Download - MECH 261/262 Measurement Lab (& Statistics)

MECH 261/262

Measurement Lab (& Statistics)January 6, 2010

Final Exams and Midterm Testsin 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 completedwork sheets handed in by the student at the end of that final exam or midterm test. Worked outsolutions will be added eventually but it has taken me six years to compile this lot in somesemblance of order so don’t expect miracles. Furthermore some questions are (very) bad andsome answers are wrong. Unfortunately that’s life but at least no attempt is made here to coverup such mistakes and assurances are given that most of these are caught before they accrue tothe detriment of a student’s final grade. In a nutshell, any small consensus of students’ answersto a particular question, that differ from mine, triggers a re-solution by me and independentlyby 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 immediatelyafter Q.1,2,3 on the 04-02-19 midterm. Notice how Q.3 is solved and due to ambiguity of thatquestion, 3 out of the possible 6 multiple choices were deemed to be correct, acceptable answers.(FD1)/Exams/ExaMT781.tex

1

keyName:-

Student Number:-

(circle one) MECH 261 262

Final Exam 10-04-29

(FD3)MLXA043

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?

?

100Nm

=

= 13.5mm

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

keyName:-

Student Number:-


Final Exam 10-04-29

(FD3)MLXB043

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

... a four lead bridge

0.422V

0.417V

0.4125V

13

Oil

Weight

0.5kg Piston

5. Find mass of weight necessary to exert pressure

mmof 1MPa on oil in dead weight pressure tester.

13.030 kg

What is that in psi (pounds/sq. in)?145.05 psi

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.

v

17mm

20

Express the air stream velocity

in ft./s, mph, m/s or kmh.8.845m/s or 31.844kmh

29.02’/s or 19.79mph

144.7 (hard way)

key

Student Number:-

MECH 262

Final Exam 10-04-29

(FD3)STXA043

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

, taken at random at both A sample of 50 airline passengers on the Montreal-Toronto run2.

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

on a 5-card deal? 0.024% or 1 in 4165hand

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 to0.6At? First, find the time constant Ct. Set t = Ct and note that for the 3 minute interval X = 1.9−1.8

1.9 = 0.11.9 .

eT − 1eT

−X =e( t

Ct) − 1

e( tCt

)−X = 0 →

(1− 0.1

1.9

)e3/Ct − 1 = 0 → Ct = 55.486min.

With Ct in hand solve for X = (1.9− 0.6)/1.9.

e( tCt

) − 1

e( tCt

)− 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 samestress magnitude in tension and compression as the maximum shear stress, i.e., on the surface of the bar at ro.

Γro

J− τ = 0 → 100r

πr4/2− 207000000 = 0 → r3 =

200207000000π

→ r = 13.5mm

3 Problem 3.

The red-white lead insulation identifies the couple as Fe-con type J and the 82F temperature must be convertedto Celsius.

(82− 32)59

=2509

= 27.777C

Couple output at 82F= 27.77C with reference to 0C is calculated with linear interpolation between 20C and 40C.

1.019 +2509 − 2040− 20

(2.058− 1.019) = 1.4230555 . . .

Similarly, couple output at 125C, with reference to 0C, is interpolated between 120C and 140C.

6.359 +125− 120140− 120

(7.457− 6.359) = 6.6335

The difference between these, hence the required reading, is

6.6335− 1.4230555 . . . = 5.210444 . . .mV

4 Problem 4.

This problem involves substituting Vs = 2V, R2 = R3 = 100Ω, RRTD = 246.05Ω (the tabulated resistance at390C), respectively, into Eqs, (9.11), (9.12) and (9,13). The equations below were derived from “scratch” circuitanalysis. 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 textproduce negative voltage on the terminals shown as positive. First a two-wire solution.

Vo =Vs(RRTD −R2)2(RRTD + R2)

=2(246.05− 100)2(246.05 + 100)

= 0.422049V

Next for three wires.

Vo =Vs(RRTD −R2)

2(RRTD + 2RL + R2)=

2(246.05− 100)2(246.05 + 4 + 100)

= 0.417226V

And finally for four wires.

Vo =Vs(RRTD −R2)

2(RRTD + 4RL + R3)=

2(246.05− 100)2(246.05 + 8 + 100)

= 0.412512V

Notice that inclusion of the lead wire resistances successively lowers the sensor resistance RRTD by not countingthe lead wire resistance as part of the sensor.

5 Problem 5.

Pressure is force f divided by area a = πr2 of piston. Force in this case is weight (w + 0.5)g of mass m of 0.5kgpiston plus the applied weight. Required pressure p is 1000000Pa.

p =f

a=

mg

a=

(w + 0.5)9.810.00652π

→ w =106(0.00652)π

9.81− 0.5 = 13.0303kg (≡ 127.8273N)

As regards converting 1MPa to psi, the conversion table handed out in class gives 1psi=6.894× 103Pa so 106Pa isjust the reciprocal

10.006894

= 145.054psi

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× 2.2π(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 = 70Fand 29in-Hg. Find velocity. Taking R the air gas constant as 53.3lbs(force)-ft/lbs(mass)R and water density as62.3lbs/ft3 at T = 70F write the energy equation. ρf is air density, ρm is manometer fluid density, v is air flowvelocity, 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.

rhofv2

22− ρmgh =

pv2

2RT− ρmgh =

2929.22

(14.696)144v2

2(53.3)(70 + 460)− 0.8(62.3)32.16

17 sin 20

12(25.4)= 0

Solving produces |v| = 29.02ft/s=19.79mph=8.845m/s=31.844kmh.

End of Measurement Part

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

P (x) =λxe−λ

λ!

The probability of seeing less than two is

P (x < 2) = P (0) + P (1) =λ0e−2.6

0!+

λ1e−2.6

1!= 0.26738

The probability of seeing more than five is

P (x > 5) = 1− (P (0) + P (1) + P (2) + P (3) + P (4) + P (5)

= 1−(

2.60e−2.6

0!+

2.61e−2.6

1!+

2.62e−2.6

2!+

2.63e−2.6

3!+

2.64e−2.6

4!+

2.65e−2.6

5!

)= 0.049037

8 Question 2.

A sample of 50 airline passengers with mean mass of µ = 68kg and standard deviation σ = 4.3kg can be foundwith Eqs. (6.46) to have a mean confidence interval

µ = x± zα/2σ√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 0.05indicating 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 directlyon v = n− 1 and headed with the “tail” fractions α/2. The pertinent equation is Eq. (6.51).

µ = x± tα/2S√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

→ ±2.49kg

which is clearly greater than the prescribed tolerance on the mean.

3

10 Question 4.

There are only 13 Fours-of-a-Kind in an ordinary 52 card deck that offers

52C5 = 2598960

possible 5-card combinations. However the fifth card may be any of the remaining 48 having “removed” the 4 thatmake up the ones of same denomination. Therefore of the 2598960 possible hands 13×48 are Fours-of-a-Kind. Theprobability of being dealt one, without intermediate removal of cards like when dealing to more than one receiver,is

13× 482598960

→ 0.024%

or 1 chance among 4165 hands, on average.

4

keyName:-

Student Number:-


Midterm Test 10-03-01, 40min.

MLT03a

+-

5V

+

-

R 1R 2

R 3R 4

M

1. A four active arm strain gauge bridge with 300 SR4 gauges is shown. Gauge factor = 2.

R i

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.

-60kN

+60kN

a) Label the spaces i,ii,iii,iv properly with R ,R ,R ,R .1 2 3 4

i

ii

iii

iv

iii

iiiv

b) What is the compressive stress and strain on the transducer block?

c) The 4 resistance values are

R 1 R 2 R 3 R 4

d) If R what is i

8 the reading of M?

R 1

R 3R 2

R 4

299.565 300.130 300.130 299.565

150MPa 724.6

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

8

-8

2

0 t

V=kt2

0.5ms

1ms

78.5 V/kN

keyName:-

Student Number:-



MLT03c

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?

5mV 20dB

300

10k 10

2k

?mV

48.302mV

2. The number $AD7649 is mapped into a 24-bit register shown below. "$" means base 16

or "hexadecimal". Show its binary image.

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

Interpret this as a 2’s complement number

and write its equivalent decimal magnitude.

+

-5409207

3. Design a low-pass 1st order Butterworth filter with R =7.2k R =22k f =40Hz.c1 2

C= What is the DC gain? And at 120Hz?

CR 1

R 2

3.055/+9.701dB180.9nF 1.019/+0.159dB

keyName:-

Student Number:-

MECH 262


STT03e

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?

2. Given the points A(3,3,-4), B(7,8,-5), C(9,4,0), D(10,7,-2) fin

four point system.

d a) The centroid of this

b) Its "inertia" matrix

I xx

I yy

I zz

-I xy -I xz

-I yz

-I yz-I xz

-I xy

numerical elements

c) Write the characteristic equation.

d) Given the eigenvalues 967.766, 343.713, 624.521 write a of equations whose

solutions are the direction numbers of the best fitting plane’s normal vector.

94.4%

4:-29:22:-11

pair

Use the top 2 rows of the

inertia matrix.

207736832-1151680 +1936 -2 3 =0

-459.766e -184e -304e =01 2 3

-184e -271.766e +88e =01 2 3

508 -184 -204

88

88

-164 696

732-204

MECH 261/262

Measurement Lab & StatisticsMidterm Test Solutions 2010

March 2, 2010

1 Question 1 Monday

1 a) −i− = R1 or R4, −ii− = R2 or R3, −iii− = R4 or R1, −iv− = R3 or R2

The compressive gauges (↓) are oriented in the direction of the applied, compressive force. Thetensile, Poisson gauges (↑) are oriented in the direction normal to the applied force.

1 b)

σ =F

A=

60000N

0.022m2= 150MPa, ε =

σ

E=

150× 106Pa

207× 109Pa= 724.6µε

1 c)

R1 = R4 =R0

1 + FGε=

300

1 + 2× 724.6× 10−6 = 299.565Ω

R2 = R3 = R0(1 + FGµ× ε) = 300× (1 + 2× 0.3× 724.6× 10−6) = 300.130Ω

Note that µ is Poisson’s ratio, ε is strain while µε has been used to mean “micro-strain”.

1 d) Calculate the voltage difference between the junction of resistors R1 and R2 and that betweenresistors R3 and R4. Vs is the power supply voltage of 5V. M is the meter reading.

M = Vs

(R2

R1 + R2

− R4

R3 + R4

)= 5

(300.130− 299.565

300.130 + 299.565

)= 4.711mV

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) = kt2 so v(t)2 = k2t4. TheRMS voltage is given as VRMS = 2V .

VRMS =

√∫ 1

0

(k2t4)dt =

√k2t5]10

5=

k√5

= 2V, k = 2√

5

Notice that peak-to-peak goes as much negative as positive. So V 12PtP = k = 2

√5 ≈ 4.472V. The

signal is bi-polar so VPtP ≈ 8.944V.

1

3 Question 1 Wednesday

Input Vi to the amplifier is given in terms of the given output of the source Vs = 5mV, its outputimpedance Rs = 300Ω and the input impedance of the amplifier Ri = 10kΩ.

Vi = Vs

(Ri

Rs + Ri

)Output Vo of the amplifier depends on the gain G and Vi. Note that GdB = +20 is given.

Vo = GVi, G = 10GdB/20 = 10

Finally the meter reading VL is determined by Vo, the meter input impedance RL = 2kΩ and theamplifier output impedance Ro = 10Ω. Putting it all together

VL = GVs

(Ri

Rs + Ri

) (RL

RL + Ro

)= 10× 5

(10

10.3

) (2

2.01

)= 48.302mV


The hexadecimal number$AD7649

translates to straight binary as

[1010][1101][0111][0110][0100][1001]

Since the leading bit is [1] this is a negative, 2’s complement number. Reversing all the bits andadding [1] produces the positive equivalent of the negative number

[0101][0010][1000][1001][1011][0111]

Converting back to base-16 gives$5289B7

This translates to

5× 165 + 2× 164 + 8× 163 + 9× 162 + 11× 161 + 7× 160

= 5242880 + 131072 + 32786 + 2304 + 176 + 7 = (−)5409207


To get the required capacitance C, rearrange Eq. (3.26) in Wheeler & Ganji.

C =1

2πfcR2

=1

2π × 40× 22000= 180.9nF

2

The DC gain Go is just the ratio G2 : G1. It is then expressed in dB.

Go =R2

R1

=22000

7200= 3.055., GodB = 20 log10 Go = +9.702dB

To get the gain at 120Hz, use Eq. (3.27) to get the magnitude ratio |G : Go|.∣∣∣∣ G

Go

∣∣∣∣ =1

2πfCR2

=1

2π × 120× 22000=

1

3

So G120Hz = 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 concernsto odds of there being 6 bad ”eggs” among a lot of 55 given a defect expectation/mean probabilityof 5% or 1 in 20. Count up the “exactlys” from 1 to 5 and subtract from unity.

P0 = 55C0 × 0.050 × 0.9555 = 0.0595

P1 = 55C1 × 0.051 × 0.9554 = 0.1723

P2 = 55C2 × 0.052 × 0.9553 = 0.2449

P3 = 55C3 × 0.053 × 0.9552 = 0.2277

P4 = 55C4 × 0.054 × 0.9551 = 0.1558

P5 = 55C1 × 0.055 × 0.9550 = 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 ahomogenizing coordinate on the left.

A(3, 3,−4), B(7, 8,−5), C(9, 4, 0), D(10, 7,−2) → G4 : 29 : 22 : −11 →(

29

4,22

4,−11

4

)The last 3 coordinates are subtracted from 4 times the original, 4 given coordinates. That wayone 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.

Ixx = (−10)2 + (−5)2 + 102 + (−9)2 + (−6)2 + 112 + 62 + 32 = 508

Iyy = (−17)2 + (−5)2 + (−1)2 + (−9)2 + 72 + 112 + 112 + 32 = 696

Izz = (−17)2 + (−10)2 + (−1)2 + 102 + 72(−6)2 + 112 + 62 = 732

Ixy = (−17)(−10) + (−1)10 + 7(−6) + 11(6) = 184

Ixz = (−17)(−5) + (−1)(−9) + 7(11) + 11(3) = 204

Iyz = (−10)(−5) + 10(−9) + (−6)11 + 6(3) = −88

Then populate the inertia matrix, subtract the diagonal λ-matrix and take the determinant to getthe characteristic equation.

|[I]| =

∣∣∣∣∣∣ 508− λ −184 −204

−184 696− λ 88−204 88 732− λ

∣∣∣∣∣∣ = 207736832− 1151680λ + 1936λ2 − λ3 = 0

This produces three eigenvalues. Since the first is of greatest magnitude it belongs to the eigen-vector (normal) to the best fitting plane, the one that minimizes the sum of least squares distanceswith respect to the 4 given points.

λ1 = 967.77, λ2 = 343.71, λ3 = 624.52

4

The test, as distributed, contained a misprinted first eigenvalue of λ1 = 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 = [e1 e2 e3]

> equationsproduces

−459.77e1 −184e2 −204e3 = 0−184e1 −271.766e2 +88e3 = 0

Solving homogeneously and normalizing on e1 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, againnormalizing on the constant term.

1− 0.0525x + 0.0572y + 0.0668z = 0

(MECH261/262)MLSCTSol10.tex

5

keyName:-

Student Number:-


Final Exam 08-04-14

(FD1)MLXA84N

R = C=2

1. Design a lowpass Butterworth filter with a gain of 0dB, corner frequncy 30Hz and input

resistor R =10k .1

R 1

R 2

C10k 530.5nF

Hint:- The unit of capacitance is "Farad" (F).

Resistors are big; capacitors are small.741

2. This is a "1st order" filter. What is the gain in dB if the input signal frequency is 60Hz?

G dBIf it is desired to reduce (attenuate) a 60Hz (noise) signal down to 1%

=

n= 6,7,8

about -7dbof its original amplitude or less, what is the smallest even number n

(order of the filter) that can do this?

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

p

2Material is assumed

to be steel, E=200GPa.

keyName:-

Student Number:-


Final Exam 08-04-14

(FD1)MLXB84N

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? Cu

Con.

±7.934mV

Cu

23.7C

192.18C

-320.24C

Why might you consider the negative

reading result to be unreliable? It’s off scale.

The table is in 50C increments.

0 20 40 60 80 100 120

0

3

2

1

0.2

0.6

1

C

R

R@25C

V

V@25C (2.1V)

(130 )

0.25

5. Calibrate the thermistor using the three given points and equation

1/T=A+BlnR+C(lnR)3

A=B=

C=

6. Here’s a "pissing" tank of water.

Where does the stream land?

s=

0

H=1m

h=0.3m

s

917mm

H 2 O

for reference only

4. A Copper-Constantan thermocouple reads 7.934mV when the reference junction is

Hint: A three-point calibration.

(7.5,2)0.004073451

-0.19807153460.5343453399

Find

key

Student Number:-

MECH 262

Final Exam 08-04-14

(FD1)STXA84N

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?

What’s the probability you’ll have exactly one frustrating wait during 4 consecutive weeks?

What’s the chance of getting caught at least once in 5 weeks?

0.857375

0.171475

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 efficiencyfor the mean

for the entire batch? If it’s desired to get the confidence interval up to

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.

98%, how many motors should be tested?0.57

16

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 R2 = R1 = 10kΩ because anycapacitor C in parallel with feedback resistor R2 has XC → ∞. Then Eq. (3.26), p.65, Wheeler & Ganji (W+G)specifies the corner frequency fc as

fc =1

2πCR2

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 the60Hz noise is down to 45% of its input level. fc = 30, f = 60.

G160Hz =

1√1 +

(ffc

)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.

G660Hz =

1√1 +

(ffc

)2n=

1√4097

→ 0.0156 (−36.12dB)

G860Hz =

1√1 +

(ffc

)2n=

1√65537

→ 0.0039 (−48.16dB)

3 Problem 3.

The pressure transducer diaphragm is of thickness t = 1mm, radius ρ = 100mm and sustains a material tensilestress σ = 200MPa when a p = 3.92MPa pressure is applied to the instrument. The Young’s modulus specified asE = 200GPa is a “red herring”, i.e., is not a required specification. The governing design equation, derived in theweb-based notes can be solved for h, the height of the spherical “bubble”.

p

σ= 4

th

ρ2 + h2→ h2 − 4σt

ph + 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, flatdiaphragm.

4 Problem 4.

Given a Copper-Constantan (Type T, blue-red) thermocouple whose reference junction is at 23.7C and whoseoutput 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.

V23.7 = V20 +Tactual − 20

40− 20(V40 − V20) = 0.789 +

23.7− 2040− 20

(1.611− 0.789) = 0.94107mV

Notice that in the top diagram in Fig. 1 since the output is positive and the reversed polarity, the reference junction

23.7C

+7.934mV

Cu

Con

23.7C

-7.934mV

Cu

Con

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− 8.2359.286− 8.235

(200− 180) = 192.18C

In the second case the output of the measuring junction is negative, the same polarity as the reference junction, sothe 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 estimateby backward extrapolation.

−250 +−6.993− (−6.181)−5.603− (−6.181)

50 = −320.24C

There are two reasons why the result is unreliable. Firstly, it is inadvisable to use a thermocouple outside itsprescribed, 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 coefficientsA,B,C in Eq. (9.15), p.285, W+G that relates thermistor temperature T to its resistance R.

1T

= A + B lnR + C(lnR)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 − 17.5

= 0

A + B ln 130 + C(ln 130)3 − 125

= 0

A + B ln 32.5 + C(ln 32.5)3 − 160

= 0

2

Simultaneous solution provides the following unique result.

A = 0.534345, B = −0.198072, C = 0.00407345

6 Problem 6.

The water has a static head of 1 − 0.3 = 0.7m from the liquid surface in the tank to the horizontally emergingstream. First, this initial velocity v is calculated.

12v2

x = H − h→ vx =√

2g(H − h) =√

2× 9.81× 0.7 = ±3.7059m/s

Then the time to fall vertically through h = 0.3m is calculated.

t =√

2h/g =√

2× 0.3/9.81 = ±0.24731s

Finally, the distance s traveled horizontally at vx is

s = vxt = 3.7059× 0.24731 = 0.9165m

Notice the ± results for vx and t are inevitable when one uses the energy equation. Since the two kinetic energydegrees 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 Pd = 0.05. No delays in threeconsecutive weeks, n = 3, is given by the exact, single term contribution of

P 3d0 =n CrP

rd (1− Pd)n−r =3 C0 × 0.050 × (1− 0.05)3 = 0.857

or about an 86% probability of experiencing three consecutive weeks with no interruption of your trip. To experienceone bad trip in four weeks, n = 4, is still a single term calculation.

P 4d1 =4 C1 × 0.054 × (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 onedelay is still a one term problem. It is the complement of zero delays in five weeks, n = 5.

P≥1d5 = 1−5 C0 × 0.050 × 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 20minutes so the arrival rate per period is i = 7 events per period and we wish to assess the probability that therewill be a line-up of at least 5 people in any given 20 minute period. The probability, like in the binomial case of inProblem 7, of exactly i in line in any period is

Pi =e−λλi

i!

so the complement of “up-to-four-in-line” is represented by the following summation.

1− e−7

(70

0!+

71

1!+

72

2!+

73

3!+

74

4!

)= 0.827

. . . 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 motorsthat exhibit a mean efficiency x = 91% with a sample standard deviation S = ±0.8%. Be careful, this efficiency isthe measured quantity. Do not confuse it with the confidence interval. The tolerance on the mean at this level ofconfidence is obtained with Table 6.6, p.144, W+G under the column α/2 = 0.025 in the row v = 9. The value atthe intersection, i.e., tα/2 = 2.262. Therefore the tolerance on the efficiency is

Tr = ±tα/2S√

n=

2.262× 0.8√10

= ±0.57%

To determine the sample size necessary to get the confidence interval ≥98% one examines the column downwardunder α/2 = 0.010 and looks for the first value ≤2.262. This occurs in the row v = 15, n = 16 at tα/2 = 2.602.

4

keyName:-

Student Number:-



(FD1)MLT83c

+-

5V

+

-

R 1R 2

R 3R 4

M

1. A four active arm strain gauge bridge with 300 SR4 gauges that have gauge factor of 2

is loaded to a nominal maximum 1000 . What is the resistance of each

gauge? R R R R1 2 3 4

2. What is the sensitivity of this system if ...R i

R i 8 R =10k R =2ki i

3. A 0 to 10V ADC returns a reading of 0123456789

1 0 0 1 1 0 1 1 1 0If this instrument is perfect

what is the highest

and lowest

possible voltage this could represent?

Convert this to a Gray coded integer.

0123456789

1 0 1 1

Now to a 2’s complement number. 0123456789

Finally to a 6-digit 10’s complement integer.

0 1 1 0 0 1 0 0 1 0111

299.4 300.6 300.6 299.4

100101

3 7 8999

:-

6.085V

6.075V

10 V/ 9.709 8.696V/ V/

keyName:-

Student Number:-



(FD1)MLT83e

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

MV =100Vdcs

+

R 1 R 2

R i

6. Choose the resistors R and R to make a 20:11 2

voltage divider, i.e., reduce the 100V supply

to 5V across R when the meter is not there,

i.e., R .8i

2

You can use resistors of any

value but all are rated at 0.25W, maximum.

R 1 R 2

If R =20k what will the meter read? (Forget about the small increase in

power due to decreased overall

resistance.)

i

Bonus: The new Jeep Cherokee uses

16.5L/100 km. What’s that in miles per

U.S. gallon?

36.1k 1.9k

4.586V

14.26

keyName:-

Student Number:-


(FD1)STT83g MECH 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?

What is the probability of this happening? What is the probability that I

complete my trip without experiencing any two-card failures?

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"

a= a= a=

b= b= b=

Find the linear least squares best fit coefficients

2

a and b for each using a simple minimization of

y (not a normal distance squared

minimization)

9. Which type is most sensitive? Least?

Calculate the standard deviation of the five measurements for each thermocouple

from their respective best fit line. T

E

J

What desireable property does

type J have?

1 is most likely

0.3697

0.366

0.04657

-0.2628

0.06727

-0.2835

0.054095

-0.104298

E T

0.05503

0.05932

0.01787It’s the most

nearly linear.

MECH 261/262

Measurement Lab & StatisticsMidterm Test Solutions 2008

March 26, 2008

1 Question 1

If gauge factor is 2 and strain on all 4 gauges is 0.001 and their original resistance is 300Ω theup/down change will be ±(2×0.001×300) = ±0.6Ω. I.e., R1 = R4 = 299.4Ω, R2 = R3 = 300.6Ω.

2 Question 2

If the meter input impedance Ri is infinite and Vs = 5V the sensitivity is given by(R2

R1 + R2

+R4

R3 + R4

)Vs

1000µε= 10

µV

µε

If Ri = 10kΩ and 2kΩ, respectively, the following seven simultaneous equations, based on Kirch-hoff’s node (current) and branch (voltage) laws, are solved simultaneously for V1 and V2.

i1 − i2 − ii = 0, i4 − i3 − ii = 0, 5− V1 − 299.4i1 = 0, V1 − 300.6i2 = 0

V1 − V2 −Riii = 0, 5− V2 − 300.6i3 = 0, V2 − 299.4i4 = 0

This results in

V1 − V2 = 2.504854369− 2.495145631→ 9.708738µV

µε

and

V1 − V2 = 2.504347828− 2.495652172→ 8.695656µV

µε

3 Question 3

10V are divided among 1023 quanta. Each quantum is worth

10000

1023= 9.775171065mV

A count of 1001101110 represents 29 + 26 + 25+E= 622.

62210000

1023= 6080.156403± 1

2

10000

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 tobinary 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, uponnegation, the 9’s complement number . . . 999377. Note how the leading most significant 9’s replaceleading 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 log10 G.

log10 G = 2.5→ 10log10 G = G = 102.5 = 316.227766

Since gain-bandwidth product is 25MHz

fc =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 attaina (temperature) rise of (θ(t)− t(0)) according to the following special solution of the differentialequation

et/τ − 1

et/τ− θ(t)− t(0)

θ(∞)− θ(0)= 0

2

The statement of specification says that at t = 1

e1/τ − 1

e1/τ− 36.95− 20

37− 20= 0

This gives the required time constant as τ = 0.1715576139s. The simulator is set up with a constanttemperature bath at 40C and a “trigger” that measures the time interval t until the sensor outputsignal reached 39.5C and we wish to know the calibration time limit tc at or below which the thetimer must trigger if the sensor is “good”. We now know τ from the specification.

etc/τ − 1

etc/τ− 39.5− 20

40− 20= 0

The solution of this second expression gives tc = 0.6328553571s.

6 Question 6

The first equation says that the voltage drop across the second resistor is 5% of the total voltagedrop across both resistors in series.

R2

R1 + R2

− 1

20= 0

The second equation says that the total voltage drop across the two resistors in series is the sumof the drop in each resistor due to the common current i that flows in both.

iR1 + iR2 − 100 = 0

The third equation says that the greater power dissipation in the resistor of higher value, R1 mustnot exceed 0.25W.

i2R1 −1

4= 0

This gives i = 1/380A, R1 = 36100Ω, R2 = 1900Ω and a check with the third equation establishesthat the power loss in R1 is indeed 1

4W. A 2000Ω/V meter on the 0-10V scale has an input

impedance of 10× 2000 = 20kΩ. Putting this in parallel with R2 gives an effective R2

R∗2 =

11

1900+ 1

20000

=380000

219= 1735.159817Ω

The meter reading is

VsR∗2

R1 + R∗2

=100× 1735.159817

36100 + 1735.159817= 4.586104105V

The reason for bringing R1 up to maximum allowable power is that in this way the outputimpedance of the voltage divider is minimized. This is a fundamental design principle. It’s espe-cially important if we’ve got a meter with low input impedance. 2000Ω/V is just about as low onecan 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 17.11218605mpG, “big gallon”.

8 Question 7

This is clearly a binomial distribution problem. Since the 10% card failure rate is the same foreach card and independent of each other, a failure occurs when both fail, one immediately afterthe other. Therefore Pµ = 0.01. To have exactly r failures in n tries

P (r : n) =n CrPrµ(1− Pµ)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 C0 × (0.01)0(0.99)100 = 0.3660323413

P (1 : 100) =100 C1 × (0.01)1(0.99)99 = 0.3697296376

P (2 : 100) =100 C2 × (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 thetime 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 ycan be input after specifying that number n. The computation of coefficients a and b are doneconventionally, 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 iscomputed.

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 bT 0.05503354 0.04657249 −0.2628484E 0.05932262 0.06727248 −0.28345J 0.01786823 0.054095 −0.1042984

The values of a are sensitivity expressed in mV/C. The type E couple is clearly the most sensitiveand the type T is the least sensitive among the three. The type J couple is the most linear. Thismay not be quite so obvious from the graph in Fig. 2.

10

0

5

mV

20 60 100 140 180

C

Notice how hard it is to

estimate linearity

without calculating

standard deviation

from line of best fit.

(FD1)LinBF83d

Figure 2: Linearized Thermocouple Characteristics

(MECH261/262)MLSCTSol08.tex

5

keyName:-

Student Number:-


Final Exam 07-12-19

(FD1)MLXA7C2

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?

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. m/s

57.4 Manometer fluid is alcohol, SpG=0.8.

30

±1/2bit=2.8 104.1±5.65.6

19.52

(See diagram on MLXA7C1.)

keyName:-

Student Number:-


Final Exam 07-12-19

(FD1)MLXA7C1

0

Disc rotation

The row of stationary sensors read "1" on black patches,

... "0" on clear ones.

sensors

011(0/1)10

keyName:-

Student Number:-


Final Exam 07-12-19

(FD1)MLXB7CS

+-

5V

+

-

R 1R 2

R 3R 4

Assume all gauges have increased or

or decreased by the same value R.

V

4. What does the meter read with all gauges at 1000 micro-strain if they

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 1 R 2

R 3 R 4

299.4

299.4

300.6

300.6

V=

Gauge factor=2.

10mV

5. Find the a) maximum and b) estimated uncertainty in the meter reading

a) b)

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?

What are the minimum pressures that can be detected accurately?

0-10Atm 0-100Atm

0-10Atm 0-100Atm

OIL

gauge weightpiston assembly

cyl. dia.

screw

8.045mm 2.544mm

0.4762At

48.25kPa

4.762At

482.5kPa

±20.0 V±20.2mV, ±0.2% full scale

if V =5V±10mV, R =300 ±0.001 , R=??±10s o

keyName:-

Student Number:-

Final Exam 07-12-19, PM, 3hrs. Page 3 of 3

(FD1)STXA7Cs

7. Twenty random numbers have been sorted into five bins.

MECH 262

frequency

0 2 4 6 8 10

Taking this as a probability mass

function express the "expected value" and the probability represented by

each bin.

4.4

0-2 2-4 4-6 6-8 8-107/20 1/20 4/20 7/20 1/20

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.

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.

4.00034mm ±0.00051mm

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 1.192mV40C 2.419mV80C 4.983mV

100C 6.317mV120C 7.683mV

Linear interpolation between 20 and 40C produces

220

(2.419− 1.192) + 1.192 = 1.315mV

Therefore the measured output is6.317− 1.315 = 5.002mV

On either side of 100C we get

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 26 = 64 quanta or “counts” from 0 to 63. Each quantum “patch” subtends 2π/64 radiansor 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 1 0

0 1 0

0,1

0,1 1,0 1,0

010011 or 010100 = a count of 19 or 20 = 101.25 or 106.875

æ2.8125 so the reading might be returned as being anywhere

between 98.4375 or 109.6875 an uncertainty span of 11.25 .

2MLX7Cs

This comes to 104.0625 æ5.625 .

Figure 1: Gray to Binary Conversion

For the specified input the straight binary output is 010011 or 010100, a count of 19 or 20 that might be interpretedliterally as 101.25 or 106.875, ±2.8125. So the reading might actually be anywhere between 98.4375 and109.6875, 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 measuredby the Pitot tube.

pdynam =ρairV

2

2= ρmanogh = pimpact

First the air density is calculated.

ρair =po

RT=

743760 × 101300

287× (273 + 19)= 1.1817kg/m3

Then the manometer pressure in Pa is determined.

pimpact = pmano = 800× 9.81× 0.05742

= 225.24N/m2

The first equation can now be solved having taken into account the density of the manometer fluid (800kg/m3)and inclination (30 → h = 57.4mm × sin 30).

V =

√2× 225.24

1.1817= 19.52m/s

4 Problem 4.

Since the all gauges are at nominal maximum strain of 1000µε 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 appearsat the upper corner of the bridge R1 and R4 must be the “down” gauges while the greater voltage drops mustoccur across the “up” gauges, R2 and R3. The voltage V12 is given by

VsR2

R1 + R2= 5

300.6600

= 2.505

Similarly V34 is

VsR4

R3 + R4= 5

299.4600

= 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

V =VsR2

R1 + R2− VsR4

R3 + R4=

Vs

2Ro[(Ro + ∆R)− (Ro −∆R)] =

Vs∆R

Ro

Now we can write∂V

∂Vs=

∆R

Ro,

∂V

∂∆R=

Vs

Ro,

∂V

∂Ro=

Vs∆R

R2o

andwV s = 0.01, w∆R = 0.00001, wRo = 0.001, Ro = 300, Vs = 5, ∆R = 0.6

So maximum uncertainty is ∣∣∣∣wV s∆R

Ro

∣∣∣∣ +∣∣∣∣w∆RVs

Ro

∣∣∣∣ +∣∣∣∣wRo

Vs∆R

R2o

∣∣∣∣ = 20.2µV

2

and the estimated uncertainty is√(wV s∆R

Ro

)2

+(

w∆RVs

Ro

)2

+(

wRoVs∆R

R2o

)2

= 20.0µV

If one calculates individual contributions to uncertainty it becomes obvious that uncertainty due to Vs is 20× 10−6

while that due to ∆R and Ro are 16×10−6 and 1

3×10−7, respectively. That’s why there’s so little difference betweenmaximum 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

πr2=

5.25× 9.81πr2

=

1013250Pa10132500Pa

Solving for r for both cases gives diameters

d1 = 8.045mmd2 = 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 wecan measure down to

121

= 0.4762Atm ≡ 1013.2521

= 48.25kPa

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

• 7 in the 0-2 range with a mean of 1,



• 7 in the 6-8 range with a mean of 7 and

• 1 in the 8-10 range with a mean of 9.

so(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 theprobabilities are

P0−2 =720

, P2−4 =120

, P4−6 =420

, P6−8 =720

, P8−10 =120

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” of5%. These are, respectively,

• zr = 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.

• zl = 1.645, using linear interpolation between the values z = −1.64 → 0.4495 and z = −1.65 → 0.4505. Notethat the 45% integral area has a 5% left hand tail.

From the change-of-variable definition

z =xi − x

σ

one gets two equations.

x + 1.28166 . . . σ − 400.1 = 0x− 1.645σ − 3.9995 = 0

that make x = 4.000343mm and σ = 0.0005125mm.

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

1 2

1 2

1 2 3

3

1 2 3 4

4

1 2 3 4 5

5

X

3/51 48/51

2x(3/50)

3x(3/49)

44/50

40/49

4x(3/48)

5x(3/47)

36/48

32/47

3STX7Cs

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 failurenode [×], shows the probability of not getting a pair as

4851

+4450

+4049

+3648

+3247

=67584195755

or about 34.52%. That means the player stands to win 65.47% of the time while the house wins only 34.52% of thetime. 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 apair thus.

351

+4851

650

+4851

4450

949

+4851

4450

4049

1248

+4851

4450

4049

3648

1547

=128171195755

One may check as follows.67584195755

+128171195755

= 1

5

keyName:-

Student Number:-



(FD1)MLT7A3

1. Given the series of four 1:10 inverting amplifiers shown below, specify:-

a) The value of R in each case,

when the input voltage v =8mV and R =22kOhm.3

R =

R =

1

3

1 b) The value of R such that the output voltage v =10Vo

iR 2R 1 R 2R 1

-

+ v o

R 3

2

-

+v i-

+

R 2R 1R 2R 1

-

+

2.2k

15.4k

2. A 10-bit 0 to +10V (unipolar) ADC is connected to measure v .o When v =10V it reads

a full-count of 10 ones 1111111111. a) What decimal number does that correspond to?

b) If a strain guage bridge shows an out of balance voltage of 8mV at =0.1%

then what is v and the count when =0.00075?o

and

o

1023

1011111111(binary)

7.5V

+-

10V

+

-

R 1R 2

R 3R 4

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 1 R 2

R 3 R 4

Assume all gauges have increased or

or decreased by the same value R.

299.76

299.76

300.24

300.24

M

keyName:-

Student Number:-



(FD1)MLT7A5

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= b= What is the sensitivity according to this

calibration?

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.

-2 2

+5y=5-(5/4)x

2

6. Given an under-the-tongue thermometer with time constant =4s

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?

Hint:- y-y

y -yi

ie=(1-e )

-t/

14.55s

3.651

0.0775 0.4833

0.0775mV/kg

keyName:-

Student Number:-


(FD1)MST7Bb

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 random distribution.uniform

MECH 262

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

frequency

0 2 4 6 8 10

(or bimodal)

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

needs 10 sheets how many sheets should you buy in order that

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?

11(Poisson)

about 0.5%

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 Toward teens Toward kids$375000 $400000 $225000

Hint:- Bayes’ theorem?

Is there another way to get the

answer? Using a binomial model will

give 13 sheets at 3.4% risk.

MECH 261/262

Measurement Lab (& Statistics)October 29, 2007

Mid-Term Test 07-10-29

1 Question 1., Four Operational Amplifiers

• Since all four inverting op-amps have |G| = 10 and R2 = 22kΩ

G = −R2

R1

therefore R1 = 2.2kΩ.

• Since the first three stages get us up to (0.008V.)(−G1)(−G2)(−G3) = −8V., the last stagemust have a gain of −10

8. Therefore

R2

R1 + R3

=22

2.2 + R3

=10

8

and R3 = 15.4kΩ.

2 Question 2., Unipolar 10V. ADC

• A 0-10V. ADC (unipolar) returns a positive integer. There is no sign bit. The binary number

111111111 = 29 + 28 + 27 + 26 + 25 + 24 + 23 + 22 + 21 + 20

= 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 1023.

On the other hand one may simply say that this is 210 − 1 = 1023.

• If a linear instrument, of which a strain gauge bridge is a fair approximation, produces a fullcount of 1023, representing 10V., at ε = 0.1% = 0.001 then at ε = 0.00075 we get

0.00075

0.00110V. = 7.5V.

and 0.75(1023) = 767.25 that gets rounded down to a count of 767. Encoded in 10-bits thisis

1011111111 = 29 + 0(28) + 27 + 26 + 25 + 24 + 23 + 22 + 21 + 20

= 512 + 0 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 767

1

3 Question 3., Four (Active) Arm Strain Gauge Bridge

Since the higher (+) voltage is above and the unstrained resistance is 300Ω, R1 = R4 = R − ∆Rand R2 = R3 = R + ∆R. From

R2

R1 + R2

Vs = V12 andR4

R3 + R4

Vs = V34

one may take the difference V12 − V34 = 8mV. and solve for ∆R, given Vs = 10V.

8mV. =300 + ∆R

60010 − 300 − ∆R

60010

So ∆R = 0.24Ω thusR1 = R4 = 299.76Ω and R2 = R3 = 300.24Ω

Don’t forget to put up-arrows on R2 ↑ and R3 ↑ and down-arrows on R1 ↓ and R4 ↓.(FD1)mltSol7A3.tex

2

MECH 261/262

Measurement Lab (& Statistics)October 31, 2007


1 Question 4., Load Cell Linear Fit

• The purpose here is to evaluate

a =n

∑xiyi −

∑xi

∑yi

n∑

x2i − (

∑xi)2

, b =

∑yi − a

∑xi

n

Using the following table

•

xi yi xiyi x2i

20 1.7 34 40060 5.8 348 3600

100 7.9 790 10000∑xi = 180

∑yi = 15.4

∑xiyi = 1172

∑x2

i = 14000

• Therefore, with n = 3,

a =3× 1172− 180× 15.4

3× 14000− 1802= 0.0775, b =

15.4− 0.0775× 180

3= 0.4833

• The static sensitivity is just the slope, a, with the appropriate units of output/input or0.0775mV/kg.

2 Question 5., Root Mean Square of Parabolic Wave Form

The function y = 5 − 54x2 describes the first quarter wave. Therefore we need to integrate the

square of this from x = 0 to x = 2, take the square-root and divide by the interval 0 → 2.

∫ 2

0

(5− 5

4x2

)2

dx = 25x +5

16x5 − 25

6x3

]2

0=

80

3

Taking the square-root and dividing by 2 produces the RMS value of this wave form amplitude.√80

5/2 =

2√

30

3= 3.651

1

3 Question 6., First Order Thermometer Response

Fromy − yi

ye − yi

= 1− e−t/τ

we use y = 39.0C because the precision, ±0.5C, of the thermometer is such that when it reacheswithin 0.5C of the feverish patient’s temperature it will have, for all intents and purposes, reachedstable, constant reading. Of course te = 39.5C and ti = 20C while τ = 4s. Substituting

39.0− 20

39.5− 20= 1− e(−t/4) so − t

4= loge

1

38therefore t = 14.55s.

(FD1)mltSol7A5.tex

2

MECH 262(Measurement Lab &) Statistics

November 2, 2007


1 Question 7., Bins and Distributions

The list of given numbers can be sorted as follows.

1.97 7.901.74 7.400.74 7.310.66 5.56 7.250.46 5.49 7.080.32 5.30 6.670.31 3.05 4.81 6.29 8.41

They fall into the following bins.

| 0 ≤ · · · < 2 | 2 ≤ · · · < 4 | 4 ≤ · · · < 6 | 6 ≤ · · · < 8 | 8 ≤ · · · < 10 |

2 Question 8., Plywood and Poisson or Binomial

2.1 Try Poisson

This is a Poisson’s distribution problem with mean λ = 110

= 0.1. We wish to buy enough sheetsof plywood so that our chance of getting 10 “clear” ones is 95% or better.

• The probability of getting exactly 0 knotty sheets if we buy only 10 is

P (x) =e−λλx

x!=

e−0.1λ0

0!= 0.9048374180

. . . not quite good enough.

• If we buy 11 sheets, we can tolerate 0 or 1 defective ones and still have 10 or more goodones.

e−0.1λ0

0!+

e−0.1λ1

1!= 0.9048374180 + 0.09048374180 = 0.995321159

or about a 0.47% probability of 2 bad sheets in 10.

1

• Well, that expresses the probability of getting 9 good sheets out of 10. If we buy 11 sheetsand 10% are bad we have 1.1 bad sheets. Never mind the fraction of a sheet. This is statisticsso λ becomes 0.11. What’s the probability of getting 2 or more bad sheets in 11?

1−(

e−0.11λ0

0!+

e−0.11λ1

1!

)= 0.0056241098

If we buy 11 sheets our chances of having 10 good ones is about 0.56%. As one might expect,getting 2 bad sheets in 11 is slightly greater than getting 2 bad in 10

2.2 Try Binomial

To buy 10 good sheets with a 0.1 probability of getting a bad one every time we pick is expressedby the probability of exactly 1 “bad egg” in the 10 we buy.

10 good in 10 = 10C0(0.1)0(0.9)10 = 0.3486784401

We’re playing “Russian roulette” with 4 of 6 chambers loaded. Let’s try 11 sheets.

10 good in 11 = 11C0(0.1)0(0.9)11 + 11C1(0.1)1(0.9)10 = 0.6971568802

Better but still way short of the of the 5% risk of getting fewer than 10 good sheets that we decidedto tolerate. Now for 12.

10 good in 12 = 12C0(0.1)0(0.9)12 + 12C1(0.1)1(0.9)11 + 12C2(0.1)2(0.9)10 = 0.8891300223

We’re getting there. 13 sheets should do it.

10 good in 13 = 13C0(0.1)0(0.9)13 + 13C1(0.1)1(0.9)12

+13C2(0.1)2(0.9)11 + 13C3(0.1)3(0.9)10 = 0.9658382791

That did it. If we buy 13 sheets our risk is down to about 3.4%.

3 Question 9., Junk Food and Bayes

This one is easy. Below appears the amount that should be allotted to each age group. The firstrepresent adult targeted advertising, the second is aimed at teenagers and the third towards kids.

10000000.6× 0.5

(0.6× 0.5) + (0.4× 0.8) + (0.3× 0.6)= $375000

10000000.4× 0.8

(0.6× 0.5) + (0.4× 0.8) + (0.3× 0.6)= $400000

10000000.3× 0.6

(0.6× 0.5) + (0.4× 0.8) + (0.3× 0.6)= $225000

(FD1)mstSol7Bb.tex

2

keyName:-

Student Number:-


Final Exam 07-04-25

(FD1)MLX74yA

1. Given the bridge circuit Fig. 1, find the total current that must be supplied by the power

supply. What is the reading of the millivoltmeter? 2. Which of the

applications illustrated in Fig. 2 might produce the state of the bridge in Fig.

The state of strain corresponds, approximately, to what fraction of full scale?

5V0V

R 1

R 2 =300.1

=299.9R 2 =300.1

R 1 =299.9

R i 8

Fig. 1

A

B

-

C

1?

Fig. 2

3. If R =10k what is the reading on the millivoltmeter? i

Is this reading too high or too low?

By about what percent?

T T

-T -T-F

F

mV1.618

low

%2.91

(50/3)mA

A,B not C

1/3

(5/3)mV

keyName:-

Student Number:-


Final Exam 07-04-25

(FD1)MLX74yB

R 1 R 2

R 3

5V

0V

R

R

R RTD

+ -

4. A two-wire platinum RTD bridge has v =5V, R =R =R =100s 1 2 4 . Each of the two lead wires

contribute R =R =0.1leadA leadB . When the sensing element is immersed in boiling water at

100C, estimate the temperature measurement error incurred

if one applies Eq. (9.11), p.283, literally. C

What’s v in this case?o V

v o

v s

+0.52

-411.1m 139.16 -vs- 139.36

100C -vs- 100.52C

5. A type J thermocouple produces a measured

output of 22.000mV with a reference junction

assumed to be at 0C. Later, it’s found that it

was at 4C. What’s the true hot junction

temperature? C This is C

above below the one originally recorded.

406.48 3.69

u o

11.7mm

6. A U-tube manometer filled with alcohol (SpG=0.8) is con-

nected to a Pitot-static tube that "looks into" a free stream

wind velocity u in an air atmosphere at 100kPa and 20C.

What is the wind velocity? m/s o

leadA

leadB

12.42

keyName:-

Student Number:-

Final Exam 07-04-25

(FD1)STX74yC

75 75.3

7. Automotive pistons are manufactured and found to have a mean

diameter of 75mm and a standard deviation of ±0.13mm.

If this product has an acceptable diameter range of ±0.1mm

about the mean, how many acceptable pistons might be expected

in a lot of 100000? 55880 If oversize pistons for

rebuilt engines are nominally 75.3mm in diameter and may

also vary ±0.1mm, how many of these may be recovered

from this same lot? 6110

8. A wholesaler buys an initial quantity of 50 of these pistons whose manufacturer claims to

be of average diameter 75mm. The client decides to continue using this supplier if it may

be concluded, with a confidence level of 9 times out of 10, that in the long run the mean

diameter with prove not to differ from that claimed. Between what diameter limits

=x=mm mm should the measured diameter limits be in order to

continue buying?

74.9698 75.0302

Use the population standard deviation as claimed above.

keyName:-

Student Number:-

Final Exam 07-04-25

(FD1)STX74yD0

9. A 5-card "flush" means all 5 cards are of the same "suite", i.e., all hearts, all clubs, all

diamonds or all spades. The denomination does not matter. What is the probability of

drawing 5 cards in a row, all from the same suite of (initially) 13 cards, from an initially

complete deck of 52?

What is the probability of drawing a 5-card straight? In this case the

the sequence can contain all of the suits but after the 5th card is drawn the player must

have a run, e.g., 7H,5C,9S,6D,8C. Furthermore "wrap-around" is permitted, e.g.,

K,A,2,3,4 is O.K. The diagram below, followed by the one on the next page, may help.

(MECH261-2)Straight

8/49

8/51

8/50

8/518/51

8/51

8/50 8/50 4/508/50

8/50 8/50 8/50 8/50 4/50

8/498/494/49

4/49 4/494/494/49

8/49

8/49 4/49

4/494/49

8/49 4/49

4/494/49

4/49

4/49

4/494/49

(30720)/(5997600)=0.5122048%

11880/5997600=0.1980792%30720/5997600=0.5122048%

keyName:-

Student Number:-

Final Exam 07-04-25

(FD1)STX74yD1

4/484/484/484/484/484/484/488/488/48 4/48 8/48 4/48 4/48 4/48 4/48 8/48 4/48 8/48 4/48 4/48 4/48

2007 Measurement Final Answers and Solutions(FD1)FXW07Sol82l.tex

February 12, 2008

1 Problem 1.

a) Total bridge resistance when Ri →∞ is, by inspection, 300Ω. Therefore

I =E

R=

5000mV300Ω

=503

mA

b) Designate V1 the voltage between R1 and R2 and V2 that between R3 and R4.

V1 =R2

R1 + R2Vs =

300.1600

5 = 2.500833V

V2 =R4

R3 + R4Vs =

299.9600

5 = 2.499167V

Therefore millivoltmeter reads 1.667mV.

2 Problem 2.

a) Since the gauge pairs are loaded symmetrically, i.e., δR = ±0.1Ω, the pairs must be in tension/compression ofidentical magnitude like on the torque cell (A) or the cantilever (B). There are two “Poisson” gauges on (C) thatwould sustain, nominally, only 30% of the strain sustained by the two “direct” (tensile/compressive) gauges. Sothe answer is (A), (B), not (C).b) δR = 0.1Ω corresponds to about 1/3 of nominal, maximum strain 1000µε at an assumed gauge factor of 1(unity). So 1/3 is the fraction of full scale.

3 Problem 3.

In the following i1, i2, i3, i4, ii are currents flowing through the resistors with the corresponding subscripts. Vs = 5Vand V1 and V2 are the respective voltages between R1 = 299.9Ω and R2 = 300.1Ω and between R3 = 300.1Ω andR4 = 299.9Ω. Note that here Ri = 10kΩ. Write the following seven linearly independent equations.

5− V1 − 299.9i1 = 0, V1 − 300.1i2 = 0, V1 − V2 − 10000ii = 0, 5− V2 − 300.1i3 = 0

V2 − 299.9i4 = 0, i1 − i2 − ii = 0, i4 − i3 − ii = 0

The complete solution gives

i1 = 0.008333414266, i2 = i3 = 0.008333252454, i4 = 0.008333414266, ii = 0.1618122983× 10−6

All these currents are in Amperes and

V1 = 2.500809061, V2 = 2.499190939

in Volts. The difference is V1 − V2 = 1.618122mV. Recall from 1 b), above, with Ri →∞, V1 − V2 = 1.666667mV,a difference of 0.048544mV that corresponds to a reduction in signal (sensitivity) of about 2.9%.

1

4 Problem 4.

Looking up RRTD in Table 9.3, p.282 under 100C one notes 139.16Ω but the bridge is completed with

R3 = RLead(A) + RRTD + RLead(B) = 0.1 + 139.16 + 0.1 = 139.36Ω

An RRTD reading of 139.36Ω produces

100 + 10139.36− 139.16143.01− 139.16

= 100 + 100.23.85

= 100.52C

which is 0.52C too high. vo can be found as

RRTD = R2vs − 2vo

vs + 2vo

or139.36 = 100

5− 2vo

5 + 2vo→ vo = −411.096mV

5 Problem 5.

The apparent temperature for a Type J (iron/constantan) thermocouple with an assumed reference junction tem-perature of 0C, that is actually at 4C, can be obtained by interpolating directly from Table 9.2, p.277.

400 + 5022− 21.846

24.607− 21.846= 402.79C

The output of 22mV would be greater had the reference junction been at 0C.

22 +421.019 = 22.2038mV

Therefore the actual temperature was

400 + 5022.2038 − 21.84624.607− 21.846

= 406.48C

or 3.69C above that originally assumed.

6 Problem 6.

The Pitot tube measures impact (velocity) pressure of a free stream directly by stagnating the flow into its openingthat faces the stream.

ρalcoholgh =ρairV

2

2Since the manometric fluid is alcohol with specific gravity 0.8 the left hand side of the energy equation thatrepresents specific potential energy is

ρalcoholgh = 0.8× 1000kgm3

× 9.81Nkg× 11.7

1000m = 91.8216

Nm2

(Pa)

Air density is calculated with the perfect gas law.

ρair =P

RT, P = 100000Pa, T = 20 + 273 = 293K

Since I only recall

Rair = 53.3ft#f

#mR

2

we need the following multiplier

12”/ft39.37”/m

9.81N/kgf

2.2#f/kgf 2.2#m/kgm 9R5K

= 5.382169164

to convert R = 286.9J/kgK.

ρair =100000

286.9× 293= 1.1897kg/m3

V 2 =91.8216× 2

1.1897= 154.357 → V = 12.42m/s

7 Problem 7.

a) Refer to the Gaussian normal distribution plotted to the left of the question. Means of standard 75mm andoversize 75.3mm piston diameters are shown. Augment the diagram by indicating a band ±0.1mm on either sideof these two means, i.e., at 74.9mm, 75.1mm, for the standard pistons and 75.2mm and 75.4mm for the oversizeones. Using Table 6.3, p.136 and computing the standardized variable z for the 75.1mm size limit first as

z =75.1− 75

0.13≈ 0.77 → 0.2794

where 0.13mm is the given population standard deviation σ. This shows that half of the area under the curvefrom 75 → 75.1mm is 27.9%. Since the tolerance is bilaterally symmetric one expects 2× 0.2794 → 55.88% of theproduction lot pistons are acceptable as “standard”. This is about 55880 out of the 100000 total batch of pistons.b) The band comprising acceptable oversize pistons is obtained by subtracting the integral under the standardizedcurve from 75 to 75.2mm from that from 75 to 75.4mm.

75.4− 760.13

≈ 3.08 → 0.4990

75.2− 750.13

≈ 1.54 → 0.4382

This represents about 6.08% or 6080 pistons.

8 Problem 8.

Useσx =

σ√n

, z =x− µ

σx

to get

±‖|x| − |µ|‖ =zσ√

n= 1.645

0.1350

→ ±0.0302mm

So the sample lot of piston diameters should fall between

74.9698mm ≤ x ≤ 75.0302mm

Note that 1.645 is 45% of the area under the z-bell curve, i.e., half of 90%. The sample size was n = 50 pistonsand σ = 0.13mm from Problem 7.

3

9 Problem 9.

In drawing five card poker hands from a deck of 52, without replacement, one obtains multiple paths of probabilitiesof success at each stage that must be added as numerator products of the stages. The denominator is just theproduct 51 × 50 × 49 × 48 = 5997600. The first card doesn’t matter because the required pattern can be filled inany sequence. It supplies a multiplier of 52 to both the numerator and the denominator. The flush is easy becauseonce the suite -clubs, hearts, diamonds or clubs- has been established by the first card dealt, the probability ofrepeating it becomes

1251× 11

50× 10

49× 9

48=

118805997600

≈ 0.1980792%

This serves as a nice “warm-up” for the straight that makes use of a chequer-board tree to illustrate the problem.It shows the ways in which the straight can be “filled” at any intermediate stage as one proceeds from root toleaf and has the possibility of success by obtaining any one of four cards in any one of the open columns, beforeproceeding to the next stage (card draw). For example the left-most path contributes 84 = 4096 to the numerators.All 21 paths yield

4096 + 2048 + 2048 + 1024 + 1024 + 1024 + 1024 + 2048 + 1024 + 2048 + 1024 + 1024

+2048 + 2048 + 1024 + 1024 + 1024 + 1024 + 1024 + 1024 + 1024 = 30720

With the same denominator, one obtains

307205997600

≈ 0.5122048%

So it is about two-and-a-half times easier to get an arbitrary straight than an arbitrary flush.

4

SS,H,D,C

5-card flush

SS

SS S

SS SS

SS SS S

12/51

11/50

10/49

9/48

p=12x11x10x9

51x50x49x48

=33/16660

about 0.198%

<1 chance in 500

5CFl792

keyName:-

Student Number:-



(FD1)MLT721A

2. Do 4219-2146=2073 in 2s complement binary arithmetic. Express the result as

a) A 16 bit binary number and b) As a string of 4 hexadecimal digits.

+

=

0

1

-4V

-2V

0

4V

2V

0

1. A cycle of a peculiar 3 harmonic wave form is shown below.

What is the RMS voltage?

0

3. Given the Gray coded 12-bit number

below, express it as a binary integer.

What is that in decimal?

Hint:- study the logic block below

or the section on arithmetic on the

website.

0 0 1 0 1 0 1 0 0 0 01

g

b

b i-1

i-1

i

0 0 1 0 0 1 1 0 0 0 0 0

608

CA

B

A B C

0 0 0

0 1 1

1 0 1

1 1 0

g 0

b 0b 11

g 6

-2146

2073

4219

0 8 9

0 0 0 0 0 0 0 0 0

0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1

1 11 1 1 1 1 1 1 1 1 1

1 1 1 1

Hex

(2 6 )/3=1.633V

The waveform consist of two

half-cycles. Each contaning

three isosceles triangular

saw-tooth forms spanning

1/6 cycle. Two have peak

amplitude of 2V, one of 4V.

keyName:-

Student Number:-



(FD1)MLT722B

1. Make a 10:1 voltage divider by choosing R and R so that the meter reads exactly 500mV

when the source voltage is 5V. The circuit consisting of R , R and R

effective resistance of 10000R .1 2 i must have an

o

V =5V

1 2

R =10k

s R

R 2

1

i

R =1.7o

0 1000mV

R =

R =

1

2

15299.83

2048.439546

2. An inverting 741 Op.Amp. has a gain

of G=100 and an input impedance

Z = 1000 .i Find f , R and R .c 1 2

f = R = R =219.90kHz

c

v i

R 2R 1

-

+ v o

meter

3. A lowpass amplifier with corner

frequency f c of 1kHz produces

an output amplitude of 0.1V with

an input sine wave signal of 3kHz.

What is the output amplitude

voltage for a 20kHz sine wave signal

input with the same amplitude as the

3kHz one?

1kHz3kHz

20kHz

Gdb

f =c

3dB

vo =0.1V @ 3kHz

vo = ?

Rolloff is 24dB

per octave.

v @20kHz=o52 V

1000 100k

key

(FD1)STT73bC

Midterm Test 07-03-02 ,40min.

MECH 262

Name:-

Student Number:-

2. 180 seats are sold for an aircraft with 175 seats. If 1 in 20 passangers are on average

"no shows" what is the most likely number of empty seats that will remain?

What is the probability of that occuring?

4

13.5% Hint:- 180Cr

3. Tellers in a bank must, on average, serve 20 customers per hour. What is the probability

that a) Exactly 25 customers arrive in a given hour?

b) There will be less than 10 customers arriving during a given hour?

Hint:- Poisson; did you try 6.29?

1. Sometimes we need a linear fit that minimizes the sum of squares of y-deviations

subject to the line passing throgh the origin.

Write the relations for the coefficients in the

equation y=ax+b.

a= i=1 ... nx y

x i2

i i

(2,1)

(4,5)

(6,8)

x

y

O

For the data at right

a=

b=b=

5/4

00

4.46%

0.50%

2007 Measurement Midterm Answers and Solutions with Maple(FD1)MT07Sol882.tex

August 28, 2007

The two exam sheets (for Measurement only) with answers follow the solution of question 1 on the 07-02-26 midterm;the one about root-mean-square voltage.

0 11/3 2/3 t

2V

4V

v=6tv=12t

(MECH261-2)Vrms3Tr

Figure 1: Midterm Test 07-02-26

Consider Fig. 1. The composite waveform shown in the actual question can be reduced to these three elementsbecause if all harmonics of a cyclic wave are represented in their correct proportional contribution a simpler problemcan be solved to give an equivalent result. Each element spans 1

3 cycle so we double the contribution of the smalltriangle and include the larger just once. Don’t forget to square v(t).

Vrms =

√2∫ 1

3

0

(6t)2dt +∫ 1

3

0

(12t)2dt = 2

√23

The solutions for questions 2 and 3 on the 07-02-26 midterm are omitted. Binary arithmetic and Gray to binarycode conversion are covered adequately in the notes posted on the website.Turning to the 07-02-28 midterm, the solution to question one is summarized in the following Maple worksheet.

> restart:> eq1:=R1+1/(1/R2+1/Ri)-10000*Ro;eq2:=(1/(1/R2+1/Ri))/((Ro+R1+1/(1/R2+1> /Ri)))-.1;

eq1 := R1 +1

1R2

+1Ri

− 10000Ro

eq2 :=1

(1

R2+

1Ri

)

Ro + R1 +1

1R2

+1Ri

− 0.1

> eq3:=subs(Ri=10000,Ro=1.7,eq1);eq4:=subs(Ri=10000,Ro=1.7,eq2);

eq3 := R1 +1

1R2

+1

10000

− 17000.0

eq4 :=1

(1

R2+

110000

)

1.7 + R1 +1

1R2

+1

10000

− 0.1

> R2:=solve(eq3,R2);

R2 := −10000. (R1 − 17000.)R1 − 7000.

> eq4;

1

1/

((−0.0001000000000 (R1 − 7000.)

R1 − 17000.+

110000

)1.7 + R1 +1

−0.0001000000000 (R1 − 7000.)R1 − 17000.

+1

10000

)− 0.1

> R1:=solve(eq4);

R1 := 15299.83000> R2;

2048.439546

Problem 1. Ms2Mt722, Measurement MECH 261/262 midterm 07-02-28. 10:1 voltage divider was dealt withabove.

Question 2 just requires some reading from the text. On page 47 it says that an inverting Op-Amp has an inputimpedance of ≈ R1. So if Zi = 1000Ω, R1 = 1000Ω. To get a gain of G = 100 it is noted that for an invertingOp-Amp G = R2/R1 so R2 = 100kΩ. Finally we are told

fc =GBP

G, GBPnon−inv = 10MHz, GBPinv =

R2

R1 + R2

so

fc =100kΩ

1kΩ + 100kΩ= 9.90kHz

Finally the solution to question 3 on the 07-02-28 midterm is summarized below. Converting decades to octavesrequires the ratio log 10

log 2 as a multiplier.> restart:> Oct13:=evalf(log(3/1)/log(2));Oct120:=evalf(log(20/1)/log(2));Oct320:> =evalf(log(20/3)/log(2));

Oct13 := 1.584962501Oct120 := 4.321928095Oct320 := 2.736965594

> Oct120-Oct13;

2.736965594> dB1:=3;dB3:=24*Oct13+dB1;dB20:=24*Oct120+dB1;

dB1 := 3dB3 := 41.03910002dB20 := 106.7262743

> G1:=evalf(10^(dB1/20));

G1 := 1.412537545> G3:=evalf(10^(dB3/20));

G3 := 112.7080669> G20:=evalf(10^(dB20/20));

G20 := 216927.0523> vo3:=.1:vi:=vo3*G3;vo1:=vi/G1;vo20:=vi/G20;

vi := 11.27080669vo1 := 7.979120081

vo20 := 0.00005195666732Problem 3.16:- A lowpass Butterworth filter has a corner frequency of 1kHz. An input of a 3kHz harmonic

sustains an output voltage vo3=0.1V. The filter has a 24bD rolloff in its stopband. Find vo20, the output voltage

2

for a 20kHz pure sinusoidal input signal with the same amplitude as the 3kHz wave that was attenuated to 0.1V.vi is the input signal amplitude for all waves of the same amplitude and form. vo1 is the output amplitude of the1kHz signal. (MECH261-2)dBperOctave, 07-02-25.

> restart:with(combinat):

Warning, the protected name Chi has been redefined and unprotected

> a:=(2*1+4*5+6*8)/(2^2+4^2+6^2);

a :=54

A least squares linear fit that is forced to go through the origin requires only one degree of freedom, i.e., a, theslope of the line. This is computed as the sum of the products of the point coordinates divided by the sum of thesquares of the x-coordinates. This is question 1 on the 2007 MECH 262 midterm test.

> numbcomb(180,7)*.05^7*.95^173;

0.1180760941> numbcomb(180,8)*.05^8*.95^172;

0.1343892387> numbcomb(180,9)*.05^9*.95^171;

0.1351751406> numbcomb(180,10)*.05^10*.95^170;

0.1216576265

This is an iterative solution to the ”rotten egg” problem to find the most likely (highest probability) number in abinomially distributed sample of 180 ”events” with a probability of 1/20. This is a unimodal distributionfunction, i.e., it has a single peak or maximum. We see this is the probability of 9 ”no-shows” at about 13.5%.Sorry, I subtracted 175-(180-9)=2, the wrong answer. It should have been 4 empty seats.This is question 2 on the2007 MECH 262 midterm test.

> P25:=exp(-20)*20^25/25!;evalf(P25);

P25 :=5120000000000000000000

236682282155319e(−20)

0.04458764909> P0:=evalf(exp(-20)*20^0/0!);P1:=evalf(exp(-20)*20^1/1!);P2:=evalf(exp> (-20)*20^2/2!);P3:=evalf(exp(-20)*20^3/3!);P4:=evalf(exp(-20)*20^4/4!)> ;P5:=evalf(exp(-20)*20^5/5!);P6:=evalf(exp(-20)*20^6/6!);P7:=evalf(exp> (-20)*20^7/7!);P8:=evalf(exp(-20)*20^8/8!);P9:=evalf(exp(-20)*20^9/9!)> ;

P0 := 0.2061153622 10−8

P1 := 0.4122307244 10−7

P2 := 0.4122307244 10−6

P3 := 0.2748204829 10−5

P4 := 0.00001374102415P5 := 0.00005496409659P6 := 0.0001832136553P7 := 0.0005234675866P8 := 0.001308668966P9 := 0.002908153258

> P0to9:=P0+P1+P2+P3+P4+P5+P6+P7+P8+P9;

P0to9 := 0.004995412306> P10:=evalf(exp(-20)*20^10/10!);

P10 := 0.005816306518> P0to9+P10;

3

0.01081171882

The bank problem is suited to a Poisson distribution model. We get about 4.46% probability that exactly 25customers will arrive in an hour long period if the average number of arrtivals is 20 per hour. The probabilitythat less than 10 will arrive is computed as the sum of the probability of exactly 0,1,2,3,4,5,6,7,8,9 arrivals.Unfortunately added the probability of exactly 10 arraivals as well to get about 1.08% probability. The answershould have been about 0.5%.(MECH261-2)MStMt73b.mws. Recomputed 07-03-03.

4

Name:-

Student Number:-

key

MECH 261/262

Final Exam 06-04-21

(FD1)MLX64uA

Measurement Question 1.

A 12-bit bi-polar ADC is scaled to measure ±5V-dc

What is its static sensitivity? (Value and units stated clearly, please)0.4094 bits/mV

Given this instrument, what measurement does the count ...

± 10 9 8 7 56 4 3 2 1 0

1 0 0 1 0 1 1 0 0 1 1 1 represent?

± 10 9 8 7 56 4 3 2 1 0

± 10 9 8 7 56 4 3 2 1 0

± 10 9 8 7 56 4 3 2 1 0

to help with your rough work

empty 12-bit registers

-4.125V

Name:-

Student Number:-

key

MECH 261/262

Final Exam 06-04-21

(FD1)MLX64uB

Measurement Questions 2 and 3.

7.05V

8.2

Question 2. If the frequency response of an amplifier is stated as "2dB down at 200Hz" what

is the actual magnitude of a voltage signal measured as 5.6V at this frequency?

Question 3. An instrument transducer is specified as having an output impedance of 10 .

It consists of a themocouple voltage source with an internal impedance of 1.8 .

The "black box" transducer model is shown below. What is the value of the

unkown impedance Z?

Z

1.8

V o (open circuit voltage)

Name:-

Student Number:-

key

MECH 261/262

Final Exam 06-04-21

(FD1)MLX64uC

Measurement Question 4.

A thermocouple can be represented as a first order system as regards its

A particular example of such an instrument has a time constant of 10s and is subject to a

temperature rise of constant rate equal to 10C per minute. What is the time delay between

any given instantaneous reading and the time at which it occurred?

What is the error inherent in any instantaneous reading?

Time

Output

0th order behaviour

actual behaviour

10s

10/6=1.6667C

time response.

T

Name:-

Student Number:-

key

MECH 262

Final Exam 06-04-21

(FD1)STX64uD

Statistics Question 1.A

A

A

A

A

A

A

A

A

A A

B

B

B

B

B

B

BB

B

B

B

BB

1

48/51

24/25

3/493/49

2/49

1/16

3/51

1/25 3/50

3/49

2/49

1/241/24

48/49

1/16

1/24 1/24

1/24 1/24

This is the decision tree for a "full house"

probability on drawing 5 cards from

(initially) a deck of 52 playing

of any kind and 2 of any

other kind. Say, 3 Jacks

and 2 sevens. Of course

a pair of Aces and 3 sixes

would be equally

valid.

Multiply the numbers along each of the 10

equally likely branches and add up the result.

P=

cards such that you get 33/50

2/49

1/16

1/16

0.00144057623

Name:-

Student Number:-

key

MECH 262

Final Exam 06-04-21

(FD1)STX64uE

Statistics Question 2.

A sample n=100 of eggs have a mean x=300g measured weight and a st on

S=±15g.

±7.5g

In a sample of 100 eggs from the same population how many of these might one

expect to deviate no more than

±30g

±75g

±1.5g

... from the 300g mean? Assume a normal dstribution of weight about the mean with the

same standard deviation.

andard deviati

(estimated number)

(estimated number)

(estimated number)

(estimated number)

38

95

100

8

Name:-

Student Number:-

key

MECH 262

Final Exam 06-04-21

(FD1)STX64uF


The following is data resulting from a loadcell calibration.

x (kg.) INPUT y (mV.) OUTPUT

-1.5

4.34

9.52

14.64

19.20

26.60

29.55

0

5

10

15

20

25

30

A linear calibration y-regression fit yields the relation

y=1.0525x-1.166071429

Compute the seven y-data deviations from this line.

1

2

3

4

5

6

7

Now compute the mean deviation x

and the standard deviation of these deviations from this mean

S

Based on Thompson’s method, pp.149-150 and table 6.8 in Wheeler

and Ganji, which, if any, of these deviations qualify as outliers?

Do not confuse x with x.In the first instance I’m using standard notation for sample mean.

The other just means data on an x/y graph.

-0.333928571

0.243571429

0.161071429

0.01857143

-0.68392857

1.45357143

-0.85892857

0.000000001

0.7654531026

6 =1.453>1.3097 = S7

O P

Q

R

C

O(0,0)

P(2,0)

Q(2,1)

R(1,2)

r

C(0.939726079,0.924164211)

r=1.22539915

x =

y =C

C

Name:-

Student Number:-

key

MECH 262

Final Exam 06-04-21

(FD1)STX64uG

r=


Assuming you have digested the circle fitting

discussed in the last two lectures, study the

attached two pages. Using the coordinates

of centroid G and the three eigenvector direction

numbers, figure out the circle centre coordinates and

radius magnitude.

G1:1.25:0.75:3.5 -13.89011176

-13.66009144

7.390510948

e =

0.939726079

0.924164211

1.22539915

keyName:-

Student Number:-



(FD1)MLT621A

1. Given a "semi-circular" and a sinusoidal wave form of the same peak amplitide A

what is the ratio of their RMS amplitudes (voltage)?

A

A

A1.10

1.15

1.20

1.25

1.30

(check one)

2. Do (-176)+(-333)=(-509) in 2s complement binary arithmetic. Express the result as

a) A 2s complement, 12 bit binary number and b) As a string of 6 hexadecimal digits.

(-176)

(-333)

(-509)

+

=

1 1 1 0 1 0 1 0 0 0 01

1 1 1 0 1 0 1 1 0 0 1 1

110000111 0000

E 0 3 Hex.

3. Sometimes change in frequency

is expressed in decades , increase or

decrease by factors of 10 rather than

factors of 2 or octaves. How many

octaves per decade?

Express a roll-off of 6dB/octave as

dB/decade.

3.322

19.932

keyName:-

Student Number:-



top

rightfront

F

-F1. A force transducer consists of a 4-arm SR4 (nominal

300ohms each) strain gauge bridge mounted on a solid

steel cylinder compression element 10mm dia. and

20mm high. What force F does this element sustain at

1000 micro-strain in the direction shown? 16.26kN

2. If Poisson’s ratio for the strain gauges’ metal sensing

Gwires is 0.3, what is the gauge factor F ? 1.6

A B3. What is the bridge output voltage V -V if the axial

compressive strain is 1000 micro-strain and the supply

10.406mVvoltage is 10V?

10V0V

A

B

R w

R x

R y

R z

R w

R x

R y

R z

300.144ohms

299.52ohms

300.144ohms

299.144ohms

This is to help you

get part marks.

(FD1)MLT63aB

key

(FD1)STT63cC

Midterm Test 06-03-03 ,40min.

MECH 262

Name:-

Student Number:-

0.51. A probability density function is given by a sine wave function

P(x)=[sin(x)]/2 in the interval x=0 to 180deg.

Compute the mean, , the variance, v, and the standard deviation, .

2. 180 seats are sold for an aircraft with a 175 passanger capacity. If, on average,

1 passanger in 20 is a "no-show", what is the probability that "overbooking" will occur?

I.e., more than 175 passangers show up.

3. If a barber can process 5 customers per hour what is the probability that someone will

have to wait beyond the one-hour period in which he/she arrived if, on average, the

barber processes 40 customers during an 8-hour shift?38.4%

37%

90 = /2 (2/4)-2 0.6836673905=0.4674

(express as %)

(express as %)

keyName:-

Student Number:-


Final Exam, MECH 261/262, 05-04-27, 2 hours

1. Uncertainty analysis:- Given an end loaded cantilever displacement transducer

that obeys the relationship =2L 3h(L-C)

3

find the % bilateral tolerance uncertainty

if L=200mm 0.1mm+

=0.001 0.00001+

h=4mm 0.01mm

C=20mm 0.1mm

+

+

... ... in ...C L

h% uncertainty =1.037%

(FD1)MLX541A

2. A chromel-constantan thermocouple junction is placed in an oven. The reference

junction is in a brine-ice bath at an equilibrium tempreature of -3C. If the measured

voltage output is exactly 40mV what is the temperature in the oven, according to linear

interpolation of the table on p.277?

V

-3C40mV

deflection =7.407mm

TT= 535.05C

3. Refer to the table , the figure

and the equation on the next page.

A 100ohm at 0C resistance thermometer

behaves according to the table and is connected as shown.

R is adjusted to make V =0 at 0C2 o . Find V at 390C if a) the lead resistances are R =0

and b) R =2ohms. All three lead resistances are identical, R .L

L

L

o

Express V to theo

nearest microVolt.

a)

b)

-0.422049V

-0.417226V

keyName:-

Student Number:-


Final Exam, MECH 261/262, 05-04-27, 2 hours

4. An 8-bit digital displacement transducer returns the reading shown below. It is coded

to represent a straight binary (unsigned) integer but it is Gray coded and must be decoded

using the digital circuit shown. Put the readable binary result in the second register.7 6 5 4 3 2 1 0

1 0 1 0 1 0 1 0

3

71 1 0 0 1 1 0 0

5. Air of density =1.2kg/m flows at velocity V=20m/s in a duct of internal diameter3

D=0.2m. Find the Reynolds number Re. I looked for air viscosiy but all I could find was

stated in old Imperial units, i.e., at room temperature µ= 0.038x10-5

slugs/(ft-sec) !

Note that a slug is 32.16 pounds, there are 2.2 pounds/kg, 25.4mm/inch and 12 in/ft.

Re= 2.6x105

Re= VD/

6. You’ve been hired by MTC to track down possible leakage from their 750V(DC) Metro

"3rd rail". All you’ve got are 22kOhm resistors and a crappy 10k Ohm input impedence, 10V

full scale meter. Using the circuit you designed (below), can you detect if the supply

drops to 700V if you can read the scale to ±0.1V?

....100x22kOhm resistors

...

750V

(700V)

V10k

V@750V

V@700V

yes

no

2.36

2.20

(FD1)MLX541B

keyFinal Exam, MECH 262, 05-04-27, 1 hour additional

(FD1)STX541C

7. Given the distribution function f(x), find x and f(x) at the mean, mode and median

values. Make appropriately precise estimates where analysis bogs down.

x

1

2

0

0 2 4 6

f f=x /82 x f

mean

median

mode

0

42 2

4

6

(x-4) +f -4=0

2.7821

0.968045.75011

4 2

4.1188 1.9965

Name

Student Number

8. A sample of 10 eggs has an average weight of 71.5g with standard deviation of 0.5g.

Examine a claim of 71g average weight for the population at a LoC of 99%.

Is < or > 71g?

What about 15 eggs averaging 71.4g with S=0.49g?

71.5 - Is < or > 71g?71.5 -0.542 0.377

9. Running the 15-egg test 13 times reveals an average standard deviation S= 0.483g.

Estimate the probable range of the population standard deviation at a LoC of

<<

95% .

LoC=Level of Confidence

0.3605g 0.797g

keyName:-

Student Number:-


1. A cheap 1000ohm input impedence voltmeter is used to measure the terminal voltage

of a battery with an internal resistance of 0.5ohm. The reading is 1.4V.

Estimate the reading error in mV. Is it too high or too low? mV

V

0.7 low

2. Answer the six questions below.

10k-ohm

10k-ohm

In dB that is dB

The gain is G=

10k-ohm

dB

The gain is G=

In dB that is

10k-ohm

In dB that is dB

The gain is G=

100k-ohm

100k-ohm

3. The RMS voltage of the asymmetric sawtooth waveform is 7V. Find the peak voltage.

V

V

0t

20ms 10ms

T/2 (hint)

Midterm Test part 1, 05-02-28, 1 hour

(FD1)MLT522A

1

0

10

+20

1/10

-20

These are inverting op. amps. but use output

magnitude, ignore sign change.

7 3 = 12.124

keyName:-

Student Number:-


Midterm test part 2, 05-03-02, 1 hour

4. Represent 379 in binary, octal and hexadecimal.10

0123456789

Octal Hexadecimal0 1 0 1 1 1 1 0 1 1 573 17B

Represent its 2’s and 10’s complement.

0123456789

1

11 10

1 1 1 0 1 0 0 0 0 1 0 2’s complement 10’s complememt... 999621

Fill all squares with 1 or 0.

5. Do a linear fit on the four given points, i.e., find a and b in y=ax+b such t hat

[y -(ax +b)]i i2

is minimized.

Express a and b as reduced fractions like 3/4, not like 12/16.

a=

b= (1,1)

(2,3)

(3,4)

(5,4)

x

y

24/35

39/35

(FD1)MLT52bB

6. Label the gauges on the bridge A,B,C,D where means tensile and means compressive.

If E=207GPa and =0.207GPa and =0.3, compute the measured out-of-bal ance

Signal V in mV. Gauges are 300ohm steel SR4’s mounted on a steel block.

V

+5V

100kN

30mm

20mm

-100kN

BA

CD

A

C

B

D

V= mV2.617485

max.

keyMidterm Test part 3, 05-03-04, 1 hour

MECH 262

7. In a general poplation of non-smokers lung cancer is the cause of death in 1 out of

a 1000 deaths . A random sample of 250 smokers who died reveals 1 lung cancer death.

Based on binomial analysis, what is the probability that the frequency of lung cancer

death among smokers is no more frequent than among smokers? Express your answer

as a decimal fraction p, 0<p<1. p= 0.21296626

Note that these are the sort of data that the upcoming anti-tobacco class-action

lawsuits will be concerned with thereby making many lawyers very, very rich.

8. Magna Corp. manufactures a great many automotive pistons of mean diameter 100 mm.

The standard deviation is æ 10 microns. Assuming normal Gaussian distribution, what

percentage of production can be assigned to "oversize" stock, i.e., to recondition worn

engines by reaming out the cylinders to slightly greater diameter and and fitting oversize

pistons in the range +15 to +20 microns? % to oversize stock4.4

(FD1)STT53cC

9. Customers arrive at an average rate of 50 per hour. The sole cashier checks o ut

shoppers in exactly 1 minute. What is the probability that during any given one minute

interval there are more than 2 customers at the check out? Express the probability as a

percentage. % of the time there are more than 2 waiting.5.234463

Name:-

Student Number:-

keyName:-

Student Number:-


Final Exam 04-04-29, 2hrs. 261, 3hrs. 262

g g g g g g g g

b b b b b b b b 01234567

7 6 5 4 3 2 1 0 8-bit binary Gray code digital signal.

A 0-10v "flash" ADC outputs an

Express the straight binary number

output from the Gray-to-binary

convertor.

= 0

= 1

(FD1)MLX443A

This represents the decimal integer 135 109 91 75

that corresponds to 3.569v 5.294v 2.941v 4.275v

= true

= false

Question 1.

Topics 1,2 &6

keyName:-

Student Number:-



v

300 typ.

+

Output impedance 75 150 225 300 375 450 525 600 1200

Using Thevenin’s theorem, find the output

impedance of a standard four resistor

strain gage bridge circuit. All arms

contain elements of identical resistance.

9v

What is the current drain on the battery?

120mA 60mA 40mA 30mA 24mA

20mA 17mA 15mA 7.5mAnone ofthese

none ofthese

Question 2.

Topics 3 & 5

(FD1)MLX443B

keyName:-

Student Number:-



A photodiode may be used in an absolute or incremental displacement transducer.

A light emitting diode (LED) " " "

A thermocouple is used together with a bridge circuit.

A thermistor may be used in a bridge circuit.

A bimetallic strip may be used as a digital temperature sensor.

A bimetallic strip may be used as an analog temperature sensor.

A strain gage torque/angular displacement transducer may be modelled as

a 1st order system.

An advantage of an advanced system like LabView is that analog sensors like

thermocouples and strain gages become unnecessary.

A digital to analog convertor may be used in a system to measure voltage.

Question 3.

Topics (miscellaneous)

(FD1)MLX443C

= true

= false

keyName:-

Student Number:-



(FD1)MLX443D

N

S

0 Lt

0 M

GPS satellite

earth

r=6375km

Alt.=40250km

45 NLt

75 WLg

Question 4.

Topics 15 & 18

180 -

Montréal is at approximately 45N-75W. Using the satellite position over the Equator

and on the plane of Greenwich, the 0-180 degree meridian compute the distance

between the satellite and Montréal at this instant.

=44731.181km

=45114.424km

=45888.298km

=46019.532km

This 3-digit precision is enough to resolve a difference of

1m "on the ground". True

FalseHint: Don’t forget how to find Cartesian

coordinates from a spherical frame.

(x,y,z)

Hint: Subtract vectors.

plane of orbit

path of orbit

Montréal

keyName:-

Student Number:-


Final Exam 04-04-29, 2hrs. 261, 3hrs. 262Question 5.

Topic 22

(FD1)MLX443E

water

waterv 1

v 2

The pipe diameter is 600mm. The orifice (obstruction) meter is of diameter 100mm.

Estimate the mass flow in kg/s.

100mm 600mm

With C =1, estimate the rate of fluid

mass flow in the pipe shown above.d

200mm

Use g=9.81 m/s , =1000 kg/m .2 3

1.401kg/s 1.441kg/s 11.113kg/s 11.001kg/s 14.007kg/s 14.407kg/s

keyName:-

Student Number:-



(FD1)MLX443F

Question 6.

Topic 22

Reading instruments and making conversions are essential in obtaining measurements.

It reads in inches (in) of mercury (Hg). The corresponding reading of a thermometer

on the wall was 22.5C. What was the density of air in my office?

An aneroid (desk) barometer was photographed in my office.

1.9923kg/m 1.2057kg/m 1.2093kg/m

1.2104kg/m 1.2187kg/m 1.2216kg/m

Take mercury density as 13590kg/m

Absolute zero is -273.3C

1in = 0.0254m

3 3 3

333

3

30

31

29

28

"Hg 30.255

This is a facsimile of the photograph

that was very hard to read and made

candidates unhappy.

MECH 2004 Measurement Final Answers andSolutions

Ques. 6 only, more to come(FD1)FX04Sol79x.tex

September 26, 2007

1 Aneroid Barometer, Given and Find

Just for fun this question will be done backwards. I.e., it will be assumed that the air densityρ = 1.2104kg/m3 and temperature T = 22.5C are given and one must find the barometer readingin inches of mercury (′′Hg).

2 Solution using Imperial Units

Because I recall the specific gas constant for air as R = 53.3ft#(f)/[#(m)R]. First we do thetemperature conversion to absolute Fahrenheit, (R), called degrees Rankine.

TR = (22.5 + 273.3)9

5

The perfect gas law saysp

ρ= RT

so the density in #(m)/ft3 is required.

1.2104× 2.2

(39.37/12)3= 0.075404816

Notice the use of the necessary conversion factors between kg and #, ′′/m and ′′/ft. Finally oneobtains the air pressure in #(f)/′′2 (psi) as follows.

p =0.075404816× 53.3× (22.5 + 273.3)9

5

144= 14.86053625

Dividing the absolute pressure by 14.696psi, a standard atmosphere, one gets 1.011195988Atm.Knowing that this is equivalent to 760mmHg or 29.92”Hg the barometer reading should be

1.011195988× 29.92 = 30.255

an that is what the meter says. Now do the problem as asked, using the barometer reading to getρ in kg/m3.

(FD1)FX04Sol79z.tex.tex

1

keyName:-

Student Number:-



(FD1)MLX443G

Question 7.

Topic 10

100mm

At 1000 this solid cylindrical steel torsion bar torque transducer sustans a torque of

Use E=2.07x10 N/m11 2

40.64kNm 4.064kNm 20.42kNm 25.18kNm 36.74kNm

=

keyName:-

Student Number:-


Final Exam 04-04-29, 2hrs. 261, 3hrs. 262Question 8

Topic 12

(FD1)MLX443H

Regard the four thermocouple/millivoltmeter circuits as systems with two voltage sources

and an instrument of infinite input impedance.

mV

mV

mV

mV

100C

100C

100C

100C

0C 0C

20C

20C

+

-

+ -

+

-

+

-

+

-

+ -

- +

- +

-A--B-

-C-

-D-

Rank the circuit outputs from 1 (lowest reading)

to 4 (greatest absolute reading value).

-A-

-B-

-C-

-D-

1 2 3 4

20C 20C

10C

10C

Hint:- Assume output

=0mV at 0C.

keyName:-

Student Number:-

Question 9.

Topics 7,8

(FD1)STX443I

Final Exam 04-04-29, 3hrs. 262

Given five points (2,1),(3,2),(4,3),(5,5),(6,4), find the line of best least-squares

normal distance fit.

Which of the points are on this line, in the order presented above?

(2,1) (3,2) (4,3) (5,5) (6,4)

Carefully plot the straight line and points on the graph below. Indicate points with (+)

and G, the centroid with (*).

1 2 3 4 50

0

1

2

3

4

5

6

(Statistics)

keyName:-

Student Number:-


(Statistics)

Question 10.

Topic 1

(FD1)STX443J

Volkswagen offers three types of "Golf". (CL, GLS, GT)

(2L, 1.8T, VR6, 1.9TD)Some of these are available with four engine choices.

All three car models are available with either manual or automatic transmission. (5Sp, AT)

Not all five colours are available with all models.

(Bk=black, Bl=blue, Yw=Yellow, Gr=green, Gy=Grey)

You can’t get a grey CL or one with a 1.8T or VR6 engine.

The GLS is available in all possible combinations.

There are

26 33 512 616 1000 1331

....

.... different ways to configure your Golf.

CL

GLS

GT

2 8 6 9 5 A k l w r yThis table may help

70472

There are no 2L GT’s and you can’t get a blue or green one.

This is the tree for ...

CL GLS GT

keyName:-

Student Number:-


(Statistics)

Topic 10

Question 11.

(FD1)STX443K

The correlation coefficient of the line fit in Question 9., specified by is ...min

max

1/16 1/17 1/18 1/19 1/20 1/21 1/22 1/23 1/24

The best coefficient possible for any fit is 0

keyName:-

Student Number:-


(Statistics)

Question 12.

(FD1)STX443L

Topic 6

CTM claims that the 24 bus at Sherbrooke and McGill College arrives at the time scheduled

within a standard deviation of plus/minus one minute. What is the probability that any given

bus will arrive within this two-minute interval if arrivals are normally distributed?

If the interval between buses is 15 minutes, what is the probability the bus you’re waiting

for won’t arrive? Assume you got there exactly on time.

What’s the probability, upon your arriving at the stop exactly on

time, that the bus has already passed?

Assume the claim is valid.

0 0.5 0.68260.3413 0.4772 0.9544

0 0.5 0.68260.3413 0.4772 0.9544

0 0.5 0.68260.3413 0.4772 0.9544

Hint:- See "Appendix 3" in notes.

If table doesn’t show the

interval you need, extrapolate.

The bus is late, not early.

1. A thermistor is

a) A semiconductor device

where k is constant.

.

c) Is a resistor.

d) Is used to measure strain.

T F

2. An 8-bit differential ADC with an input voltage range of 10v+

should be augmented with a differential input instrumentation

amplifier of gain x10 x10 x10 x10 x10-2 -1 0 1 2

Answer by blackening the

(They are only

available with

in order to achieve maximum resolution

when measuring a signal expected to vary between -30 and +50v?

3. What is the precision in counts/volt achieved in 2., above?2 2 2 2 2 20 1 2 3 4 5 2 6

(FD1)MLT42sA

key

right rectangles.b) Has a resistance -vs- temperature

behaviour given by R=k T

Name:-

Student Number:-


Midterm Test 04-02-19, 80min. 261, 120min. 262

a)

b)

c)

d)

settings of x10 )æn

2. An 8-bit differential ADC with an input voltage range of 10v+

should be augmented with a differential input instrumentation

amplifier of gain x10 x10 x10 x10 x10-2 -1 0 1 2

(They are only

available with+n

in order to achieve maximum resolution

when measuring a signal expected to vary between -30 and +50v?

3. What is the precision in counts/volt achieved in 2., above?2 2 2 2 2 20 1 2 3 4 5 2 6

keyName:-

Student Number:-



settings of x10 )

Q 3.:- Raw:- Input span is +10/-10v=20v. Output span is 255 counts.

20

Since 12 is equidistant from 2 and 2 , either would be acceptable.3 4

With amplifier:- Input span is 200v. Output span is still 255 counts.

Therefore 255 =1.225counts/v. Again, round down.

200

Therefore 255 =12.25counts/v. Always round down in integer arirhmetic.

Since 1=2 , this too is acceptable.0

The -30v to +50v doesn’t change anything.

(80/200)255=1.225

80

(FD1)MLT42sA1

Name:-

Student Number:-


key

(FD1)MLT42sB

4. Represent the number 57 in binary

in the top 8-bit field, its one’s

complement in the next and its

negative, two’s complement,

= "one"

= "zero"

+ 6 5 4 3 2 1 0

5. What is the shear modulus G of aluminum whose Young’s modulus

E=0.70x10 Pa and whose Poisson’s ratio is µ=0.33?11

Choose the closest value. x1011Pa0.40 0.35 x10

11Pa 0.30 x10

11Pa

0.25 x1011Pa 0.20 x10

11Pa 0.15 x10

11Pa

6. What is the peak-to-peak voltage of a 220v AC RMS domestic

power supplied by Hydro-Qu bec? 120v 220v 320v 420v

520v 620v 720v

(sinusoidal)


in the bottom one.


right rectangles.

Name:-

Student Number:-


keyMidterm Test 04-02-19, 80min. 261, 120min. 262Answer by blackening the

right rectangles.

(FD1)MLT42sC

7. The linearized first order temperature measuring system refered to

on the included sheet of explanation is sitting at 20C and is plunged

into boiling water at 100C, suddenly, at t=0.

The initial voltage output was 1.6mv. After a long time the output is 4.3mv.

Find k =k and specify the units.0

80/2.7 2.7/80 120/5.9 5.9/120 -2.7/120 -80/2.7

v/C C/mv mv/C

8. The exercise above is repeated but this time a reading at t=1.4s

yields an output of 2.9mv. Find the time constant and coefficient c.

c= 4.918 6.213 8.043 5.675 k/c= 1.894 5.221 -2.227

Name:-

Student Number:-


key


Midterm Test 04-02-19, 120min. 262

STATISTICS 9. Given the 4 points (1,1),(2,1),(3,3),(4,2) do a best least squares sum fit

minimizing the sum of x .2

neatly.

a’= -10/11 10/11 -11/10 11/10

b’= 1 2 3 -1 -2 -3

I.e., Find a’ and b’ and plot results below,

-3

0

4

4x

(FD1)STT42sD

and drawing a straight line.

right rectangles

Name:-

Student Number:-


key



STATISTICS 10. Using the results from Question 9, compute the variance, v, and the

standard deviation, , of y in y=a’x+b’. Use n=4 because we are

refering these to the line which we now know. The first entry in the

i

1 1.1-1=0.1 1 +0.9

2

3

4

1

3

2

0.81

y’=a’x+b’

2.2-1=1.2

3.3-1=2.2

4.4-1=3.2

-0.2

+0.8

-1.2

0.04

0.64

1.44

2.93

4

the right rectangle.

(FD1)STT42sE

v= 0.417 0.189 0.733 0.629

= 0.718 0.856 0.337 0.504

table below is done to make your job easy.

Name:-

Student Number:-


key



STATISTICS

the right rectangle.

11. A lazy student attempts this test by randomly choosing to fill

rectangles for all previous questions except Question 4 which

requires multiple filling. All questions are of equal value so consider these

nine questions, 1,2,3,5,6,7,8,9,10 will yield a perfect score of 10 each =90.

What is his expected score based on the law of large numbers?

Falses are not deducted from Trues in T/F questions.

If a question has 2 parts, each is worth 5 marks.

Do not include this question.

*

* I don’t think the ladies will mind my political incorrectness.

5 10 20 25 30 35 40 50

Sorry about the lack of a 4th statistics question and for leaving out binomial, Poisson’s

distribution, rotten eggs, cancer and all that stuff. You’ll see it on the final. I Think

we’ve all had enough by now and somebody’s gotta grade it.

(FD1)STT42sF

Download - MECH 261/262 Measurement Lab (& Statistics)

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