Solution Manual For Introduction To Engineering Experimentation, 3rd Edition

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Solution Manual For Introduction To Engineering Experimentation, 3rd Edition offers step-by-step solutions to help you understand tough concepts with ease.

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1.1
CHAPTER 1
1.1 (a) The weight of a person is 175 pounds which is equivalent to 778 Newtons.lbf175
*lbf
N
1
4482.4 = 778 Newtons
(b) The magnetic force of 1,200 dynes is equivalent to 0.0027 pounds or 0.012 Newtons.dyn1200
*dyn
N
1
10 5− = 0.012 Newtonsdyn1200
*dyn
N
1
10 5− *N
lbf
4482.4
1 = 0.0027 pounds
(c) The force exerted by the spring is 10 Newtons which is equivalent to 1,000,000 dynes or
2.248 poundsN10
*N
dyn
5
10
1 − = 1,000,000 dyneN10
*N
lbf
4482.4
1 = 2.248 pounds
1.2 (a) The tire pressure is 30 psig which is equivalent to 2.069 bars or 2.041 atm.
Note: 1 psig is considered to be = 1 psipsig30
*psi
atm
696.14
1 = 2.041 atmpsig30
*psi
kPa
1
895.6 *kPa
bar
100
1 = 2.069 bars
(b) The compressed air at 10 bars is equivalent to 145 psig or 9.87 atm or 295.3 inches of
mercury or 102 meters of waterbars10
*bar
kPa
1
100 *kPa
psi
895.6
1 = 145 psibars10
*bar
kPa
1
100 *kPa
psi
895.6
1 *psi
atm
696.14
1 = 9.87 atmbars10
*bar
kPa
1
100 *kPa
psi
895.6
1 *psi
atm
696.14
1 *atm
cmHG
1
76 *cm
in
54.2
1 = 295.3 in of Hgbars10
*bar
kPa
1
100 *kPa
psi
895.6
1 *psi
atm
696.14
1 *atm
cmWater
1
2.1033 *cm
m
100
1 = 102 meters water
(c) 50 cm of water is equivalent to 0.0490 bars or 0.711 psi or 1.45 inches of mercurycmWater50
*cmWater
atm
2.1033
1 *atm
kPa
1
325.101 *kPa
bar
100
1 = 0.0490 barscmWater50
*cmWater
atm
2.1033
1 *atm
kPa
1
325.101 *kPa
psi
895.6
1 * = 0.711 psicmWater50
*cmWater
atm
2.1033
1 *atm
cmHG
1
76 *cm
in
54.2
1 = 1.45 in of Hg

1.2
1.3 (a) The household energy use of 750 kWh is equivalent to 2,559,241.71 Btu or 645,007
kCal or 2,700,000,000 jouleskWh750
*kWh
kJ
1
3600 *kJ
J
1
1000 *J
Btu
1055
1 = 2,559,241.71 BtukWh750
*kWh
kJ
1
3600 *kJ
J
1
1000 *J
cal
186.4
1 *cal
kcal
1000
1 = 645,007 kcalkWh750
*kWh
kJ
1
3600 *kJ
J
1
1000 = 2,700,000,000 J
(b) The gas water heater uses 50,000 Btu or 52,750,000 joules or 12,601.5 kcal or
38,908,386 ft.lbfBtu50000
*Btu
J
1
1055 = 52,750,000 joulesBtu50000
*Btu
J
1
1055 *J
cal
186.4
1 *cal
kcal
1000
1 = 12,601.5 kcalBtu50000
*Btu
lbfft
00128507.0
.1 = 38,908,386 ft.lbf
(c) The amount of heat required is 250 kCal which is equivalent to 992 Btu or 250,000 Cal
or 1,046,500 jouleskcal250
*kcal
cal
1
1000 *cal
J
1
186.4 *J
Btu
1055
1 = 992 Btukcal250
*kcal
cal
1
1000 = 250,000 calkcal250
*kcal
cal
1
1000 *cal
J
1
186.4 = 1,046,500 J

1.3
1.4 (a) The automobile rating of 150 hp is equivalent to 112 kW or 82,500 ft.lbf/sec or
106.02 Btu/sechp150
*hp
W
1
7.745 *W
kW
1000
1 = 112 kWhp150
*hp
s
lbfft
1
.
550 = 82,500 ft.lbf/sechp150
*hp
W
1
7.745 *W
s
Btu
04.1055
1 = 106.02 Btu/sec
(b) The truck rating of 400 kW is equivalent to 536 hp or 295,025 ft.lbf/sec or 379 Btu/seckW400
*kW
W
1
1000 *W
hp
7.745
1 = 536 hpkW400
*kW
W
1
1000 *W
hp
7.745
1 *hp
s
lbfft
1
.
550 = 295,025 ft.lbf/seckW400
*kW
W
1
1000 *W
s
Btu
04.1055
1 = 379 Btu/sec
(c) The water heater rating of 40,000 Btu/hr is equivalent to 11.72 kWhr
Btu
40000
*hr
Btu
W
1
293.0 *W
kW
1000
1 = 11.72 kW

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1.4
1.5 (a) 50 oF is equivalent to 10 oC
(9
5 )(50 – 32) = 10 oC
(b) 150 oC is equivalent to 302 oF
(5
9 )(150 oC) + 32 = 302 oF
(c) The water temperature increase of 40 oC is equivalent to a change of 40 K or 72 oF or 72
oR
(40 oC) *C
K


1
1 = 40 K
(40 oC) *C
F


1
8.1 = 72 oF
(40 oC) *C
R


1
8.1 = 72 oR
(d) The air temperature change of 30 oF is equivalent to a change of 16.7 K or 16.7 oC or 30
oR
(30 oF) *F
K


1
556.0 = 16.7 K
(30 oF) *F
C


1
556.0 = 16.7 oC
(30 oF) *F
R


1
1 = 30 oR

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1.5
1.6 (a) 4 gallons is equivalent to 15.1 liter or 15,142 cm3 or 0.535 ft3gal4
*gal
m
1
0037854.0 3 *33
10
1
m
Liter
− = 15.1 litergal4
*gal
m
1
0037854.0 3 *36
3
10
1
m
cm
− = 15,142 cm3gal4
*gal1
0037854.0 *3
3
02832.0
1
m
ft = 0.535 ft3
(b) 10 liters is equivalent to 2.64 gallons or 10,000 cm3 or 0.353 ft3Liter10
*liter
m
1
10 33− *3
0037854.0
1
m
gal = 2.64 gallonsLiter10
*liter
m
1
10 33− *36
3
10
1
m
cm
− = 10,000 cm3Liter10
*liter
m
1
10 33− *3
3
02832.0
1
m
ft = 0.353 ft3
(c) 5 ft3 is equivalent to 37.4 gallons or 141,600 cm3 or 142 liters3
5 ft
*3
13368.0
1
ft
gal = 37.4 gallons3
5 ft
*3
3
1
02832.0
ft
m *36
3
10
1
m
cm
− = 141,600 cm33
5 ft
*3
3
1
02832.0
ft
m *33
10
1
m
liter
− = 142 liters
1.7 The air gas constant of 53.34 ft.lbf/lbm.oR is equivalent to 0.0685 Btu/lbm.oR or 287
joules/kg.K or 0.0686 kcal/kg.KRlbm
lbfft

.
34.53
*lbfft
Btu
.1
00128507.0 = 0.0685 Btu/lbm.oRRlbm
lbfft

.
34.53
*lbfft
Btu
.1
00128507.0 *Btu
J
1
1055 *kg
lbm
4536.0
1 *K
R


9
5
1 = 287 J/kg.KRlbm
lbfft

.
34.53
*lbfft
Btu
.1
00128507.0 *Btu
J
1
1055 *J
cal
186.4
1 *cal
kcal
1000
1 *kg
lbm
4536.0
1 *K
R


9
5
1 =Kkg
kcal

0685.0

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1.6
1.8 The universal gas constant is 1.986 Btu/lb mole.oR which is equivalent to 1.986 kCal.kg
mole.K or 1,545 ft.lbf/lb mole.oR or 8,314 joules/kg mole.KRlbmole
Btu

986.1
*Btu
J
1
1055 *J
cal
186.4
1 *cal
kcal
1000
1 *kg
lbm
4536.0
1 *K
R


9
5
1 = 1.986 kCal.kg mole.KRlbmole
Btu

986.1
*Btu
lbfft
00128507.0
.1 = 1,545 ft.lbf/lb mole.oRRlbmole
Btu

986.1
*Btu
J
1
1055 *kg
lbm
4536.0
1 *K
R


9
5
1 = 8,314 J/kg mole.K
1.9 The thermal conductivity is 200 W/m.oC or 116 Btu/hr/ft/oF or 0.048 kcal/sec.m.oCCm
W

200
*Cm
W
Ffthr
Btu


7307.1
..
1 = 116 Btu/hr/ft/oFCm
W

200
*W
s
Btu
04.1055
1 *Btu
J
1
1055 *J
cal
186.4
1 *cal
kcal
1000
1 = 0.048 kcal/sec.m.oC
1.10 The thermal conductivity is 50 Btu/hr.ft.oF or 86.54 W/m.oC or 20.7 Cal/sec.m.oCFfthr
Btu
.
50
*Ffthr
Btu
Cm
W


.
1
7307.1 = 86.54 W/m.oCFfthr
Btu
.
50
*Ffthr
Btu
Cm
W


.
1
7307.1 *W
s
Btu
04.1055
1 *Btu
J
1
1055 *J
cal
186.4
1 = 20.7 cal/sec.m.oC

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2.1
CHAPTER 2
2.1 (a) The mechanical speedometer measures vehicle speed by measuring the angular
velocity of the wheels. The angular velocity of the wheels (through some gears) causes a
flexible cable to rotate. This cable causes a magnet to rotate inside a metal cup creating a
circumferential drag on the cup (a torque). The drag is sensed by allowing the cup to rotate
less than a complete turn against the resistance of a torsional spring. The cup is connected
to a pointer which can be compared to a stationary scale.
The sensing element is the magnet rotating inside the metal cup creating drag. The signal
modification system is the system that allows the cup to rotate against a torsional spring.
The indicator is the comparison of the pointer to a stationary scale.
(b) The fuel level inside a fuel tank is measured with a mechanical float which follows the
surface of the fuel (see Ch. 10). The position of the float is sensed with a connected arm
which rotates a rotary variable resistor (angular potentiometer, see Ch. 8). The resistance
of the variable resistor is sensed by passing applying a voltage to the resistor and
measuring the current with an electromechanical gage. The sensing element is the float.
The signal modification system consists of the angular potentiometer and the indicator is the
electromechanical gage.
(c) Most of these devices use a variable resistance device called a thermistor. The
resistance of the thermistor is a strong function of it’s temperature. The resistance is
sensed by passing applying a voltage to the device and sensing the current, which will be a
function of temperature. The current is then converted to a digital form which is then output
to the display. In this case, the thermistor is the sensing device. The signal modification
system is quite complicated including the creation of the current and the conversion to
digital form. Finally, the indicator is the liquid crystal display.
Wheel
Gears
Flex
Cable
Rot.
Magnet
Cup Pointer

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2.2
2.2 True Value = 0.5000 inches
Determination of Bias Error.
First, the average of readings must be calculated:
Average of Readings =in4821.0
48214.0
10/)4822.04820.04822.04821.04822.04820.04821.04821.04824.04821.0(
=
=
+++++++++
Bias Error = Average of Readings -True Value= −
= −
= −
0 4821 0 5000
0 01786
0 0179
. .
.
. in
Determination of Maximum Precision Error:
Maximum Precision Error =Largest difference between a single
reading and the Average of Readings


 

= −
=
0 4824 0 4821
0 0003
. .
. in
2.3 Readings:
20.2, 20.2, 20.6, 20.0, 20.4, 20.2, 20.0, 20.6, 20.0, 20.2 (lb)
First determine the average of the reading:
Average W = 20.2 lb
For bias error,
Bias Error = Average Value -True Value
= 20.24-20.0
= 0.2 lb
For maximum precision error, we need the reading with the greatest deviation from
the average reading (20.6 lb). Therefore,
Maximum Precision Error = 20.6 -20.2 = 0.4 lb
2.4 (a) Intrusive; The thermometer causes a loading error.
(b) Non-intrusive; The photography does not affect the speed of the bullet at any time.
(c) Non-intrusive; Optical thermal radiation device would yield a non-intrusive measurement
as long as it is insulated from the furnace.
(d) Non-intrusive; The speed of the car is unaffected by waves measured by the radar gun.

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2.3
2.5 (a) A single conducting wire induces a small magnetic field around itself and if it is
alternating current, will induce an alternating magnetic in the clamp on ammeter. The clamp
on ammeter will have a negligible effect on the current in the wire and for all practical
purposes is non-intrusive.
(b) The orifice meter (see Ch.10) measures fluid flowrate by obstructing the flow in pipe and
measuring the resultant pressure drop. The pressure drop is significant and this device is
intrusive.
(c) This device passes a beam of infrared radiation through the gases which absorb some of
the radiation (see Ch. 10). This measurement has no effect on the composition of the gases
and negligible effect on the gas temperature. It is non-intrusive.
(d) This device measures rotational speed by shining a pulsing light on a mark on the shaft
and adjusting the pulsing rate until the mark appears stationary (see Ch. 8). The light has
negligible effect on the rotation of the shaft and is non-intrusive.
2.6 (a) Bias Error; The output will consistently deviate from the true value.
(b) Precision Error; The speedometer output shows data scatter.
(c) If the difference is consistent with time, then it is a bias error - either of calibration or
spatial error. If the difference varies with time, it is precision.
2.7 (a) In most cases, this error will by systematic since repeated measurements at the
same time will produce the same error. However, if measurements are made over a long
period of time and the temperature varies randomly, the error will be random.
(b) This error will be the same each time the measurement is made by the same person
with the same procedure and hence is systematic. However, if the measurement is made by
several people using different procedures, it may appear to be random.
(c) This error will be the same each time the measurement is made and is always
systematic.
2.8 (a) This error is usually considered systematic if the readings are all made at the same
ambient temperature. However, if the readings are taken over a period of time and the
ambient temperature varies randomly, then the error will appear random.
(b) This is always random since the fields normally vary in a random manner.
(c) Since this is a malfunction, it is not predictable in occurrence so it would be considered
random.
2.9 (a) This error is usually considered random even if the readings are all made at the
same conditions since corn-growing conditions are highly dependent on various factors.
(b) The deterioration of asphalt/concrete in a highway is a combination of factors that needs
further analysis since the same-grade concrete will deteriorate at a same rate which gives a
systematic error, but given the various conditions at which the different portions of the
highway is exposed to the elements, each section will have a random error.
(c) The variation of height of the same type of tree in an orchard qualifies as a random error
since the height of each tree is highly dependent various growth factors.
(d) The variation of drying time of concrete columns of a highway is subjected to both
systematic and random error; only if the drying conditions are constant for all columns and
the concrete grade is uniform throughout the highway will the error be limited to systematic.

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2.4
2.10 (a) The variation in access for a popular website would usually be considered a
random error since it is dependent on unknown factors.
(b) The variation in average access per day of a popular website would also be considered
a random error since it is due to uncontrollable factors.
(c) The variation in the rider-ship of a bus or train line would usually be considered a
random error since it is dependent on unknown factors.
(d) The variation in the rider-ship of a bus or train would also be considered a random error
since it is due to uncontrollable factors.
2.11 Resolution or readability does not necessarily give any information about accuracy so
we cannot make any statement about accuracy. However, the digital device can be read to
only 1 part in 999 of the full scale reading. It may be possible to interpolate between
divisions on the analog device giving an effective resolution that is better.
2.12 The span is 50 - 0 = 50 m/s.
2.13 (a) The span is 50 - 5 = 45 psig
(b) 70 cm vacuum is taken to be 70 cm of mercury which is equivalent to 93 kPa. Thus, the
span is 200 - 93 = 107 kPa
(c) The span is 150 - 10 = 140 kPa
2.14 Device (D) would be the best. Device (C) is really the closest in its range. However,
measurement errors might cause device (C) to be over range for some measurements
producing meaningless results.
2.15 Manufacturer Accuracy = 2.0% of full scale
= 0.02(30V)
= 0.6V
% uncertainty of accuracy =  (0.6V/5V)(100) = 12%
The resolution of the device is 0.1 Volts. With a reading of 5V,
% uncertainty of resolution = (0.1V/5V)(100) = 2% (or 1%)
2.16 (a) The maximum reading for each range will be 2.999, 29.99, 299.9 and 2999. and the
resolution uncertainty will be 1 in the least significant digit. So the resolution uncertainty will
be 0.001V, ).01V, 0.1V and 1V for the three ranges. This could also be viewed as 0.0005V,
0.005V,0.05V and 0.5V
(b) The uncertainties will be 2% of full scale. This is .02*3 for the lowest scale or 0.06V.
Similarly for the higher ranges, the uncertainties will be 0.6V, 6V and 60V.
(c) The resolution uncertainty is negligible compared to the accuracy. Hence we can use the
results of part (b). For the 30 V range the relative uncertainy will be 0.06/25 = 2.4%. For
the higher ranges, the uncertainties are 24% and 240%.

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2.5
2.17 Since the device reads 0.5 psi when it should read zero, it has a zero offset of 0.5 psi
which will affect all readings. Zero offset is not a component of accuracy. The accuracy
specification of 0.2% of full scale gives an uncertainty of 0.00250 = 0.1 psi. This means
that we can have an expected error in any reading of 0.50.1 psi. For an applied pressure of
20 psi, the reading would be expected to be in the range 20.4 to 20.6 psi.
We can reduce the expected error by either adjusting the zero (if possible) or by subtracting
0.5 psi from each reading. It may be possible to reduce the error due to the accuracy
specification by a calibration of the gage.
2.18 With 2V reading when leads are shorted together,
Error 1 = 2V
Error 2 = 4%(100) = 4V
 Maximum Total Error = +6/-2V = (6V/80V)(100) = +7.5%
With 0V reading when leads are shorted together,
Error = 4%(100) = 4V
 Maximum Percent Error = (4V/80V)(100) = 5%
2.19 The range of both temperature gages will allow the intended 300 C measurement. The
uncertainty for each of the two gages is 2% of its span; this gives an uncertainty of 8 C and
18 C for the temperature gage with the smaller and larger span respectively. Thus, the
temperature gage with range of 100 C to 500 C should be selected since there is smaller
uncertainty.
2.20 The sensitivity is output/input. This is (125-5)/1000-100) = 0.1333 mV/kPa.
2.21 The sensitivity is output/input. This is (150-10)/100-10) = 1.556 mV/psi
2.22 The relationship between pressure and temperature is given by:PV mRT
P mRT
V
=
=
The sensitivity is given by:dP
dT
mR
V
P
T R m and V cons ti
i
= = ; , , tan
 We see the sensitivitydP
dT is proportional to the initial pressure and is changed when the
initial filling pressure is changed.
2.23 (a) The sensitivity from A to C is not a constant and gets smaller from A to C.
(b) If a high degree of sensitivity is required, use A-B. For most purposes, B-C would not
be recommended due to the sensitivity approaching zero at C.
2.24 Usually, the maximum output increases proportionally with increasing range so the
sensitivity will be unchanged.
2.25 Installing an amplifier will increase the sensitivity.

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2.6
2.26 (a) For this device, the output is P and the input is Q. So the sensitivity is simply the
derivative of P with respect to Q. Solving for P we get:2






= C
Q
P
then2
2
C
Q
dQ
Pd =

(b) The sensitivity increases with flowrate and with pressure drop.
(c) This device is best for values of P which are high relative to the design value. At 10% of
the design Q, the pressure drop will be only 1% of the design pressure drop.
2.27 Ideal sensitivity = 0.1 cm/N
Input span = 200 N
 Ideal output span = Input span  Ideal Sensitivity
= 200N  0.1cm/N
= 20 cm
Actual sensitivity = 0.105 cm/N
Actual output span = Input span  Actual sensitivity
= 200N  0.105cm/N
=21 cm
 %error of output span =( )
Actual Ideal
Ideal
− 100( )
= −
=
21 20
20 100
5%

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2.7
2.280
20
40
60
80
100
120
Gage Reading - psi
0 20 40 60 80 100
True Pressure - psi
Plot of Data for Problem 2.13
The best fit line is:
Reading = 2.9 + 0.978 (True Pressure) psi
True Deviation (Readings - Best Fit) (psi)
Pressure
(psi) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5
20 -2 -3 -2 -2 -3
40 -3 -2 -3 -3 -3
60 -3 -4 -3 -4 -2
80 -1 -1 -2 -2 -1
100 0 -1 -1 0 1
80 3 2 3 3 3
60 1 1 1 0 0
40 1 0 1 1 2
20 2 2 2 2 2
0 2 1 1 3 1
The accuracy can be evaluated from the extreme deviation values. The values are +3 and -
4 psi. For a span of 100 psi, these values result in an accuracy of +3% and -4% of full scale.
The repeatability can be evalued from Table P2.13 in the problem statement. The maximum
deviation for a given measurand and direction is 2 psi. The repeatability can then be
expressed as 1% of full scale.
Hysteresis is the maximum difference between the up and down readings for any
measurand value in the same calibration cycle. The maximum difference is 5 psi, occuring
several times (Cycle 2, 20 psi is an example). This can be expressed as 5% of span.-4
-3
-2
-1
0
1
2
3
Reading minus Best Fit - psi
0 20 40 60 80 100
True Pressure - psi
Deviation Data for Problem 2.13

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2.8
2.29-200
0
200
400
600
800
1000
1200
Gage Reading -kPa
0 200 400 600 800 1000
True Pressure - kPa
Plot of Data for Problem 2.14
Best Fit
Line
Best Fit Line:
Reading = -4.5 + 1.017 (True Pressure)
Deviations based on difference between readings and best fit line.
True Deviations (Readings - Best Fit) (N)
Force
(N) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5
200 -7 -8 -5 -6 -7
400 -11 -13 -11 -10 -9
600 -10 -9 -9 -13 -10
800 -2 -3 -4 -6 -4
1000 10 10 9 10 8
800 7 5 7 7 8
600 0 -3 -3 -3 -2
400 -3 1 0 -1 1
200 4 2 3 1 6
0 15 17 14 16 15
The accuracy can be determined from the deviation data. The maximum positive deviation
is 17 Pa and the largest negative deviation is -13 Pa. For a span of 1000 Pa, this translates
to an accuracy of +1.7% and -1.3% of full scale. Note: this accuracy only applies when the
readings are corrected using the above curve fit to the data.
The repeatability can be evaluated from the deviations given in Table P2.14. This occurs at
the 200 Pa "down" reading and has a value of 5 Pa. This translates to 0.5% of full scale.
The hysteresis error is given by the maximum deviation between the up and the down
readings for any value of the measurand in one cycle. The value is 14 Pa which occurs at
400 Pa in Cycle 2.-15
-10
-5
0
5
10
15
20
Reading minus
0 200 400 600 800 1000
True Pressure -k Pa
Deviation Data for Problem 2.14Best Fit - kPa

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2.9
2.30-10
0
10
20
30
40
50
Output - mV
0 20 40 60 80 100
Force - N
Plot of Data for Problem 2.15
Best Fit
Line
Equations of the best fit line:
mV = -0.27 + 0.408 Force
or Force = 0.662 + 2.45 mV
True Deviations (Readings - Best Fit ) (mV)
Force
(N) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5
20 -0.05 0.04 0.07 -0.06 0.00
40 -0.09 -0.05 -0.14 -0.20 -0.12
60 -0.14 -0.17 -0.05 -0.04 -0.17
80 -0.04 0.01 0.05 -0.07 0.07
100 0.26 0.29 0.33 0.19 0.30
80 -0.06 -0.08 -0.06 0.00 0.03
60 -0.05 -0.11 -0.18 -0.08 -0.15
40 -0.08 -0.18 -0.10 -0.16 -0.11
20 -0.07 -0.02 0.09 -0.02 -0.05
0 0.28 0.25 0.35 0.32 0.20
The accuracy is determined from the maximum and minimum deviation values. From the
deviation table, these values are +0.35 mv and -0.20 mV. For an output span of 40 mV, this
corresponds to +0.9% and -0.5% of full scale.
The repeatability can be determined directly from Table P2.15 in the problem statement.
The maximum deviation for any reading (taken in the same direction of varying force) is 0.16
mV at 20 N (down reading). This can be stated as 0.2% of full scale.-0.2
-0.1
0
0.1
0.2
0.3
0.4
Reading minus Best Fit - mV
0 20 40 60 80 100
Force - N
Deviation Data for Problem 2.15

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2.10
2.31 Static calibration refers to calibration processes where the measurand is changed
slowly while allowing the device to come to equilibrium. On the other hand, dynamic
calibration refers to processes of a more complicated nature often used for devices where
the measured variable is changing rapidly.
Types of recommended calibration:
(a) Oral thermometer- Static: always allowed to come into equilibrium.
(b) Pressure gage used in water line: Static for most purposes. Dynamic effects need more
sophisticated instruments.
(c) Car speedometer: Static and Dynamic. Dynamic effects are important during
acceleration.
2.32 In this case, temperature is the measurand and since it is varying over time, technically
a dynamic measurement. However, since the room temperature varies so relatively slowly
at 6 C/h, we can consider its measurement to be static. The reading will be an accurate
representation of room temperature because the time compared to the time constant.
2.33 In this case, temperature is the measurand and since it is varying over time, technically
a dynamic measurement. However, since the room temperature varies so relatively slowly
at 1 F/h, we can consider its measurement to be static. The reading will be an accurate
representation of room temperature because the time compared to the time constant.
2.34 Both thermometers are dipped in ice water; the smaller one will reach equilibrium
faster. The time constant is mc/hA. Since m is proportional to volume, the volume to surface
area will be smaller for the smaller thermometer and hence the smaller thermometer will
have the smaller time constant.

Loading page 17...

2.11
2.35 (a) The concepts of time constant, response time, rise time, and settling time do not
apply to zero order systems due to their instantaneous responses. All of the concepts
would have values approaching zero.
(b) First order systems as shown in Response A of Figure 2.8 (b) can most effectively
be represented by the concept of time constant in Eq. 2.3:y
y e at t
e
t
= − = =
−
1 0 632


.
The preceding equation is a numerical specification of the transient response of the
first order system. Although the time constant is the most appropriate concept, the
response time and rise time concepts can also be used.
(c) For overdamped second order systems also shown in Response A of Figure 2.8(b),
the time constant concept of Eq. 2.3 is not really applicable. Therefore, use of the
response time and rise time terms is more appropriate occurring wheny
y e values are
0.95 and 0.1 to 0.9, respectively.
(d) Underdamped second order systems as in Response B of Figure 2.8(b) are
oscillatory responses which can best be represented by the settling time concept.
This concept is the time until the amplitude of the oscillations are less than fraction of
the equilibrium response.
2.36 (a) The light intensity is changing slowly relative to the response of the meter. Each
time a measurement is made, it is essentially a static measurement. This is not a dynamic
measurement.
(b) The cycling occurs at 1500/60 = 25 times per second. A pressure transducer with a
response time of 2 seconds will just measure some kind of average value and will not
respond to the pressure variations. The response time must be much shorter than the time
for each cycle (1/25 = 0.04 sec) to show the variation in pressure. The response time should
probably be less than 1% of 0.04 seconds or 0.0004 seconds.
(c) No, it would not measure the power accurately, It is possible to construct a scenario in
which the power if off every time the power is measured. It would be necessary to measure
the power every minute or more frequently to get a reasonable measurement. There are
better ways to measure the power consumption. For example, the on-off time of the heater
can be measured over time and separately the power consumption when it is on.

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2.12
2.37 The response is shown below:0
3000
Time
rpm
reading
Since the responses for the tachometer resembles that of Response B of Figure
2.8(b), the measuring system may be characterized as second order. To eliminate
oscillations, the damping should be increased.
2.38 Assuming this to be a first order system,y
y e
e
t
= − −
1

where t = 5 sec; and = 2 sec
We want y = 0 at t = 0. Hence, define y = T − Ti = T − 20.
Then ye = T − Ti = 80 − 20 = 60
This results inT e
−
− = − −20
80 20 1 5 2
Solving, we get T = 75.1C
2.39 Assuming this to be a first order system,y
y e
e
t
= − −
1

where t = 4 sec; and = 2 sec
We want y = 0 at t = 0. Hence, define y = T − Ti = T − 75.
Then ye = T − Ti = 180 − 75 = 105
This results in2
4
1
75180
75 −
−=
−
− e
T
Solving, we get T = 165.8 F
2.40 Since this is a first order system,y
y e
e
t
= − −
1

where ye = 80 − 20 = 60 C; and  = 4 sec.
Since the rise time is the time it takes y/ye to increase from 0.1 to 0.9,0 1 1 0 42
0 9 1 9 21
9 21 0 42 8 79
4 1
4 2
2 1
. . sec
. . sec
. . . sec
= − → =
= − → =
= − = − =
−
−
e t
e t
t t t
t
t

Therefore the rise time is 8.79 seconds and the 90% response time is 9.21 seconds.

Loading page 19...

Loading page 20...

2.14
2.42 (a) For an input of 20 lb, the nominal output voltage is calculated below:( )
Output mV= − =
20
100 0 30 6
(b) Linearity Uncert.=  0.1%(full scale)( )
( )
=  = 
=  = 
0 001 30 0 03
0 03
6 100 0 5%
. .
. .
mV mV
mV
mV of reading
Hysteresis Uncert. =  0.08%(full scale)( )
( )
readingof
mV
mV
mVmV
%40.0100
6
024.0
024.0300008.0
=

=
==
Repeatability Uncert. = 0.03%(full scale)( )
( )
=  = 
=  = 
0 0003 30 0 009
0 009
6 100 0 15%
. .
. .
mV mV
mV
mV of reading
Zero Balance Uncert. = 2%(full scale)( )
( )
readingof
mV
mV
mVmV
%10100
6
6.0
6.03002.0
=

=
==
Temp Effect Uncert. on Zero:
Max. temp. variation = +25, -20F
Max. + error :( )
( )
readingof
mV
mV
mVFmV
%25.0100
6
015.0
015.0)25(3000002.0
==
==
Max. - error  0.012 mV or 0.20% of reading
Temp effect uncert. on span: same values as uncertainty on zero
Summary Of Results:
Type Uncertainty, mV Uncertainty, % of
reading
Linearity 0.03 0.5
Hysteresis 0.024 0.4
Repeatability 0.009 0.15
Zero Balance 0.6 10
Temp Eff. (Zero) +0.015/−0.012 +0.25/−0.20
Temp Eff. (Span) +0.015/−0.012 +0.25/−0.20

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2.15
2.43 (a) For an input of 20 lb, the nominal output voltage is calculated below:( )
mVOutput 8.424
01000
200 =
−
=
(b) Linearity Uncert.=  0.1%(full scale)( )
( )
readingof
mV
mV
mVmV
%5.0100
8.4
024.0
024.024001.0
=

=
==
Hysteresis Uncert. =  0.08%(full scale)( )
( )
readingof
mV
mV
mVmV
%40.0100
8.4
0192.0
0192.0240008.0
=

=
==
Repeatability Uncert. = 0.05%(full scale)( )
( )
readingof
mV
mV
mVmV
%25.0100
8.4
012.0
012.0240005.0
=

=
==
Zero Balance Uncert. = 1%(full scale)( )
( )
readingof
mV
mV
mVmV
%5100
8.4
24.0
24.015(2401.0
=

=
==
Temp Effect Uncert. on Zero:
Max. temp. variation = +15, -10C
Max. + error :( )
( ) readingof
mV
mV
mVCmV
%15.0100
8.4
0072.0
0072.0)15(2400002.0
==
==
Max. - error  0.0048 mV or 0.1% of reading
Temp effect uncert. on span: same values as uncertainty on zero
Summary Of Results:
Type Uncertainty, mV Uncertainty, % of
reading
Linearity 0.024 0.5
Hysteresis 0.0192 0.4
Repeatability 0.012 0.25
Zero Balance 0.24 5
Temp Eff. (Zero) +0.0072/−0.0048 +0.15/−0.1
Temp Eff. (Span) +0.0072/−0.0048 +0.15/−0.1

Loading page 22...

3.1
CHAPTER 3
3.1 Output = 5 Volts = Vo
Input = 5 V = 510-6 volts = ViGain G V
V
G G dB
o
i
dB
= = =  =
= = =
−
5
5 10 10
20 20 10 120
6
6
10 10
6
log log
3.2 GdB=60dB
Vi=3mV=310-3 volts
GdB=60dB=20log10G
3dB= log10G
 G = 103
 G = Vo/Vi
Vo = GVi = 103(310-3)
= 3 volts
3.3 Eq. 3.2 applies. For G =10, GdB = log10(10) = 20. Similarly, for G = 100 and
500, the decibel gains are 40 and 54.
3.4 The circuit resembles Fig. 3.9 (a). For this problem we want the voltage drop
across the resistor Rs to be 0.01xVs. The current in the loop is)/( iss RRVI +=
and the voltage drop across the resistor is Vdrop = IsxRs.
Combining these:120)120/()/(01.0 ississs RVRRRVV +=+=
. Solving for Ri, we get
Ri = 11,880 .
3.5 The circuit resembles Fig. 3.9(a). The input voltage, Vi, is IxRi. The current is
Vs/(Rs+Ri). Combining, Vi=RixVs/(Rs+Ri). In the first case:
0.005=5x106xVs/(Rs+5x106) For the second case:
0.0048=10,000xVs/(Rs+10,000) These can be solved simultaneously to give
Rs = 416 .

Loading page 23...

3.2
3.6 a) From Eq. 3.14,G R
R
R
R
R
R
= +
= +
=
1
100 1
99
2
1
2
1
2
1
Since R1 and R2 typically range from 1k to 1M, we arbitrarily choose:
R2=99k
 R1 = 1k
b) f = 10 kHz = 104 Hz
GPB = 106 Hz for 741
G = 100
From Eq. 3.15,f GPB
G
Hz Hzc = = =
10
100 10
6
4
This is the corner frequency so signal is -3dB from dc gain.
dc gain = 100 = 40dB. Gain at 104 Hz is then 37 dB.
From Eq. 3.16,

= − 

 

 = − 

 

 = − = −− −
tan tan1 1
4
4
10
10 4 45
f
fc


Loading page 24...

Loading page 25...

3.4
3.8G R
R
R
R
= = +
=
1000 1
999
2
1
2
1
Selecting R2 = 999 k, R1 can be evaluated as 1 k.
Since GBP = 1MHz for the A741C op-amp and G = 1000 at low
frequencies,
GBP = 1MHz = 1000(Bandwidth)
 Bandwidth = 1 kHz = fc
If f = 10 kHz and fc = 1 kHz, we must calculate the number of times fc
doubles before reaching f.f f
x
c
x
x
 =
 =
 =
2
1000 2 10000
3 32.
Now the gain can be calculated knowing that for each doubling the gain
decreases by 6dB (i.e. per octave)Gain dB dB
dB
( ) log . ( )= −
=
20 1000 3 32 6
40
10
From Eq. 3.16,
= − 

 


−
tan 1 f
fc= − 

 


= −
−
tan
.
1 10000
1000
84 3

Loading page 26...

3.5
3.9 G = 100 (Actually -100 since signal inverted)
Input impedance = 1000  R1
From Eq. 3.17,G R
R
R
R k
= −
− =
 =
2
1
2
2
100 1000
100 
Since GPBnoninv = 106 Hz, from Eq. 3.18,( )
GPB R
R R GPB
Hz
inv noninv= +
= +
= 
2
1 2
6
5
100000
1000 100000 10
9 9 10.
From Eq. 3.15,f GPB
G kHzc = =  =
9 9 10
100 9 9
5
. .
3.10 G = 10 (Actually -10 since output inverted)
Input impedance = 10 k = 10000   R1
From Eq. 3.17,G R
R
R
R k
= −
− =
 =
2
1
2
2
10 10000
100 
Since GPBnoninv = 106Hz, from Eq. 3.18,( )
GPB R
R R GPB
kHz
inv noninv= +
= +
=
2
1 2
6100000
10000 100000 10
909
From Eq. 3.15,f GPB
G kHzc = =  =
0 909 10
10 90 9
3
. .

Loading page 27...

3.6
3.11 (a) 10 = 2N. N = ln10/ln2 = 3.33
(b) dB/decade = NxdB/octave = 3.33x6 = 20 dB/decade
3.12
The gain of the op-amp itself is
Vo=g(Vp − Vn) [A]
Vp is grounded so Vp = 0 [B]
The current through the loop including Vi, R1, R2, and Vo isI V
R
V V
R R
L
i o
= = −
+

 1 2
Vn can then be evaluated as( )
V V I R V R V V
R R
n i L i
i o
= − = − −
+
1
1
1 2
[C]
Substituting [C] and [B] into [A]( )
V g V R V V
R R
o i
i o
= − + −
+


 


1
1 2
Rearranging:( ) ( )
V R R gR V R R R g
V
V G R g
R R gR
o i
o
i
1 2 1 1 2 1
2
1 2 1
+ + = − − +
= = −
+ +
Noting the g is very largeG R
R
= − 2
1

Loading page 28...

3.7
3.13 The complete circuit is as follows,
For a loading error of 0.1%, the voltage drop across Rs should be
900.001 = 0.09 V. The current through Rs is then:I V
R AR
R
s
S
s
= = =
0 09
10 0 009
. .
IRs also flows through R1 and the combination of R2 and Ri. For R2 and Ri,
we have:V IR
R R R k
R
o
i
= = =
+












=
+












=
10 0 009 1
1 1 0 009 1
1 1
100
1124
2 2
2
. .

The voltage drop across R1=90−0.09−10= 79.91VR V
I
1
79 91
0 009 8879 0= = =
.
. . 

Loading page 29...

3.8
3.14
a) If we ignore the effects of Rs and R0, we can use Eq. 3.19:V
V
R
R R
R
R
R
o
i
= +
= +
 =
2
1 2
2
2
2
8
120 100000
7142 9. 
(If we include Rs and R0 ,the value of
R2 is 7193 , less than 1%
different.)
b)I V
R
V
R R As
= = + = + =
 1 2
120
100000 7142 9 0 00112
. . (neglecting load
effects)
 P = I2R = (0.0012)2(100000 + 7142.9) = 0.13 W
c)I V
R
A
A = =
+ +
+










= + +
=

120
0 5 100000 1
1
7142 9
1
10
120
0 5 100000 7092 2
0 00112
6
.
.
. .
.
Voltage drop across line load resistorV I R V smallA= =  =0 00112 0 5 0 00056. . . ( )
3.15 If f1 = 7600 Hz and f2 = 2100 Hz then the following equation may be used,
f2  2x = f1 where x = # octaves
Substituting,2100 2 7600
2 3 619
2 3 619
1856
 =
=
=
 =
x
x
x
x octaves
.
log log .
.Rs
R1
R2 R0
IA
120V

Loading page 30...

3.9
3.16 fc = 1kHz = 1000Hz , Butterworth
Rolloff = 24 dB/octaveA V
f kHz Hz
f kHz Hz
out1
1
2
0 10
3 3000
20 20000
=
= =
= =
.
Since Rolloff = 24 dB/octave = 6n dB/octave,
 n = 4
From Eq. 3.20,( )
( )
G
f fc
n1
1
2
2 4
1
1
1
1 3000 1000
0 01234
=
+
=
+
=

.
A A
G V A in1in
1out
1
2
010
0 01234 81= = = =
.
. .
From Eq. 3.20,( )
G2 2 4
1
1 20000 1000
0 00000625
=
+
=

.
A G A mVout in2 2 2 0 00000625 81 0 051= = =. ( . ) .
3.17 Using Eq. 3.2,)6.5/(log202 10 oV=− . Solving, Vo = 4.45
3.18 We want a low-pass filter with a constant gain up to 10 Hz but a gain of
0.1 at 60 Hz. Using Eq. 3.20:( )
( )
G
f fc
n
n
=
+
=
+
1
1
0 1 1
1 60 10
1
2
2
.
Solving for n, we get 1.28. Since this is not an integer, we select n = 2. With this
filter, the 10 Hz signal will be attenuated 3 dB. If this is a problem, then a higher
corner frequency and possibly a higher filter order might be selected.

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Solution Manual For Introduction To Engineering Experimentation, 3rd Edition

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