 3 = 0, V 2 299.4i 4 = 0 V 1 V 2 = 2.504854369 2.495145631 9.708738 µv µɛ V 1 V 2 = 2.504347828 2.495652172 8.695656 µv µɛ 3 Question 3 10V are divided among 1023 quanta. Each quantum is worth 10000 1023 = 9.775171065mV A count of 1001101110 represents 2 9 + 2 6 + 2 5 +E= 622. 622 10000 1023 = 6080.156403 ± 1 10000 2 1023 = 4.88759mV The reading could represent anything from 6085.043988 to 6075.268817mV. 1

(MECH261)BGB83a 1 0 0 1 1 0 1 1 1 0 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 Figure 1: 10-Bit Binary-to-Gray-to-Binary Conversion 1001101110 is converted to 1101011001 in Gray code. The converter, followed by reconversion to binary is illustrated in Fig. 1. In order to convert 1001101110 to 2 s complement reverse the bits to 0110010001 and add 1 to get... 1110110010010. Note the most significant leading 1 s continue to the left indefinitely, implicitly, like most significant 0 s in a positive number. Since the original count was 622 it becomes, upon negation, the 9 s complement number... 999377. Note how the leading most significant 9 s replace leading 1 s in 2 s complement binary. Adding 1 gives the 10 s complement... 999378. 4 Question 4 A gain of +50dB corresponds to 50 = 20 log 10 G. Since gain-bandwidth product is 25MHz log 10 G = 2.5 10 log 10 G = G = 10 2.5 = 316.227766 f c = GBP G = 25000kHz 316.227766 = 79.05694151kHz 5 Question 5 The fundamental first order response to a step input change (θ( ) θ(0)) after time t is to attain a (temperature) rise of (θ(t) t(0)) according to the following special solution of the differential equation e t/τ 1 e t/τ θ(t) t(0) θ( ) θ(0) = 0 2

The statement of specification says that at t = 1 e 1/τ 1 36.95 20 e 1/τ 37 20 This gives the required time constant as τ = 0.1715576139s. The simulator is set up with a constant temperature bath at 40C and a trigger that measures the time interval t until the sensor output signal reached 39.5C and we wish to know the calibration time limit t c at or below which the the timer must trigger if the sensor is good. We now know τ from the specification. = 0 e tc/τ 1 39.5 20 e tc/τ 40 20 = 0 The solution of this second expression gives t c = 0.6328553571s. 6 Question 6 The first equation says that the voltage drop across the second resistor is 5% of the total voltage drop across both resistors in series. R 2 R 1 + R 2 1 20 = 0 The second equation says that the total voltage drop across the two resistors in series is the sum of the drop in each resistor due to the common current i that flows in both. ir 1 + ir 2 100 = 0 The third equation says that the greater power dissipation in the resistor of higher value, R 1 must not exceed 0.25W. i 2 R 1 1 4 = 0 This gives i = 1/380A, R 1 = 36100Ω, R 2 = 1900Ω and a check with the third equation establishes that the power loss in R 1 is indeed 1 W. A 2000Ω/V meter on the 0-10V scale has an input 4 impedance of 10 2000 = 20kΩ. Putting this in parallel with R 2 gives an effective R 2 The meter reading is R 2 = 1 1 1900 + 1 20000 = 380000 219 = 1735.159817Ω V s R 2 R 1 + R 2 = 100 1735.159817 36100 + 1735.159817 = 4.586104105V The reason for bringing R 1 up to maximum allowable power is that in this way the output impedance of the voltage divider is minimized. This is a fundamental design principle. It s especially important if we ve got a meter with low input impedance. 2000Ω/V is just about as low one can get with a commercial volt/ohm meter. 3

7 Bonus L/U.S.gal. is 0.355(128/12)=3.7866.... Miles /100km is 62.13711922. The product is 235.2925581. This divided by 16.5 gives 14.26015504mpg. Since the imperial gal. is 6/5U.S.ga., that s 17.11218605 mpg, big gallon. 8 Question 7 This is clearly a binomial distribution problem. Since the 10% card failure rate is the same for each card and independent of each other, a failure occurs when both fail, one immediately after the other. Therefore P µ = 0.01. To have exactly r failures in n tries P (r : n) = n C r P r µ(1 P µ ) n r In all cases n = 100 so let s start with r = 0 and work up until P (r : n) starts to diminish. P (0 : 100) = 100 C 0 (0.01) 0 (0.99) 100 = 0.3660323413 P (1 : 100) = 100 C 1 (0.01) 1 (0.99) 99 = 0.3697296376 P (2 : 100) = 100 C 2 (0.01) 0 (0.99) 98 = 0.1848648188 From this we conclude that the most likely is one double failure per trip at almost 37% of the time but close behind is the chance 36.6% chance of completing the trip without embarrassment. 9 Questions 8 and 9 Here we see an opportunity to write a small program where up to 10 combinations of x and y can be input after specifying that number n. The computation of coefficients a and b are done conventionally, without even bothering to zero on the mean. After a line has been fit, the y- deviations from the fit line of all y-values are squared and added and their standard deviation is computed. 100 DIM X(10),Y(10),XY(10),X2(10) 110 INPUT N 120 SX=0:SY=0:SXY=0SX2=0 130 FOR I=1 TO N 140 INPUT X(I),Y(I):SX=SX+X(I):SY=SY+Y(I):SXY=SXY+X(I)*Y(I):SX2=SX2+X(I)^2 150 NEXT I 160 D=N*SX2-SX^2:A=(N*SXY-SX*SY)/D:B=(SX2*SY-SX*SXY)/D 170 SSD=0 180 FOR I=1 TO N 190 SSD=SSD+(Y(I)-A*X(I)-B)^2: 200 NEXT I 210 LPRINT SQR(SSD)/(N-1),A,B:STOP:END 4

Thermocouple type σ a b T 0.05503354 0.04657249 0.2628484 E 0.05932262 0.06727248 0.28345 J 0.01786823 0.054095 0.1042984 The values of a are sensitivity expressed in mv/c. The type E couple is clearly the most sensitive and the type T is the least sensitive among the three. The type J couple is the most linear. This may not be quite so obvious from the graph in Fig. 2. 10 Notice how hard it is to estimate linearity without calculating standard deviation from line of best fit. mv 5 (FD1)LinBF83d 0 20 60 100 140 180 C Figure 2: Linearized Thermocouple Characteristics (MECH261/262)MLSCTSol08.tex 5

1. A type "E" thermocouple is used as a sensor to accurately control the temperature of an oil bath to 100C. Assuming that the reference junction is held at 22C, what is the output of this system? 5.002mV If this signal can be read to ±1 V, estimate what is the smallest change in temperature that can be detected. ±0.015C 2. A transparent circular disc is encoded with a 6-level Gray code on 6 concentric tracks. Express the angular resolution of this device. I.e., what is the change in angle (degrees) that can be detected? What is the reading depicted? ±1/2bit=2.8 5.6 104.1±5.6 (See diagram on MLXA7C1.) 3. A pitot tube connected to an inclined manometer gives a reading of 57.4mm at 30. If the room temperature is 19C and the barometer reads 743mm Hg estimate the velocity of the air approaching. 19.52 m/s 57.4 Manometer fluid is alcohol, SpG=0.8. (FD1)MLXA7C2 30 Final Exam 07-12-19 (circle one) MECH 261 262

Disc rotation 0 sensors 011(0/1)10 The row of stationary sensors read "1" on black patches,... "0" on clear ones. (FD1)MLXA7C1 Final Exam 07-12-19 (circle one) MECH 261 262

4. What does the meter read with all gauges at 1000 micro-strain if they have unstrained resistance of 300 Assume all gauges have increased or or decreased by the same value R. a) Show with up/down arrows which of the resistors in the 4 arm bridge have inceased/decreased. b) Calculate their respective resistances. Gauge factor=2. R 1 299.4 R 2 R 3 300.6 R 4 V= 10mV 300.6 299.4 5. Find the a) maximum and b) estimated uncertainty in the meter reading if V =5V±10mV, R =300 ±0.001, R=??±10 s o a) ±20.2mV, ±0.2% full scale b) ±20.0 V R 2 R 4 + V - - + 5V R 1 R 3 6. A dead-weight tester is meant to calibrate two types of Bourdon pressure gauge. One type measures 0-10Atm. the other 0-100Atm. If the two piston assemblies can be tailored to both have a mass of exacltly 250g and there are wieghts of up to 5kg mass what are the two required cylinder diameters? 0-10Atm 8.045mm 0-100Atm 2.544mm What are the minimum pressures that can be detected accurately? 482.5kPa 0-10Atm 0.4762At 0-100Atm 4.762At gauge weight 48.25kPa piston assembly Final Exam 07-12-19 cyl. dia. screw OIL (FD1)MLXB7CS (circle one) MECH 261 262

7. Twenty random numbers have been sorted into five bins. Taking this as a probability mass function express the "expected value" 4.4 and the probability represented by each bin. 0-2 7/20 2-4 1/20 4-6 4/20 6-8 7/20 8-10 1/20 frequency 0 2 4 6 8 10 8. Exhaustive sampling experiments show that 10% of a production run of precision 4mm dia. steel drill rod is 4.001mm or greater while 5% is 3.9995mm of less. Assuming normal or Gaussian distribution, find the mean and standard deviation of this population. 4.00034mm ±0.00051mm 9. In an EUS carnival there are gambling games to collect money for worthy causes. At one of the booths you are dealt 5 cards starting with a full deck of 52. You bet $1 that there will be at least one pair dealt to you. If there is, you win $0.25. What is the profit margin for the "house"? Express your answer as a percentage of the total amount bet as being 100%. Base your answer on statistical probability. 0.345-0.25=0.195, 19.5% Win/loss odds are 128171/195755 -vs- 67584/195755. (FD1)STXA7Cs Final Exam 07-12-19, PM, 3hrs. Page 3 of 3 MECH 262

2007 Measurement Final Answers and Solutions (FD1)FXF07Sol7B3.tex December 27, 2007 1 Problem 1. Below are the relevant tabulations for a type E (chromel-constantan) thermocouple at 20, 40, 80, 100 and 120C. 20C 40C 80C 100C 120C 1.192mV 2.419mV 4.983mV 6.317mV 7.683mV Linear interpolation between 20 and 40C produces Therefore the measured output is On either side of 100C we get 2 (2.419 1.192) + 1.192 = 1.315mV 20 6.317 1.315 = 5.002mV 6317 4983 = 1334µV/20C 66.7µV/C and 7683 6317 = 1366µV/20C 68.3µV/C Taking the average gives 67.5µV/C and the reciprocal is about ±0.015C/µV. 2 Problem 2. A 6-level code contains 2 6 = 64 quanta or counts from 0 to 63. Each quantum patch subtends 2π/64 radians or 360/64 = 5 5 8 degrees. Angular resolution is therefore ±1/2 bits or ±2.8125 degrees. The reading depicted is, starting with the most significant, innermost, bit 011 10. The indicates this may be either a 0 or a 1. Notwithstanding, Fig. 1 shows the following. 0 1 1 0,1 1 0 2MLX7Cs 0 1 0 0,1 1,0 1,0 010011 or 010100 = a count of 19 or 20 = 101.25 or 106.875 Figure 1: Gray to Binary Conversion æ2.8125 so the reading might be returned as being anywhere between 98.4375 or 109.6875 an uncertainty span of 11.25. For the specified input the straight binary output is 010011 or 010100, a count of 19 or 20 that might be interpreted This comes to 104.0625 æ5.625. literally as 101.25 or 106.875, ±2.8125. So the reading might actually be anywhere between 98.4375 and 109.6875, an uncertainty span of 11.25. This comes 104.0625 ± 5.626. 1

3 Problem 3. This problem is solved by equating the dynamic (velocity) pressure to the impact (stagnation) pressure measured by the Pitot tube. p dynam = ρ airv 2 = ρ mano gh = p impact 2 First the air density is calculated. Then the manometer pressure in Pa is determined. ρ air = p 743 o RT = 760 101300 287 (273 + 19) = 1.1817kg/m3 p impact = p mano = 800 9.81 0.0574 2 = 225.24N/m 2 The first equation can now be solved having taken into account the density of the manometer fluid (800kg/m 3 ) and inclination (30 h = 57.4mm sin 30 ). 2 225.24 V = = 19.52m/s 1.1817 4 Problem 4. Since the all gauges are at nominal maximum strain of 1000µɛ and the gauge factor is 2 the up gauges go to 300Ω + 2 0.001 300Ω = 300.6Ω Similarly the down gauges decrease by the same amount and go to 299.4Ω. Since the + -sign for voltage appears at the upper corner of the bridge R 1 and R 4 must be the down gauges while the greater voltage drops must occur across the up gauges, R 2 and R 3. The voltage V 12 is given by Similarly V 34 is V s R 2 R 1 + R 2 = 5 300.6 600 = 2.505 V s R 4 R 3 + R 4 = 5 299.4 600 = 2.495 Therefore the meter V reads 2.505 2.495 = 0.01 or 10mV. 5 Problem 5. Stating the result for Problem 4 a little more formally Now we can write and So maximum uncertainty is V = V sr 2 R 1 + R 2 V sr 4 R 3 + R 4 = V s 2R o [(R o + R) (R o R)] = V s R R o V = R, V s R o V R = V s R o, V = V s R R o Ro 2 w V s = 0.01, w R = 0.00001, w Ro = 0.001, R o = 300, V s = 5, R = 0.6 w V s R R o + w R V s R o + w Ro V s R = 20.2µV R 2 o 2

and the estimated uncertainty is (wv ) 2 ( ) 2 ( ) 2 s R w R V s wro V s R + + = 20.0µV R o R o R 2 o If one calculates individual contributions to uncertainty it becomes obvious that uncertainty due to V s is 20 10 6 while that due to R and R o are 1 6 10 6 and 1 3 10 7, respectively. That s why there s so little difference between maximum and estimated uncertainties. As it stands, uncertainty is about ±0.2% of full-scale out of balance voltage. A lot could be gained with a better power supply. 6 Problem 6. For reference, a standard atmosphere is 101.325kPa while a bar is 100kPa. Applying Pascal s Principle { W 5.25 9.81 1013250Pa = πr2 πr 2 = 10132500Pa Solving for r for both cases gives diameters d 1 = 8.045mm d 2 = 2.544mm } Since the 0.25kg piston assembly weight is always there and this is 1 21 of the total, maximum weight of 5.25kg we can measure down to 1 1013.25 = 0.4762Atm = 48.25kPa 21 21 with the 10Atm gauge and 10 times that with the 100Atm gauge. 7 Problem 7. Statistics: MECH 262 only. Although we don t know the individual values of the 20 numbers we can conclude that there are so 7 in the 0-2 range with a mean of 1, 1 in the 2-4 range with a mean of 3, 4 in the 4-6 range with a mean of 5, 7 in the 6-8 range with a mean of 7 and 1 in the 8-10 range with a mean of 9. (7 1 + 1 3 + 4 5 + 7 7 + 1 9)/20 = 4.4 Expected value means mean in this case, i.e., 4.4. Given this distribution and randomly choosing any number between 0 and 10 produces the conclusion that the probabilities are P 0 2 = 7 20, P 2 4 = 1 20, P 4 6 = 4 20, P 6 8 = 7 20, P 8 10 = 1 20 3

8 Problem 8. Statistics: MECH 262 only. Refering to Table 6.3 on p.136 we can find z that corresponds to a right tail area of 10% and a left tail area of 5%. These are, respectively, z r = 1.28166..., using linear interpolation between the values z = 1.28 0.3997 and z = 1.29 0.4015. Note that the 40% integral area has a 10% right hand tail. z l = 1.645, using linear interpolation between the values z = 1.64 0.4495 and z = 1.65 0.4505. Note that the 45% integral area has a 5% left hand tail. From the change-of-variable definition one gets two equations. z = x i x σ that make x = 4.000343mm and σ = 0.0005125mm. x + 1.28166... σ 400.1 = 0 x 1.645σ 3.9995 = 0 9 Problem 9. Statistics: MECH 262 only. The tree for two-of-a-kind from a straight five card deal from a full 52 card deck is shown in Fig. 2. 1 3/51 48/51 1 2 44/50 3STX7Cs 2x(3/50) 1 2 3 40/49 3x(3/49) 1 2 3 4 36/48 4x(3/48) 1 2 3 4 5 32/47 5x(3/47) 1 2 3 4 5 X Figure 2: Probability:- Two-of-a-Kind It shows two ways to get the answer. Following the right-hand branch at every juncture, all the way to the failure node [ ], shows the probability of not getting a pair as 48 51 + 44 50 + 40 49 + 36 48 + 32 47 = 67584 195755 or about 34.52%. That means the player stands to win 65.47% of the time while the house wins only 34.52% of the time. If it pays out only 25 cents per loss it gets to make a profit of 19.5% on every dollar bet... in the long run. The other, long way to solve this problem is to evaluate every way a player may draw a pair. That means adding 4

up the probability products along the 15 paths that end in [1], [2], [3], [4] or [5] to get the probability of drawing a pair thus. 3 51 + 48 6 51 50 + 48 44 9 51 50 49 + 48 44 40 12 51 50 49 48 + 48 44 40 36 15 51 50 49 48 47 = 128171 195755 One may check as follows. 67584 195755 + 128171 195755 = 1 5

1. Given the series of four 1:10 inverting amplifiers shown below, specify:- a) The value of R 1 in each case, b) The value of R such that the output voltage v 3 o =10V when the input voltage iv =8mV and R 2 =22kOhm. R 1 R 2 R 1 R 2 R 1 R 2 R 1 R 2 R 1 = 2.2k - - - R 3 - v i + + + + R = 15.4k v o 3 2. A 10-bit 0 to +10V (unipolar) ADC is connected to measure v o. When v o =10V it reads a full-count of 10 ones 1111111111. a) What decimal number does that correspond to? 1023 R 2 b) If a strain guage bridge shows an out of balance voltage of 8mV at =0.1% then what is v o and the count when =0.00075? (binary) M + R 1 and 7.5V 1011111111 3. When meter M reads 8mV as indicated and the 4 strain gauges have unstrained resistance of 300 a) Show with up/down arrows which of the resistors in the 4 arm bridge have inceased/decreased. b) Calculate their respective resistances. R 4 (FD1)MLT7A3 - - R 3 R 1 299.76 R 2 300.24 + R 3 300.24 R 4 299.76 10V Assume all gauges have increased or or decreased by the same value R. Midterm Test 07-10-29, 40min. (circle one) MECH 261 262

4. A weigh-scale is calibrated with 3 accurate weights of mass 20kg, 60kg and 100kg. The corresponding strain gauge force transducer bridge outputs are 1.7mV, 5.8mV and 7.9mV. Find the linear coefficients in y=ax+b where y is output and x is the applied weight. a= 0.0775 b= 0.4833 What is the sensitivity according to this calibration? 0.0775mV/kg 5. Given a parabolic wave-form consisting of a periodic series of upward "lumps" of amplitude 5 and wave-length 4, find the RMS amplitude value. 3.651 y=5-(5/4)x 2 +5 6. Given an under-the-tongue thermometer with time constant =4s -2 2 and a patient with a temperature of 39.5C. Assuming a 1st order response and a precision of +0.5C, how long should one wait to get an accurate reading if the thermometer was at 20C to begin with? 14.55s Hint:- y-y i y e -y i (FD1)MLT7A5 -t/ =(1-e ) Midterm Test 07-10-31, 40min. (circle one) MECH 261 262

7. Sort the following 20 random numbers into 5 bins. Label the bin boundaries and construct the appropriate bar chart. It most closely represents a 0.46,0.66,5.30,1.97,5.49,6.29,4.81,0.74,7.08,3.05 0.32,7.40,0.31,7.31,8.41,7.25,6.76,5.56,7.90,1.74 8. You re building a plywood boat. The sheets must be clear of knots. The best marine ply you can find guarantees that no more than 1 sheet in 10 has knots. If the boat uniform (or bimodal) needs 10 sheets how many sheets should you buy in order that frequency you run, at worst, a 5% (1 in 20 tries) risk of having to go back to the store? If you buy this amount, what is the actual risk (in % probability) that you won t have 10 good sheets of plywood? answer? Using a binomial model will give 13 sheets at 3.4% risk. about 0.5% random distribution. 0 2 4 6 8 10 11(Poisson) Is there another way to get the 9. Yucky Junk Food, Inc. made a survey of a population consisting of 60% adults, 40% teens and 30% children. 50% of adults like the stuff. So do 80% of teens and 60% of the kids. Targeting the potential consumers how much of $1M advertising should they allot to each age group? Toward adults $375000 Toward teens $400000 Toward kids $225000 Hint:- Bayes theorem? (FD1)MST7Bb Midterm Test 07-11-02, 40min. MECH 262