CHAPTER 1
1.1 (a) The weight of a person is 175 pounds which is equivalent to 778 Newtons.
4.4482 N
175lbf *
= 778 Newtons
1lbf
(b) The magnetic force of 1,200 dynes is equivalent to 0.0027 pounds or 0.012 Newtons.
10 5 N
1200dyn *
= 0.012 Newtons
1dyn
10 5 N
1lbf
*
= 0.0027 pounds
1dyn
4.4482 N
(c) The force exerted by the spring is 10 Newtons which is equivalent to 1,000,000 dynes or
2.248 pounds
1dyn
10 N * 5
= 1,000,000 dyne
10 N
1lbf
10 N *
= 2.248 pounds
4.4482 N
1200dyn *
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 psi
1atm
30 psig *
= 2.041 atm
14.696 psi
6.895kPa
1bar
30 psig *
*
= 2.069 bars
1 psi
100kPa
(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 water
100kPa
1 psi
*
= 145 psi
10bars *
1bar
6.895kPa
1atm
100kPa
1 psi
*
*
= 9.87 atm
10bars *
1bar
6.895kPa 14.696 psi
1atm
100kPa
1 psi
76cmHG
1in
*
*
*
*
= 295.3 in of Hg
10bars *
1bar
6.895kPa 14.696 psi
1atm
2.54cm
1atm
100kPa
1 psi
1033.2cmWater
1m
10bars *
*
*
*
*
= 102 meters water
1bar
6.895kPa 14.696 psi
1atm
100cm
(c) 50 cm of water is equivalent to 0.0490 bars or 0.711 psi or 1.45 inches of mercury
1atm
101.325kPa
1bar
*
*
= 0.0490 bars
50cmWater *
1033.2cmWater
1atm
100kPa
1atm
101.325kPa
1 psi
*
*
* = 0.711 psi
50cmWater *
1033.2cmWater
1atm
6.895kPa
1atm
76cmHG
1in
*
*
= 1.45 in of Hg
50cmWater *
1033.2cmWater
1atm
2.54cm
1.1
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 joules
3600kJ 1000 J
1Btu
750kWh *
*
*
= 2,559,241.71 Btu
1kWh
1kJ
1055 J
3600kJ 1000 J
1cal
1kcal
*
*
*
= 645,007 kcal
750kWh *
1kWh
1kJ
4.186 J 1000cal
3600kJ 1000 J
*
= 2,700,000,000 J
750kWh *
1kWh
1kJ
(b) The gas water heater uses 50,000 Btu or 52,750,000 joules or 12,601.5 kcal or
38,908,386 ft.lbf
1055 J
= 52,750,000 joules
50000 Btu *
1Btu
1055 J
1cal
1kcal
*
*
= 12,601.5 kcal
50000 Btu *
1Btu
4.186 J 1000cal
1 ft.lbf
= 38,908,386 ft.lbf
50000 Btu *
0.00128507 Btu
(c) The amount of heat required is 250 kCal which is equivalent to 992 Btu or 250,000 Cal
or 1,046,500 joules
1000cal 4.186 J
1Btu
250kcal *
*
*
= 992 Btu
1kcal
1cal
1055 J
1000cal
250kcal *
= 250,000 cal
1kcal
1000cal 4.186 J
250kcal *
*
= 1,046,500 J
1kcal
1cal
1.2
1.4 (a) The automobile rating of 150 hp is equivalent to 112 kW or 82,500 ft.lbf/sec or
106.02 Btu/sec
745.7W
1kW
150hp *
*
= 112 kW
1hp
1000W
ft.lbf
550
s
= 82,500 ft.lbf/sec
150hp *
1hp
Btu
1
745.7W
s
*
150hp *
= 106.02 Btu/sec
1hp
1055.04W
(b) The truck rating of 400 kW is equivalent to 536 hp or 295,025 ft.lbf/sec or 379 Btu/sec
1000W
1hp
400 kW *
*
= 536 hp
1kW
745.7W
ft.lbf
550
1000W
1hp
s
*
*
= 295,025 ft.lbf/sec
400kW *
1kW
745.7W
1hp
Btu
1
1000W
s
*
= 379 Btu/sec
400kW *
1kW
1055.04W
(c) The water heater rating of 40,000 Btu/hr is equivalent to 11.72 kW
Btu 0.293W
1kW
*
*
= 11.72 kW
40000
Btu
hr
1000W
1
hr
1.3
1.5 (a) 50 oF is equivalent to 10 oC
5
( )(50 – 32) = 10 oC
9
(b) 150 oC is equivalent to 302 oF
9
( )(150 oC) + 32 = 302 oF
5
(c) The water temperature increase of 40 oC is equivalent to a change of 40 K or 72 oF or 72
o
R
1 K
= 40 K
(40 oC) *
1C
1.8 F
= 72 oF
(40 oC) *
1C
1
.8 R
= 72 oR
(40 oC) *
1C
(d) The air temperature change of 30 oF is equivalent to a change of 16.7 K or 16.7 oC or 30
o
R
0.556 K
(30 oF) *
= 16.7 K
1 F
0.556C
(30 oF) *
= 16.7 oC
1 F
1 R
= 30 oR
(30 oF) *
1 F
1.4
1.6 (a) 4 gallons is equivalent to 15.1 liter or 15,142 cm3 or 0.535 ft3
0.0037854m 3
1Liter
* 3 3 = 15.1 liter
4 gal *
1gal
10 m
0.0037854m 3
1cm 3
4 gal *
* 6 3 = 15,142 cm3
1gal
10 m
0.0037854
1 ft 3
4 gal *
*
= 0.535 ft3
3
1gal
0.02832m
(b) 10 liters is equivalent to 2.64 gallons or 10,000 cm3 or 0.353 ft3
10 3 m 3
1gal
10 Liter *
*
= 2.64 gallons
1liter
0.0037854m 3
1cm 3
10 3 m 3
* 6 3 = 10,000 cm3
10 Liter *
1liter
10 m
3
3
1 ft 3
10 m
*
= 0.353 ft3
10 Liter *
1liter
0.02832m 3
(c) 5 ft3 is equivalent to 37.4 gallons or 141,600 cm3 or 142 liters
1gal
5 ft 3 *
= 37.4 gallons
0.13368 ft 3
5 ft 3 *
0.02832m 3
1cm 3
*
= 141,600 cm3
1 ft 3
10 6 m 3
5 ft 3 *
0.02832m 3
1liter
* 3 3 = 142 liters
3
1 ft
10 m
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.K
ft.lbf
0.00128507 Btu
53.34
*
= 0.0685 Btu/lbm.oR
lbm R
1 ft.lbf
ft.lbf
0.00128507 Btu 1055 J
1lbm
1 R
*
*
*
*
= 287 J/kg.K
53.34
5
lbm R
1 ft.lbf
1Btu
0.4536kg
K
9
0.00128507 Btu 1055 J
1lbm
1 R
ft.lbf
1cal
1kcal
*
*
*
*
*
*
=
53.34
5
lbm R
1 ft.lbf
1Btu
4.186 J 1000cal 0.4536kg
K
9
kcal
0.0685
kg  K
1.5
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.K
1lbm
Btu
1055 J
1cal
1kcal
1 R
*
*
*
*
*
= 1.986 kCal.kg mole.K
1.986
5
lbmole R
1Btu
4.186 J 1000cal 0.4536kg
K
9
Btu
1 ft.lbf
*
= 1,545 ft.lbf/lb mole.oR
1.986
lbmole R 0.00128507 Btu
1lbm
1 R
Btu
1055 J
*
*
*
= 8,314 J/kg mole.K
1.986
5
lbmole R
1Btu
0.4536kg
K
9
1.9 The thermal conductivity is 200 W/m.oC or 116 Btu/hr/ft/oF or 0.048 kcal/sec.m.oC
Btu
1
W
hr. ft. F
*
= 116 Btu/hr/ft/oF
200
W
mC
1.7307
mC
Btu
1
W
1055 J
1cal
1kcal
s
*
*
*
*
= 0.048 kcal/sec.m.oC
200
mC 1055.04W
1Btu
4.186 J 1000cal
1.10 The thermal conductivity is 50 Btu/hr.ft.oF or 86.54 W/m.oC or 20.7 Cal/sec.m.oC
W
1.7307
Btu
mC = 86.54 W/m.oC
*
50
Btu
hr. ft  F
1
hr. ft  F
W
Btu
1.7307
1
Btu
1055 J
1cal
mC *
s
50
*
*
*
= 20.7 cal/sec.m.oC
Btu
hr. ft  F
1055.04W
1Btu
4.186 J
1
hr. ft  F
1.6
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.
Rot.
Magnet
Gears
Wheel
Flex
Cable
Pointer
Cup
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.
2.1
2.2
True Value = 0.5000 inches
Determination of Bias Error.
First, the average of readings must be calculated:
Average of Readings =
(0.4821 0.4824 0.4821 0.4821 0.4820 0.4822 0.4821 0.4822 0.4820 0.4822) / 10
 0.48214
 0.4821in
Bias Error = Average of Readings -True Value
 0.4821  0.5000
  0.01786
  0.0179 in
Determination of Maximum Precision Error:
 Largest difference between a single 
Maximum Precision Error = 

 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.
2.2
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.
2.3
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%.
2.4
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
mRT
P
V
The sensitivity is given by:
dP mR Pi


; R, m, and V cons tan t
dT
V
Ti
dP
is proportional to the initial pressure and is changed when the
 We see the sensitivity
dT
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.
2.5
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
dP 2Q
Q 
then
P   

dQ C 2
C 
(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 =
21  20
100
20
 5%
Actual  Ideal
100
Ideal

2.6
2.28
Plot of Data for Problem 2.13
Gage Reading - psi
120
100
80
60
40
20
0
0
20
40
60
True Pressure - psi
80
100
The best fit line is:
Reading = 2.9 + 0.978 (True Pressure) psi
Deviation (Readings - Best Fit) (psi)
Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5
-2
-3
-2
-2
-3
-3
-2
-3
-3
-3
-3
-4
-3
-4
-2
-1
-1
-2
-2
-1
0
-1
-1
0
1
3
2
3
3
3
1
1
1
0
0
1
0
1
1
2
2
2
2
2
2
2
1
1
3
1
Reading minus Best Fit - psi
True
Pressure
(psi)
20
40
60
80
100
80
60
40
20
0
Deviation Data for Problem 2.13
3
2
1
0
-1
-2
-3
-4
0
20
40
60
True Pressure - psi
80
100
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.
2.7
2.29
Plot of Data for Problem 2.14
Gage Reading -kPa
1200
1000
Best Fit
Line
800
600
400
200
0
-200
0
200
400
600
800
True Pressure - kPa
1000
Best Fit Line:
Reading = -4.5 + 1.017 (True Pressure)
Deviations based on difference between readings and best fit line.
True
Force
(N)
200
400
600
800
1000
800
600
400
200
0
Deviations (Readings - Best Fit) (N)
Reading minus
Best Fit - kPa
20
Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5
-7
-8
-5
-6
-7
-11
-13
-11
-10
-9
-10
-9
-9
-13
-10
-2
-3
-4
-6
-4
10
10
9
10
8
7
5
7
7
8
0
-3
-3
-3
-2
-3
1
0
-1
1
4
2
3
1
6
15
17
14
16
15
Deviation Data for Problem 2.14
15
10
5
0
-5
-10
-15
0
200
400
600
800
True Pressure -k Pa
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.
2.8
1000
2.30
Plot of Data for Problem 2.15
50
Output - mV
40
Best Fit
Line
30
20
10
0
-10
0
20
40
60
Force - N
80
100
Equations of the best fit line:
or
mV = -0.27 + 0.408 Force
Force = 0.662 + 2.45 mV
Deviation Data for Problem 2.15
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
Reading minus Best Fit - mV
0.4
0.3
0.2
0.1
0
-0.1
-0.2
0
20
40
60
Force - N
80
100
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.
2.9
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.
2.10
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:
t
y
 1  e   0.632 at t  
ye
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
y
values are
response time and rise time terms is more appropriate occurring when
ye
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.
2.11
2.37
The response is shown below:
rpm
reading
3000
0
Time
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,
t
y
 1  e  where t = 5 sec; and = 2 sec
ye
We want y = 0 at t = 0. Hence, define y = T  Ti = T  20.
Then ye = T  Ti = 80  20 = 60
5
T  20
2
This results in 80  20  1  e
Solving, we get T = 75.1C
2.39 Assuming this to be a first order system,
t
y
 1  e  where t = 4 sec; and = 2 sec
ye
We want y = 0 at t = 0. Hence, define y = T  Ti = T  75.
Then ye = T  Ti = 180  75 = 105
4
T  75
2
This results in 180  75  1  e
Solving, we get T = 165.8 F
2.40 Since this is a first order system,
t
y
 1  e  where ye = 80  20 = 60 C; and  = 4 sec.
ye
Since the rise time is the time it takes y/ye to increase from 0.1 to 0.9,
0.1  1  e
t
0.9  1  e
4
t
4
 t 1  0.42 sec
 t 2  9.21sec
t  t 2  t1  9.21  0.42  8.79 sec
Therefore the rise time is 8.79 seconds and the 90% response time is 9.21 seconds.
2.12
2.41 Since this is a first order system,
t
y
 1  e  where ye = 180  75 = 105 F; and  = 4 sec.
ye
Since the rise time is the time it takes y/ye to increase from 0.1 to 0.9,
0.1  1  e
t
0.9  1  e
t
4
 t1  0.42 sec
4
 t 2  9.21sec
t  t 2  t1  9.21  0.42  8.79 sec
2.13
2.42 (a) For an input of 20 lb, the nominal output voltage is calculated below:
20
Output 
30  6mV
100  0
(b)
Linearity Uncert.=  0.1%(full scale)
 0.00130mV   0.03 mV
0.03mV
100  0.5% of reading
6 mV
Hysteresis Uncert. =  0.08%(full scale)
 0.000830mV   0.024 mV
 0.024mV
100   0.40%of reading

6mV
Repeatability Uncert. = 0.03%(full scale)
 0.000330 mV   0.009mV

0.009mV
100  0.15% of reading
6 mV
Zero Balance Uncert. = 2%(full scale)
 0.02 30mV   0.6mV

 0.6mV
100  10%of reading
6mV
Temp Effect Uncert. on Zero:
Max. temp. variation = +25, -20F
Max. + error :
 0.0000230mV ( 25 F )  0.015mV
0.015mV
100  0.25%of reading

6mV

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)
+0.25/0.20
+0.015/0.012
Temp Eff. (Span)
+0.25/0.20
+0.015/0.012
2.14
2.43 (a) For an input of 200N, the nominal output voltage is calculated below:
200
 24  4.8mV
Output 
1000  0
(b)
Linearity Uncert.=  0.1%(full scale)
 0.001 24mV   0.024mV
 0.024mV
100  0.5%of reading
4.8mV
Hysteresis Uncert. =  0.08%(full scale)
 0.000824mV   0.0192mV
 0.0192mV
100  0.40%of reading

4.8mV
Repeatability Uncert. = 0.05%(full scale)
 0.0005 24mV   0.012mV

 0.012mV
100  0.25%of reading
4.8mV
Zero Balance Uncert. = 1%(full scale)
 0.0124mV (15  0.24mV

 0.24mV
100  5%of reading
4.8mV
Temp Effect Uncert. on Zero:
Max. temp. variation = +15, -10C
Max. + error :
 0.0000224mV (15C )  0.0072mV


0.0072mV
100  0.15%of reading
4.8mV
 0.0048 mV or 0.1% of reading
Max. - error
Temp effect uncert. on span: same values as uncertainty on zero
Summary Of Results:
Type
Linearity
Hysteresis
Repeatability
Zero Balance
Temp Eff. (Zero)
Temp Eff. (Span)
Uncertainty, mV
0.024
0.0192
0.012
0.24
+0.0072/0.0048
+0.0072/0.0048
2.15
Uncertainty, % of
reading
0.5
0.4
0.25
5
+0.15/0.1
+0.15/0.1
CHAPTER 3
3.1 Output = 5 Volts = Vo
Input = 5 V = 510-6 volts = Vi
5
V
Gain  G  o 
 10 6
6
Vi 5  10
GdB  20 log10 G  20 log10 10 6  120dB
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 I  Vs /(Rs  Ri )
and the voltage drop across the resistor is Vdrop = IsxRs.
Combining these:
0.01 Vs  Vs /(Rs  Ri )  Rs  Vs /(120  Ri )120 . 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 .
3.1
3.6
a) From Eq. 3.14,
R
G  1 2
R1
100  1 
R2
R1
R2
R1
Since R1 and R2 typically range from 1k to 1M, we arbitrarily choose:
R2=99k
 R1 = 1k
99 
b) f = 10 kHz = 104 Hz
GPB = 106 Hz for 741
G = 100
From Eq. 3.15,
GPB 10 6 Hz
fc 

 10 4 Hz
G
100
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,
 10 4 
f

   tan 1     tan 1  4     45 
4
 fc 
 10 
3.2
3.7
G  1
R2
R1
100  1 
R2
R1
R2
R1
Selecting R1 = 1k, R2 can be evaluated as 99k..
99 
Since GBP = 1 MHz = 100(Bandwidth)
 bandwidth = 10 kHz = fc
Gain will decrease 6dB from DC value for each octave above 10 kHz.
The phase angle can be determined from Eq. 3.16,
f
   tan 1 
 fc 
f(Hz)
0
5k
10k
100k
0
-26.6
-45
-84.3

3.3
3.8
G  1000  1 
R2
R1
R2
R1
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.
fc  2 x  f
999 
1000  2 x  10000
 x  3.32
Now the gain can be calculated knowing that for each doubling the gain
decreases by 6dB (i.e. per octave)
Gain(dB)  20 log 10 1000  3.32(6dB)
 40dB
From Eq. 3.16,
f
   tan1 
 fc 
 10000 
  tan 1

 1000 
 84.3 
3.4
3.9
G = 100 (Actually -100 since signal inverted)
Input impedance = 1000  R1
From Eq. 3.17,
R
G 2
R1
R2
1000
 R2  100k
Since GPBnoninv = 106 Hz, from Eq. 3.18,
R2
GPBinv 
GPBnoninv
R1  R2
100 
 
100000
10 6
1000  100000
 9.9  10 5 Hz
From Eq. 3.15,
GPB 9.9  10 5
fc 

 9.9 kHz
G
100

3.10
G = 10 (Actually -10 since output inverted)
Input impedance = 10 k = 10000   R1
From Eq. 3.17,
R
G 2
R1
R2
10000
 R2  100k
Since GPBnoninv = 106Hz, from Eq. 3.18,
R2
GPBinv 
GPBnoninv
R1  R2
10 
 
100000
10 6
10000  100000
 909kHz
From Eq. 3.15,
GPB 0.909  10 3
fc 

 90.9kHz
G
10

3.5
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)
Vp is grounded so Vp = 0
[A]
[B]
The current through the loop including Vi, R1, R2, and Vo is
V  Vi  Vo
IL 
 R R1  R2
Vn can then be evaluated as
R V  Vo 
[C]
Vn  Vi  ILR1  Vi  1 i
R1  R2
Substituting [C] and [B] into [A]

R V  Vo  
Vo  g Vi  1 i

R1  R2 

Rearranging:
Vo R1  R2  gR1   Vi  R1  R2  R1 g
Vo
R2 g
G 
Vi
R1  R2  gR1
Noting the g is very large
R
G 2
R1
3.6
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:
VR
0.09
IRS  s 
 0.009 A
Rs
10
IRs also flows through R1 and the combination of R2 and Ri. For R2 and Ri,
we have:








1
1 



 0.009
Vo  10  IR  0.009
1
1 
 1
 1

 



 R2 100k 
 R2 Ri 
R2  1124
The voltage drop across R1=900.0910= 79.91V
V 79.91
R1  
 8879.0
I 0.009
3.7
3.14
a) If we ignore the effects of Rs and R0, we can use Eq. 3.19:
Vo
R2
Rs

Vi R1  R2
8
R2

120 100000  R2
R1
120V
 R2  7142.9
R2
(If we include Rs and R0 ,the value of
IA
R2 is 7193 , less than 1%
different.)
V
Vs
120
b) I 


 0.00112A (neglecting load
 R R1  R2 100000  7142.9
R0
effects)
 P = I2R = (0.0012)2(100000 + 7142.9) = 0.13 W
c)
3.15
V
120


R


1
0.5  100000  

1
1 


 7142.9 10 6 
120

0.5  100000  7092.2
 0.00112A
Voltage drop across line load resistor
V  IAR  0.00112  0.5  0.00056V (small )
If f1 = 7600 Hz and f2 = 2100 Hz then the following equation may be used,
f2  2x = f1
where x = # octaves
Substituting,
2100  2 x  7600
IA 

2 x  3.619
x log 2  log 3.619
 x  1856
.
octaves
3.8
3.16
fc = 1kHz = 1000Hz , Butterworth
Rolloff = 24 dB/octave
A1out  0.10V
f1  3kHz  3000Hz
f2  20kHz  20000Hz
Since Rolloff = 24 dB/octave = 6n dB/octave,
n=4
From Eq. 3.20,
1
G1 
2n
1  f1 fc 

1
1  3000 1000
2 4
 0.01234
A
010
.
 81
. V  A 2 in
 A 1in  1out 
G1
0.01234
From Eq. 3.20,
1
G2 
24
1  20000 1000
 0.00000625
 A2out  G2 A2 in  0.00000625(8.1)  0.051mV
3.17 Using Eq. 3.2,  2  20 log10 (Vo / 5.6) . 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:
1
G 
2n
1  f1 fc 
0.1 
1
1  60 10 
2n
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.
3.9
3.19 We want a low-pass filter with a constant gain up to 100 Hz but an
attenuation at 1000 Hz of 20log10 0.01= -40 dB (G = 0.01). Using Eq. 3.20:
1
G 
2n
1  f1 fc 
0.01 
1
1  1000 100 
2n
Solving for n, we get n = 2
With the selected corner frequency, the 100 Hz signal will be attenuated 3dB. If
this were to be a problem, a higher corner frequency would be required and also
a higher order filter.
3.20 fc = 1500 Hz
f = 3000 Hz
a) For a fourth-order Butterworth filter
n=4
From Eq. 3.20,
1
G
2n
1  f fc 

1
1  3000 1500
2 4
 0.0624  6.24%
 24dB
b) For a fourth-order Chebeshev filter with 2 dB ripple width
n=4
f
3000

2
fc 1500
From Fig. 3.18 we see that for n = 4 and f/fc = 2,
G(dB) = 34dB
Frequency Ratio
c) For a fourth-order Bessel filter
n=4
f
3000

2
fc 1500
From Fig. 3.20 we see that for n = 4 and f/fc = 2,
G(dB) = 14 dB
Frequency Ratio
3.10
3.21
n 1
G 1
fc  12kHz
R1  1000
At dc, Eqs. 3.21 and 3.17 are equivalent. Since we require no gain, set R1 = R2.
Thus, R1 = R2 = 1000
From Eq. 3.26, we can calculate C,
1
fc 
2R2C
1
2 (1000)C
 C  0.013 F
12kHz 
3.22 It would not be possible to solve problem 3.16 using a simple Butterworth
filter based on the inverting amplifier. This is because R1 would have to be on the
order 10 M. Such a resistance is higher than resistances normally used for
such circuits because it is on the order of various capacitive impedances
associated with the circuit. The signal should first be input to an amplifier with a
very high input impedance such as a non-inverting amplifier and the signal then
passed through a filter.
3.23
n4
G 1
fc  1500 Hz
f  25 kHz
From Eq. 3.20,
G

1
1  f fc 
2n
1
1  25000 1500
2 4
 130
.  10 5
GdB = 20log10(1.310-5)
3.24 Vin  Deflection  V / div  4.3  2  8.6V
3.11
3.25 Range  Maximum Deflection  V / div  8  100 mV  800mV
3.26 The visual resolution is on the order of the beam thickness (for thick beams
is may be on the order of ½ the beam thickness since one can interpolate within
the beam. Taking the resolution as the beam thickness, the fractional error in
reading is 0.05/1 = 0.05 (5%). In volts the resolution is 0.05x5 mV = 0.25 mV.
3.21
n 1
G 1
fc  12kHz
R1  1000
At dc, Eqs. 3.21 and 3.17 are equivalent. Since we require no gain, set R1 = R2.
Thus, R1 = R2 = 1000
From Eq. 3.26, we can calculate C,
1
fc 
2R2C
1
2 (1000)C
 C  0.013 F
12kHz 
3.22 It would not be possible to solve problem 3.16 using a simple Butterworth
filter based on the inverting amplifier. This is because R1 would have to be on the
order 10 M. Such a resistance is higher than resistances normally used for
such circuits because it is on the order of various capacitive impedances
associated with the circuit. The signal should first be input to an amplifier with a
very high input impedance such as a non-inverting amplifier and the signal then
passed through a filter.
3.12
3.23
n4
G 1
fc  1500 Hz
f  25 kHz
From Eq. 3.20,
G

1
1  f fc 
2n
1
1  25000 1500
2 4
 130
.  10 5
GdB = 20log10(1.310-5)
= 97.7 dB
3.13
CHAPTER 4
4.1
Remainder
147 2
73 2
36 2
18 2
1
1
0
92
0
42
22
1
12
0
0
1
0
From the remainders, we get the 8-bit number as follows:
N2 = 1 0 0 1 0 0 1 1
4.2
Remainder
145 2
72 2
1
36 2
0
18 2
92
0
42
0
1
22
0
12
0
0
1
From the remainders, we get the 8-bit number as follows:
N2 = 1 0 0 1 0 0 0 1
4.3
Remainder
4.1
1149 2
574 2
287 2
1
0
143 2
1
71 2
35 2
1
1
17 2
1
8 2
4 2
1
0
22
0
12
0
0
1
From the remainders, we get the 12-bit number as follows:
N2 = 0 1 0 0 0 1 1 1 1 1 0 1
Note that the MSB must be zero since we have only eleven remainders.
4.4
Remainder
872 2
436 2
218 2
109 2
54 2
27 2
13 2
62
32
12
0
0
0
0
1
0
1
1
0
1
1
From the remainders, we get the 12-bit number as follows:
N2 = 0 0 1 1 0 1 1 0 1 0 0 0
Note that the MSB must be zero since we have only eleven remainders.
4.2
4.5 We first find the 8-bit number for +121:
Remainder
121 2
60 2
30 2
1
0
15 2
0
7 2
3 2
1
1
12
0
1
1
The 8-bit number for +121 is 01111001.
To find the 8-bit number for -121, we first invert the 8-bit number for +121:
10000110.
Then we add 1 to obtain the 8-bit number for -121: 10000111.
4.6 We first find the 8-bit number for +101:
Remainder
101 2
50 2
25 2
12 2
62
32
12
0
1
0
1
0
0
1
1
The 8-bit number for +121 is 01100101.
To find the 8-bit number for -121, we first invert the 8-bit number for +121:
10011010.
Then we add 1 to obtain the 8-bit number for -121: 10011011.
4.3
4.7
Remainder
891 2
445 2
1
222 2
1
111 2
55 2
0
1
27 2
1
13 2
1
1
62
32
0
12
1
0
1
The 2's complement binary equivalent is thus,
001101111011
4.8
Remainder
695 2
347 2
1
173 2
1
86 2
43 2
1
0
21 2
1
10 2
52
1
22
12
0
0
1
0
1
The 2's complement binary equivalent is thus,
001010110111
4.4
4.9 To find the decimal value for 10010001, we must first subtract 1 from the
LSB thus giving:
10010000
Next, we invert the 1's and 0's as follows:
01101111
Finally, we evaluate the decimal:
 N10  027   126   125   02 4   123   12 2   121   120 
 111
 N10  111
4.10 To find the decimal value for 10001001, we must first subtract 1 from the
LSB thus giving:
10001000
Next, we invert the 1's and 0's as follows:
01110111
Finally, we evaluate the decimal:
 N10  027   126   125   12 4   023   12 2   121   120 
 119
 N10  119
4.11 The following table presents the maximum decimal number versus the
number of bits for simple binary:
No. Bits
Max. Dec. No. Simple Binary
12
13
14
15
16
212 -1 = 4095
213 -1 = 8191
214 -1 = 16383
215 -1 = 32767
216 -1 = 65535
Consequently, 15 bits are needed to represent 27541 in simple binary. For a
two's complement binary number, the MSB will be zero so 16 bits will be
required.
4.5
4.12 The following table presents the maximum decimal number versus the
number of bits for simple binary:
No. Bits
Max. Dec. No. Simple Binary
12
13
14
15
16
212 -1 = 4095
213 -1 = 8191
214 -1 = 16383
215 -1 = 32767
216 -1 = 65535
Consequently, 14 bits are needed to represent 12034 in simple binary. For a
two's complement binary number, the MSB will be zero so 16 bits will be
required.
4.13 The following table presents the maximum decimal number versus the
number of bits for simple binary:
No. Bits
8
9
10
11
12
Max. Dec. No. for Simple Binary
28 -1 = 255
29 -1 = 511
210 -1 = 1023
211 -1 = 2047
212 -1 = 4095
Consequently, 10 bits are needed to represent 756 in simple binary. However, for
-756, an additional bit is required to represent the sign. Hence, 11 bits will be
required. The representation of -756 in 2's complement binary is 10100001100.
4.14 The following table presents the maximum decimal number versus the
number of bits for simple binary:
No. Bits
8
9
10
11
12
Max. Dec. No. for Simple Binary
28 -1 = 255
29 -1 = 511
210 -1 = 1023
211 -1 = 2047
212 -1 = 4095
Consequently, 10 bits are needed to represent 534 in simple binary. However, for
-534, an additional bit is required to represent the sign. Hence, 11 bits will be
required. The representation of -534 in 2's complement binary is 11011101010.
4.6
4.15
N = 12
Vru = 8V
Vrl = -8V
Vin = input voltage
(a) By Eq. B in Fig. 4.7:
 V  Vrl  N 
D0  int  in
2 
 Vru  Vrl  
 4.2   8  12 
 int 
2 
 8   8 

 int3123.2




 3123
(b) By Eq. B in Fig. 4.7:
  5.7   8 12 
2 
D0  int 
 8   8

 int588.8
 589
(c) Since 10.9V falls outside the input range, Do will have the maximum
output:
D0  212  1  4095
(d) Since -8.5V falls outside the input range, D0 will take the minimum
value:
D0  0
4.7
4.16
N = 12
Vru = 8V
Vrl = -8V
Vin = input voltage
(a) By Eq. B in Fig. 4.7:
 V  V  
D0  int in rl 2 N 
 Vru  Vrl  
  2.4    8  12 
 int 
2 
  8    8 

 int2662.4
 2662
(b) By Eq. B in Fig. 4.7:
  6.3    8  12 
D0  int 
2 
  8    8 

 int 3660 .8
 3660
(c) Since 11V falls outside the input range, Do will have the maximum
output:
D0  212  1  4095
(d) Since 9.2V falls outside the input range, D0 will take the maximum
value:
D0  212  1  4095
4.8
4.17
N8
Vru  10V
Vrl  0V
Vin  input voltage
a) By Eq. B in Fig. 4.7:
 V  Vrl  N 
D0  int  in
2 
 Vru  Vrl  
 5.75  0 8 
2 
 int 

 10  0
 int147.2
 147
b) The input is below the input range. Hence the output will be 0,
D0 = 0
c) Since 11.5V falls outside the input range, the output will be the
maximum possible. The output will be:
D0  28  1  255
d) By Eq. B in Fig. 4.7:
 0  0  8 
D0  int 
2 
 10  0 
0
4.9
4.18
N 8
V ru  15V
V rl  0V
Vin  input voltage
a) By Eq. B in Fig. 4.7:
 Vin  Vrl  N 
D0  int 
2 
 Vru  Vrl  
  6.42  0 8 
2 
 int 

 15  0
 int109.6
 109
b) The input is below the input range. Hence the output will be 0,
D0 = 0
c) By Eq. B in Fig. 4.7:
 Vin  Vrl  N 
D0  int 
2 
 Vru  Vrl  
 12  0 8 
2 
 int 

 15  0
 int204.8
 204
d) By Eq. B in Fig. 4.7:
  0  0 8 
D0  int 
2 
 15  0 
0
4.10
4.19 We need Equation A of Figure 4.7 to solve this problem.
(a) When the 1.5V signal is amplified with a gain of 10, it becomes 15V which
exceeds the input range of the A/D converter (it is saturated). According to Figure
4.7, the maximum output is 2N/2-1 = 112/2-1 = 2047
(b) With the gain of 10, the input becomes 8V. The output, in decimal, is then:
V  Vrl N  2 N
 8  ( 10) 12  212
 int
2 
2 
 1638
Do  int  in
10  ( 10)
 2
Vru  Vrl
 2
(c) When amplified, -1.5V results in an input to the A/D converter which is below
the input range (it is saturated). The largest negative output is –2N/2 = -2048
(d)With the amplifier, this voltage results in an input to the A/D of –8V. The output
is then:
 V  Vrl N  2 N
  8  (10) 12  212
Do  int  in
2 
 int 
2 
 1638
 10  (10)
 2
Vru  Vrl
 2
4.20 We need Equation A of Figure 4.7 to solve this problem.
(a) When the 5.2V signal is amplified with a gain of 10, it becomes 52V which
exceeds the input range of the A/D converter (it is saturated). According to Figure
4.7, the maximum output is 2N/2-1 = 112/2-1 = 2047
(b) When the 1.5V signal is amplified with a gain of 10, it becomes 15V which
exceeds the input range of the A/D converter (it is saturated). According to Figure
4.7, the maximum output is 2N/2-1 = 112/2-1 = 2047
(c) When amplified, -5.2V results in an input to the A/D converter which is below
the input range (it is saturated). The largest negative output is –2N/2 = -2048
(d) When amplified, -1.5V results in an input to the A/D converter which is below
the input range (it is saturated). The largest negative output is –2N/2 = -2048
4.21
N  16
Vru  5V
Vrl  0V
Vin  136
. V
From Eq. 4.1:
V  V 
Input Resolution Error  0.5 ru N rl Volts
 2

 5  0
 0.5 16 Volts
 2 
 3.815  10 5Volts
The quantization error (as a percent reading) for an input of 1.36V is:
 3.815  10 5Volts 

 100  0.0028%
136
. Volts


4.11
4.22
N  12
Vru  5V
Vrl  0V
Vin  2.45V
From Eq. 4.1:
V  V 
Input Resolution Error  0.5 ru N rl Volts

 2
50
 0.5 12 Volts
 2 
 6.104  10 4 Volts
The quantization error (as a percent reading) for an input of 2.45V is:
  6.104  10 4 Volts 

100  0.00025%
2.45Volts


4.23 Quantization error is computed using equation 4.1.
Vru  Vrl
2N
For 8 bits this becomes
8  (8)
input res error  0.5
 0.0313V and this is 0.42% of the 7.5V input.
28
For 12 bits this becomes
8  (8)
input res error  0.5
 0.00195V and this is 0.026% of the 7.5V input.
212
For 16 bits this becomes
8  (8)
input res error  0.5
 0.000122 V and this is 0.0016% if the 7.5V input.
216
input res error  0.5
4.12
4.24 Quantization error is computed using equation 4.1.
Vru  Vrl
2N
For 8 bits this becomes
10  (10)
input res error  0.5
 0.0391V and this is 0.49% of the 8V input.
28
For 12 bits this becomes
10  (10)
input res error  0.5
 0.00244V and this is 0.031% of the 8V input.
212
For 16 bits this becomes
10  (10)
input res error  0.5
 0.000153V and this is 0.0019% if the 8V input.
216
input res error  0.5
4.25
N  12
Vru  8V
Vrl  8V
Vin  4.16V
From Eq. 4.1,
V  V 
Input Resolution Error  0.5 ru N rl Volts
 2

 8   8 
 0.5
Volts
12

 2
 1996
 10 3Volts
.
The quantization error (as a percent reading) for an input of -4.16V is:
 1996
.
 10 3Volts 

 100  0.048%
4.16Volts


4.13
4.26
N  12
V ru  5V
V rl  5V
Vin  2.46V
From Eq. 4.1,
V  V 
Input Resolution Error  0.5 ru N rl Volts
 2

 5    5 
Volts
 0.5
 212 
 1.221  10  3 Volts
The quantization error (as a percent reading) for an input of -2.46V is:
  1.221 10 3 Volts 
100  0.050%

 2.46Volts


4.27 Since the signal from the transducer varies between 15mV (0.015V) and
the A/D converter input range is 10V, we can select a gain of 100 which will
yield an input of 1.5V. A gain of 100 is chosen such that the amplified signal is
not saturated (i.e. greater than the input range).
The quantization error from Eq. 4.1 is as follows:
V  V 
Quantization Error  0.5 ru N rl Volts
 2

 10   10  
 0.5

212


 2.44  10 3 Volts
The transducer voltage is 3.75mV but after a gain of 100 it becomes
0.375V. Thus, the quantization error as a percent reading is as follows:
 2.44  10 3Volts 

 100  0.651%
0.375Volts 

If the transducer output were attenuated by a factor of 2/3 (to 10 mV) the
gain could be set to 1000 without saturating the A/D converter. The 3.75
mv output would then become 3.7510-3(2/3)1000 = 2.5 V at the input
to the A/D converter and the resolution error would be reduced to
0.098%.
4.14
4.28 The amplified input must not exceed 10V.
When amplified, the maximum input will be 0.075V, 0.75V and 3.75V for gains
of 10, 100 and 500 respectively. So we can use the maximum gain of 500.
The input resolution error is:
10  (10)
input res error  0.5
 2.44mV
212
This is 0.033% of the maximum 7.5V input.
4.29 The amplified input must not exceed 10V.
When amplified, the maximum input will be 0.1V, 1V and 5V for gains of 10, 100
and 2000 respectively. The gain of 2000 will saturate the amplified signal and
become greater than the input limit of 10V. Thus the maximum gain that can be
used is 100.
The input resolution error is:
10  (10)
input res error  0.5
 2.44 mV
212
This is 0.024% of the maximum 10V input.
4.30
N8
Vru  5V
Vrl  0V
D in  Digital input  32
The output range will be divided into increments of:
5
incr.  8  0.0195
2
An input of 32 would then give an output of:
Vout  32  0.0195  0.625V
4.15
4.31
N  12
Vru  10V
Vrl  0V
Din  Digital input  45
The output range will be divided into increments of:
10
incr.  12  0.00244
2
An input of 45 would then give an output of:
Vout  45  0.00244  0.110V
4.32
The reference voltage increment is:
 input span   10 
V  
   12   0.0024414

 2 
2N
Trial digital output (D0)
100000000000
010000000000
011000000000
010100000000
010110000000
010101000000
010101100000
010101010000
010101001000
010101000100
010101000010
010101000011
(2048)
(1024)
(1536)
(1280)
(1408)
(1344)
(1376)
(1360)
(1352)
(1348)
(1346)
(1347)
D0V
Pass/Fail
5.0
2.5
3.75
3.12
3.44
3.28
3.36
3.32
3.30
3.2901
3.286
3.289
F
P
F
P
F
P
F
F
F
F
P
P
Actual digital output
010000000000
010000000000
010100000000
010100000000
010101000000
010101000000
010101000000
010101000000
010101000000
010101000010
010101000011
The output is 010101000011 or 1347 in decimal.
4.16
(1024)
(1024)
(1280)
(1280)
(1344)
(1344)
(1344)
(1344)
(1344)
(1346)
(1347)
4.33 The voltage increment is V = 16/212 = 0.00390625 V. Since this converter
is offset binary, the expected input for a given digital output Do is 0.0039806Do 8.0. The expected input for the final output is 4.2 V.
Trial Do
VDo - 8.
Pass/Fail
Actual Digital Output
100000000000 (2048)
110000000000 (3072)
111000000000 (3584)
110100000000 (3328)
110010000000 (3200)
110001000000 (3136)
110000100000 (3104)
110000110000 (3120)
110000111000 (3128)
110000110100 (3124)
110000110010 (3122)
110000110011 (3123)
0.0
4.0
6.0
5.0
4.5
4.25
4.125
4.1875
4.21875
4.203125
4.195313
4.199219
P
P
F
F
F
F
P
P
F
F
P
P
100000000000 (2048)
110000000000 (3072)
110000000000 (3072)
110000000000 (3072)
110000000000 (3072)
110000000000 (3072)
110000100000 (3104)
110000110000 (3120)
110000110000 (3120)
110000110000 (3120)
110000110010 (3122)
110000110011 (3123)
The final output is 110000110011 in binary or 3123 in decimal. The same result
is obtained from Eq. B in Figure 4.7.
4.34
non-linearity error
ADC span error
ADC zero
amplifier gain
MUX crosstalk
quantization
aperture
drift
bias
bias
bias
bias
precision
precision
precision
precision
4.35 How many channels available?
How many bits does the ADC output?
Is there programmable gain and what values of gain are possible?
Maximum number of samples per second?
Is there capability for automatic timing of data taking?
Is there a simultaneous sample and hold capability?
Is there a capability for digital input?
Is there a capability for frequency input? (not discussed in chapter)
Is there a capability for analog output?
What range of input voltages will not permanently damage the system?
What software is available for the system?
If intended for a harsh environment, how durable is the package?
4.17
CHAPTER 5
5.1
The ramp function appears as the following:
T = 0.02 sec
o = 2f
= 2(1/T)
= 2(1/0.02)
= 100 rad/sec
f(t) = 100t
f(t) = 100t  2
0  t  0.01
0.01  t  0.02
From Eq. 5.5,
2 T
b n   f (t ) sin n  o tdt
T 0
Thus,
0.02
2  0.01
b1 
100 t sin(100t )dt   (100 t  2 ) sin(100t )dt 

0
0
.01


0 .02 

 
 100 3.1831  10 3  3.1831  10 3

 0.6366
b2 
0.02
2  0.01
100 t sin( 2  100t )dt   (100 t  2 ) sin( 2  100t )dt 


0.01
0 .02  0

 
 100 15916
.
 10 3  15916
.
 10 3

 0.3183
ao is the average over the period, Eq. 5.3
1 T
a o   f t dt
T 0
Thus,
0.02
1  0.01
a0 
100
t
cos(
0
)
dt

0.01 (100t  2) cos(0)dt 
0.02 0

 
 50 5  10 3  5  10 3

0
From Eq. 5.4,
5.1
an 
2
T

a1 
T
0
f (t ) cos n o tdt
0.02
2  0.01
100 t cos(100t )dt   (100 t  2) cos(100t )dt 

0
0
.01


0.02 

 
 100 2.023  10 3  2.026  10 3

0
a2 
0.02
2  0.01
100t cos( 2  100t )dt   (100t  2) cos( 2  100t )dt 


0.01
0.02  0
0
Without actually evaluating the values for ao, a1, and a2, we could have found
that they were each equal to zero since f(t) is an odd function.
5.2
The ramp function appears as the following:
T = 0.04 sec
o = 2f
= 2(1/T)
= 2(1/0.04)
= 50 rad/sec
f(t) = 50t
f(t) = 50t  2
0  t  0.02
0.02  t  0.04
From Eq. 5.5,
2 T
bn  0 f (t ) sin n o tdt
T
Thus,
0.04
2 0.02
b1 
t ) dt  0.02 (50t  2) sin( 50
t ) dt
50t sin( 50

0
0.04
 506.3662  103   6.3662  103 


 0.6366
b2 
2
0.04

0.02
0
t ) dt  0.02 (50t  2) sin( 2  50
t ) dt
50t sin( 2  50
0.04
5.2

 50 3.1831103    3.1831103 
 0.3183
ao is the average over the period,Eq. 5.3
1 T
a o   f t dt
T 0
Thus,
0.04
1 0.02
a0 
50
t
cos(
0
)
dt

(50t  2) cos(0)dt


0.02
0.04 0
 25  0.01    0.01 
0
From Eq. 5.4,
2 T
a n   f (t ) cos n o tdt
T 0
0.02
0.04
2
a1 
50t cos( 50
t ) dt  0.02 (50t  2) cos( 50
t ) dt

0
0.04
 50 4.053  103   4.053  103 




0
a2 
2
0.04

0.02
0
t )dt  0.02 (50t  2) cos(2  50
t )dt
50t cos(2  50
0.04

0
Without actually evaluating the values for ao, a1, and a2, we could have found
that they were each equal to zero since f(t) is an odd function.
5.3
5.3
0 t 1
V(t) = 25t
T = 1sec
o = 2f
= 2(1/T)
= 2(1/1)
= 2 rad/sec
25
0
By Eq. 5.5.
2 T
b n   f (t ) sin n  o tdt
T 0
Thus,
2 1
b1   25 t sin( 2t )dt
1 0
= 2(-3.9789)
= -7.9578
2 1
25 t sin( 2  2t )dt
1 0
= 2(-1.9894)
= -3.9788
b2 
ao is the average, Eq. 5.3
1 T
a o   f t  cos n  o tdt
T 0
Thus,
1 1
a 0   25 t cos( 0 )dt
1 0
= 12.5
2 T
f t  cos n o tdt
T 0
2 1
a1   25 t cos( 2t )dt
1 0
=0
and an 
2 1
25 t cos( 2  2t )dt
1 0
=0
a2 
5.4
1
2
5.4
0t 2
V(t) = 10t
T = 2sec
o = 2f
= 2(1/T)
= 2(1/2)
= 1 rad/sec
10
0
By Eq. 5.5.
2 T
bn  0 f (t ) sin n o tdt
T
Thus,
2 2
b1  010t sin( 
t ) dt
2
= -6.3662
2 2
 10t sin( 2  t )dt
2 0
= -3.1831
b2 
ao is the average, Eq. 5.3
1 T
ao  0 f  t  cos notdt
T
Thus,
1 2
a0  010t cos( 0) dt
2
= 0.5(20)
= 10
and an 
2 T
f t  cos n o tdt
T 0
2 2
10t cos(t )dt
2 0
=0
a1 
2 2
10t cos( 2  t )dt
2 0
=0
a2 
5.5
2
4
time
-0.01
-0.0098
-0.0096
-0.0094
-0.0092
-0.009
-0.0088
-0.0086
-0.0084
-0.0082
-0.008
-0.0078
-0.0076
-0.0074
-0.0072
-0.007
-0.0068
-0.0066
-0.0064
-0.0062
-0.006
-0.0058
-0.0056
-0.0054
-0.0052
-0.005
-0.0048
-0.0046
-0.0044
-0.0042
-0.004
-0.0038
-0.0036
-0.0034
-0.0032
-0.003
-0.0028
-0.0026
-0.0024
-0.0022
-0.002
-0.0018
-0.0016
-0.0014
-0.0012
-0.001
-0.0008
-0.0006
-0.0004
-0.0002
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0.0016
0.0018
0.002
0.0022
0.0024
0.0026
0.0028
0.003
0.0032
0.0034
0.0036
0.0038
0.004
0.0042
0.0044
0.0046
0.0048
0.005
0.0052
0.0054
0.0056
0.0058
0.006
0.0062
0.0064
0.0066
0.0068
0.007
0.0072
0.0074
0.0076
0.0078
0.008
0.0082
0.0084
0.0086
0.0088
0.009
0.0092
0.0094
0.0096
0.0098
0.01
f(t) n>
-1.0000
-0.9800
-0.9600
-0.9400
-0.9200
-0.9000
-0.8800
-0.8600
-0.8400
-0.8200
-0.8000
-0.7800
-0.7600
-0.7400
-0.7200
-0.7000
-0.6800
-0.6600
-0.6400
-0.6200
-0.6000
-0.5800
-0.5600
-0.5400
-0.5200
-0.5000
-0.4800
-0.4600
-0.4400
-0.4200
-0.4000
-0.3800
-0.3600
-0.3400
-0.3200
-0.3000
-0.2800
-0.2600
-0.2400
-0.2200
-0.2000
-0.1800
-0.1600
-0.1400
-0.1200
-0.1000
-0.0800
-0.0600
-0.0400
-0.0200
0.0000
0.0200
0.0400
0.0600
0.0800
0.1000
0.1200
0.1400
0.1600
0.1800
0.2000
0.2200
0.2400
0.2600
0.2800
0.3000
0.3200
0.3400
0.3600
0.3800
0.4000
0.4200
0.4400
0.4600
0.4800
0.5000
0.5200
0.5400
0.5600
0.5800
0.6000
0.6200
0.6400
0.6600
0.6800
0.7000
0.7200
0.7400
0.7600
0.7800
0.8000
0.8200
0.8400
0.8600
0.8800
0.9000
0.9200
0.9400
0.9600
0.9800
1.0000
ao =
0
f(t)sinwt
1
2.65359E-06
0.061537253
0.12032233
0.176140739
0.228796872
0.278117339
0.323951517
0.366171967
0.404674727
0.439379478
0.470229576
0.497191956
0.520256918
0.539437779
0.554770412
0.56631266
0.574143647
0.578362966
0.579089776
0.576461797
0.570634205
0.561778455
0.550081016
0.535742036
0.518973944
0.5
0.479052791
0.456372692
0.432206294
0.406804811
0.380422475
0.353314924
0.325737592
0.297944123
0.270184791
0.242704958
0.215743575
0.18953172
0.164291194
0.140233179
0.117556965
0.096448751
0.077080528
0.059609054
0.044174911
0.030901674
0.019895175
0.011242869
0.005013325
0.001255809
0
0.001255809
0.005013325
0.011242869
0.019895175
0.030901674
0.044174911
0.059609054
0.077080528
0.096448751
0.117556965
0.140233179
0.164291194
0.18953172
0.215743575
0.242704958
0.270184791
0.297944123
0.325737592
0.353314924
0.380422475
0.406804811
0.432206294
0.456372692
0.479052791
0.5
0.518973944
0.535742036
0.550081016
0.561778455
0.570634205
0.576461797
0.579089776
0.578362966
0.574143647
0.56631266
0.554770412
0.539437779
0.520256918
0.497191956
0.470229576
0.439379478
0.404674727
0.366171967
0.323951517
0.278117339
0.228796872
0.176140739
0.12032233
0.061537253
2.65359E-06
b1 =
0.636411395
f(t)sin2wt
2
-5.30718E-06
-0.122831626
-0.238747029
-0.34604144
-0.443217317
-0.529010205
-0.602404449
-0.662643891
-0.709237464
-0.741959702
-0.760846263
-0.766184661
-0.758500506
-0.738539597
-0.707246305
-0.665738758
-0.615281351
-0.557255192
-0.49312709
-0.424417719
-0.352669606
-0.279415566
-0.206148202
-0.13429104
-0.065171858
1.32679E-06
0.060161165
0.114398436
0.161975758
0.202337364
0.235114788
0.260128459
0.277385206
0.287071823
0.289544888
0.285317102
0.275040508
0.259486972
0.239526396
0.216103147
0.190211238
0.162868796
0.135092395
0.107871788
0.082145597
0.058778482
0.038540264
0.022087455
0.009947587
0.002506663
0
0.002506663
0.009947587
0.022087455
0.038540264
0.058778482
0.082145597
0.107871788
0.135092395
0.162868796
0.190211238
0.216103147
0.239526396
0.259486972
0.275040508
0.285317102
0.289544888
0.287071823
0.277385206
0.260128459
0.235114788
0.202337364
0.161975758
0.114398436
0.060161165
1.32679E-06
-0.065171858
-0.13429104
-0.206148202
-0.279415566
-0.352669606
-0.424417719
-0.49312709
-0.557255192
-0.615281351
-0.665738758
-0.707246305
-0.738539597
-0.758500506
-0.766184661
-0.760846263
-0.741959702
-0.709237464
-0.662643891
-0.602404449
-0.529010205
-0.443217317
-0.34604144
-0.238747029
-0.122831626
-5.30718E-06
b2 =
-0.317891435
f(t)coswt
1
1
0.978066034
0.952429807
0.923349576
0.89109595
0.8559502
0.818202551
0.778150429
0.736096709
0.692347943
0.647212597
0.6009993
0.554015108
0.506563799
0.458944213
0.411448625
0.364361185
0.317956412
0.272497763
0.228236274
0.185409288
0.14423927
0.104932719
0.067679178
0.032650354
-6.63397E-07
-0.03014006
-0.057653845
-0.082448283
-0.104450206
-0.123607202
-0.139887686
-0.153280856
-0.163796518
-0.171464804
-0.176335769
-0.178478877
-0.177982378
-0.174952575
-0.169512995
-0.161803461
-0.151979073
-0.140209102
-0.126675809
-0.111573192
-0.09510566
-0.077486657
-0.058937237
-0.039684589
-0.019960535
0
0.019960535
0.039684589
0.058937237
0.077486657
0.09510566
0.111573192
0.126675809
0.140209102
0.151979073
0.161803461
0.169512995
0.174952575
0.177982378
0.178478877
0.176335769
0.171464804
0.163796518
0.153280856
0.139887686
0.123607202
0.104450206
0.082448283
0.057653845
0.03014006
6.63397E-07
-0.032650354
-0.067679178
-0.104932719
-0.14423927
-0.185409288
-0.228236274
-0.272497763
-0.317956412
-0.364361185
-0.411448625
-0.458944213
-0.506563799
-0.554015108
-0.6009993
-0.647212597
-0.692347943
-0.736096709
-0.778150429
-0.818202551
-0.8559502
-0.89109595
-0.923349576
-0.952429807
-0.978066034
-1
a1 =
1.39888E-16
f(t)cos2wt
2
-1
-0.972271768
-0.929838618
-0.87398817
-0.806199982
-0.728112768
-0.641489579
-0.548181607
-0.450091346
-0.34913579
-0.247210365
-0.146154254
-0.047717735
0.046467885
0.134917249
0.216314369
0.289532139
0.353647637
0.407953028
0.451961946
0.48541132
0.508258735
0.520675445
0.523035292
0.515899825
0.5
0.476214903
0.445547975
0.409101276
0.368048355
0.323606299
0.277007554
0.229472106
0.182180592
0.136248882
0.092704644
0.052466359
0.016325177
-0.01507003
-0.041224142
-0.061803601
-0.076640428
-0.085732402
-0.089239439
-0.087476288
-0.080901731
-0.070104551
-0.055786596
-0.038743329
-0.019842294
0
0.019842294
0.038743329
0.055786596
0.070104551
0.080901731
0.087476288
0.089239439
0.085732402
0.076640428
0.061803601
0.041224142
0.01507003
-0.016325177
-0.052466359
-0.092704644
-0.136248882
-0.182180592
-0.229472106
-0.277007554
-0.323606299
-0.368048355
-0.409101276
-0.445547975
-0.476214903
-0.5
-0.515899825
-0.523035292
-0.520675445
-0.508258735
-0.48541132
-0.451961946
-0.407953028
-0.353647637
-0.289532139
-0.216314369
-0.134917249
-0.046467885
0.047717735
0.146154254
0.247210365
0.34913579
0.450091346
0.548181607
0.641489579
0.728112768
0.806199982
0.87398817
0.929838618
0.972271768
1
a2 =
-2.66454E-17
5.6
5.5
The spreadsheet is shown at the left.
The values determined are b1 = 0.6364,
b2 = -0.3179, ao = 0, a1 = 0, and a2 =0.
These compare to exact respective values
of 0.6366, -0.3183, 0, 0 and 0.
note: The integration is from -T/2 to T/2
which is equivalent to 0 to T
time
f(t)
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.4
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
0.5
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.6
0.61
0.62
0.63
0.64
0.65
0.66
0.67
0.68
0.69
0.7
0.71
0.72
0.73
0.74
0.75
0.76
0.77
0.78
0.79
0.8
0.81
0.82
0.83
0.84
0.85
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1
f(t)sinwt
1
n>
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
5
5.25
5.5
5.75
6
6.25
6.5
6.75
7
7.25
7.5
7.75
8
8.25
8.5
8.75
9
9.25
9.5
9.75
10
10.25
10.5
10.75
11
11.25
11.5
11.75
12
12.25
12.5
12.75
13
13.25
13.5
13.75
14
14.25
14.5
14.75
15
15.25
15.5
15.75
16
16.25
16.5
16.75
17
17.25
17.5
17.75
18
18.25
18.5
18.75
19
19.25
19.5
19.75
20
20.25
20.5
20.75
21
21.25
21.5
21.75
22
22.25
22.5
22.75
23
23.25
23.5
23.75
24
24.25
24.5
24.75
25
bo =
0
0.015697617
0.062666564
0.140535869
0.248689682
0.386270928
0.552186385
0.745113172
0.963506604
1.205609381
1.469462057
1.752914735
2.053639925
2.369146504
2.696794692
3.033811974
3.377309882
3.724301543
4.071719906
4.416436545
4.755280941
5.085060141
5.402578676
5.704658653
5.988159891
6.25
6.487174298
6.696775446
6.876012704
7.022230693
7.132927562
7.205772459
7.238622205
7.229537071
7.176795585
7.078908254
6.934630146
6.742972238
6.503211473
6.214899449
5.877869697
5.49224348
5.058434088
4.577149582
4.049393962
3.476466739
2.859960895
2.201759236
1.504029131
0.769215658
3.31699E-05
-0.800544682
-1.629296443
-2.482765809
-3.357276003
-4.248945501
-5.153705051
-6.0673159
-6.985389162
-7.90340623
-8.816740142
-9.720677803
-10.61044296
-11.48121983
-12.32817724
-13.14649321
-13.9313798
-14.6781082
-15.38203377
-16.03862113
-16.64346894
-17.19233448
-17.68115762
-18.10608444
-18.46348991
-18.75
-18.96251265
-19.09821786
-19.15461651
-19.12953802
-19.02115657
-18.82800588
-18.54899256
-18.18340764
-17.7309366
-17.19166748
-16.56609727
-15.85513639
-15.06011123
-14.18276475
-13.22525512
-12.19015235
-11.08043291
-9.899472362
-8.651035991
-7.339267499
-5.968675728
-4.544119506
-3.070790643
-1.554195141
-0.000132679
a1 =
12.5
f(t)sin2wt
2
0
0.031333282
0.124344841
0.276093192
0.481753302
0.734731029
1.026819963
1.348397346
1.688654941
2.035859953
2.377640471
2.701289338
2.994079945
3.243587149
3.438006352
3.566463781
3.619311102
3.588397793
3.467315073
3.251605736
2.938934849
2.529217044
2.024696981
1.429980448
0.752014566
1.65849E-05
-0.814648221
-1.678638001
-2.576852526
-3.492694581
-4.408370071
-5.305221481
-6.164088622
-6.965689901
-7.691016885
-8.321734472
-8.840578812
-9.231744957
-9.481256326
-9.577308257
-9.510578283
-9.27449628
-8.865468299
-8.283048635
-7.530055615
-6.612627561
-5.540216457
-4.325517995
-2.984337864
-1.535395321
-6.63397E-05
1.597930252
3.232899034
4.877581018
6.503606793
8.081982278
9.583598821
10.97975875
12.24270709
13.34615969
14.26581822
14.97986207
15.46940789
15.71892759
15.71661638
15.45470303
14.92969558
14.14255658
13.09880334
11.80852967
10.28634711
8.551244936
6.62636985
4.538727408
2.318808986
0.000149264
-2.381179375
-4.787127939
-7.17827867
-9.514489938
-11.75556765
-13.86195198
-15.79540774
-17.5197065
-19.00128838
-20.20989173
-21.11913911
-21.70706873
-21.95660093
-21.85593071
-21.39883804
-20.584909
-19.41966263
-17.91457919
-16.08702791
-13.96009349
-11.56230248
-8.927252545
-6.093149079
-3.102255558
-0.000265359
a2 =
-7.955142436
f(t)coswt
1
0
0.249506683
0.496057357
0.73671546
0.968583214
1.188820748
1.394664905
1.583447619
1.752613769
1.899738408
2.022543266
2.118912441
2.18690719
2.224779729
2.230985968
2.204197111
2.143310048
2.047456475
1.916010702
1.748596079
1.545090019
1.305627575
1.030603538
0.720673056
0.376750744
8.29247E-06
-0.408129426
-0.84598973
-1.311658984
-1.802990874
-2.317616101
-2.852953428
-3.406222039
-3.97445515
-4.554514807
-5.143107808
-5.736802659
-6.332047489
-6.925188845
-7.512491253
-8.090157466
-8.654349286
-9.201208865
-9.726880369
-10.22753189
-10.69937751
-11.13869937
-11.5418697
-11.90537258
-12.22582542
-12.5
-12.72484295
-12.89749561
-13.01531306
-13.0758823
-13.0770395
-13.01688612
-12.89380385
-12.70646836
-12.45386165
-12.13528299
-11.75035842
-11.29904864
-10.78165531
-10.19882571
-9.551555701
-8.841190915
-8.069426234
-7.238303469
-6.350207267
-5.407859233
-4.41431028
-3.372931225
-2.28740166
-1.161697123
-7.46322E-05
1.192943387
2.412586701
3.653856342
4.911545068
6.180259129
7.454441254
8.728394754
9.996308641
11.25228365
12.49035906
13.70454017
14.88882635
16.03723947
17.14385266
18.2028192
19.20840143
20.15499954
21.03718011
21.84970426
22.58755526
23.24596546
23.82044244
24.30679421
24.70115336
25
b1 =
-3.973642941
f(t)cos2wt
2
0
0.248028679
0.484291607
0.697332452
0.876306885
1.011271633
1.093453595
1.115492984
1.071655024
0.958005351
0.77254501
0.515301769
0.188375372
-0.204064713
-0.655829492
-1.158808051
-1.70311102
-2.277257404
-2.868401329
-3.462594423
-4.045078733
-4.600604433
-5.113765944
-5.569349685
-5.952686292
-6.25
-6.448747807
-6.537941148
-6.50844306
-6.353234181
-6.067641495
-5.649524319
-5.099412855
-4.420595458
-3.619151734
-2.703929616
-1.686465613
-0.580848562
0.596471693
1.826928171
3.090129564
4.364197377
5.626141825
6.852270085
8.018619734
9.101409602
10.07749977
10.92485213
11.62298273
12.15339711
12.5
12.64947109
12.59159894
12.31956588
11.83017746
11.12403085
10.20561775
9.083358284
7.769563897
6.280328131
4.635345769
2.857662038
0.973354855
-0.988845569
-2.997994267
-5.021419532
-7.025253721
-8.97499824
-10.83611328
-12.57462234
-14.15772097
-15.55437913
-16.73592611
-17.67660752
-18.35410376
-18.75
-18.85019853
-18.64526498
-18.13070091
-17.30713671
-16.18043971
-14.76173389
-13.06732927
-11.11856034
-8.94153454
-6.566793469
-4.028891047
-1.365894254
1.381186341
4.169027782
6.952677952
9.686280053
12.32382665
14.81993092
17.13060224
19.21401285
21.0312425
22.54698787
23.73022407
24.55480626
25
b2 =
-4.22194E-05
-4.21777E-05
5.7
5.6
The spreadsheet is shown at the left.
The values determined are b1 = -7.955,
b2 = -3.974, a0 = 12.5, a1 = 0, and a2 =0.
These compare to exact respective values
of -7.9578, -3.9788, 12.5, 0 and 0.
5.7
If the function is made into an odd function, i.e. v(t) = -v(-t), then it can be
represented with sine functions only. This can be accomplished by adding the
function from t = -0.1 to t = 0 having the form 20t + 200t2. The original function plus
this new function form one cycle of a function on which a Fourier analysis can be
performed.
Voltage
0.6
0.4
2
v(t) = 20t - 200t
0.2
0
-0.2
2
v(t) = 20t + 200t
-0.4
-0.6
-0.1
-0.05
0
0.05
0.1
Time - sec
The resulting Fourier series will properly represent the original function in the 0 0.1 second interval but will be incorrect outside that interval.
5.8
If the function is made into an odd function, i.e. v(t) = -v(-t), then it can be
represented with sine functions only. This can be accomplished by adding the
function from t = -0.2 to t = 0 having the form 20t + 100t2. The original function plus
this new function form one cycle of a function on which a Fourier analysis can be
performed.
The resulting Fourier series will properly represent the original function in the 0 0.2 second interval but will be incorrect outside that interval.
5.8
5.9
The function can be represented with only cosine terms if it is an even function,
f(t) = f(-t). One such function is shown below.
v(t)
4
3
2
1
0
-1
-2
-3
-4
-6
-4
-2
0
2
4
6
time - seconds
The resulting Fourier will represent the original function correctly in the 0 - 1.5
second interval but will be incorrect outside that interval.
5.10
The function can be represented with only cosine terms if it is an even function,
f(t) = f(-t). One such function is shown below.
The resulting Fourier will represent the original function correctly in the 0 - 2 second
interval but will be incorrect outside that interval.
5.9
5.11
f(t) 1.2
1
first 20 terms
0.8
b0 , b 1 , b2 , b3 terms
0.6
0.4
0.2
0
-0.2
0
0.1
0.2
0.3
0.4
0.5
0.6
time - sec
The first four terms do not do a very good job of representing this square waves.
This is why square waves are a sensitive test of the frequency response of
instruments. For reference, the sum of the first 20 terms is also shown.
5.10
f(t)
3
2.871651566
1.183250021
0.424079962
2.12396463
5.422294952
7.803153495
7.336129722
4.403878647
1.359405907
0.421978514
1.707936728
3.17016518
2.498103016
-0.738323335
-4.637804818
-6.620225193
-5.696273486
-3.310949009
-2.003290419
-3.093180742
-5.482581588
-6.652700577
-4.957472928
-1.143869125
2.274093196
3.226047986
1.890628905
0.471250995
1.148357663
4.058389982
7.125272795
7.899143154
5.746285898
2.416680299
0.477612381
1.025127872
2.735401501
3.109028387
0.783125743
-3.177231726
-6.176592366
-6.370160297
-4.274984328
-2.261606546
-2.387137882
-4.525560184
-6.482143137
-5.988753492
-2.77980659
1.124988499
3.188687984
2.58487515
0.879896973
0.558205579
2.72435974
6.057407815
7.96355213
6.890027853
3.715191016
0.961924643
0.543338635
2.072309777
3.257116489
2.023572337
-1.553351437
freq (Hz)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
F(f)
2.99963853703405
3.03107808323142-0.953489727174751i
3.13972600731118-2.19264191621691i
3.39446291040749-4.35277576822253i
4.1506503807292-10.3318395080083i
42.0778018981402-315.139900191749i
1.15673145040832+14.5606087209i
2.0687931824973+7.94063600672844i
2.45177094174856+5.81805890632637i
2.73216871318734+4.80571676035284i
2.99947307192228+4.25004076361989i
3.29223988807091+3.94021835272683i
3.63963995929184+3.79303154623195i
4.07528245235743+3.77711044930519i
4.64861413905354+3.89155708187997i
5.44394983322112+4.16524610793015i
6.62445094289296+4.67557346031329i
8.55830664201133+5.61403491112103i
12.2949154415459+7.54174889789134i
22.5086297516064+12.9766095685749i
162.608034374744+88.4852822222177i
-29.8880606599922-15.4066520489326i
-13.4508975188424-6.58827818481786i
-8.58753125233437-4.00709570329827i
-6.2617659309384-2.78971922963066i
-4.90140821032262-2.08882384099306i
-4.01063677914423-1.63756127310175i
-3.38347692063223-1.32532271690295i
-2.91901176398376-1.09808332141237i
-2.56197898616181-0.926376517565872i
-2.2795792213038-0.792797515774453i
-2.05111784768694-0.686426317762479i
-1.86289456542791-0.600079203753877i
-1.70548044113495-0.528846341613602i
-1.57217514308172-0.469264699141709i
-1.45808779197661-0.418826334322632i
-1.35956589935354-0.375673663607068i
-1.2738277338382-0.338403845548055i
-1.1987183134243-0.305939362053351i
-1.13254310168028-0.277440124902437i
-1.07395199924385-0.252242399286549i
-1.02185674147784-0.229815493691055i
-0.975370998471055-0.209730489103945i
-0.933766222021456-0.191637292692324i
-0.896438617865371-0.175247552311383i
-0.862884109708393-0.16032176510352i
-0.83267913164779-0.146659432058179i
-0.805465730442839-0.134091454589227i
-0.780939895538502-0.122474201673354i
-0.758842335060168-0.111684835739446i
-0.738951125784759-0.101617596738633i
-0.721075813698939-9.21808224173024E-0
-0.705052648416223-8.32945390417479E-0
-0.69074071219551-7.48884975129065E-00
-0.67801876123611-6.69005595738468E-00
-0.666782639208102-5.92753608008256E-0
-0.656943154719748-5.19631934414062E-0
-0.648424338491339-4.49190644708364E-0
-0.641162014451039-3.81018935315351E-0
-0.635102633248227-3.14738224842301E-0
-0.630202327871022-2.49996136833994E-0
-0.626426159944596-1.86461181921036E-0
-0.62374753243364-1.23817982669434E-00
-0.622147750372017-6.17629078533588E-0
-0.621615716175162
-0.622147750372011+6.17629078533999E-0
|F(f)|
2.999638537
3.177511134
3.829563679
5.519876388
11.13439743
317.9366259
14.60648329
8.205705663
6.313556065
5.528079183
5.201892463
5.134604576
5.256811519
5.556481838
6.062493706
6.854623617
8.108288197
10.23533099
14.42369308
25.98135503
185.1243312
33.62530442
14.97771857
9.476418574
6.855089055
5.32794402
4.332068132
3.633785406
3.118720356
2.724318223
2.41350561
2.162929845
1.957158965
1.785592895
1.64071449
1.517048288
1.410514139
1.318011479
1.237143681
1.166030403
1.103176833
1.047380715
0.997665005
0.953228309
0.91340785
0.877651443
0.845496023
0.816551016
0.790485326
0.767017074
0.745905424
0.726944037
0.709955785
0.69478847
0.681311328
0.669412172
0.658995055
0.649978342
0.642293144
0.63588203
0.63069799
0.626703607
0.623870414
0.622178407
0.621615716
0.622178407
5.12
Amplitude
time
0
0.007874016
0.015748031
0.023622047
0.031496063
0.039370079
0.047244094
0.05511811
0.062992126
0.070866142
0.078740157
0.086614173
0.094488189
0.102362205
0.11023622
0.118110236
0.125984252
0.133858268
0.141732283
0.149606299
0.157480315
0.165354331
0.173228346
0.181102362
0.188976378
0.196850394
0.204724409
0.212598425
0.220472441
0.228346457
0.236220472
0.244094488
0.251968504
0.25984252
0.267716535
0.275590551
0.283464567
0.291338583
0.299212598
0.307086614
0.31496063
0.322834646
0.330708661
0.338582677
0.346456693
0.354330709
0.362204724
0.37007874
0.377952756
0.385826772
0.393700787
0.401574803
0.409448819
0.417322835
0.42519685
0.433070866
0.440944882
0.448818898
0.456692913
0.464566929
0.472440945
0.480314961
0.488188976
0.496062992
0.503937008
0.511811024
400
300
200
100
0
0 3 6 9 12 15 18 21 24 27 30 33 36
Frequency - Hz
Note: peaks occur at the frequencies
in the generating function
5.11
0.519685039
0.527559055
0.535433071
0.543307087
0.551181102
0.559055118
0.566929134
0.57480315
0.582677165
0.590551181
0.598425197
0.606299213
0.614173228
0.622047244
0.62992126
0.637795276
0.645669291
0.653543307
0.661417323
0.669291339
0.677165354
0.68503937
0.692913386
0.700787402
0.708661417
0.716535433
0.724409449
0.732283465
0.74015748
0.748031496
0.755905512
0.763779528
0.771653543
0.779527559
0.787401575
0.795275591
0.803149606
0.811023622
0.818897638
0.826771654
0.834645669
0.842519685
0.850393701
0.858267717
0.866141732
0.874015748
0.881889764
0.88976378
0.897637795
0.905511811
0.913385827
0.921259843
0.929133858
0.937007874
0.94488189
0.952755906
0.960629921
0.968503937
0.976377953
0.984251969
0.992125984
1
-5.253975531
-6.654266847
-5.256118151
-2.889965405
-2.029518013
-3.540399016
-5.894110744
-6.556461971
-4.297832112
-0.340474383
2.69406209
3.091251797
1.527448962
0.397006012
1.593257113
4.748004319
7.520282281
7.676642507
5.088936023
1.849089024
0.397420617
1.35138986
2.990907747
2.861768829
0.048127139
-3.938796782
-6.460992062
-6.074200387
-3.779732533
-2.081814452
-2.70291604
-5.01886925
-6.625611204
-5.526256294
-1.965143306
1.747018651
3.262000087
2.250726831
0.63688536
0.800648776
3.376056779
6.63168196
7.995961579
6.353922911
3.045588561
0.66618074
0.74951434
2.422132071
3.239542613
1.447528332
-2.374073117
-5.771156984
-6.567937388
-4.774921543
-2.535053219
-2.159292329
-4.024989253
-6.2332661
-6.334502788
-3.565174975
0.422714993
2.999867304
-0.623747532433636+1.23817982669474E-0
-0.626426159944614+1.86461181921076E-0
-0.630202327871017+2.49996136834509E-0
-0.635102633247936+3.14738224842586E-0
-0.641162014451052+3.81018935315351E-0
-0.648424338491355+4.49190644708368E-0
-0.656943154719751+5.19631934414049E-0
-0.666782639208101+5.92753608008274E-0
-0.678018761236112+6.69005595738419E-0
-0.690740712195535+7.4888497512885E-00
-0.705052648416258+8.32945390418962E-0
-0.721075813698932+9.21808224173266E-0
-0.738951125784765+0.101617596738646i
-0.758842335060181+0.111684835739462i
-0.780939895538509+0.122474201673366i
-0.805465730442842+0.134091454589245i
-0.832679131647796+0.146659432058196i
-0.862884109708437+0.160321765103543i
-0.896438617865513+0.175247552311575i
-0.933766222021401+0.19163729269233i
-0.975370998471051+0.209730489103939i
-1.02185674147784+0.229815493691042i
-1.07395199924384+0.252242399286547i
-1.13254310168027+0.277440124902439i
-1.19871831342429+0.305939362053357i
-1.27382773383837+0.338403845547929i
-1.35956589935359+0.375673663607144i
-1.4580877919766+0.418826334322665i
-1.57217514308172+0.469264699141715i
-1.70548044113496+0.528846341613596i
-1.86289456542791+0.600079203753877i
-2.05111784768693+0.686426317762493i
-2.2795792213038+0.792797515774465i
-2.56197898616183+0.926376517565844i
-2.91901176398385+1.09808332141246i
-3.3834769206321+1.32532271690313i
-4.01063677914423+1.63756127310176i
-4.90140821032263+2.08882384099306i
-6.2617659309384+2.78971922963067i
-8.58753125233436+4.00709570329832i
-13.4508975188424+6.58827818481792i
-29.8880606599922+15.4066520489327i
162.608034374744-88.4852822222182i
22.5086297516063-12.9766095685749i
12.2949154415458-7.54174889789138i
8.55830664201125-5.6140349111211i
6.62445094289294-4.6755734603133i
5.44394983322108-4.16524610793018i
4.64861413905351-3.89155708187998i
4.07528245235733-3.77711044930534i
3.63963995929178-3.79303154623196i
3.29223988807091-3.94021835272679i
2.99947307192227-4.25004076361991i
2.73216871318728-4.80571676035299i
2.45177094174854-5.81805890632638i
2.06879318249724-7.94063600672859i
1.15673145040826-14.5606087209i
42.0778018981418+315.139900191749i
4.15065038072927+10.3318395080083i
3.39446291040754+4.35277576822282i
3.13972600731119+2.19264191621688i
3.03107808323142+0.953489727174949i
-1.28-52.1414i
5.12
0.623870414
0.626703607
0.63069799
0.63588203
0.642293144
0.649978342
0.658995055
0.669412172
0.681311328
0.69478847
0.709955785
0.726944037
0.745905424
0.767017074
0.790485326
0.816551016
0.845496023
0.877651443
0.91340785
0.953228309
0.997665005
1.047380715
1.103176833
1.166030403
1.237143681
1.318011479
1.410514139
1.517048288
1.64071449
1.785592895
1.957158965
2.162929845
2.41350561
2.724318223
3.118720356
3.633785406
4.332068132
5.32794402
6.855089055
9.476418574
14.97771857
33.62530442
185.1243312
25.98135503
14.42369308
10.23533099
8.108288197
6.854623617
6.062493706
5.556481838
5.256811519
5.134604576
5.201892463
5.528079183
6.313556065
8.205705663
14.60648329
317.9366259
11.13439743
5.519876388
3.829563679
3.177511134
52.15710876
f(t)
0
0.000787402
0.001574803
0.002362205
0.003149606
0.003937008
0.004724409
0.005511811
0.006299213
0.007086614
0.007874016
0.008661417
0.009448819
0.01023622
0.011023622
0.011811024
0.012598425
0.013385827
0.014173228
0.01496063
0.015748031
0.016535433
0.017322835
0.018110236
0.018897638
0.019685039
0.020472441
0.021259843
0.022047244
0.022834646
0.023622047
0.024409449
0.02519685
0.025984252
0.026771654
0.027559055
0.028346457
0.029133858
0.02992126
0.030708661
0.031496063
0.032283465
0.033070866
0.033858268
0.034645669
0.035433071
0.036220472
0.037007874
0.037795276
0.038582677
0.039370079
0.04015748
0.040944882
0.041732283
0.042519685
0.043307087
0.044094488
0.04488189
0.045669291
0.046456693
0.047244094
0.048031496
0.048818898
0.049606299
0.050393701
0.051181102
freq (Hz)
8
-1.001264893
-5.77992304
2.175730456
0.389538306
-1.472531294
5.106322592
-0.756970292
-7.636525619
2.942428177
5.890432359
-3.355176234
-1.021855252
1.348628445
-4.101026101
2.068165004
6.617226701
-4.756626851
-5.318021241
4.790059281
1.246670585
-1.767720112
3.055511354
-2.769472158
-5.140169668
6.153693982
4.081876412
-6.189160753
-0.894451937
2.562990265
-2.257007351
2.83276874
3.487761877
-6.9245361
-2.337642887
7.253700679
-0.074514159
-3.474443875
1.925282318
-2.368860985
-1.966088397
6.985042837
0.345388418
-7.74206003
1.559736732
4.206625937
-2.16427335
1.601552938
0.839333256
-6.393058266
1.584988948
7.522342121
-3.340637899
-4.494187063
2.939058053
-0.817230892
-0.273494243
5.334752963
-3.159098978
-6.601372449
5.122836396
4.161054415
-4.083169508
0.300925461
0.301581782
-4.083368667
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
420
430
440
450
460
470
480
490
500
510
520
530
540
550
560
570
580
590
600
610
620
630
640
F(f)
7.99937
8.00425+0.196478i
8.01896+0.393915i
8.04374+0.593295i
8.079+0.795648i
8.12536+1.00209i
8.18363+1.21383i
8.25491+1.43224i
8.34059+1.65891i
8.4425+1.89567i
8.56297+2.14473i
8.70502+2.4088i
8.8726+2.69124i
9.07099+2.99638i
9.30727+3.3299i
9.59131+3.69944i
9.93715+4.11572i
10.3656+4.5943i
10.9086+5.15889i
11.6189+5.84784i
12.5885+6.728i
13.9972+7.92843i
16.2474+9.73722i
20.4723+12.9624i
31.5848+21.1016i
154.285+108.645i
-28.1429-20.869i
-7.99645-6.23935i
-2.03189-1.66704i
0.979711+0.845228i
2.92885+2.65441i
4.41931+4.20714i
5.72223+5.72158i
7.00261+7.35412i
8.40265+9.26959i
10.0975+11.704i
12.3757+15.0773i
15.8372+20.2899i
22.0866+29.7746i
37.574+53.3399i
150.575+225.3i
-66.0264-104.242i
-25.4567-42.4607i
-15.0649-26.5856i
-10.2921-19.2491i
-7.54606-14.9862i
-5.7612-12.1763i
-4.50915-10.1681i
-3.5842-8.64881i
-2.87522-7.4496i
-2.31683-6.47102i
-1.8681-5.65069i
-1.50207-4.94742i
-1.20027-4.33284i
-0.949682-3.78672i
-0.740847-3.29419i
-0.566788-2.84402i
-0.422255-2.42751i
-0.303287-2.03776i
-0.206866-1.66916i
-0.130696-1.31704i
-0.073063-0.977412i
-0.0327292-0.646777i
-0.00885656-0.321957i
-0.000945956
-0.00885656+0.321957i
|F(f)|
7.99937
8.006661081
8.028629304
8.065590626
8.118084549
8.186920025
8.273160415
8.378236719
8.503965189
8.652708882
8.827475403
9.032147621
9.27177456
9.553070328
9.885014358
10.28003326
10.75574736
11.33813282
12.06696731
13.00753896
14.27362309
16.08669046
18.94179136
24.23094883
37.98522231
188.6997542
35.03623812
10.14261806
2.628231218
1.293925815
3.95272751
6.10166599
8.091995667
10.15478349
12.51118802
15.45778517
19.50597154
25.73901602
37.07215538
65.24530947
270.985093
123.3931929
49.50711686
30.55724692
21.82785315
16.77883226
13.47047539
11.12307023
9.362072637
7.985200699
6.873267134
5.951478395
5.170413808
4.496014962
3.903990812
3.376468871
2.899947999
2.463961056
2.060206014
1.681930036
1.323508899
0.98013898
0.647604577
0.322078792
0.000945956
5.13
5.13
Amplitude
time
300
250
200
150
100
50
0
0
70 140 210 280 350 420 490 560 630
Frequency - Hz
Spikes in FFT occur at the frequencies
of the generating function. The
spread at the peak frequencies
is due to insufficient resolution of
signal (sampling rate only about
three times maximum frequency.)
0.051968504
0.052755906
0.053543307
0.054330709
0.05511811
0.055905512
0.056692913
0.057480315
0.058267717
0.059055118
0.05984252
0.060629921
0.061417323
0.062204724
0.062992126
0.063779528
0.064566929
0.065354331
0.066141732
0.066929134
0.067716535
0.068503937
0.069291339
0.07007874
0.070866142
0.071653543
0.072440945
0.073228346
0.074015748
0.07480315
0.075590551
0.076377953
0.077165354
0.077952756
0.078740157
0.079527559
0.080314961
0.081102362
0.081889764
0.082677165
0.083464567
0.084251969
0.08503937
0.085826772
0.086614173
0.087401575
0.088188976
0.088976378
0.08976378
0.090551181
0.091338583
0.092125984
0.092913386
0.093700787
0.094488189
0.095275591
0.096062992
0.096850394
0.097637795
0.098425197
0.099212598
0.1
4.160002467
5.123842099
-6.600583844
-3.160351437
5.334672217
-0.272744699
-0.817830481
2.939295569
-4.493355663
-3.341770928
7.52183541
1.586382956
-6.393176027
0.838490071
1.602157717
-2.164487132
4.20599503
1.560899841
-7.741862601
0.343941211
6.985420913
-1.96518981
-2.36954195
1.925441863
-3.473960722
-0.075621455
7.253797831
-2.33624433
-6.925203276
3.486878699
2.833587469
-2.25712012
2.562583776
-0.893458914
-6.189499769
4.080622999
6.154638085
-5.139391821
-2.770464269
3.055620507
-1.767318949
1.245812955
4.790561336
-5.316986349
-4.757795366
6.61664527
2.06932917
-4.101200109
1.34817893
-1.021115184
-3.355753576
5.889653364
2.943737309
-7.636214285
-0.758265265
5.106638981
-1.472011193
0.388865998
2.176305109
-5.779396071
-1.002615148
7.999999861
-0.0327292+0.646777i
-0.073063+0.977412i
-0.130696+1.31704i
-0.206866+1.66916i
-0.303287+2.03776i
-0.422255+2.42751i
-0.566788+2.84402i
-0.740847+3.29419i
-0.949682+3.78672i
-1.20027+4.33284i
-1.50207+4.94742i
-1.8681+5.65069i
-2.31683+6.47102i
-2.87522+7.4496i
-3.5842+8.64881i
-4.50915+10.1681i
-5.7612+12.1763i
-7.54606+14.9862i
-10.2921+19.2491i
-15.0649+26.5856i
-25.4567+42.4607i
-66.0264+104.242i
150.575-225.3i
37.574-53.3399i
22.0866-29.7746i
15.8372-20.2899i
12.3757-15.0773i
10.0975-11.704i
8.40265-9.26959i
7.00261-7.35412i
5.72223-5.72158i
4.41931-4.20714i
2.92885-2.65441i
0.979711-0.845228i
-2.03189+1.66704i
-7.99645+6.23935i
-28.1429+20.869i
154.285-108.645i
31.5848-21.1016i
20.4723-12.9624i
16.2474-9.73722i
13.9972-7.92843i
12.5885-6.728i
11.6189-5.84784i
10.9086-5.15889i
10.3656-4.5943i
9.93715-4.11572i
9.59131-3.69944i
9.30727-3.3299i
9.07099-2.99638i
8.8726-2.69124i
8.70502-2.4088i
8.56297-2.14473i
8.4425-1.89567i
8.34059-1.65891i
8.25491-1.43224i
8.18363-1.21383i
8.12536-1.00209i
8.079-0.795648i
8.04374-0.593295i
8.01896-0.393915i
8.00425-0.196478i
5.14
f(t)
0
0.015748031
0.031496063
0.047244094
0.062992126
0.078740157
0.094488189
0.11023622
0.125984252
0.141732283
0.157480315
0.173228346
0.188976378
0.204724409
0.220472441
0.236220472
0.251968504
0.267716535
0.283464567
0.299212598
0.31496063
0.330708661
0.346456693
0.362204724
0.377952756
0.393700787
0.409448819
0.42519685
0.440944882
0.456692913
0.472440945
0.488188976
0.503937008
0.519685039
0.535433071
0.551181102
0.566929134
0.582677165
0.598425197
0.614173228
0.62992126
0.645669291
0.661417323
0.677165354
0.692913386
0.708661417
0.724409449
0.74015748
0.755905512
0.771653543
0.787401575
0.803149606
0.818897638
0.834645669
0.850393701
0.866141732
0.881889764
0.897637795
0.913385827
0.929133858
0.94488189
0.960629921
0.976377953
0.992125984
1.007874016
1.023622047
freq (Hz)
0
1.690774167
5.238447449
7.443918867
6.318412902
2.876850305
0.222647147
0.749221098
3.981736088
7.005283935
7.093391792
4.166608731
0.864150321
0.163995956
2.697507191
6.18009984
7.471359139
5.406896893
1.848334133
0.004591385
1.538256705
5.066353187
7.407433155
6.450436593
3.058331535
0.289935903
0.641639999
3.796295983
6.90920474
7.173312557
4.350461209
0.986146234
0.114125951
2.520741356
6.035836099
7.489686775
5.57128903
2.010550782
0.018354295
1.391155225
4.891035519
7.361991345
6.57584762
3.24150648
0.36569745
0.541670459
3.610742511
6.805389478
7.244850525
4.532843315
1.114910103
0.073159251
2.346985654
5.885974943
7.498856897
5.731221306
2.177026887
0.04125503
1.249829939
4.712923755
7.307704713
6.694338886
3.425926592
0.44974627
0.44955728
3.425530045
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
10.5
11
11.5
12
12.5
13
13.5
14
14.5
15
15.5
16
16.5
17
17.5
18
18.5
19
19.5
20
20.5
21
21.5
22
22.5
23
23.5
24
24.5
25
25.5
26
26.5
27
27.5
28
28.5
29
29.5
30
30.5
31
31.5
32
F(f)
476.25
-3.75741-0.092231i
-3.78098-0.185729i
-3.82107-0.281831i
-3.87892-0.381998i
-3.95643-0.487931i
-4.0563-0.601634i
-4.18228-0.725621i
-4.33958-0.86311i
-4.53544-1.01836i
-4.78015-1.19724i
-5.08862-1.40806i
-5.48327-1.66315i
-5.99918-1.98164i
-6.69433-2.395i
-7.67189-2.95904i
-9.13438-3.78315i
-11.5411-5.11522i
-16.2032-7.66263i
-28.9632-14.577i
-204.081-109.07i
36.5439+20.6991i
16.0018+9.58982i
9.92636+6.28493i
7.02244+4.69157i
5.32486+3.74964i
4.21392+3.12478i
3.43221+2.6781i
2.85361+2.34151i
2.40908+2.07773i
2.05767+1.86463i
1.77351+1.6882i
1.53951+1.53921i
1.3439+1.41125i
1.17832+1.29979i
1.03666+1.2015i
0.914375+1.11389i
0.80799+1.03507i
0.714826+0.963563i
0.632766+0.898192i
0.56013+0.838029i
0.495564+0.782312i
0.437965+0.730424i
0.386425+0.681855i
0.340196+0.636169i
0.29864+0.593005i
0.261244+0.552052i
0.227553+0.513046i
0.197192+0.475753i
0.169851+0.439984i
0.145236+0.405558i
0.123121+0.372317i
0.1033+0.340127i
0.0855925+0.308866i
0.0698582+0.278423i
0.0559598+0.248692i
0.0437916+0.219586i
0.0332556+0.191019i
0.0242768+0.162911i
0.016783+0.135188i
0.0107256+0.107779i
0.00605607+0.0806229i
0.00274071+0.0536478i
0.000764668+0.0267928i
0.000116199
0.000764668-0.0267928i
|F(f)|
476.25
3.7585418
3.785538934
3.83144942
3.897684292
3.986403769
4.10067472
4.244760516
4.424580596
4.648362409
4.927800487
5.279837731
5.729949198
6.317994758
7.109857885
8.222761938
9.886815559
12.6238847
17.92371582
32.42461847
231.3986159
41.99892103
18.65535448
11.74874325
8.445442121
6.512598111
5.246081568
4.353422228
3.691308591
3.18129351
2.776841887
2.448541803
2.1769838
1.948767244
1.754392221
1.586904599
1.441121984
1.31309472
1.19976242
1.098699995
1.007987213
0.926064654
0.851664582
0.783741361
0.721418266
0.663958417
0.610745314
0.561245549
0.515000584
0.471630451
0.430779279
0.392146311
0.355467672
0.32050629
0.287053192
0.2549102
0.223910061
0.193892221
0.164709918
0.136225785
0.108311363
0.080850034
0.053717762
0.02680371
0.000116199
5.15
5.14
400
300
Amplitude
time
200
100
0
0
5
10
15
20
25
30
Frequency - Hz
10 Hz is double function frequency since
squaring makes negative portion of sine wave
positive leading to an apparent doubling of the
frequency. Spike at f = 0 is due to the fact that
the average value of the funcion is not zero
(there is a DC component of the signal.
1.039370079
1.05511811
1.070866142
1.086614173
1.102362205
1.118110236
1.133858268
1.149606299
1.165354331
1.181102362
1.196850394
1.212598425
1.228346457
1.244094488
1.25984252
1.275590551
1.291338583
1.307086614
1.322834646
1.338582677
1.354330709
1.37007874
1.385826772
1.401574803
1.417322835
1.433070866
1.448818898
1.464566929
1.480314961
1.496062992
1.511811024
1.527559055
1.543307087
1.559055118
1.57480315
1.590551181
1.606299213
1.622047244
1.637795276
1.653543307
1.669291339
1.68503937
1.700787402
1.716535433
1.732283465
1.748031496
1.763779528
1.779527559
1.795275591
1.811023622
1.826771654
1.842519685
1.858267717
1.874015748
1.88976378
1.905511811
1.921259843
1.937007874
1.952755906
1.968503937
1.984251969
2
6.694092366
7.307830518
4.713308442
1.250126617
0.041196172
2.176665567
5.730883344
7.498847049
5.88630209
2.347354792
0.073237512
1.114626915
4.532454042
7.244706192
6.805620237
3.611140276
0.541876547
0.365526022
3.241112121
6.575585942
7.362098315
4.891414678
1.391464667
0.018314983
2.010198165
5.570941079
7.489657254
6.036151628
2.521117408
0.114223423
0.98587723
4.350068303
7.173150051
6.909419172
3.796693991
0.64186268
0.289782455
3.05794033
6.450160397
7.407521027
5.066725888
1.538578154
0.004571715
1.847991082
5.406539806
7.471310017
6.18040298
2.697889236
0.1641124
0.86389616
4.166213154
7.09321151
7.005481516
3.982133364
0.749459828
0.222512056
2.876463212
6.318122866
7.443987427
5.238812782
1.691106834
2.11246E-08
0.00274071-0.0536478i
0.00605607-0.0806229i
0.0107256-0.107779i
0.016783-0.135188i
0.0242768-0.162911i
0.0332556-0.191019i
0.0437916-0.219586i
0.0559598-0.248692i
0.0698582-0.278423i
0.0855925-0.308866i
0.1033-0.340127i
0.123121-0.372317i
0.145236-0.405558i
0.169851-0.439984i
0.197192-0.475753i
0.227553-0.513046i
0.261244-0.552052i
0.29864-0.593005i
0.340196-0.636169i
0.386425-0.681855i
0.437965-0.730424i
0.495564-0.782312i
0.56013-0.838029i
0.632766-0.898192i
0.714826-0.963563i
0.80799-1.03507i
0.914375-1.11389i
1.03666-1.2015i
1.17832-1.29979i
1.3439-1.41125i
1.53951-1.53921i
1.77351-1.6882i
2.05767-1.86463i
2.40908-2.07773i
2.85361-2.34151i
3.43221-2.6781i
4.21392-3.12478i
5.32486-3.74964i
7.02244-4.69157i
9.92636-6.28493i
16.0018-9.58982i
36.5439-20.6991i
-204.081+109.07i
-28.9632+14.577i
-16.2032+7.66263i
-11.5411+5.11522i
-9.13438+3.78315i
-7.67189+2.95904i
-6.69433+2.395i
-5.99918+1.98164i
-5.48327+1.66315i
-5.08862+1.40806i
-4.78015+1.19724i
-4.53544+1.01836i
-4.33958+0.86311i
-4.18228+0.725621i
-4.0563+0.601634i
-3.95643+0.487931i
-3.87892+0.381998i
-3.82107+0.281831i
-3.78098+0.185729i
-3.75741+0.092231i
5.16
5.15
f(t)
w(n)
0
0.003922
0.007843
0.011765
0.015686
0.019608
0.023529
0.027451
0.031373
0.035294
0.039216
0.043137
0.047059
0.05098
0.054902
0.058824
0.062745
0.066667
0.070588
0.07451
0.078431
400
0.082353
0.086275
350
0.090196
300
0.094118
0.098039
250
0.101961
200
0.105882
0.109804
150
0.113725
0.117647
100
0.121569
50
0.12549
0.129412
0
0.133333
1
0.137255
0.141176
0.145098
0.14902
0.152941
0.156863
0.160784
0.164706
0.168627
0.172549
0.176471
0.180392
0.184314
0.188235
0.192157
0.196078
0.2
0.203922
0.207843
0.211765
0.215686
0.219608
0.223529
0.227451
0.231373
0.235294
0.239216
0.243137
0.247059
0.25098
0.254902
0.258824
0.262745
2
2.560006
2.049354
1.267592
1.01408
1.52654
2.316328
2.507956
1.46111
-0.733176
-3.214141
-4.811389
-4.724188
-2.996915
-0.492563
1.619046
2.537596
2.252438
1.453049
1.00319
1.32937
2.124775
2.561975
1.875572
-0.066945
-2.584947
-4.526322
-4.928061
-3.585873
-1.176101
1.146607
2.423491
2.417133
1.666602
1.053186
4 7 10
1.169499
1.906853
2.530531
2.193828
0.55459
-1.912745
-4.124828
-4.999925
-4.094401
-1.867182
0.593791
2.211444
2.525707
1.89191
1.160509
1.058599
1.68139
2.426268
2.412491
1.112244
-1.22202
-3.622363
-4.936983
-4.503123
-2.541379
-0.024061
1.899749
2.562415
2.110914
1.317452
1.004639
1.466526
2.264766
0
0.000152
0.000607
0.001366
0.002427
0.003791
0.005455
0.007421
0.009685
0.012247
0.015105
0.018258
0.021703
0.025438
0.029462
0.033771
0.038364
0.043237
0.048387
0.053812
0.059507
0.06547
0.071697
0.078184
0.084927
0.091922
0.099165
0.106651
0.114376
0.122335
0.130524
0.138937
0.147569
0.156415
0.16547
13 16 19
0.174728
0.184184
0.193831
0.203664
0.213677
0.223864
0.234219
0.244735
0.255406
0.266226
0.277187
0.288284
0.29951
0.310857
0.322319
0.333889
0.345559
0.357324
0.369175
0.381106
0.393109
0.405177
0.417302
0.429477
0.441696
0.45395
0.466231
0.478534
0.490849
0.50317
0.515489
0.527798
0.540091
fxw
f
F(f) [w/o Hann]
|F| w/o Hann
|F| with Hann
0
0 9.520899079.52089907615234
9.520899
-1.83056267609769E-002-2.55671310739411E-003i
0.018483
-1.92929477549381E-002-5.29251415784474E-003i
0.020006
0.000389
1 9.543982259.54398225491964+0.592611312 9.562363
0.001244
2 9.614245209.61424520583671+1.195398829 9.688276
-2.10542353890507E-002-8.4144408297074E-003i
0.022673
0.001731
3 9.734840289.73484028645024+1.819243935 9.903371
-2.37887838246068E-002-1.2193253877418E-002i
0.026732
0.002461
4 9.911418849.91141884239111+2.476548448 10.21614
-2.78364257427318E-002-1.70183637239435E-002i
0.032627
-3.37659173848057E-002-2.34926070300524E-002i
0.041134
0.005787
5 10.152818610.1528186286056+3.182300831 10.63987
0.012637
6 10.472243210.4722432772704+3.955609767 11.19441
-4.25464464212656E-002-3.26097854105798E-002i
0.053606
0.01861
7 10.889251310.8892513310703+4.822078601 11.90917
-5.58960315401776E-002-4.61122117038859E-002i
0.072462
0.014151
8 11.433174811.4331748967985+5.817720476 12.82823
-7.70384765327427E-002-6.7268570002083E-002i
0.102274
-0.008979
9 12.149236112.1492361449759+6.995824011 14.01947
-0.112499212438008-0.10272860632803i
0.152346
-0.048549
10 13.110136713.1101367239048+8.439837570 15.59187
-0.176882453591918-0.167488142711216i
0.243597
-0.087844
11 14.439751514.4397515416454+10.28958945 17.73082
-0.307691721662172-0.300397592619051i
0.430015
-0.102528
12 16.366695816.3666958002919+12.80043742 20.77787
-0.622619657668311-0.62453151177262i
0.88187
-0.076236
13 19.363289919.3632899198369+16.49655896 25.43764
-1.63437812057132-1.68129435121605i
2.344769
-0.014512
14 24.586074124.5860741810464+22.65665294 33.4335
-7.84124173583645-8.26794992168868i
11.39491
0.054678
15 35.813492335.8134923608497+35.44998523 50.39154
87.90976257063+95.0359566135189i
129.4603
0.097352
16 76.571811176.5718111545271+80.81435654 111.3293
-123.626530071246-137.091701774767i
184.6013
42.1659316649976+47.9534504903429i
63.8553
0.097389
17 -230.88462-230.884626101933-257.6306950 345.9498
0.070309
2.92746337045364+3.40598118951088i
4.491186
W/O Hann 18 -40.364497-40.3644971813686-47.12744822 62.0507
With Hann
0.053983
19 -20.437092-20.4370925215447-24.62623781 32.00197
0.825664436417024+0.975311646376146i
1.277871
0.079107
20 -12.881498-12.8814983153593-15.71012701 20.31603
0.340305830822219+0.400551621763411i
0.525595
200
0.169916497289174+0.190699988080646i
0.255418
0.139109
21 -8.9361985-8.93619857309896-10.70686338 13.94606
180 9.814375
0.183686
22 -6.5363485-6.5363485821225-7.3210720565
9.53740986802364E-002+9.129516981575E-002i
0.132027
0.146639
23 -4.9436730-4.94367304927676-4.712516500
5.8003808947335E-002+3.2183343037416E-002i
0.066334
160 6.829913
3.76062146503254E-002-1.32939552229795E-002i
0.039887
-0.005685
24 -3.8303918-3.83039184025816-2.482549237
140 4.564532
-0.237613
25 -3.0311093-3.03110938233169-0.394208898
2.61373517575831E-002-6.01240390550676E-002i
0.06556
120 3.056636
-0.448851
26 -2.4567276-2.45672767371197+1.734190159 3.007146
2.03740186151045E-002-0.123478638584714i
0.125148
100
-0.525582
27 -2.0603546-2.06035469896185+4.088731135 4.578513
1.99086055927278E-002-0.230227913098095i
0.231087
80
2.81278831041657E-002-0.447811711847271i
0.448694
-0.410138
28 -1.8245704-1.82457044481194+6.919989928 7.156488
60 10.79395
-0.143879
29 -1.7622992-1.7622992003391+10.649120028
6.04791096851053E-002-0.999899356943792i
1.001727
40 16.24983
0.201099756437104-2.98393755897354i
2.990706
0.14966
30 -1.9375816-1.9375816315704+16.133897144
20 25.67786
1.56257067251059-19.9557484735226i
20.01683
0.336712
31 -2.5486481-2.54864816647615+25.55106335
0.356694
32 -4.3482349-4.34823496977889+46.72382435
-9.58854860337493+105.66285785475i
106.097
0 46.92572
11.3840548649404-109.93060570717i
110.5185
0.260682
33 -13.832795-13.8327950067868+146.3309259 146.9833
1 4 7 10 13 16
19 22 25 28 31 34 37 40 43 46 49 52 55 58
0.174271
34 17.960861917.9608619432907-177.51364875
178.42
-2.78806322628668+23.9918851781958i
24.15334
22 25 28 31 34 37 40 43 46 49 52 55 58 61 64
f- Hz
-0.413763790828728+3.21955914445288i
3.246038
0.204344
35 6.689927556.68992755636645-60.119310614 60.49038
0.351211
36 4.657744424.65774442124189-37.637320394 37.92443
-0.148295585182181+1.05574525769422i
1.06611
-7.2408954752263E-002+0.475968363231916i
0.481445
0.490495
37 3.862244553.86224455141116-28.003883728 28.26897
-4.17275908629139E-002+0.255032170127355i
0.258423
0.446804
38 3.465450813.46545081260024-22.603845539 22.86795
0.118503
39 3.243458873.24345887698067-19.121698033 19.39483
-2.66982980903635E-002+0.152523557935908i
0.154843
-0.428195
40 3.111594943.11159494425029-16.672758350 16.96063
-1.83655556549138E-002+9.84632358361419E-002i
0.100161
-0.966113
41 3.031068593.03106859208115-14.845493033 15.15177
-1.33249828364367E-002+6.72463349626715E-002i
0.068554
-1.223657
42 2.981788532.98178853615779-13.422285969 13.7495
-1.00713535814055E-002+4.79543194593184E-002i
0.049
-1.045735
43 2.952423152.9524231503951-12.2771179547 12.62713
-7.86257082364794E-003+3.5385368836958E-002i
0.036248
-0.497092
44 2.936170662.93617066994831-11.331873184 11.70609
-6.30127032602193E-003+2.6842739725767E-002i
0.027572
0.164591
45 2.928751242.92875124347351-10.535493731
10.935
-5.16040054871906E-003+2.08318603641251E-002i
0.021462
0.637524
46 2.927371512.92737151068435-9.8531576022 10.27882
-4.30323252236245E-003+1.64782843206885E-002i
0.017031
-3.64377954386447E-003+1.32468666149042E-002i
0.013739
0.756474
47 2.930154912.93015491346819-9.2602725197 9.712799
-3.12595756926322E-003+1.07974504814177E-002i
0.011241
0.588113
48 2.935811482.93581148129834-8.7389566933 9.218913
-2.71202994918288E-003+8.90668982668203E-003i
0.00931
0.374054
49 2.943438062.94343806341119-8.2758818976 8.783738
0.353454
50 2.952393092.95239309139856-7.8609003594 8.397046
-2.3759130018119E-003+7.42375796275876E-003i
0.007795
-2.09914886887435E-003+6.2442674219692E-003i
0.006588
0.58102
51 2.962215662.96221566708471-7.4861417538 8.050903
-1.86840242115474E-003+5.29438835853882E-003i
0.005614
0.866964
52 2.972571922.97257192746298-7.1454024122 7.739054
-1.67386020180755E-003+4.52088748614375E-003i
0.004821
0.890632
53 2.983218692.98321869596771-6.8337218855 7.456497
-1.50818005108094E-003+3.88471468076608E-003i
0.004167
0.423883
54 2.993978372.99397837173733-6.5470829461 7.199181
-0.480387
55 3.004721283.0047212897412-6.28219489807 6.963787
-1.36578626395333E-003+3.3567760594796E-003i
0.003624
-1.24238693268426E-003+2.91508897056632E-003i
0.003169
-1.467696
56 3.015353143.01535314609505-6.0363343304 6.747569
-2.060212
57 3.025805913.02580591790716-5.8072262500 6.548235
-1.13463733893784E-003+2.54283040222814E-003i
0.002784
-1.03990127150382E-003+2.22697496402402E-003i
0.002458
-1.933989
58 3.036031233.03603123173695-5.5929540941 6.363853
-1.122516
59 3.045995473.04599547175924-5.3918907278 6.192784
-9.5607920606733E-004+1.95732925840776E-003i
0.002178
-8.814828854118E-004+1.72583733657927E-003i
0.001938
-0.010922
60 3.055676133.05567613943102-5.2026449081 6.033628
-8.14742602125932E-004+1.52607447223154E-003i
0.885722
61 3.065059123.06505912359628-5.0240193008 5.885181
-7.54737839044773E-004+1.35287364765499E-003i
1.226202
62 3.074136633.07413663960986-4.8549772280 5.746401
-7.00544819071779E-004+1.2020468243561E-003i
1.03614
63 3.082905663.08290566458054-4.6946160907 5.61638
0.662902
3.091366743.0913667435362-4.54214594532929i
-6.51396436445562E-004+1.07017474256731E-003i
-6.06651355273534E-004+9.54446846780328E-004i
0.51788
3.099523073.09952307497051-4.39687209809818i
Remaining
rows not shown for
0.77403
3.107379803.107379808229-4.25818086119435i
-5.65769967673952E-004+8.52538259817063E-004i
presentation
purposes
-5.28295534871413E-004+7.6251442421007E-004i
1.223179
3.114943503.11494350247595-4.12552781514903i
|F|
t
5.17
5.16
The function has frequencies of 5 and 20 Hz. The minimum sampling rate would
then be 40 Hz to avoid aliasing. Sampling at 30 Hz would not produce aliases for
the 5 Hz signal but would for the 20 Hz signal. The alias frequency can be
evaluated using the folding diagram in Section 5.1.
fN = 30/2 = 15 Hz. f/fN = 20/15 = 1.3333. From the folding diagram, fa/fN = 0.666.
Thus fa = 0.66615 = 10 Hz. This is the difference between the sampling rate and
the signal frequency.
5.17
The function has frequencies of 10 and 15 Hz. The minimum sampling rate would
then be 30 Hz to avoid aliasing. Sampling at 50 Hz would not produce false aliases
for either signals. The alias frequency can be evaluated using the folding diagram in
Section 5.1.
5.18
The signal in problem 5.11 has frequencies of 5 and 20 Hz. The minimum sampling
rate to avoid aliasing would then be twice the maximum frequency, or 40 Hz in this
case. We will evaluate the alias frequencies using the folding diagram in Section
5.1.
For a sampling rate of 5 Hz, fN = 2.5 Hz. For the 5 Hz signal, f/fN = 5/2.5 = 2.
Reading from the folding diagram, fa/fN = 0. So the alias frequency will be 0 Hz (i.e.
DC). For the 20 Hz signal, f/fN = 20/2.5 = 8. This is over the range of Figure A.1, but
if the construction of the diagram is examined, it will be noticed that fa/fn = 0 again.
So again, fa = 0. Both of these signals are integer multiples of the sampling rate, so
the dc alias is to be expected.
5.19
The maximum frequency in the signal of problem 5.11 is 20 Hz. The sampling rate
should exceed twice this value, or 40 Hz.
5.18
5.20
The signal has frequencies of 250 and 400 Hz. The minimum sampling rate to avoid
aliasing would then be twice the maximum frequency, or 800 Hz in this case. We
will evaluate the alias frequencies using the folding diagram in Section 5.1.
For a sampling rate of 400 Hz, fN = 200 Hz. For the 250 Hz signal, f/fN = 250/200 =
1.25. Reading from the folding diagram, fa/fN = 0.75 So the alias frequency will be
0.75200 = 150 Hz. For the 400 Hz signal, f/fN = 400/200 = 2. The value of fa/fN = 0.
So fa = 0, or dc.
5.21
The signal has frequencies of 50 and 250 Hz. The minimum sampling rate to avoid
aliasing would then be twice the maximum frequency, or 500 Hz in this case. We
will evaluate the alias frequencies using the folding diagram in Section 5.1.
For a sampling rate of 200 Hz, fN = 100 Hz. This is adequate to avoid aliasing for the
50 Hz signal. For the 250 Hz signal, f/fN = 250/100 = 2.5. Reading from the folding
diagram, fa/fN = 0.5 So the alias frequency will be 0.5100 = 50 Hz.
5.22
The maximum frequency in the signal of problem 5.11 is 400 Hz. The sampling rate
should exceed twice this value, or 800 Hz.
5.19
time
0.00025
0.00275
0.00525
0.00775
0.01025
0.01275
0.01525
0.01775
0.02025
0.02275
0.02525
0.02775
0.03025
0.03275
0.03525
0.03775
0.04025
0.04275
0.04525
0.04775
0.05025
0.05275
0.05525
0.05775
0.06025
0.06275
0.06525
0.06775
0.07025
0.07275
0.07525
0.07775
0.08025
0.08275
0.08525
0.08775
0.09025
0.09275
0.09525
0.09775
0.10025
f(t)
6.816737266
2.897171027
2.897333587
6.817124503
1.273976625
5.193861135
5.193891276
1.2744186
6.817832692
2.898162376
2.898471966
6.818158801
1.27501086
5.194999313
5.194882292
1.275513565
6.818927586
2.899153195
2.899609814
6.819192565
1.276044565
5.196136959
5.195872775
1.276608001
6.820021945
2.900143483
2.900747132
6.820225795
1.277077741
5.197274072
5.196862726
1.277701907
6.821115771
2.901133239
2.901883918
6.821258492
1.278110386
5.198410652
5.197852144
1.278795283
6.822209063
term 1
2.771638978
-1.14806041
-1.148030991
2.771626792
-2.77165421
1.148097184
1.147994217
-2.771611559
2.771669442
-1.148133957
-1.147957442
2.771596326
-2.771684673
1.148170731
1.147920668
-2.771581092
2.771699903
-1.148207504
-1.147883893
2.771565858
-2.771715134
1.148244277
1.147847118
-2.771550623
2.771730363
-1.14828105
-1.147810343
2.771535388
-2.771745592
1.148317822
1.147773567
-2.771520152
2.771760821
-1.148354595
-1.147736792
2.771504916
-2.771776049
1.148391367
1.147700016
-2.771489679
2.771791277
term 2
4.045098287
4.045231437
4.045364578
4.045497711
4.045630835
4.045763952
4.04589706
4.046030159
4.046163251
4.046296334
4.046429408
4.046562475
4.046695533
4.046828583
4.046961624
4.047094657
4.047227682
4.047360699
4.047493707
4.047626707
4.047759699
4.047892682
4.048025657
4.048158624
4.048291582
4.048424532
4.048557474
4.048690408
4.048823333
4.04895625
4.049089158
4.049222059
4.049354951
4.049487834
4.049620709
4.049753576
4.049886435
4.050019285
4.050152128
4.050284961
4.050417787
amplitude
5.23
8
7
6
5
4
3
2
1
0
0
0.02
0.04
0.06
0.08
0.1
0.12
time - seconds
This signal when sampled at the sampling rate of
400 Hz will produce a DC alias of the second term (with
amplitude 4.05) and a alias of the first term
with frequency 150 Hz.
5.20
5.24
Use Appendix A-3. fN = 2000/2 = 1000 Hz. f/fN = 3500/1000 = 3.5. From Figure A.1,
fa/fN = 0.5. Thus fa = 0.5*1000 = 500 Hz.
5.25
Use Appendix A-3. fN = 3000/2 = 1500 Hz. f/fN = 5000/1500 = 3.333. From Figure
A.1, fa/fN = 0.666. Thus fa = 0.666*1500 = 9990 Hz.
5.26
Use Appendix A-3. fN = 1500/2 = 750 Hz. f/fN = 1000/750 = 1.333. From Figure A.1,
fa/fN = 0.666. Thus fa = 0.666*750 = 500 Hz.
5.27
Use Appendix A-3. fN = 4000/2 = 2000 Hz. f/fN = 3000/2000 = 1.5. From Figure A.1,
fa/fN = 0.5. Thus fa = 0.5*2000 = 1000 Hz.
5.28
The dynamic range of an A/D converter can be determined from Eq. 5.18. Since the
16 bit converter is bipolar, one bit is used for the sign and the dynamic range is
determined from 15 bits. The dynamic range is then:
dynamic range = 20 log10(215) = 90.3 dB
5.29
The dynamic range can be determined from Eq. 5.18. For the 14 bit unipolar
converter, the dynamic range is:
dynamic range = 20 log10(214) = 84.3 dB
5.21
5.30
The required attenuation is the dynamic range of the A/D converter. For the 12 bit
bipolar converter, 1 bit is used for the sign and the dynamic range is:
dynamic range = 20 log10(211) = 66.2 dB
A second order Butterworth filter will attenuate the signal at a rate of 12 dB octave.
Hence, 66.2/12 = 5.5 octaves will be required. The maximum frequency will then be
given by Eq. 5.20:
fm = 1000025.5 = 452 kHz.
This is a rather high frequency and will require a sampling rate greater than 900
kHz. It will probably be better to use a higher order filter. For a fourth order filter, fm
will be only 68 kHz.
5.31
From Eq. 5.18,
dynamic range = 20 log10(2N) dB
= 20 log10(211) dB
= 66.2 dB
Since 1 bit for sign with bipolar.Next choose a fourth order Butterworth filter
which attenuates the signal 46 = 24 dB/octave.
From Eq. 5.19, we can find the number of octaves required for attenuation:
Dynamic Range
Noct 
Filter Attenuation Rate
66.2 dB

24 dB / oct
 2.75 octaves
Choose fc = 500 Hz and we can find fm from Eq. 5.20:
fm  fc 2 N
= 50022.75
= 3363 Hz
From Eq. 5.16, we find the actual minimum sampling frequency:
fs = 2(3363 Hz) = 6274 Hz, which is less than our maximum sampling rate of
10,000 samples per second.
The corner frequency should also be chosen to be equal to fc and thus 500
Hz.
oct
5.22
5.32
From Eq. 5.18,
dynamic range = 20 log10(2N) dB
= 20 log10(2(81)) dB
= 42.14 dB
A first order Butterworth filter has a filter attenuation rate of 16 = 6
dB/octave
Thus, from Eq. 5.19,
Dynamic Range
Noct 
Filter Attenuation Rate
4214
. dB

6 dB / oct
 7.02 octaves
If we choose fc = 100 Hz, from Eq. 5.20:
fm  fc 2 N
= 10027.02
= 13015.01 Hz
oct
From Eq. 5.16, the minimum sampling rate is fs = 26031 Hz.
5.23
CHAPTER 6
6.1 a) Data arranged in bins with width 2mm
Bin
(cm)
48.9-49.09
49.1-49.29
49.3-49.49
49.5-49.69
49.4-49.89
49.9-50.09
50.1-50.29
50.3-50.49
50.5-50.69
b)
No. Meas.
3
Number of
Measurements
1
2
2
0
1
1
2
0
1
No. meas. 10, bin width 2 mm
2
1
0
48.7 48.9 49.1 49.3 49.5 49.7 49.9 50.1 50.3 50.5 50.7
Length range - cm
b)
6.2 a) Data arranged in bins with width 2in
No. Meas.
Bin
(in)
48.1-50.0
50.1-52.0
52.1-54.0
54.1-56.0
56.1-58.0
58.1-60.0
60.1-62.0
62.1-64.0
64.1-66.0
68.1-70.0
Number of
Measurements
0
0
0
0
0
8
2
0
0
0
8
No of meas. 10, bin
width 2 in
7
6
5
4
3
2
1
0
48.1
50.1
52.1
54.1
56.1
58.1
60.1
62.1
Length range - in
6.1
64.1
66.1
68.1
6.3
Bin
80-84.9
85-89.9
90-94.9
95-99.9
100-104.9
105-109.9
110-114.9
115-119.9
120-124.9
No.
0
1
2
0
2
2
3
1
0
3
Total no. meas. = 12
No.
Bin width 5 psi
Meas.
2
1
0
80
85
90
95
100
105
110
115
120
125
Pressure range - psi
6.4
Bin
8-8.49
8.5-8.99
9-9.49
9.5-9.99
10-10.49
10.5-10.99
No.
0
3
3
4
2
0
No of meas. 12, bin width 0.5 bar
No.
Meas.
4
3
2
1
0
8
8.5
9
9.5
10
Pressure range - bar
6.2
10.5
11
6.5
S
Using the data from problem 6.1:
Mean:
n
x  x 2    x n
x
x 1
 i
n
i 1 n
49.3  50.1  48.9  49.2  49.3  50.5  49.9  49.2  49.8  50.2
x
10
x  49.6cm
Median:
Arranging data in ascending order:
48.9, 49.2, 49.2, 49.3, 49.3, 49.8, 49.9, 50.1, 50.2, 50.5
49.3  49.8
 Median 
 49.6cm
2
Standard Deviation:
( x i  x )2
i 1 ( n  1)
n

 ( 49.3  49.64 )

(50.1  49.64)
( 48.9  49.64 )
   


S
(10  1)
(10  1)
(10  1)


2
2
2
S  0.53 cm
Modes: 49.2cm, 49.3cm
6.3
1
2
6.6
Using the data from problem 6.2:
Mean:
n
x  x      xn
x
x 1 2
 i
n
i 1 n
59.3  60.0  58.8  59.1  59.2  60.4  59.8  59.3  59.8  60.3
x
10
x  59.6in
Median:
Arranging data in ascending order:
58.8, 59.1, 59.2, 59.3, 59.3, 59.8, 59.8, 60.0, 60.3, 60.4
59.3  59.8
 Median 
 59.6in
2
Standard Deviation:
( xi  x ) 2
S 
i 1 ( n  1)
n
1
 (58.8.1  59.6) 2 (59.1  59.6) 2 (59.2  59.6) 2
2

   

S 
(10  1)
(10  1)
(10  1)


S  0.54in
Modes: 59.3in,59.8in
6.4
6.7
x
110  104  106  94  92  89  100  114  120  108  110  115
12
x  105 psi
Median:
Arranging data in ascending order
89, 92, 94, 100, 104, 106, 108, 110, 110, 114, 115, 120
 Median 
106  108
2
Median  107
Standard Deviation:
S
( x i  x )2

i 1 ( n  1)
n
1
 (110  105.2)2 (104  105.2)2 (106  105.2)2
2
S


 
(12  1)
(12  1)
(12  1)


S  9.7 psi (10psi )
Mode:
110psi
6.5
6.8
x
9.5  9.3  9.4  8.9  8.8  8.7  9.0  9.8  10.2  10  9.5  9.9
12
x  9.4bar
Median:
Arranging data in ascending order
8.7, 8.8, 8.9, 9.0, 9.3, 9.4, 9.5, 9.5, 9.8, 9.9, 10.0, 10.2
9.4  9.5
 Median 
2
Median  9.45
Standard Deviation:
S
( x i  x )2

i 1 ( n  1)
n
1
 (8.7  9.4) 2 (8.8  9.4) 2 (8.9  9.4) 2
2
   
S 


(12  1)
(12  1)
 (12  1)

S  0.50bar 1bar
Mode:
9.5bar
6.9 Probability of having a 6 and a 3 in tossing two fair dice:
1 1 2
 
 0.056
 5.6%
6 6 36
There are two ways of getting a 6 and a 3 - 6,3 and 3,6
P(6 and 3)  2 
6.10 Probability of having a 4 and a 2 in tossing two fair dice:
1 1 2
 
 0.056  5.6%
6 6 36
There are two ways of getting a 4 and a 2 - 4,2 and 2,4
P(64and 2)  2 
6.11
Probability of an undergraduate electrical engineering student to be a
woman:
P  0.15  0.8  0.120
 12%
6.6
6.12
Probability of an undergraduate biology student to be a woman:
P  0.55  0.85  0.468  46.8%
6.13
Probability of all three components being defective:
P  0.03  0.03  0.03  27  10 6 or 0.0027%
6.14
Probability of all three components being defective:
P  0.02  0.02  4  10 4 or 0.04%
6.15 Binomial distribution
all 5 > 12 oz.; p=0.99, n=5, r=5
 n  5
5!
     
1
 r   5  5! 0!
P(5)  1 0.995 (1  0.99)0  0.951
all 5 < 12 oz.; p=0.99, n=5, r=0
 n   5  5!
     
1
 r   0  0!5!
P(0)  1  0.99 0 (1  0.99) 5  10 5
6.16 Binomial distribution
all 5 > 8 oz.; p=0.98, n=5, r=5
 n  5 5!
  
1
 r  5 5!0!
P(5)  1  0.985 (1  0.98) 0  0.904
all 5 < 12 oz.; p=0.98, n=5, r=0
 n   5  5!
     
1
 r   0  0!5!
P(0)  1  0.980 (1  0.98) 5  3.2  10 9
6.17 Binomial distribution
all 6 > 3000 hours; p=0.9, n=6, r = 6
 n   6  6!
     
1
 r   6  6!0!
P (6)  1 0.9 6 (1  0.9)0  0.531
6.7
6.18 Binomial distribution
all 6 > 3600 hours; p=0.95, n=6, r = 6
 n   6  6!
  
1
 r   6  6!0!
P(6)  1  0.95 6 (1  0.95) 0  0.735
6.19 Binomial distribution
Success is failure before 1000 hours. We want probability of 1 or 2 failures.
p=0.2, n=2, r=1 and r=2
 2  2!
 2  2!
  
  
2
1
1  1!1!
 2  2!0!
P(1)  2  0.21(1  0.2)1  0.32 P (2)  1 0.2 2 (1  0.2)0  0.04
The probability of 1 or 2 is then P(1)+P(2) = 0.36
6.8
6.20
Probability distribution function:
3x 2
f ( x) 
2 x3
35
0
(a) f(x) satisfies the requirement of a probability distribution function
because:
 f ( x )  0    x  

 P(   x  ) 
 f ( x )dx 

3

2
3

2
3
 f ( x )dx   f ( x )dx   f ( x )dx

2
3
3
3x
x
27  8 

    1
dx 
35
35 2 35  35 
2

(b) Expected (mean) value of x:

3
3x 2
   xf ( x )dx   x
dx
35

2
3
3
3x 3
3 x4
3
dx 


81  16
35
4
35
140
2
2
.
 1393
(c) Variance of population:

 2   ( x   )2 f ( x )dx

3
3x 2
  (x  )
dx 
35
2
2
3


2

3

( x 2  2 x   2 )
2
3
3
3x 2
dx
35
3x 4
3x 3
3x 2
3
dx  2  
dx   2 
dx 
x5
35
35
35
5
35

2
2


3
3 5  25  194
.  2.774
5  35
  1666
.
6.9
 
3
2
 2 2   2
6.21
Probability distribution function:
x3
f ( x) 
1 x  3
20
0
(a) f(x) satisfies the requirement of a probability distribution function
because:
 f ( x)  0    x  
 P(  x  ) 

1


3

1
3
 f ( x)dx   f ( x)dx   f ( x)dx   f ( x)dx
4 3
3
x3
1 x
1

(81  1)  1
dx  
4 20 1 80
1 20

(b) Expected (mean) value of x:

3
x3
 xf ( x)dx   x 20 dx


3
1
3
4
1 x5
1
x
 243  1
  dx 

5 20 1 100
1 20
 2.42
(c) Variance of population:

 2   ( x   ) 2 f ( x ) dx

3
  (x   )2
1
3
x3
x3
dx   ( x 2  2 
x   2 ) dx
20
20
1
3
3 4
3 3
x5
x
x
dx  2   dx   2  dx
1 20
1 20
1 20

1 x6
 [
20 6
3
1
3
3
 2 x 4
2   x5


]
5 1
4 1
1 1
2  2.42
(2.42) 2
[ (729  1) 
(243  1) 
(81  1)]
20 6
5
4
1
 [121.333  234.256  117.128]
20
 0.210

  0.458
6.10
6.22
Probability of the following cases of problem 6.20:
2
0
(a) for x  0
P ( x  0) 

f ( x )dx 

0


2

3x 2
x3
dx 
35
35
1
(b) for 0  x  1 P ( 0  x  1) 

0
6.23

0
f ( x )dx 
0

x
2
2
1
3x
x3
1
dx 

 0.0286 or
35
35 0 35

x
f ( x )dx 


0dx  0
  x  2

x

3x 2
x3
8
dx


2 35
35 35
2  x  3
3
3x 2
dx  1

35
2
f ( x )dx
8
 0.2286 or 22.86%
35
Cumulative distribution of random variable x:
F( x ) 

2
3 x 
F(-2) = 0 , F(0) = 0.229 and F(3) = 1
6.11
2.86%
6.24
Binomial distribution can be used because of the satisfactory/ unsatisfactory
outcome of the process.
 n
 n
n!
P (r )    p r (1  p )n  r ,   
 r  r !(n  r )!
r
In this case: p=0.95
n=4
(a) All four parts be satisfactory:
 4
P ( 4)    (0.95)4 (1  0.95 )0
 4
 1  0.8145  1  0.8145 or 8145%
.
(b) For at least two parts to be satisfactory, we should
calculate the probability that 2,3 and 4 parts be satisfactory:
 4
 4
4!
P( 2)    ( 0.95 )2 (1  0.95 )2
6
 
 2
 2 (3 !)(1!)
 6  0.9025  0.0025
 0.0135
 4
P(3)    (0.95)3 (1  0.95)1
 3
 4
4!
4
 
 3 ( 3 !)(1!)
 4  0.8574  0.05
 0.1715
Probability of having at least two satisfactory parts:
 P (2)  P (3)  P (4)
  0.0135  0.1715  0.8145   0.9995
6.25
or
99.95%
We want the probability that at most 2 computers will fail. The is the
probability that 0, 1 or 2 failures. Define “success” as a computer failure. Then
p=0.1. Then we want the probability that 0, 1 or 2 will fail.
 20 
P (0)    0.10 (1  0.1)20  0.121577
 0
 20 
P (1)    0.11(1  0.1)19  0.27017
 1
 20 
P (2)    0.12 (1  0.1)18  0.28518
 2
P(2 or less) = .121577 + 0.27017 + 0.28518 = 0.677
6.12
6.26
Define success = failure. P = 0.1
We want the probability that 2, 3, 4, or 5 will fail.
P(2-5) = P(2)+P(3)+P(4)+P(5)
for example:
 20
P (2)    (0.1) 2 (1  0.1)18  0.2852;
 2
 20
P (3)    (0.1) 3 (1  0.1)17 .1901;
 2
 20
P (4)    (0.1) 4 (1  0.1)16  0.098;
 4
 20
20 !
 190
 
 2  2! 18 !
 20
20 !
 1140
 
 3  3 ! 17 !
 20
20 !
 4845
 
 4  4 ! 16 !
 20
 20
20 !
P (5)    (0.1) 5 (1  0.1)158  0.032;   
 15504
 5
 5  5 ! 15 !
P(2-5) = 0.2852+.1901+.098+.032 = 0.597
6.27
(a) For all 6 parts to be satisfactory, using binomial distribution:
 6
 6
6!
P( 6 )    ( 0.95 )6 (1  0.95 )0
1
 
 6
 6 6 ! 0 !
 0.7351
 73.51%
(b) For at least two parts to be satisfactory, we should find the sum of
probabilities for 2,3,4,5 and 6 parts to be satisfactory:
P(number of successes  2)  1  P(0)  P (1)
 6
P(0)    (0.95)0 (1  0.95)6
 0
 6
6!
1
 
 0 0 ! 6 !
 156
.  10 8
 6
P(1)    (0.95)1(1  0.95)5
 1
 6
6!
6
 
 1 1! 5 !
 6  0.95  (3.15  10 7 )  178
.  10 6
so P(# of success  2)  100%
6.13
Probability of one or more power failure = 0.05
Probability of no power failure (success) = 0.95
Binomial distribution will be used here:
6.28
 n
 n
n!
P (r )    p r (1  p )n  r ,   
r
 r  r !(n  r )!
a- No power failure in three months: n=3, r=3
 3
P(3)    (0.95)3 (1  0.95)0  0.857  85.7%
 3
b- Exactly one month with power failure in four months: n=4, r=3
 4
P( 3)    (0.95)3 (1  0.95 )1
 3
 4  0.8574  0.05  0.1715
 4
4!
4
 
 3 3 !(4  3 )!
 17.15%
c- At least one power failure in the nest five months: n=5
 5
P (0)    (0.95) 0 (0.05) 5  3.125  10 7
 0
 5
P (1)    (0.95)1 (0.05) 4  2.969  10 5
 1
 5
5!
5
 
 1 1! 4 !
 5
P (2)    (0.95) 2 (0.05) 3  1128
.
 10 3
 2
 5
5!
 10
 
 2 2! 3 !
 5
P (3)    (0.95) 3 (0.05) 2  0.0214
 3
 5
5!
 10
 
 3 3 ! 2!
 5
P(4)    (0.95) 4 (0.05)1  0.2036
 4
 5
5!
5
 
 4 4 ! 1!
4
 P(i )  0.226
Chance of at least on power failure in 5 months.
i 0
6.29
Binomial distribution can be used:
success = a failure n=16 r=0 p=0.01
 n
 n
n!
P (r )    p r (1  p )n  r ,   
r
 r  r !(n  r )!
 16
P ( r  0)    (0.01) 0 (0.99)16
 0
P ( r  0)  0.85
 16
16 !
1
 
 0  0 ! 16 !
 85% chance of no failures
6.14
6.30
Binomial distribution
(a) p=0.05, n=100, r = 2, 5, 10
 n  100  100!
   
 
 4950
P (2)  4950  0.05 2 (1  .05)98  0.081
 r   2  2!98!
 n  100  100!
   
 
 75287520
P (2)  75287520  0.05 5 (1  .05)95  0.180
 r   5  5!95!
 n  100 
100!
   
 
 1.731x1013 P ( 2)  1.731x10 13  0.05 10 (1  .05 ) 90  0.0167
r
10
10
!
90
!

  

Note: One cannot normally compute 100! with a hand calculator – it is a very large number. However
100!/(2!98!) can be rewritten 100x99x98!/(2!98!) =100x99/2!, which can be readily computed.
(b) We want the probability of 0 or 1 failing to be 0.99. i.e. P(1)+P(2)=0.99
 n  10  10!
 n  10  10!
     
     
1
 10
 r   0  0!10!
 r   1  1!9!
P(0)  P(1)  1 p 0 (1  p)10  10  p1(1  p)9  0.99
This must be solved by trial and error for p. Using a spreadsheet, the answer is
p=0.0155
6.31
Binomial distribution
(a) p=0.03, n=100, r = 1, 4, 15
 n  100 100!
  
 100
 r   1  1!99!
P(1)  100  0.031 (1  .03) 99  0.147
 n  100 100!
  
 3921225
 r   4  4!96!
P(4)  3921225 0.034 (1  .03) 96  0.171
 n  100 
100!
  
 2.533  1017 P (15)  2.533 x1017  0.0315 (1  .03) 85  2.729  10 7
 r   15  15!85!
Note: One cannot normally compute 100! with a hand calculator – it is a very large number. However
100!/(1!99!) can be rewritten 100x99x98!/(1!99!) =100/1!, which can be readily computed.
(b) We want the probability of 0 or 1 failing to be 0.99. i.e. P(1)+P(2)=0.99
 n  10  10!
 n  10  10!
  
  
1
 10
 r   0  0!10!
 r   1  1!9!
P(0)  P(1)  1 p 0 (1  p)10  10  p1 (1  p) 9  0.99
This must be solved by trial and error for p. Using a spreadsheet, the answer is
p=0.0155
6.15
6.32 We are looking at the probability that more than 175 passengers will show up. This
can be solved as a binomial distribution problem. Consider success that a passenger
shows up, so p=0.95. We then want more than 175 successes out of 180 trials.
 n 180 
180!
180  179  178  177  176!
 
 

 42.297  10 6
176! x 4!
 r 176  176! x 4!
 n 180 
180!
180  179  178  177!
 
 

 955,860
177! x 3!
 r 177  176! x3!
 n 180 
180!
180  179  178!
 
 

 16110
178! x 2!
 r 178  178! x 2!
 n 180 
180!
180  179!
 
 

 180
179! x1!
 r 179  179! x1!
 n 180 
180!
 
 
1
 r 180  180! x 0!
P (176)  42.297  10 6 x 0.95176 (1  0.95) 4  0.03174
P (177)  955860x 0.95177 (1  0.95)3  0.01363
P (178)  16110x 0.95178 (1  0.95)2  0.004363
P (179)  180 x 0.95179 (1  0.95)1  0.0009263
P (180)  1x 0.95180 (1  0.95)0  0.00009778
P(r>175) = P(176)+ P(177)+ P(178)+ P(179)+ P(180) =
0.03174+0.01363+0.004363+0.0009263+0.00009778 = 0.0508
6.16
6.33 We are looking for the probability that there will 5 or less defective components.
Consider success to be a defective component then p = 0.05, n=55 and r = 0,1,2,3,4,5
 55  55!
 1 P (0)  1x 0.05 0 (1  .05)55  0.0595
r=0:   
 0  0!55!
r=1:
r=2:
r=3:
r=4:
r=5:
 55  55!
  
 55
P (1)  55 x 0.051(1  .05)54  0.1725
1  1!54!
 55  55!
  
 1485 P (2)  1485x 0.05 2 (1  .05)53  0.2449
 2  2!53!
 55  55!
  
 26235 P (3)  26235x 0.05 3 (1  .05)52  0.2277
 3  3!52!
 55  55!
  
 341055 P (0)  341055x 0.05 4 (1  .05)51  0.1558
 4  4!51!
 55  55!
  
 3478761 P (5)  3478761x 0.05 5 (1  .05)50  0.0836
 5  5!50!
So the probability of 5 or less defective components is
0.0595+0.1725+0.2449+0.2277+0.1558+0.0836 = 0.944. This is the probability that
there are 50 or more good components.
6.34 This is a Poisson distribution problem.  = 40/8 = 5 visits/hour. The probability of
more than 5 visits in an hour is 1 – [P(5)+P(4)+ P(3)+P(2)+P(1)+P(0)].
P (0)  e 5 5 0 / 0!  6.74 x10 3 P (1)  e 5 51 / 1!  .03369 P (2)  e 5 5 2 / 2!  0.0842
P (3)  e 5 5 3 / 3!  0.1404 P ( 4)  e 5 5 4 / 4!  0.1755 P (5)  e 5 5 5 / 5!  0.1755
So P(x>5) = 1  (6.74 x10 3  0.03369  0.0842  0.1404  0.1755  0.1755) =0.384. The
probability that there will be more than 5 visits in an hour is 0.384
6.35 This can be solved as either a binomial distribution (p=0.001, n=4, r=1) or as a
Poisson distribution. For the poisson distibution, the expected occurrence () will be
4/1000 = 0.004. We are then looking for the probability of x = 1.
P (1)  e 0.004 .0041 / 1!  3.984x10 3 .
6.36 This is a Poisson distribution. The expected number of crashes, , is (1/3)5=1.666.
Then, P (0)  e 1.666 1.666 0 / 0!  0.189
6.37 This is a Poisson distribution problem although it can be done as a binomial
distribution. The expected frequency () of defects will be (1/50)x10 = 0.2 for the typical
10 m2 kitchen. The probability of 1 or more defects is 1-P(0).
P(0)  e 0.2 0.20 / 0!  0.818 P(x1) = 1-0.818 = 0.182. If solved as a binomial distribution
, p= 0.02, n=10. For r = 0,
10  10!
  
 1 P (0)  1x.020 (1  .02)10  0.818 . So P(x1) = 1-0.818 = 0.182
 0  0!10!
6.17
6.38 Poisson distribution. The fact that a bulb has failed does not affect the probability
of other failures. We need to know what the probability of two or more failures during the
day. The probability of 2 or more is:
P(x2) = 1 – P(0) – P(1).  = 2.
P(0)  e  x / x!  e 2 20 / 0!  0.13533
P (1)  e   x / x!  e 2 21 / 1!  0.2707
P(x2) = 1 – 0.1355 – 0.2707 = 0.5938
6.39 Poisson distribution. The expected number of failures in 50 calls will be (5/100)x50
= 2.5.
(a) The probability of exactly 5 failures is:
P (5)  e   x / x!  e 2.5 2.5 5 / 5!  0.0668
(b) P(5 or less) = P(0)+P(1)+P(2)+P(3)+P(4)+p(5) Similar to part (a)
P(5 or less) = 0.08208+0.2052+0.2565+0.2137+0.1336+.0668 = 0.958
(c) P(more than 5) = 1 – P(5 or less) = 1 – 0.958 =0.042
6.40 Poisson distribution. The average value of customers in that 1 hour is 20 = .
(a) P (25)  e 20 20 25 / 25!  0.0446
(b)P(20 to 25) = P(20)+ P(21)+ P(22)+ P(23)+ P(24)+
P(25)=0.0888+0.0846+0.0769+0.0669+.0557 = 0.417558
10
(c)P(10 or less) = i 0 P ( i ) = 0.010812
(d) P(x>10) = 1 – P(x10) = 1 – 0.010812 = 0.989
6.41 Poisson distribution.  =3 for 1 sheet
(a) P (25)  e 3 310 / 10!  0.00081
(b) P(0) = 0.049787
(c)  =3/6 = 0.5 for a single board. P (25)  e 0.5 0.51 / 1!  0.3033
(d) P(more than 1) = 1 – P(1) – P(0) = 1 – 0.3033 - 0.6065 = 0.0902
6.42 For P(1or more) = 0.01, P(0) = 0.99 = e   0 0! Solving for  we get  = 0.01
defects per board or 0.06 defects per sheet.
6.18
6.43
The area from 100 to 100.5 is 0.2. From Table 6.3, z =
0.52
0.52 
110.5  100

20%
20%
,   0.96
Probability of error greater than 0.75 Volts:
100.75  100
z
 0.78125
0.96
99.5
100
100.5
From the normal distribution curve(Table 6.3) for
z = 0.7812 P(z) = 0.2826
28.26%
 for error greater than 100.75 or less than 99.25
(1000.75) we will have:
P (z )  2  (0.5  0.2826 )  0.4348
 43.48%
6.44
The area from 110 to 110.5 is 0.25. From Table 6.3, z =
0.67
0.67 
110.5  110

,   0.75
Probability of error greater than 1 Volts:
111  110
z
 1.33
0.75
From the normal distribution curve(Table 6.3) for
z = 1.33 P(z) = 0.4082
 for error greater than 111 or less than 109 (1101)
we will have:
P ( z )  2  (0.5  0.4082)  0.1836  18.36%
6.19
100
100.75
6.45
6.827  6.832
 0.5
0.01
a) z1 
From Table 6.3 area = 0.1915
 100(0.1915)(2)
= 38 readings within 0.5 cm
6.812  6.832
2
0.01
b) z1 
From Table 6.3 area = 0.4772
 100(0.4772)(2)
= 95 readings within 2 cm
6.782  6.832
5
0.01
c) z1 
From Table 6.3 area = 0.5
 100(0.5)(2)
= 100 readings within 5 cm
d) z1 
6.831  6.832
 0.1
0.01
From Table 6.3 area = 0.0398
100(0.0398)(2)
= 8 readings within 10 cm
6.20
6.46
7.76  7.75
 1.0
0.01
a) z1 
From Table 6.3 area = 0.3413
 20(0.3413)(2)
= 14 readings within 1 cm
7.77  7.75
2
0.01
b) z1 
From Table 6.3 area = 0.4772
 20(0.4772)(2)
= 19 readings within 2 cm
7.80  7.75
5
0.01
c) z1 
From Table 6.3 area = 0.5
 20(0.5)(2)
= 20 readings within 5 cm
d) z1 
7.85  7.75
 10
0.01
From Table 6.3 area = 0.5
20(0.5)(2)
= 20 readings within 10 cm
6.21
6.47 (a) The average is 71.3, the median is 70 and S = 12.62
(b) The grades according to the criterion will be:
score grade
95 A
86 B+
83 B+
79 B
79 B
78 B
75 B70 C+
70 C+
68 C+
63 C
63 C
55 C
55 C50 D
(c) There are 15 students and the division will be:
grade
A
AB+
B
BC+
C
CD
F
no. of
students
0.342
0.66
1.3785
2.247
2.8725
2.8725
2.247
1.3785
0.66
0.342
Of course, the number of students with each grade is an integer. If we round off, we will
only get 14 students total so some judgement is required for the additional student.
6.22
6.48
a)
From Table 6.3 for area0.4; z = 1.28
x  10000
  1.28 
500
x = 9,360 hrs

40%
10%

b)
z1 
7000  10000
 6  area  0.5( fromTable6.3)
500
z2 
10000  10000
 0  area  0
500
 0.5  0  0.5
50% or1000 bulbs will fail
6.49 Normal distribution problem.
9
8
7
6
5
4
3
2
1
0
(a) The probability of the weight being less than 7.9 is
the area under the curve to the left of x = 7.9. For this
value of x, z  ( x   ) /   (7.9  8.05) / .05  3.0 . From
the nomral distribution table, for z = 3.0, the area under
the curve (from the mean to the value of interest) is 0.4987. The area to the left of x =
7.9 is thus 0.5 – 0.4987 = 0.0013. So 0.13% of the cans will be rejected.
(b) For this case, z  ( x   ) /   (7.9  8.00) / .05  2.0 . For this value of z, the area
under the normal distribution curve from the mean to x=7.9 is 0.4772. So the area to the
left of 7.9 is 0.5 – 0.4772 = 0.0228. 2.28% of the cans will be rejected.
(c) for this case, z  ( x   ) /   (7.9  8.05 ) / .03  5 The table does not give a value
for z this high but for 4.9, it is given as 0.5. So the area to the left of 7.9 is 0.5 – 0.5 = 0.
For this z value the number of rejected cans is negligible. (actually 3x10-5%.
)
x
f(
7.85
7.9
7.95
8
8.05
8.1
8.15
8.2
8.25
x
6.50 (a) A 2 tolerance means that values greater than z = 2 or less than z =-2. For z =
2.0, the value from the normal distribution table is 0.4772 . The right hand tail area (for
z>2) is 0.5 – 0.4772 = 0.0228 The left hand tail has the same area. The probability of
rejection is then 2x0.0228 = 0.0456 or 4.56%.
(b) This part is best solved using the binomial distribution. For 2 rejections, we define
success as a rejection. Then for 2 rejections, r = 2, n=20 and p = 0.056. The probability
of 2 rejected parts is then 0.1705. Similarly, for r = 4 and r = 10 we get 0.0099 and
4.5x10-9 respectively.
6.23
6.51
  0.2 inches
x  7.25 ft  87 inches
d  x  x  0.3 acceptable
d
0.3
.
 15
 0.2
From Table 6.3, A  0.4332
z



P d  0.3  2  0.4332  0.8664 From Table 6.3
% rejection = 100(1-0.8664) = 13.36%
To reduce the rejection rate to 3%
1-P = 0.03  P = 0.97  z(for P/2 = 0.485) = 2.17
d
0.3
z
 
 0.14 inch
.
217

This means that the manufacturer has to cut the columns close to the standard size.
6.52
x  10.500 inches
  0.005 inches
10.520  10.500 0.020

4
0.005
0.005
P ( x  10.520)  0.5  0.5  1  100% U sin gTable 6.3
a) z 
10.485  10.500
10.500
 3
0.005
10.515  10.500
z2 
3
0.005
From Table 6.3, A = 0.4987
P (10.485  x  10.515)  2  0.4987  0.9974  99.87% From Table 6.3
b) z1 
c) From Table 6.3,
z  2.5,P ( x  2.5  x  x  2.5 )  2  0.4938  0.9876  98.76%
P ( rejection )  1  0.9876  0.012
6.24
10.520
6.53
x  25.00cm
  0.02cm
a) z 
25.02  25.00 0.02

1
0.02
0.02
P ( x  25 .02 )  0.5000  0 .3413  0.8413  84 .13% Using Table 6.3
24.95  25.00
 2.5
0.02
25.05  25.00
z2 
 2.5
0.02
From Table 6.3, A = 0.4938
P( 24.95  x  25.05)  2  0.4938  0.9876  98.76% From Table 6.3
b) z1 
c) From Table 6.3,
z  2 P ( x  2  x  x  2 )  2  0.4772  0.9544  95.44%
P ( rejection )  1  0.9544  0.046
6.54
a)  = 5000
From table 6.3 for area = 0.40
x  50,000
 128
.

5000
x = 43,600 miles
b) z1 
z2 
60,000  50,000
2
5000
z  1.28
40%
10%
50,000
area = 0.4772
70,000  50,000
 4 area = 0.5
5000
 0.5-0.4772 = 0.0228
50,000
(100,000 tires)(0.0228) = 2280 tires fail between 60,000 to 70,000 miles
c)
20,000  50,000
 6
5000
 No tires expected to have life less than 20,000
d) Major assumption: Life span of tire follow normal distribution.
6.25
60,000
70,000
6.55
 = 160 hrs
For A = 0.4, z  1.28

40%
x  3600
 128
.
160
x  3395.2
Recommend replacing bulbs after 3395 hours.
10%

6.56 x  6.686 lb, S  0.0486 lb, estimate of x  S / n  0.0486 / 10  0.01536 lb
6.57a Confidence level: 95%
1- = 0.95  = 0.05
0.5 - /2 = 0.475
  x  z
 z = 1.96 (Table 6.3)

n
(196
. )(2)
  30 
40
  30  0.620 mph with confidence of 95%
2
6.57b For a large sample, n>30,   x  z / 2 / n
For 90, 95, and 99% confidence levels, x/2 = 1.64, 1.96, and 2.58 respectively.
For 90% confidence level, the confidence interval is 1.64x0.2/(40)1/2 = 0.052 oz. For
95% and 99% confidence levels, the intervals are 0.061 oz and 0.082 oz.
6.58 In this case the sample size is small, less than 30, so we must use the tdistribution. The interval on the mean is given by:   x  t  / 2S / n . The sample size is
20 so the degrees of freedom is 19. From the t-distribution table, the values of t for  =
19 and the confidence levels of 90%, 95% and 99% (/2 = 0.05, 0.025 and 0.005) are
1.729, 2.093, and 2.861. For 90% confidence level, the confidence interval is
1.729x0.2/(20)1/2 = 0.0773 oz. For 95% and 99% confidence levels, the intervals are
0.0936 oz and 0.128 oz.
6.59 (a) The mean is 16.042 oz. and the sample standard deviation is 0.079 oz. The
standard deviation of the mean is 0.07941/(12)1/2 = 0.0229 oz.
(b) This is a t-distribution problem with = 12-1 = 11 and /2 = 0.025. From the tdistribution table, t has a value of 2.201. The confidence interval on the mean is then
t/2S/(n)1/2 = 2.201x0.0229 = 0.0504 oz.
6.26
6.60
Find 95% confidence interval on the mean
x  50,000 miles
S  5000 miles
n  100
1- = 0.95  = 0.05
z = 1.96 (From Table 6.3)
0.5-/2 = 0.475
  x  z
2

n
  50,000  196
. (5000 / 100 )
  50,000  980 miles
6.27
6.61
n  10 VCRs
x  1500 hours
S  150 hours
  0.05
Find the 95% confidence interval on the mean of the life of the VCRs.
  n  1  9 
a) 
 t   2.262 (From Table 6.6)
2
 0.025 
2

  x  t  (S
2
n)
 1500  2.262(150
10 )
 1500  107 hours
Confidence interval on the mean is  107 hours
b) For 95% confidence interval on the mean of 50 hrs:
t   S   50 hrs

n
2
(
2
 0.025)
Assuming S =150 hrs will remain the same (in reality should be recalculated and
it may change), using Table 6.6 we find n by trial and error.
n
t
15
20
25
36
2.145
2.093
2.064
2
2
t   S 
2
n
83
70
62
50
So 36 systems should be tested.
t   2 is an approximate value for n > 30, in which case the t-distribution
2
approaches normal distribution.
6.28
6.62
For this problem, we need to find a one-sided confidence interval of the form
S
x  t
  
n
t-distribution
(a) x = 41.25x106, S = 0.30x106, n=10. For 99% confidence level,  = 0.01 and = 10
– 1 = 9 from Table t = 2.821. The tolerance interval is then:
x  t
S
n
  
41.25x10  2.821
6
0.3x10 6
10
  
40.982  10 6    
41.0x106 falls into this interval so the manufacturer cannot be confident that the average
strength exceeds this limit.
(b) If we change n to 20, then  = 20 – 1 =19 and t = 2.539.
0.3x106
6
41.25x10  2.539
  
20
41.080  106    
Now the manufacturer can be confident the mean exceeds 41.0x106 psi.
6.29
6.63
n  10
x  910%
.
S  0.8%
  0.05
  n  1  9 
  0.025  t  2  2.262 (From Table 6.6)
2

  91  2.262 0.8

   x  t   S






2
n
10 
   910
.  0.57
Cut confidence interval by one half > 0.57/2 = 0.285
Assuming that n > 30, and S to remain the same:


  0.285 (  S )
z  
n
2
z 0.025  1.96
S  0.8




 z 2   

n 
0.285


2
n  30
 20 more motors should be tested.
6.30
6.64
n8
x  88.5%
S  0.5%
  0.05
  n  1  7 
  0.025 t 2  2.365 ( FromTable6.6)
2





  88.5  2.365 0.5 
   x  t  S



8
n
2
   88.5  0.42
Cut confidence interval by one half > 0.42/2 = 0.21
Assuming that n > 30, and S to remain the same:


  0.21
z  


n
2
z 0.025  1.960
(  S )
S  0.5





 z 2   
n
0.21


2
n  30
 22 more motors should be tested.
6.31
6.65
Given: Confidence level = 99%
Sheight = 2.44 in.
Shand = 0.34 in.
Find: N
Soln.: Assuming N > 30
  0.01
 2  0.005
0.5   2  0.495
 From Table 6.3
z  2.575
For height: the confidence interval:
2.44

z 2
 1 in, 2.575
 1  n  40
n
n
For sleeve length:
z 2

n
 0.4 in,
2.575
0.34
 0.2  n  18
n
In order to satisfy both conditions, 40 members should be chosen.
6.66
n  15
x  25 ppm
Find the 95% confidence interval.
S  3 ppm
  0.05   n  1  14
  0.025
2
  142 , 0.025  26.119 ( FromTable6.7)
  142 ,1.025  5.6287
(n  1) S 2
 142 ,.025
( n  1) S 2 14  32
14  32

 4.82; 2

 22.17
 14,1.025 5.6287
26.119
4.82   2  22.17 or 2.20    4.71 for 95% confidence level
6.32
6.67
The distribution governing the time interval between the arrival successive cars is
give to be
f(t,λ) = λe-λt
Where
1 1
 
 8
(a)
P (t  6)  1  e
 ( 6 )
1 e

6
8
 0.5276
(b)
P (t  15)  1  P (t  15)  1  (1  e

15
8
)  0.1534
In order to have only two arrivals in 10 minutes, if the interval between the first
(c)
two arrivals is t, mins, the interval between the next arrival should be greater than (10-t1)
Thus,
P[no more than 2 arrivals] = P[0≤t1≤10 and t2≥(10-t1)]
10
10
1
  f (t1 ) P(t 2  (10  t1 )dt1   e  t1  (e  (10t1 ) )dt1
0

1

0
10 10
e

 dt
1
0
1
 e
8
10
8

 10  1.25e 1.25  0.3581
6.68 We are given that TBF has an exponential distribution with μ = 450 hours.
Therefore
1
1
 
 450
(a)
P ( failure  300hours )  1  e
  ( 300 )
1 e

300
450
 0.4856
(b)
P ( failure  500hours )  e  ( 500)  1  e
(c)

500
450
 0.3292
P[300  failuretime  600]  P[ failure  600]  P[ failure  300]
 [1  e
600
450
300
]  [1  e 450 ]  0.7364  0.4866  0.2498
6.33
6.69 Let the pollution level, X, be in ppm. If we define y = log(X), then y is N(µ,σ),
where μ = 1.9031 and σ = 1.3010
Note: Here we have used y = log(X) rather than y = ln(X)
(a)
P[ X  90]  P[ y  log(90)]  P[Y  1.9542]
We now convert this problem to a standard normal distribution,
1.9542  1.9031
P[ y  1.9542]  P[ Z 
]  P[ Z  0.393]
1.3010
Using the standard normal distribution in Table 6.3,
P[ X  90]  P[ Z  0.393]  0.5  0.1528  0.3472
(b)
P[ X  20]  P[ y  log(20)]  P[Y  1.3010]
1.3010  1.9031
 P[ Z 
]  P[ Z  0.4628]
1.3010
 0.5  0.1782  0.3218
(c)
If the mean pollutant level is reduced to 40 ppm, then the new value of μ = log(40
= 1.6021
P[ X  20]  P[ y  log(20)]  P[Y  1.3010]
1.3010  1.6021
 P[ Z 
]  P[ Z  0.2304]
1.3010
 0.5  0.0915  0.4085
6.34
6.70 Let X be the time taken to fix the software bug. X has a mean of 200 mins, a
standard deviation of 30 mins. If y = log(X) is N(μ,σ) with μ = log(200) = 2.3010, σ =
log(30) = 1.4771
(a)
P[ X  500]  P[ y  log(500)]  P[Y  2.6990]
We now convert this problem to a standard normal distribution,
2.6990  2.3010
P[ y  2.6990]  P[ Z 
]  P[ Z  0.2694]
1.4771
 P[ X  500]  P[ Z  0.2694]  0.5  0.1062  0.3938
(b)
(c)
P[100  X  200]  P[2.000  y  2.3010]
2  2.3010
2.3010  2.3010
 P[
Z
]  P[0.2038  Z  0]
1.4771
1.4771
 0.0808
P[ X  50]  P[ y  log(50)]  P[ y  1.6990]
1.6990  2.3010
 P[ Z 
]  P[ Z  0.4076]
1.4771
 0.5  0.1582  0.3418
6.71
Chi-squared distribution.
S = 0,002 in., n = 10,  = n – 1 = 9. 95% confidence level:  = 1-0.95 = 0.05.
/2 = 0.025, 1-/2 = 0.975. From Table 6.7, 2/2 = 19.023, 21-/2 = 2.704.
(n  1)S 2
(n  1)S 2
(10  1)0.002 2
(10  1)0.002 2
2
2






19.023
2.704
 2 / 2
 12 / 2
1.89  10 6   2  1.33  10 5
or 0.00137    0.00364
6.72
Chi squared distribution.
S = 5500, n = 8,  = 7.  = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.
2/2 = 20.278, 21-/2 = 0.9893.
(n  1)S 2
 2 / 2
 
2
(n  1)S 2
 12 / 2
(8  1)5500 2
(8  1)5500 2
2
 
20.278
0.9893
10442351   2  214.0  10 6 or 3231    14360
6.35
6.73
Chi squared distribution.
S = 10, n = 12,  = 11.  = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.
2/2 = 26.757, 21-/2 = 2.6032.
(n  1)S 2
 2 / 2
2 
(n  1)S 2
 12 / 2
(12  1)10 2
(12  1)10 2
2
 
26.757
2.6032
41.11   2  422 .5 or 6.41    20.55
We cannot be 99% confident that the standard deviation is less 15 mA.
6.74
There is an error in the problem. The required variance should be 0.0004 mm2. (not m2).
(a) Chi squared distribution.
S = 0.01, n = 10,  = 9.  = 1-0.95 = 0.05, /2 = 0.025, 1-/2 = 0.975.
2/2 = 19.023, 21-/2 = 2.7004.
(n  1)S 2
 2 / 2
2 
(n  1)S 2
 12 / 2
(10  1)0.012
(10  1)0.012
2 
19.023
2.7004
4.73  10 5   2  3.333  10 4 or 6.88  10 3    0.018
(b) The maximum of the confidence interval is less than the desired variance so the part
is acceptable.
(c) S = 0.01, n = 5,  = 4.  = 1-0.95 = 0.05, /2 = 0.025, 1-/2 = 0.975.
2/2 = 11.143, 21-/2 = 0.4844.
(n  1)S 2
 2 / 2
2 
(n  1)S 2
 12 / 2
(5  1)0.012
(5  1)0.012
2 
11.143
0.4844
3.590  10 5   2  8.258  10 4
The upper end of the confidence interval exceeds the allowable so the part is not
acceptable.
6.36
6.75 There is an error in the problem. The required variance should be 0.0004 mm2.
(not m2).
(a) Chi squared distribution.
S = 0.01, n = 10,  = 9.  = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.
2/2 = 23.589, 21-/2 = 1.7349.
(n  1)S 2
 2 / 2
2 
(n  1)S 2
 12 / 2
(10  1)0.012
(10  1)0.012
2 
23.589
1.7349
3.81 10 5   2  5.188  10 4
(b) The maximum of the confidence interval is greater than the desired variance so the
part is not acceptable.
6.76 There is an error in the problem. The required variance should be 0.0004 mm2.
(not m2).
(a) Chi squared distribution.
S = 0.01,  = 1-0.99 = 0.01, /2 = 0.005, 1-/2 = 0.995.
(n  1)S 2
(n  1)S 2
2



2
2
 / 2
 1 / 2
We need to determine n so that the maximum of the confidence interval is 0.0004. We
are told that n = 10 was too small so n must be larger. The following calculations were
n
11
12
13
14
nu
10
11
12
13
chisq
2.155845
2.603202
3.073785
3.565042
upper lim
0.000463855
0.000422557
0.000390398
0.000364652
performed on a spreadsheet program.
It can be concluded that 13 measurements will make the part acceptable.
6.37
6.77
Data from problem one arranged in ascending order:
48.9, 49.2, 49.2, 49.3 ,49.3, 49.8, 49.9, 50.1, 50.2, 50.5
x  49.64
S

 x i  x 2
n 1
 0.530
  xi  x
Low po int
48.9  49.64  0.74
High po int
50.5  49.64  0.86
From the Table 6.8:
n = 10   = 1.798
S = (0.530)(1.798) = 0.95294
Since S > deviations, then:
No data should be required.
6.78
From problem 6.4:
x  105.2
S  9.71
n = 12  = 1.829 (Table 6.8)
 1  Pl arg est  P  120  105.2  14.8
 2  Psmallest  P  89  105.2  16.2
S  (9.71)(1829
.
)  17.76
Neither 1 or 2 exceeds S so No points rejected
6.38
6.79
x
y
xx
( x  x) 2
y y
( y  y) 2
( x  x)( y  y)
20
30
40
50
1.02
1.53
2.05
2.55
-33.5714
-23.5714
-13.5714
-3.5714
1127.04
555.611
184.183
12.7549
-1.5271
-1.0171
-0.4971
0.0029
51.2669
23.9745
6.7463
-0.01036
60
75
100
3.07
3.56
4.05
6.4286
21.4286
46.4286
41.3269
459.185
2155.61
0.5229
1.0129
1.5029
2.3320
1.0345
0.25711
-6
8.4110
0.27342
1.0260
2.2587
7.1717
176.8214
4535.71
3.3615
21.7050
69.7775
x  53.5714
y  2.5471
rxy 
rxy 
 (x
 ( x
i
i
 x )( y i  y )
 x )2  ( y i  y )2

12
176.8214
435.717.1717
12
rxy  0.9804
This value of rxy which is close to 1 shows a strong linear relationship.
6.39
6.80 (a)
 x  y 
n  x   x 
 x  y   x  x y 
b
n  x   x 
a
n xi y i 
i
i
2
2
i
i
2
i
i
i
i
i
2
2
i
i
xi
yi
(xi)2
(yi)2
xiyi
20
30
40
50
60
75
100
  375
1.02
1.53
2.05
2.55
3.07
3.56
4.05
  17.83
400
900
1600
2500
3600
5625
10000
 24,625
1.0404
2.3409
4.2025
6.5025
9.4249
12.6736
16.4025
 52.5873
20.4
45.9
82.0
127.5
184.2
267.0
405
1132
7(1132)  (375)(17.83)
 0.0398
7(24625)  (375)2
(24625)(17.83)  (375)(1132)
b
 0.4587
7(24625)  (375)2
a


y  3.898  10 2 x  0.4587
y  V( mV)
x  T(C)
(b)
4.5
4
3.5
3
2.5
Best Fit
Line
2
Output - mV
1.5
1
20
40
60
80
Temperature - C
6.40
100
6.81
(a)
xi(t)
yi(t)
( x i  x)
( x i  x) 2
( y i  y)
( yi  y) 2
( y i  y)( x i  x )
0.0000
0.1000
0.2000
0.3000
0.4000
0.5000
4.98
1.84
0.68
0.25
0.09
0.03
-0.2500
-0.1500
-0.0500
0.0500
0.1500
0.2500
0.0625
0.0225
0.0025
0.0025
0.0225
0.0625
3.6683
0.5283
-0.6317
-1.0617
-1.2217
-1.2817
13.4567
0.2791
0.3990
1.1271
1.4925
1.6427
-0.9171
-0.0793
0.0316
-0.0531
-0.1833
-0.3204
=18.3971
=-1.5215
=0.1750
x  0.25 sec.
y
y
rxy 
rxy 
i
N
 13117
.
Volts
 (x
 ( x
i
i
 x )( y i  y )
 x )2  ( y i  y )2
.
15215
0.1750  18.397
12

12
 0.848
Note: The relationship between Voltage and time is not linear!
(b)
xi(t)
yi(lnV)
( x i  x)
0.0000
0.1000
0.2000
0.3000
0.4000
0.5000
1.6054
0.6098
-0.3857
-1.3863
-2.4079
-3.5066
-0.2500
-0.1500
-0.0500
0.0500
0.1500
0.2500
( y i  y)
( x i  x) 2
0.0625
0.0225
0.0025
0.0025
0.0225
0.0625
=0.1750
2.5173
1.5216
0.5262
-0.4744
-1.4961
-2.5947
( yi  y) 2
6.3368
2.3154
0.2769
0.2251
2.2382
6.7324
=18.124
( y i  y)( x i  x )
-0.6293
-0.2282
-0.0263
-0.0237
-0.2244
-0.6487
=-1.780
x  0.25
y  0.912
rxy 
rxy 
 (x
 ( x
i
i
 x )( y i  y )
 x )2  ( y i  y )2
1780
.
0.175  18.124
12

12
 0.99984 , showing a closer linear relationship between
ln(V) and than between V and t.
6.41
6.82
For the best fit line, the equ. is reading=1.0525weight-1.1661
For the forced zero, reading=0.9987weight.
40
reading
30
20
y = 1.0525x - 1.1661
y = 0.9987x
10
0
-10 0
10
20
weight - lb
6.42
30
40
6.83
DelP
1.96
4.2
4.9
5.48
5.91
7.3
7.73
9
9.9
Sy.x=
rsquared=
DelP - in H2O
Voltage
0.31
0.65
0.75
0.85
0.91
1.12
1.19
1.38
1.52
0.022862 [Excel function:(steyx(yrange,xrange)]
0.99988 [Excel function: rsq(yrange,xrange)]
12
10
8
6
4
2
0
y = 6.5606x - 0.0629
0
0.5
1
1.5
2
Voltage
6.84 For the data given in Problem 6.79, we first compute the following quantities,
using x = T and y = V
 x  375,  x
i
i
2
 24265,  yi  17.86,  xi yi  1133.9,  yi  52.7712
2
We can now compute the slope, a, and the intercept, b, as
n  xi yi   xi  y i 7  1133.9  375  17.86

 0.0390
a
7  24625  (375) 2
n  xi 2  ( xi )2
b
n  xi 2  yi  (  xi )(  xi y i )
n  xi  (  xi )
2
2

14590
 0.4590
31750
Therefore the best fit is V = 0.0390T + 0.4590
(a)
The two-sided 95% confidence interval implies α/2 = 0.025 so that tα/2,n-2 = t0.025,5
= 2.571, the standard error,
S yx 
( y
 Yi ) 2
 0.2394
n2
i
The 95% two-sided confidence interval for a is
S yx  t 
,n 2
0.2394  2.571
2
(a 
)

0
.
0390

 0.0390  0.0089
24625  7  (53.17) 2
 xi 2  n x 2
or (0.0302 ≤ a ≤ 0.0479)
Similarly the 95% confidence interval for the intercept, b, is
b  S yx  t 
2
,n 2
(  xi ) 2
1
 2
n n (  xi 2  n x 2 )
6.43
1
375 2
 0.4595  2.571  0.2394  2
7 7 (24625  7  53.17 2 )
 0.4595  0.5284  (0.0688,0.9879)
(b)
When T = 70oC, V = aT + b = 3.1929V
The 95% two-sided prediction interval for the above V is
n 1
( x *  x) 2

n
 xi 2  n x 2 where
x* = 70 and y* = 3.1929V
Therefore the 95% prediction interval for V at T = 70oC is
7 1
(70  53.57) 2
3.1929  0.2394  2.571

7
24625  7  53.212
 3.1929  0.6739  (2.5191,3.8668)
y *  S yx  t 0.025,5
(c)
The upper limit of V at T = 70oC at 95% confidence level involves a one-sided
prediction interval with α = 0.05.
At α = 0.05, tα,n-2 = t0.05,5 = 2.015
Thus, the upper limit will be
7 1
(70  53.57) 2
3.1929  0.2394  2.051

7
24625  7  53.212
 3.1929  0.5281  3.7211volts
6.85 Based on the given data, we compute the following quantities, where x = ToF and
y = σ in ksi.
x
(a)
 9000,  xi 2  8180000,  y i  314.8,  xi y i  182970,  y i  10348
2
i
The independent variable is the temperature, ToF, and the dependent variable is
the tensile strength, σ (ksi). The coefficients of the linear regression model are
12  182970  9000  314.8
a
 0.0372
12  818  10 4  (9000) 2
818  10 4  314.8  9000  182.97  10 3
b
 54.099
12  818  10 4  (9000) 2
Therefore the linear regression model for the data is
σ = -0.0372T + 54.099
6.44
(b)
When the temperature is T = 670oF, the expected σ is
σexp = -0.0372x670+54.099 = 29.2056 ksi.
For a 90% two-sided confidence interval, α/2 = 0.05, n = 12.
From the tables, tα/2,n-2 = t0.05,10 = 1.812
The confidence interval for strength, σ, at T = 670oF is
 exp  S yx  t 
2
, n 2
n 1
( x *  x) 2

n
 xi 2  n x 2
where
x* = 670 and
S yx 
( y
 Yi ) 2
115.6

 3.34
10
n2
i
Therefore, 90% confidence interval for σ at T = 670oF is
13
(670  750) 2

(29.2056  3.34  1.812
)
12 818  10 4  12  750 2
 (29.2056  6.3153)  (22.89,35.52)ksi
(c)
The expected loss of strength for each 100oF increase is
Δσ = a x 100 = -0.0372 x 100 = -3.72 ksi
The 95% confidence interval for this change in strength is 100 x (95% confidence
interval for a)
The 95% confidence interval for the slope parameter, a, is
S yx  t 
,n 2
3.34  2.228
s
a
 0.0372 
818  10 4  12  750 2
 xi 2  n x
 0.0372  0.0062  (0.0434,0.0309)
Therefore, 95% confidence interval for change in strength for a 100oF change is =
(-4.34, -3.09) ksi
(d)
At T = 550oF, the expected tensile strength is
σ = a x 550 + b = -0.0372 + 550 + 54.09 = 33.641 ksi
The one-sided 95% confidence level for the strength is
6.45
(ax * b)  S yx  t 
2
, n 2
n 1
( x *  x) 2

 
n
 xi 2  n x 2
where
x* = 550, α = 0.05, n = 12
From the tables, tα,n-2 = t0.05,10 = 1.812
Then,
 0.0372  550  54.099  3.34  1.812
12  1
(550  750)

12
818  10 4  12  750 2
= 33.641-0.0051 = 33.6359 ksi
Therefore, theh minimum strength at T = 550oF at a 95% confidence level is
σmin = 33.6359 ksi
6.46
6.86
a) Best fit line for t and V
xi2
0
0.01
0.04
0.09
0.16
0.25
  0.55
xi (t)
0.0
0.1
0.2
0.3
0.4
0.5
  15.
a
4.98
1.84
0.68
0.25
0.09
0.03
  7.87
y i2
24.800
3.3856
0.4624
0.0625
0.0081
0.0009
  28.720
xiyi
0
0.184
0.136
0.075
0.036
0.015
 0.446
 x  y 
n  x   x 
n  xi y i 
i
i
2
2
i
a
yi (V)
i
6( 0.446)  (15
. )(7.87)
 8.694
6(0.55)  (15
. )2
 x  y   x  x y 
b
n  x   x 
2
i
i
i
i
2
i
b
i
2
i
(0.55)(7.87)  (15
. )(0.446)
 3.485
6( 0.55)  (15
. )2
The best fit line is therefore: y = -8.694x + 3.485
Best fit line for t and lnV
xi
0.0
0.1
0.2
0.3
0.4
0.5
  15.
x i2
0
0.01
0.04
0.09
0.16
0.25
  0.55
yi
1.6054
0.60977
-0.38566
-1.3863
-2.4079
3.5066
  5.47129
6.47
yi2
2.5774
0.37181
0.14874
1.9218
5.7982
12.2959
.
  231139
xiyi
0
0.060977
-0.077132
-0.41589
-0.96316
-1.7533
.
  31485
a
 x  y 
n  x   x 
n xi y i 
i
i
a
i
2
2
i
6( 3.1485)  (15
. )( 5.47129)
 10.175
6(0.55)  (15
. )2
 x  y   x  x y 
b
n  x   x 
2
i
i
i
i
2
i
b
i
2
i
(0.55)( 5.47129)  (15
. )( 3.1485)
.
 16319
6(0.55)  (15
. )2
Best fit line is therefore: y = -10.175x + 1.6319
b)
Standard error of the estimate for T vs V
Sy . x 
y
2
i
 b y i  a x i y i
n2
 (28.720)  (3.485)(7.87)  ( 8.694)(0.446) 
Sy . x  

62


Sy .x  11369
.
12
Standard error of the estimate for T vs ln(V)
.
)( 5.47129 )  ( 10.175)( 3.1485 ) 
 ( 23.1139)  (16319
Sy .x  

62


Sy .x  0.0378
12
The standard error for (T vs ln(V)) is much less than for (t vs V) therefore there is
less data scatter around this best fit curve.
6.48
6.87
DelP
0.05
0.07
0.09
0.12
0.15
0.17
0.19
0.21
0.23
0.25
Sy.x=
Q
2
2.35
2.7
3.12
3.5
3.72
3.85
4.1
4.35
4.45
ln(DelP)
-2.99573
-2.65926
-2.40795
-2.12026
-1.89712
-1.77196
-1.66073
-1.56065
-1.46968
-1.38629
0.105469 Sy.x=
ln(Q)
0.693147
0.854415
0.993252
1.137833
1.252763
1.313724
1.348073
1.410987
1.470176
1.492904
pred(lnQ)
0.693541
0.862349
0.988434
1.132764
1.244715
1.307509
1.363311
1.413523
1.459164
1.500996
e
0.000394
0.007934
-0.00482
-0.00507
-0.00805
-0.00621
0.015238
0.002536
-0.01101
0.008092
e/Sy.x
0.04402
0.886565
-0.53839
-0.56645
-0.89932
-0.69442
1.702752
0.283388
-1.23055
0.904233
log10(DelP)
-1.30103
-1.15490196
-1.04575749
-0.92081875
-0.82390874
-0.76955108
-0.7212464
-0.67778071
-0.63827216
-0.60205999
log10(Q)
0.30103
0.371068
0.431364
0.494155
0.544068
0.570543
0.585461
0.612784
0.638489
0.64836
0.008949
y = 0.5017x + 2.1965
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
1.6
1.4
y = 12.179x + 1.5506
1.2
1
ln(Q)
Q-cfm
Excel Sx.y function is STEYX(yrange,xrange)
0.8
0.6
0.4
0.2
0
0
0.05
0.1
0.15
0.2
0.25
0.3
-4
-3
DelP- psi
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
-0.5
2
1
e/Sx.y
log10(Q)
-1
-1
ln(DelP)
y = 0.5017x + 0.9539
-1.5
-2
0
-1
0
0
0.05
0.1
0.15
0.2
0.25
0.3
-2
log10(DelP)
DelP
The ln(Q) vs ln(DelP) is clearly the better fit than the straight day (Q vs. DelP). The fit
with log10 is different but is equivalent to the ln fit and is just as good. The standardized
residuals were computed for the ln regression analysis. The value of the standardized
residual at DelP = 0.19 is 1.702 and this point is suspect although it is not clearly an
outlier.
6.49
6.88
Data
T
90
90
90
90
90
120
120
120
120
120
rpm
2000
3000
4000
5000
6000
2000
3000
4000
5000
6000
Use the regression function in Excel under Tools/Data Analysis/regression
Regression Parameters
Meas Q
rpm
T
rpmsq
rpmxT
pred Q pred-meas
0.5
2000
90 4000000
180000 0.499563 -0.00044
0.85
3000
90 9000000
270000 0.855213 0.005213
1.13
4000
90 16000000
360000 1.124443 -0.00556
1.31
5000
90 25000000
450000 1.307253 -0.00275
1.4
6000
90 36000000
540000 1.403643 0.003643
0.5
2000
120 4000000
240000 0.495564 -0.00444
0.86
3000
120 9000000
360000 0.866214 0.006214
1.15
4000
120 16000000
480000 1.150444 0.000444
1.35
5000
120 25000000
600000 1.348254 -0.00175
1.46
6000
120 36000000
720000 1.459644 -0.00036
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.99993859
R Square 0.99987718
Adjusted R 0.99977892
Standard E 0.00530498
Observatio
10
ANOVA
df
Regression
Residual
Total
Intercept
X Variable
X Variable
X Variable
X Variable
SS
MS
4 1.145549 0.286387
5 0.000141 2.81E-05
9 1.14569
Coefficients tandard Erro
-0.369 0.038255
0.000526714 1.16E-05
-0.001133333 0.000336
-4.32143E-08
1E-09
5E-07 7.91E-08
The regression equation is:
t Stat
-9.64585
45.38555
-3.37787
-43.1045
6.322548
F
ignificance F
10176.2 5.85E-10
P-value Lower 95%Upper 95%Lower 95.0%
Upper 95.0%
0.000203 -0.46734 -0.27066 -0.46734 -0.27066
9.81E-08 0.000497 0.000557 0.000497 0.000557
0.01972
-0.002 -0.00027
-0.002 -0.00027
1.27E-07 -4.6E-08 -4.1E-08 -4.6E-08 -4.1E-08
0.001459 2.97E-07 7.03E-07 2.97E-07 7.03E-07
Q = -0.369+0.0005267T-4.3214E-8rpmsq+5E-7rpmxT
Q - l/s
2nd order fit
4
3
2
1
0
2
y = -0.0236x + 0.9779x - 2.5974
4
5
6
7
V - m/s
6.50
8
6.89
V
4
5
6
7
8
Use the plotting and trendline functions in Excel
Q
0.94
1.69
2.44
3.08
3.72
This data is non-linear and the 2nd order fit is superior. Note the points at the ends –the
linear fit is above them and the 2nd order passes through them.
Q - l/s
linear fit
4
3
2
1
0
y = 0.695x - 1.796
4
5
6
7
V - m/s
6.51
8
6.90
Velocity
0.07
0.08
0.11
0.135
0.145
0.185
0.19
0.22
0.24
0.285
0.295
Use the plotting and trendline functions
of Excel
0.4
V e lo c ity
Voltage
0.01
0.115
0.29
0.48
0.59
0.81
0.88
1.02
1.12
1.325
1.4
2
y = 0.0373x + 0.11x + 0.0698
0.3
0.2
0.1
0
0
0.5
1
1.5
Voltage
T
200
300
400
500
600
700
800
900
1000
1100
1200
1300
sigma
41.5
40.3
39.2
37.8
35
32.5
28.5
23
16
9.5
6.5
5
sigma
6.91
(a)
50
40
30
20
10
0
y = 1E-10x4 - 4E-07x3 + 0.0003x2 - 0.1061x + 53.842
0
500
1000
1500
Temp
(b) Due to the S shape of the curve, at least a 3rd order curve will be required.
(c) The 4th order curve fit is shown.
6.52
6.92
Let L1 be the length of the shelves and L2 be the length between sides of the cabinet.
The clearance between the two is  = L2 – L1. We want to be 99% confident that  is
greater than equal to zero.
S  S L22  S L21  0.06 2  0.03 2  0.067
Assume large sample size for the standard deviations so we can use the normal
distribution. We want to be sure that 99% of the shelves will not be too long. Using
Table 6.3, For 99% confidence level, A = 0.49 and z =2.33
.This value of z corresponds to
  zS  2.33  0.067  0.156 in
The longest the nominal length the shelves can be is 24 – 0.156 = 23.844 in long. Note:
We did not concern ourselves with how short the shelves can be.
6.53
CHAPTER 7
7.1 (a) Series
R
 1,
R
R  R R
1
2
w
R
 [(
1
R
1
R
2
R
R
2
2 1/ 2
w ) (
w ) ]
R R1
R R 2
1
1
2 1/ 2
 [( 0.2 * 1)  ( 0.1 * 1) ]
 0.22 
2
w
R,max

R
R

w
w
R
1
R
R R 2
1
1
 0.2  0.1  0.3 
(b) parallel
RR
R 1 2
R1  R2
R
R
R22
R12


,
R1 (R1  R2 )2 R2 (R1  R2 )2
50 2
(150)2
 0.11

100 2
150 2
 0.44

wR  [(0.11 * 0.2)2  (0.44 * 0.1)2 ]1/ 2  0.05 
wR,max  0.11*.2  0.44 * 0.1  0.07 
7.1
7.2
K
M

[tA (2gh)1/2 ]
393.00
[600 * (
0.0127 2
) * 1000(2 * 9.81 * 3.66) 1/2 ]
4
 0.610

kg
 K (dimension less)
[sec. m . kg/m 3 (m/sec 2 * m) 1/2 ]
2
1/ 2
K
K
K
K
w m )2  (
w t )2  (
w  )2  (
w n ) 2 ]
M
T

n
1
K


 155 * 10  4 kg 1
1/ 2
M
tA (2gh)
0.610
K
 
 10.16 * 10  4 sec 1
600
t
0.610
K


 4814.40m 2
A
1.266 * 10  4
0.610
K
 
 6.10 * 10  4 m 3 / kg

1000
w k  [(
K
M
K
t
K
A
K

0.610
K
K


 833.33 * 10  4 m 1
2h
2 * 3.66
( h )
A
A d  0.0127
wd ,


 0.01995 m
2
2
d
d
w A  0.01995 * 0.000025  .005 * 10  4 m 2
wA 
w K  [(1.55 * 10  4 * 0.03) 2  (10.16 * 10  4 * 1) 2  ( 4815.40 * 0.005 * 10  4 ) 2  (6.10 * 10  4 * 0.003 ) 2 ]1 / 2
w K  0.0026
So,
K= 0.610  0.003
The maximum error for this case is larger:
n
K
w K max  
w x i  10  4 [1.55 * 0.03  10.16 * 1  4815.4 * 0.005  6.10 * 0.001  833.33 * 0.00366 ]
n 1 x i
 0.037
 0.004
7.2
7.3
865
M

1/ 2
3
[tA(2gh ) ] 600 * (136
. * 10 ) * 63.36 * (2 * 32.17 * 12.02)1/ 2
lbm
 K dim ensionless
lbm
ft
2
1/ 2
sec* ft * 3 * (
* ft )
ft
sec 2
K
K
K
K
w K  [(
w M )2  ( w t )2  ( w  )2  (
w h )]1/ 2
M
t

h
K
 0.601
 10 4 [(6.94 * 0.005)2  (10 * 1)2  (440 * 10 4 * 5.45 * 10 6 )2  (249.58 * 0.01)2 ]1/ 2
 26.10 * 10 4
where
1
k
K


 6.94 * 10 4 lbm 1
1/ 2
M tA(2gh )
M
3
K
K
K
C
4
1 K
2 C
4 ft
   10 * 10 sec ,
   440 ft ,
   96.22 * 10
t
A


t
A
lbm
K
K

 249.58 * 10 4 ft 1
2h
 ( h )
A
A d  *.5 0.001
*
wd ,


 5.45 * 10 6 ft 2
wA 
2
2 * 12 12
d
d
So
K  0.601 0.003
The maximum error for this case is larger:
n
K
w K ,max  
w x 1  10 4 [6.49 * 0.05  10 * 1  (440 * 10 4 * 5.45 * 10 6 )  (249.58 * 0.01)]
i 1 x1
 0.0037
7.3
7.4 The general formula for calculation of uncertainty (Eq. 7.4) will be applied to
n
R
wR  [ (w xi 1 )2 ]1/ 2
individual cases.
x i
i 1
R 1
R
R
 a, 1  b, 1  C
(a) R1  ax 1  bx 2  cx 3
x 1
x 2
x 3
w R1  [a 2 w x21  b 2 w x22  c 2 w x23 ]1 / 2
For the general case of summation of, R   ai x i
2
wR
2 w xi 1/ 2
wR  [ a w ] and
 [ ai 2 ]
R
R
R 2
R
(b)R 2  d ( x1 )( x 2 )( x 3 ),
, d, ( x 2 )( x 3 ), 2  d ( x1 )( x 3 )
x 2
x1
2
i
2 1/ 2
xi
R 2
 d ( x1 )( x 2 )
x 3
w R 2  [d 2 ( x 2 ) 2 ( x 3 ) 2 w x21  d 2 ( x1 ) 2 ( x 3 ) 2 w x22  d 2 ( x1 ) 2 ( x 2 ) 2 w x23 ]1/ 2
For the general case of multiplication:
n
n
w
w2
R2
w R  d [ 2 w xi2 ]1 / 2 and R  [ xi2 ]1 / 2
R
i 1 x i
i 1 x i
(c)R 3  e( x 1 )( x 2 ) / x 3 ,
w R 3  e[(
R 3
x R
x R
x x
 e 2 , 3  e 1 , 3  e 1 2 2
x 3 x 2
x 3 x 3
x 1
x3
w2
x x
w2
w2
x2 2 2
x
) w x1  ( 1 ) 2 w x22  ( 1 2 2 ) 2 w x23 ]1 / 2 and w R 3 / R 3  [ x21  x22  x23 ]1 / 2
x3
x3
x1
x2
x3
x3
(d) R 4  f ( x 1 ) g ( x 2 ) h ( x 3 ) i ,
R 4
R
 fg ( x 1 ) g 1 ( x 2 ) h ( x 3 ) i , 4  fh( x 1 ) g ( x 2 ) h 1 ( x 3 ) i ,
x 1
x 2
R 4
 fi ( x 1 ) g ( x 2 ) h ( x 3 ) i 1
x 3
w R 4  f [g 2 ( x 1 ) 2( g 1) ( x 2 ) 2 h ( x 3 ) 2i w x21  h 2 ( x 1 ) 2g ( x 2 ) 2( h 1) ( x 3 ) 2i w x22  i 2 ( x 1 ) 2 g ( x 2 ) 2h ( x 3 ) 2( i 1) ]1 / 2
w
w R4
w
w
 [g 2 ( x 1 ) 2  h 2 ( x 2 ) 2  i 2 ( x 3 ) 2 ] 1 / 2
R4
x1
x2
x3
n
For the general case of R  f  x iai   multiplication
i 1
wR
n
w
w
w
 f [ a i2 R 2 ( x1 ) 2 ]1 / 2 and R  [ a i2 ( x1 ) 2 ]1 / 2
x1
R
x1
i 1
i 1
n
7.4
7.5
F  kx
( w F ) max  w k
F
F
 wx
x
k
F
 k  700 N / cm
x
F
 x  12 . 5 cm
k
( w F ) max  (18 * 12 . 5  1 . 25 * 700 )
( w F ) max  1,100 N
F 2 1/ 2
F 2
)  (w x
) ]
X
K
 [( 18 * 12 . 5 ) 2  (1 . 25 * 700 ) 2 ] 1 / 2
w F  [( w k
 903 N
F
 700 * 12 . 5  8 ,750 N ( absolute value )
wF
903
* 100  10 . 3 %( relative value )
* 100 
F
8750
7.6
F
F
 wx
k
x
F 2
F 2 1/ 2
w F  [(w k
)  (w x
) ]
k
x
F
F
 x  5 in,
 20 lbf / in
k
x
F  kx
(w F )max  w k
12.5
(w F )max

.125
F
5 * 20
w
 10.3, F .103, 10.3%
F
(w F )max  0.5 * 5  0.5 * 20  12.5 lbf ,
w F  [(0.5 * 5)2  (0.5 * 20)2 ]1/ 2
7.5
7.7
F  kx
( w F ) max  w k
F
F
 wx
x
k
F
 k  800 N / cm
x
F
 x  20 . 0 cm
k
( w F ) max  (16 * 20 . 0  1 . 25 * 800 )
( w F ) max  1,320 N
F 2 1 / 2
F 2
)  (wx
) ]
X
K
 [(16 * 20 . 0 ) 2  (1 . 25 * 800 ) 2 ]1 / 2
w F  [( w k
 1, 050 N
F
 800 * 20 . 0  16 , 000 N ( absolute value )
wF
903
* 100 
* 100  10 . 3 %( relative value )
8750
F
7.8
F 1/ 2 2
) ,  250,000,  500rad / sec
mr
Eq.7.6
 (
w
w
w
1 wm 2
[(
)  ( r ) 2  ( F ) 2 ]1 / 2

r
F
2 m
w
1 .5 2
.02 2
.5 2 1 / 2
 [(
) ]
) (
) (

2 100
20
500
1
 [25 * 10  6  10  6  10  6 ]1 / 2
2
w
 2.60 * 10 3 ,0.26%(95%confidence level )


w   1.30rad / sec (95%confidencelevel )
7.6
7.9
F 1/ 2 2
) ,  200,000,   447.21rad / sec
mr
Eq.7.6
w 1 wm 2
wr
wF 2 1 / 2
 [( )  ( ) 2  (
) ]
r
F
 2 m
w 1 .5 2
.02
1.2 2 1 / 2
 [(
)  ( )2  (
) ]
 2 120
25
600
1
 [ 4.691  10  3 ]
2
w
  2.345 * 10  3 ,0.23% (95%confidencelevel )
(

w  1.03rad / sec (95%confidencelevel )
7.10
E  (f / A) /(
L
L
)
F .L
A.L
Eq.7.6
w
wE
w
w
w
 [( F ) 2  ( L ) 2  ( A ) 2  ( L ) 2 ]1 / 2
E
F
L
A
L
5 2 1/ 2
1.5 2
1 2
.5 2
) ]
 [(
) (
) (
) (
100
100
100
100
1

[.25  1  2.25  25]1/ 2  5.3 / 100or 5.3%
100
To reduce the uncertainty in E by 50%, we have to reduce the uncertainty in
measurement of L (to less than 1/2 of the present value.)
7.7
7.11
wR  [(
R
R
wd )2  ( wT )2 ]1/ 2
d
T
We have assumed that R0 and T0 are known values with high degrees of
accuracy.
R  [R0 (1   (T  T0 )]1/ 2
R
R
 R0 (T  T0 ) and
 R0 , so

T
w R  R0 [((T  T0 )w ) 2  (wT )2 ]1/ 2
0.1
* 0.0048   0.0048 * 10 3 1 / C
100
wT  [BT2  (tST )2 ]1/ 2 t  2 ( for l arg e (  30) samples )
w  
 [0.12  (2 * 0.1)2 ]1/ 2  0.22 C
w R  100[((25  0 )*.0048 * 10 3 )2  (.0048 * 0.22)2 ]1/ 2
 0.11 ohm
while R25  100.00(1  0.0048 * 25 )  112.00 ohms.
7.8
7.12
m f
N
2
5 . 0 * 10  4 ( kg / sec)

 1 . 06 * 10  8 kg / J
1
. N .m .)
2  * 50 * 150 (
sec
kg / hr
 0 . 38 * 10  4
W
W bsfc
W m f 2
W
W
(b)
 [(
)  ( N ) 2  (  ) 2 ]1 / 2

bsfc
m f
N
(a) bsfc 
w bsfc
at 1%, the maximum tolerable uncertainty in
bsfc
 f , N and  is 1%. This implies that if we have 1% uncertainty in one, the other
m
two must have a much less uncertainty (negligible value compared to 1%.)
To have the value of
(c) If all measured parameters have the same uncertainty, then,
W
W bsfc
Wm f 2 1 / 2
W 2
 [3(
) ]  [3( N ) 2 ]  [3(
) ]
bsfc
m f
N

 0.01
Wm f WN W


 7.6 *10 3 or 0.8%
m f
N

d  Wm f  7.6 *10 3 * 5.0 *10  4  3.8 *10 6 kg / sec  14.4 g / hr!
WN  22.8RPM
W  1.14N .m
7.9
7.13
m f
N
2
2 . 5 * 10  4 ( kg / sec)

 1 . 06 * 10  9 kg / J
1
2  * 50 * 750 (
. N .m .)
sec
kg / hr
 0 . 38 * 10  5
W
W bsfc
W m f 2
W
W
(b)
 [(
)  ( N ) 2  (  ) 2 ]1 / 2

bsfc
m f
N
(a) bsfc 
w bsfc
at 1.5%, the maximum tolerable uncertainty in
bsfc
 f , N and  is 1.5%. This implies that if we have 1.5% uncertainty in one, the
m
other two must have a much less uncertainty (negligible value compared to
1.5%.)
To have the value of
(c) If all measured parameters have the same uncertainty, then,
W
W bsfc
Wm f 2 1 / 2
W 2
 [3(
) ]  [3( N ) 2 ]  [3(
) ]

bsfc
m f
N
 0.015
Wm f WN W


 8.7 *10 3 or 0.87%

m f
N
d  Wm f  8.7 *10 3 * 2.5 *10  4  2.2 *10 6 kg / sec  7.8g / hr!
WN  26.1RPM
W  1.31N .m
7.10
7.14
hysteresis
Lineariz.error
Resolution error
zero off set
repeatability
error type
systematic
systematic
random
systematic
random
 0.1C
0.2% of reading
0.05C
0.1C
0.2C
B  (0.12  [(.002)(120)]2  0.12 )1 / 2
 0.28C
Assuming that the random errors have been determined with samples>30,
P  (.05 2  .2 2 )1 / 2  0.21C
So total uncertainty
w  [ B 2  p 2 ]1 / 2  [ B 2  (tS ) 2 ]1 / 2
w  [(0.28) 2  (0.21) 2 ]1 / 2
w  0.35C
7.15
hysteresis
Lineariz.error
Resolution error
zero off set
repeatability
error type
systematic
systematic
random
systematic
random
 0.1C
0.2% of reading
0.10C
0.1C
0.1C
B  (0.12  [(.002)(100)]2  0.12 )1 / 2
 0.24C
Assuming that the random errors have been determined with samples>30,
P  (.10 2  .12 )1 / 2  0.14C
So total uncertainty
w  [ B 2  p 2 ]1/ 2  [ B 2  (tS ) 2 ]1/ 2
w  [(0.24) 2  (0.14) 2 ]1 / 2
w  0.28C
7.11
7.16
Range: 0  2 kg
Calib. Accuracy: 2% of reading (Systematic)
A/D Converter Uncertainty:
0.5Range 0.5  2
 12  0.00024 kg
2N
2
2
wm  [ wCalib
 w102 ]1 / 2
 [(0.02 *1.25) 2  (0.00024) 2 ]1 / 2
 0.025 kg
7.17
Range: 0  5 kg
Calib. Accuracy: 2% of reading (Systematic)
A/D Converter Uncertainty:
0.5Range 0.5  5
 12  0.00061 kg
2N
2
2
wm  [ wCalib
 w102 ]1 / 2
 [(0.02 * 3.20) 2  (0.00061) 2 ]1 / 2
 0.064 kg
7.12
7.18
accuracy
resolution
linearity
temp. stab.
error type
systematic
random
systematic
random
0.5% range =5 kPa
1 kPa
4 kPa
2 kPa (0  50C)
Systematic Errors: Normally, accuracy includes the error of linearity, so the only
systematic error is the accuracy. Then, B = 5 kPa.
Random Errors: The random errors are those due to resolution and temperature
stability.
Assuming more than 30 samples, t = 2
P  (12  (2)2 )1/ 2  2.2 kPa
Overall Uncertainty
w  [B 2  P 2 ]1/ 2  (5 2  2.22 )1/ 2
 5.5 kPa
or 5.5 / 500  0.011or 11%
of measured value of 500 kPa
.
7.19
accuracy
resolution
temp. stab.
error type
systematic
random
random
0.3%range =1.5 kPa
1 kPa
 1kPa(0  50c)
Random Errors: The random errors are those due to resolution and temperature
stability.
Assuming more than 30 samples, t = 2
P  (12  (1) 2 )1 / 2  1.4kPa
Overall Uncertainty
w  [B 2  P 2 ]1/ 2  (1.5 2  1.4 2 )1/ 2
 2.05kPa
or 2.05 / 250  0.0082
0.8% of measured value
7.13
7.20
Average Temperature
n
T
T
i 1
n
T  248.3 C
Random Uncertainty of Single Measurement
n
Sx  [  ( x i  x )2 / ( n  1)]1/ 2
i 1
Sx  0.55 C
Table 6.6
10  1  9  t  2.262
95% conf .
Pi  tSx  ( 2.262)( 0.55)
Pi  12
. C
Random Uncertainty of Mean
Px  tSx / n 1/ 2  (2.262)(0.55 ) / 10
Px .39 C
7.14
7.21
(a)
m av 
(b)
m
n

1.93  1.95  1.96  1.93  1.95  1.94  1.96  1.97  1.92  1.93
10
 1.94 kg
 (m
 mav ) 2 1/ 2
]
n 1
 0.01 2   0.01 2   0.02 2  .......... 1 / 2
[
]
9
 1.65 *10  2 kg
S Samp 1.70 *10  2


 0.52  10  2 kg
n
10
S samp  [
S mean
(c)
i
i
PSingle  tS samp  3.73 * 10 2 kg (95% Confidence)
t  2.262 is for   n - 1  9 degrees of freedom from Student - t Table 6.6
B single  0.015  Range  0.015  5kg  0.075kg
2
1/ 2
Total uncertainty : wsingle  ( Psin2 gle  Bsin
 0.084 (95%confidence)
gle )
(d)
Pmean  tS mean  1.18  10 2 kg (95% Confidence )
B mean  Bsingle  0.075kg
2
2
Total uncertainty : wmean  ( Pmean
 Bmean
)1 / 2  0.076(95%confidence )
As can be seen, the dominant factor is the systematic uncertainty.
7.15
7.22
(a)
(b)
S mean 
(c)
m
2.90  2.95  2.96  2.92  2.95  2.94  2.96  2.97  2.93  2.91
n
10
 2.94 lb
 (mi  mav )2 ]1/ 2
Ssamp  [
n 1
2
 0.04   0.01 2   0.02 2  .......... 1 / 2
[
]
9
 2.72 *10  4 lb
m av 
i
S Samp
n


2.72 *10  4
 0.86 *10  4 lb
10
PSingle  tS samp  6.15  10 4 lb (95% Confidence)
t  2.262 is for   n - 1  9 degrees of freedom from Student - t Table 6.6
B single  0.015  Range  0.015  10lb  0.15lb
2
1/ 2
Total uncertainty : wsingle  ( Psin2 gle  Bsin
 0.15 (95%confidence)
gle )
(d)
Pmean  tS mean  1.95  10 4 kg (95% Confidence)
B mean  Bsingle  0.15kg
2
2
Total uncertainty : wmean  ( Pmean
 Bmean
)1 / 2  0.15(95%confidence )
As can be seen, the dominant factor is the systematic uncertainty.
7.16
7.23 (a) If measurement is performed on over 50 pieces of pipe. Based on
problem 7.13, for the sample of 10
 mi  1.94kg
m av 
n
(b) Based on Problem 7.13,
(m i  mav ) 2 1/ 2

S samp  [
] = 1.65*10-2 kg
n 1
But now, n=50 so the standard deviation of the mean is:
SSamp 1.65 * 10 2
S mean 

 0.23 * 10 2 kg
n
50
(c)
PSingle  tSsamp  2 * 1.65 * 10 2 kg  3.3 * 10 2 kg (95% Confidence)
t  2 is for   30 degrees of freedom from Student - t Table 6.6
B single  0.015 * Range  0.015 * 5kg  0.075 kg
2
2
1/ 2
Total uncertaint y : w single  (Psin
 0.082 (95% confidence )
gle  B sin gle )
(d)
Pmean  tS mean  2 * 0.23  10 2  0.46  0 2 kg (95% Confidence )
B mean  Bsingle  0.075kg
2
2
Total uncertaint y : w mean  (Pmean
 Bmean
)1/ 2  0.075 (95% confidence )
As can be seen, the dominant factor is the systematic uncertainty.
7.24
S  0.1
w Total
P  2S
(Large Sample, t  2)
 2 * 0.10  0.20 kg (With 95% Confidence Level)
1.5
B 
* 12  0.18 kg
100
 [B 2  P 2 ]1/ 2
 [0.18 2  0.2 2 ]  0.27 kg (With 95% confidence level)
7.17
7.25
1 .5
* 4.5  0.07kg
100
Calib :
B1
w Total 
 0.01 * 4.5  0.045 kg
Calibration will be the only systematic uncertainty.
w 2 Total  B 2  P 2
P 2  w 2 Total - B 2
w total  .015  4.5  0.068
 0.068 2 - 0.045 2
P  0.05 kg
P
 0.027
2
This the acceptable S for the cheese blocks.
The A/D converter contributes to this precision uncertainty but there may be
other sources of precision error.
For a large sample : S Total 
7.18
7.26
 .t , where m  mass, m
  mass flow rate and t  filling time
mm
m
1
(a) t 

 4 sec

m 0.25
(b) The flowrate uncertainty is almost certainly random since it depends on
factors that vary somewhat randomly (like viscosity(
It is most likely that the time uncertainty is also random. If it were systematic, it
would likely be adjusted to close to zero by adjusting the speed of the cam. We
will treat it as random.
This means there are no significant bias errors to consider.
(c)
 m  2  m  2 
Pm  
P  
P 
 m   t t  
 m

0.5


 4.0  0.0025   0.25  0.1 
2
 Pt 2
 tPm   m
0.5
2 0.5
2
 0.0001  0.00063
0.5
 0.027 kg
Since we have not considered any systematic uncertainties, this is the total
uncertainty, wm = P
(d) The uncertainty in time has a much larger effect than the uncertainty in
flowrate (2.5% vs. 1%) therefore the approach should be to make the timing
more accurate.
Note: When the total uncertainty is computed in this problem, it doesn’t matter
whether uncertainties were classified as systematic or random since they were
combined with RSS.
7.19
7.27
 t , where m  mass, m
  mass flow ratio and t  filling time
mm
m
2
(a) t 

 4 sec
 0. 5
m
(b) The flowrate uncertainty is almost certainly random since it depends on
factors that vary somewhat randomly (like viscosity(
It is most likely that the time uncertainty is also random. If it were systematic, it
would likely be adjusted to close to zero by adjusting the speed of the cam. We
will treat it as random.
This means there are no significant bias errors to consider.
(c)
 m  2  m  2 
Pm  
P  
P 
 m   t t  

 m
0.5


 4.0  0.005   0.5  0.1 
 Pt 
 tPm   m
2 0.5
2
2 0.5
2
 0.0004  0.0025
0.5
 0.054 lbm
Since we have not considered any systematic uncertainties, this is the total
uncertainty, wm.= P.
(d) The uncertainty in time has a much larger effect than the uncertainty in
flowrate (2.5% vs. 1%) therefore the approach should be to make the timing
more accurate.
Note: When the total uncertainty is computed in this problem, it doesn’t matter
whether uncertainties were classified as systematic or random since they were
combined with RSS.
7.20
7.28
standard deviation:
S=450 kJ/kg
systematic error:
B=.01*100000=100 kJ/kg
for 95% confidence level and degrees of freedom 15-1 ,
Pi  tSx  2145
.
* 450 kJ / kg
 965 kJ / kg
P
965.25
Px  i 
 249 kJ / kg
n
15
a) w x  (B2  P 2 ) y 2  (1000 2  249 2 )1/ 2
w x  1030 kJ / kg
b) w i  (B2  Pi 2 )1/ 2  (1000 2  965 2 )1/ 2
w i  1390
kJ / kg
7.21
t=2.145
(Table 6.6)
7.29
(a) Output ( mA) is linear with input(I mH2 O) in the range;
At Depth  0 mH 2O
I  4 mA
At Depth  20 mH 2O
I  20 mA
Input span is 20 m, Output span is 16 mA
At depth  15 m
Output  4  (15/20)16  16 mA
(b)
Acccuracy 0.2% of span (systematic)
0.002  20  0.04 m H 2O
In m H 2O :
In mA :
0.002  16  0.032 mA
This is 0.032/16  0.002  0.2% of reading
Zero Balance :
2% Span (systematic)
 0.02  20  0.40 m H 2O
In mA :
0.02  16  0.3 2mA
This is 0.32/16  0.02  2% of reading
Thermal Effects :
1.5% of span (random)
0.015  20  0.3 m H 2O
In m H 2O :
0.015  16  0.24 mA
In mA :
This is 0.24/16  0.015  1.5% of reading
The following table shows the uncertainties in m, mA and % of reading:
Uncertainty
Accuracy
Zero Balance
Thermal
Effect
Total
Uncertainty
Uncertainty
Category
Systematic
Systematic
Random
Meter
Water
0.04
0.40
0.30
0.50
(c)
B Depth  [  Bi2 ]1 / 2  [0.04 2  0.4 2 ]1 / 2  0.40m water
wDepth  [ B 2  P 2 ]1 / 2
 [0.40 2  0.30 2 ]1 / 2
 0.50 m water or 0.50  16/20  0.40 mA
This will be 0.4/16  2.5%
7.22
mA
0.03
0.30
0.24
% reading at
15 m
0.26
2.6
2.0
0.40
3.3
7.30
(a) Output ( mA) is linear with input(I mH2 O) in the range;
At Depth  0 mH 2O
I  4 mA
At Depth  25 mH 2O
I  20 mA
Input span is 25 m, Output span is 16 mA
At depth  15 m
Output  4  (15/25)16  13.6 mA
(b)
Acccuracy 0.1% of span (systematic)
0.001  25  0.025 m H 2O
In m H 2O :
0.001  16  0.016 mA
In mA :
This is 0.016/16  0.001  0.1% of reading
Zero Balance :
1.5% Span (systematic)
 0.015  25  0.38 m H 2O
In mA :
0.015  16  0.24 mA
This is 0.24/16  0.015  1.5% of reading
Thermal Effects :
1% of span (random)
0.01  25  0.25 m H 2O
In m H 2O :
0.01  16  0.16 mA
In mA :
This is 0.16/16  0.01  1% of reading
The following table shows the uncertainties in m, mA and % of reading:
Uncertainty
Accuracy
Zero Balance
Thermal
Effect
Total
Uncertainty
Uncertainty
Category
Systematic
Systematic
Random
Meter
Water
0.025
0.38
0.25
0.46
(c)
B Depth  [  Bi2 ]1 / 2  [0.025 2  0.38 2 ]1 / 2  0.38m water
wDepth  [ B 2  P 2 ]1 / 2
 [0.38 2  0.25 2 ]1 / 2
 0.45 m water or 0.45  16/20  0.36 mA
This will be 0.36/16  2.3%
7.23
mA
0.016
0.24
0.16
% reading at
15 m
0.12
1.76
1.18
0.29
2.13
7.31
(a) For an un-amplified output of 20mV, considering that the device is linear,
assuming no zero offset,
F (N )
mV
VOutput 
* 3(
) * VExitation
V
FRange
VOutput 
F (N )
500N
* 3(
mV
) * 10V
V
Vout
20

 333 N
0.060 0.060
0.1
(b) Linearity Uncertainty : 0.1% FS 
* 500  0.5 N
100
or
0.060 * 0.50  0.030 mV
or 0.03/20 =0.15%
Similarly for Hysteresis and Repeatability (see table below)
0.002
* 500 * 40  20 
Max. Temperature effect: 0.002 % FS/ oC 
100
 0.20N
Vout
 0.060 F (N)
 F
Or 0.06 * 0.2  0.012 mV
Other uncertainties are shown in the table below:
Uncertainty
Linearity
Hysteresis
Repeatability
Temp. Effect
Total
(c)
Category
Systematic
Systematic
Random
Random
Total Uncertaint y :
mV
0.030
0.025
0.009
0.012
0.042
N
0.50
0.40
0.15
0.20
0.70
% output
0.15
0.12
0.05
0.06
0.21
2
2
w Total  [B Force
 PForce
]1 / 2
2
2
 BHys
B Force  [ Bi2 ]1 / 2  [B Linl
]1 / 2  0.039 mV
or
0.65N
Assuming that all random uncertainties are based on samples greater than 30 :
2
2
1/ 2
 (0.009 2  0.012 2 )1 / 2  0.015 mV
PForcel  [ P 2 i ]1 / 2  (PRe
p  PTem )
or
0.25N
2
2
 PForce
w Total  (BForce
)1 / 2  (0.039 2  0.015 2 )1 / 2  0.042 mV
or
0.70N
which is about 0.21% of output.
7.24
7.32
(a) Uncertainty due to accuracy (considered systematic)
6
* 2  0.12 lpm
For 2 lpm : 6% of flow 
100
6
* 10  0.6 lpm
For 10 lpm : 6% of flow 
100
Uncertainty due to repeatability (considered random)
1
* 2  0.02 lpm
For 2 lpm : 1% of flow 
100
1
* 10  0.10 lpm
For 10 lpm : 1% of flow 
100
(b)
Total Uncertainty: w Total  [B 2  P 2 ]1 / 2
For 2 lpm
wTotal  [0.12 2  0.02 2 ]1 / 2  0.12 lpm
For 10 lpm
wTotal  [0.6 2  0.12 ]1/2
 0.61 lpm
In both the cases, the uncertainty due to accuracy is dominant
7.33
(a) Uncertainty due to accuracy (considered systematic)
5
* 5  0.25 lpm
For 5 lpm : 5% of flow 
100
5
For 15 lpm : 5% of flow 
*15  0.75 lpm
100
Uncertainty due to repeatability (considered random)
1
For 5 lpm : 1% of flow 
* 5  0.05 lpm
100
1
For 15 lpm : 1% of flow 
*15  0.15 lpm
100
(b)
Total Uncertainty: wTotal  [B 2  P 2 ]1 / 2
For 5 lpm
wTotal  [0.25 2  0.05 2 ]1 / 2  0.25 lpm
For 15 lpm
wTotal  [0.75 2  0.152 ]1/2
 0.76 lpm
In both the cases, the uncertainty due to accuracy is dominant
7.25
7.34
Accuracy:
average:
STD Dev:
0.2 C
250 C
0.2 C
(systematic error)
(a) Random Uncertainty of Mean Value
degrees of freedom 15-1 = 14, 95% confidence.
Table 6.6  t  2.145
PX  tSx / n 1/ 2  ( 2.145 )(0.2) / 15
 0.11 C
w  ( B 2  P 2 ) y 2  ( 0.22  0.112 )1/ 2
w  0.23
C
overall accuracy of the mean value
(b) Random Uncertainty - Single Reading
Pi  tSx  ( 2.145 )( 0.2) .43
w  ( B 2  P 2 )1/ 2  (.2 2 .43 2 )1/ 2
 0.5 C
accuracy of any sin gle measurement
(c) It will have a greater effect on mean since random error is about the same,
but for single reading - the effect will not be as much because random error is
greater.
7.35
Tcorr 
Tcorr

(Ts4  Tw4 ) 
h
 135.8 K
5.669 * 10 8 * 0.9
[(600  273 )4  (545  273)4 ]
50
w
0.1 2 10 2 1/ 2
w T
w
 [(  )2  ( n )2 ]1/ 2  [(
)  ( ) ]  0.23
T

0.9
50
n

w T  0.23 * 135.8  312
. K
b
w
w T
w
0.2 2 10 2 1/ 2
 [(  )2  ( n )2 ]1/ 2  [(
)  ( ) ]  0.30
T

n
0.9
50

w T  0.30 * 135.8  40.8 K
 Tg  600  135.8
Tg  735.8  312
. / 40.8 K
7.26
7.36
10 measurements of strain
ave: 80 mV
S = 0.5 mV (Standard deviation)
t= 2.262
(deg. of freedom 10 - 1 = 9, 95% confidence)
Px  ts  2.262 * 0.5  1.13mV (duetostraingage)
15 measurements of transmitted voltage
STD = 1 mV
(Standard deviation)
t = 2.145
(deg. of freedom 15 – 1 =14, 95% confidence)
Pz  ts  2145
.
* 1  214
. mV (due to transmission )
Overall standard deviation of voltage:
m
S x  [ s i2 ]1 / 2  [.5 2  12 ]1 / 2  1.1mV
i 1
m
[ s i2 ] 2
[.5 2  12 ] 2
 20
.5 4 14
4
si / i


9 14
i 1
For 95%conf .fromTable6.6,t  2.086,Px  ts x  2.3mV
x 
i 1
m

7.37
10 measurements of strain
ave: 100 mV
S = 0.8 mV (Standard deviation)
t= 2.262
(deg. of freedom 10 - 1 = 9, 95% confidence)
Px  ts  2.262 * 0.8  1.81mV (duetostraingage)
10 measurements of transmitted voltage
STD = 1 mV
(Standard deviation)
t = 2.262
(deg. of freedom 10 - 1 = 9, 95% confidence)
Pz  ts  2.262 *1  2.26mV (duetotransmission)
Overall standard deviation of voltage:
m
S x  [  si2 ]1 / 2  [.8 2  12 ]1 / 2  1.3mV
i 1
m
[  si2 ]2
[.8 2  12 ]2

 17
x 
.8 4 14
4
 si /  i 9  9
i 1
For95%conf . fromTable6.6,t  2.110,Px  ts x  2.7mV
i 1
m
7.27
7. 38 15 measurements - degrees of freedom = 15 - 1 = 14 at 95%
confidence.
t  2.145 (Table 6.6)
Pr ecision index Sx  5 kP
P  ts x  2145
.
* 5  10.7 kPa
In Problem 7.11, there are two systematic errors, accuracy and linearity.
However, the linearity is normally included in the accuracy so the only systematic
error is the accuracy. As a result:
B=5 kPa
Overall uncertainty of the pressure measurements:
total:
w x  (Bx2  Px2 )1/ 2  (5 2  10.7 2 )1/ 2
w x  118
. kPa
7.39 Yes the uncertainty will change because it is dependent on t which is
determined from the sample size. If n = 30, t = 2.042. t would be smaller so the
uncertainty would also be smaller.
7.28
7.40 (a)
V 
V
i
 459 .9
n
 V  V  ]
2
SV  [
i
I
i
I
 1.25 V
1/ 2
n 1
 31.0 amps
n
 I  I  ]
2
SI  [
i
1/ 2
 0.63 amp
n 1
 PFi  0.79
PF 
n
 (PFi  PF )2 ]1 / 2  0.02
SPF  [
n 1
 Pi   Vi * Ii * PFi 3  19,629 Watts
P 
n
n
 P  P  ]
[
2
SP
i
1/ 2
n 1
PV  tS V  2.95 V ,
 787 Watts
PI  tSI  1.49 amp,
PPF  tSPF  0.05, and
PP  tSP  1,861Watts
where " t" has been obtained from Student - t test Table 6.6 for
  n - 1  7 for 95% confidence (
(b)
P
PP  t
(c)

SP
n
 P  V
i
i
787
8
2
 0.0025) to be 2.365
* I i * PFi 3
n
 2.365 *

n
 19,629Watts
 658 Watts
95 % confidence
B v  B v  1% of V  0.01*459.9 = 4.60 V
BI  BI  1%of I  0.01* 31.0  0.31 amp
BPF  BPF  1%of PF  0.01* 0.79  0.79 * 102
Systematic uncertainty of power (applying Eq. 7.24 and utilizing Eq. 7.6 for
simplification),
2
2
2

  P   P
 P
B P  [
BV   
BI   
BPF  ]1/ 2

  PF
  I
 V
7.29
B
BP
B2
B
 [( V ) 2  ( I ) 2  ( PF ) 2 ]1/ 2  (0.012  0.012  0.012 )1/ 2  0.017
P
V
PF
I
B P  0.017 * 19629.  334 Watts
(d)
w P  [ B P2  PP2 ]1/2  (334 2  658 2 )1/2  738 Watts
(With 95% confidence level)
most of the uncertainty is due to random effects.
7.41
El Error
freedom
Calibration
Data Acquisition
repeatedly
spatial
installation
transmission
Data Reduction
linearity
Systematic Uncertainty
.
Standard deviation Degrees of
2
---
---
--2
1
---
.5
----.3
14
----29
1
---
---
Systematic Uncertainty : Bx  (.22  22  12  12 )1/ 2  2.5 C
(.5 2 .3 2 )2
 24
[.5 4 / 14 .3 4 / 29]
from 6.6  t  2.064

Standard Deviation :
Sx  (.5 2 .3 2 )1/ 2 .58 C
P  tSx  (2.064)(.58)  12
. C
Total Uncertainty
w x  (P 2  B2 )1/ 2  (12
. 2  2.5 2 )1/ 2  2.8 C
7.30
7.42
w V  uncerta int y of volume  1 cc ( precision )
w t  uncerta int y of time   0.3
sec
V 100

 4cc / sec
t
25
Eq. 7.6
Q
wQ
w
w
 [( V )2  ( t )2 ]1/ 2
Q
V
t
1 2
.3
 [(
)  ( )2 ]1/ 2
100
25
 0.016, 16%
.
wQ  4 * 0.016  0.06 cc / sec
7.31
( precision )
7.43
Power as measured: P = V x I x PF x 3 1/2 = 460*36*0.81*3 ½ = 23,233 Watts
(a) We will take the uncertainty originated from the measurement device as
systematic uncertainty
For Voltage : B V  1 % of reading  0.01 * 460  4.60 V
For Current : B I  1% of reading  0.01 * 36  0.36 A
For PF
B PF  2% of reading  0.02 * 0.81  1.62 * 10 - 2
:
The total uncertainty due to measurement device applying Eq. 7.24 and utilizing
Eq. 7.6 for simplification):
2
2
2
 P

 P 
 P

B P  [
BV   
BI   
BPF  ]1 / 2
 V

 I

 PF

or
2
2
2
BP
B 
B 
B 
 [ V    I    PF  ]1 / 2  (0.012  0.012  0.02 2 )1 / 2  0.0245
P
 PF 
 I 
V 
B P  23,233 * 0.0245  569 Watts
(b) Assuming that a large number of tests has been performed for calculation
(n>30)of the standard deviation, S, the random uncertainties with 95%
confidence level will be:
PV  2 S V  6.00 Volts
P1  2 SI  4.00 Amp
PPF  2 SPF  0.06
The actual power is Pwr  460  35  0.81 3 0.5  22,587
and then B=0.0245x22587=553.
(c) To Calculate the total uncertainty of power measurement, we
calculate the systematic and random uncertainties separately:
need to
Systematic uncertainty, BP was calculated in part a. The random uncertainty PP
will be calculated below.
7.32
2
2
2
 P

 P 
 P

Sp  [ 
SV   
SI   
S PF  ] 1 / 2
 V

 I

 PF

or
Sp
2
2
2
S 
S 
S 
 [  V    I    PF  ]1 / 2
P
V 
 I 
 PF 
2
2
2
Sp
 3.00 
 2.00 
 0.03  1 / 2
[
 
 
 ]  0.067
P
 460 
 35.00 
 0.81 
S p  1,556 Watts with 95% confidence
Assuming a large number of measurements (n>30),
PP = 2SP = 3,112 Watts
Total uncertainty for measurement of power with 95% confidence,
w P  [ BP2  PP2 ]1/2  (5532  3,1122 )1/2  3,160 Watts
This is equivalent to 14% of the reading!
7.33
7.44
Power as measured: P = V x I x PF x 3 1/2 = 460*50*0.81*3 ½ = 32,268 Watts
(a) We will take the uncertainty originated from the measurement device as
systematic uncertainty
For Voltage : BV  1 % of reading  0.01 * 460  4.60 V
For Current : BI  1% of reading  0.01 * 50  0.5 A
For PF
BPF  1% of reading  0.01 * 0.81  0.81
:
The total uncertainty due to measurement device applying Eq. 7.24 and utilizing
Eq. 7.6 for simplification):
2
2
2
 P
  P   P

B P  [
BV   
BI   
BPF  ]1 / 2
 V
  I
  PF

or
2
2
2
BP  BV   BI   BPF  1 / 2
 ]  (0.012  0.012  0.012 )1 / 2  0.0173
 [      
 V   I   PF 
P
B P  32,268 * 0.0173  558 Watts
(b) Assuming that a large number of tests has been performed for calculation
(n>30)of the standard deviation, S, the random uncertainties with 95%
confidence level will be:
PV  2 S V  6.00 Volts
P1  2 S I  2.00 Amp
PPF  2 S PF  0.06
We will take the uncertainty originated from the measurement device as
systematic uncertainty
For Voltage : B V  1 % of reading  0.01 * 460  4.60 V
For Current : B I  1% of reading  0.01 * 49.50  0.495 A
For PF
:
B PF  1% of reading  0.01 * 0.80  0.80
The actual power is
Pwr  460  49.5  0.8  3 0.5  31,551
an then the systematic uncertainty is 0.0173x31551 = 546. W
(c) To Calculate the total uncertainty of power measurement, we need to
calculate the systematic and random uncertainties separately:
7.34
Systematic uncertainty, BP was calculated in part b. The random uncertainty PP
will be calculated below.
2
2
2
 P   P   P

Sp  [ 
SV    S I   
S PF  ]1 / 2
 V   I   PF

or
Sp
2
2
2
S  S  S 
 [  V    I    PF  ]1 / 2
 V   I   PF 
P
Sp
2
2
2
 3.00   1.00   0.03  1 / 2
 
 
 ]  0.043
[
 460   49.50   0.80 
P
Sp  1,357 Watts with 95% confidence
Assuming a large number of measurements (n>30),
PP = 2SP = 2,714Watts
Total uncertainty for measurement of power with 95% confidence,
w P  [ BP2  PP2 ]1/2  (5462  2,7142 )1/2  2,768 W
This is equivalent to 9% of the reading!
7.35
7.45 (a) Although, some of these uncertainties (repeatability and temperature
errors specifically) are usually classified as random, because they are related to
the accuracy of an instrument, we will combine them as systematic uncertainty.
As an elemental error if they are considered as random uncertainty based on a
large sample, and the uncertainty rather than the standard deviation is given, the
effect on the total uncertainty of the measurement will be the same. If we are
looking at the uncertainty of the mean, this approach will not work and we need
to separate systematic and random uncertainties.
0.5
* 100  3.5 kPa
100
0.1
BHys  0.1%of span 
* 700  0.7 kPa
100
0.1
BRe p  0.1%of span 
* 700  0.7 kPa
100
0.3
BStab  0.3%of span 
* 700  2.1 kPa
100
0.040
Bzero  0.040%of span/ C 
* 700 * 20  15  1.4 kPa
100
0.030
BSpan  0.030%of span/ C 
* 70020  15  1.05 kPa
100
4.55
BPr ess  [ B2i ]1 / 2 4.55kPa or
* 100  0.65% of span
700
Blin  0.5%of span 
(b) Assumi ng that a large number of tests has been performed for calculation of the standard deviation, S.
PP  2S p  2 * 3.5  7.0 kPa with 95% Confidence
w P  [ B 2P  PP2 ]1/ 2  [ 4.55 2  7.0 2 ]1/ 2
 8.3 kPa with 95% Confidence
The total uncertaint y is more due to pressure fluctuatio ns, i.e random unceratint y.
(c) Quantization uncertaint y, PQ , which can be considered a part of the transmissi on uncertainty
0.5Span 0.5  1400

 0.17 kPa
2N
212
0.17
This value is
* 100  2.0% of the sensor uncertaint y
8.3
Note : A unipolar A/D was assumed, For bi - polar system, the PQ will be 0.34 kPa
PQ 
(d) Uncertainty due to noise, Pn , which can be considered a part of the transmissi on uncertaint y
Pn  0.02 * Signal value
 0.02 * 550  11 kPa
Obviously, the uncertaint y due to noise is a significant part of the uncertaint y of the final result
7.36
7.46 (a) Although, some of these uncertainties (repeatability and temperature
errors specifically) are usually classified as random, because they are related to
the accuracy of an instrument, we will combine them as systematic uncertainty.
As an elemental error if they are considered as random uncertainty based on a
large sample, and the uncertainty rather than S is given the effect on the total
uncertainty of the measurement will be the same. If we are looking for the total
uncertainty in a mean, this approach will not work and we need to separate
systematic and random uncertainties.
0.5
B lin  0.5% of span 
* 100  0.5 psi
100
0.1
* 100  0.1 psi
BHys  0.1% of span 
100
0.1
* 100  0.1 psi
BRe p  0.1% of span 
100
0.3
* 100  0.3 psi
BStab  0.3% of span 
100
0.040
* 100 * 70  60   0.22 psi
B zero  0.022% of span/  F 
100
0.016
* 100 * 70  60   0.16 psi
BSpan  0.016% of span/  F 
100
0.66
BPr ess  [  B i2 ]1/ 2 0.66 psi or
* 100  0.66 % of span
100
Assumi ng that a large number of tests has been performed for calculatio n of the standard deviation, S,
(b)
PP  2S p  2 * 0.50  1.00 psi with 95% Confidence
w P  [ B 2P  PP2 ]1 / 2  [0.66 2  1.0 2 ]1 / 2
 1.20 psi with 95% Confidence
The total uncertaint y is more due to pressure fluctuatio ns, i.e random unceratint y.
Quantizati on uncertaint y, PQ , which can be considered a part of the transmissi on uncertaint y
(c)
0.5Span 200
 12  2.44 * 10 -2 psi
N
2
2
2.44 * 10 -2
This value is
* 100  3.7% of the sensor uncertaint y
0.66
Note : A unipolar DAS was assumed, For bi - polar system, the PQ will be 4.88 * 10 -2 psi
PQ 
(d)
Uncertaint y due to noise, Pn , which can be considered a part of the transmissi on uncertaint y
Pn  0.02 * Signal value
 0.02 * 80  1.60 psi
Obviously, the uncertaint y due to noise is a significan t part in the uncertaint y of the final result
7.37
7.47
(a)
P   
(b)
3000
* 165
60
 51,836 N.m/sec (watt)
The Standard deviation of power :
 2 N.   2 *
2
2

 P

 P
S P  [
S   
S  ]1 / 2
 

 

P
P
 ,



S P  [ .S    .S   ]1 / 2
2
2
 [2 * 50 * 4   (165 * 2
2
5 2 1/ 2
) ]
60
 1,260 Watts
The degrees of freedom for power (Eq. 7.28)
(c)
P 
S 
2 2
P
2
1  P 
1  [ x Si  ] 2
i 
i

Where    n  1  20  1  19 and
n
   n  1  10  1  9
P 

S P4
4
1  P
1  P 

S   
S 

   

    
4
(1,260) 4
4
1 
5 
1
3000

* 4
   2 *
165 * 2 *
19 
60 
9
60

The random uncertainty of power:
7.38
4
9
 P  9  t P  2.262 (Table 6.6) for 95% confidence
PP  t P SP
 2.262 * 1260
 2,850 Watts
The systematic uncertaint y of power :
2
2
 P
  P  1/ 2
B P  [
B   
B  ]
 
  

 [B   B  ]1/ 2
2
2
 236 Watts
(d)
Total uncertaint y of power :
w P  [B P2  PP2 ]1/ 2
 [2850   236  ]1/ 2
 2,860 Watts (95% confidence )
2
2
7.39
7.48
(a)
P   
3000
* 165
60
 51,836 N.m/sec (Watts)
 2 N.   2 *
(b) The Standard deviation of power :
2
2

  P
 P
S P  [ . S    
. S   ]1 / 2

  
 
 [S   S  ]1/ 2
2
2
 [2 * 50 * 4   165 * 2 * 5160  ]1/ 2
2
2
 1,260 Watts
(c) Random uncertainty of power, considering the large number of measuremen ts :
PP  2SP  2,520 Watts
The systematic uncertainty of power :
2
2
  P  1/ 2
 P
B   
B  ]
B P  [

  
 
 [ B    B  ]1/ 2
2
2
2
2
5   3000


* 0 .7  ] 1 / 2
 [165 * 2 *
   2
60
60  


 236 Watts
(d) Total Uncertainty of power :
w P  [BP2  PP2 ]1/ 2
 [2,520   236  ]1/ 2
2
2
 2,531 Watts
7.40
7.49 (a) The systematic uncertainty of the measurement will remain the same
as in the previous two problems:
2
2
 P
  P  1/ 2
B P  [
B   
B  ]
 
  

 [ B   (B ) 2 ]1/ 2
2
2
2
5   3000


 [165 * 2 *
* 0.7  ]1/ 2
   2
60  
60


 236 Watts
To calculate the random uncertainty:
Degrees of freedom,  = n-1 = 10-1 = 9, where n is number of measurements.
Random uncertainty of each power measurement:
PP = tSP = 2.262*0.75 = 1.70 kW
where t = 2.262 is the value for Student-t value from Table 6.6 for 95%
confidence.
For the mean power:
tS
2.262  0.75
PP  P 
 0.54 kW
n
10
(b) Total Uncertainty of a single measurement :
w P  [BP2  PP2 ]1/ 2
 [1700  236 ]1/ 2
2
2
 1,716 watt
Total uncertainty of the mean power:
Total Uncertainty of the mean power :
w P  [BP2  PP2 ]1/ 2
 [236  540 2 ]1 / 2
2
 590 watt
(c) The systematic uncertainty remains the same but the random uncertainty
depends upon the measurement process and number of measurements.
Additionally, the uncertainties in Problems 7.32 and 7.33 are for single
measurement experiment although torque and speed have been independently
measured multiple times but the averages are used for calculation of the results.
In this problem, power is measured multiple times, and the random uncertainties
are based on these measurements.
7.41
7.50
P
W
3
 340 
3  m 
 * 700 * 10 3 (Pa )

 * 10 
60
sec




  Q.
(a)

5 * 10 3 Watts
 0.793
(b)
Parameter
Q
P
W
Mean Value
340 lit/m
700 kPa
5 kW
Standard
Deviation
5.6
10
0.15
2
 

S  [ 
Si  ]1/ 2
 X i 
2
Q
QP
 P

 [
. S Q   ( .S P ) 2  ( 2 S ) 2 ]1/ 2
W
W
W

Applying Eq. 7.6 and 7.26
S

 [(
SQ
Q
.)2  (
SP 2 SW 2 1/ 2
) (
) ]
P
W
5.6 2 10 2 0.15 2 1/ 2
) (
) ]  0.037
) (
700
5
340
S  0.793 * 0.037  0.029
 [(
7.42
Degree of
Freedom
15-1
15-1
15-1
(c)
Random uncertainty : P  t S
To find t, we need to calculate degrees of freedom for  Eq. 7.28 
S 
2 2
 

2
2
1  
1
 
S

Q   
 Q  Q    P
2
2
2
  
1   
 
 
S
S



P  
W  
     W
 
 P
2
S 
4


4
1  SQ

 Q  Q




4

1
 
  P
 S P  4 
1
  

 P   W
 Sw  4 
 

 W  
4
 0.029 


0.793 

 
4
4
4
1  5.6 
1  10 
1  0.15 


 

 

14  340  14  700  14  5 
 27
Using Table 6.6, for 95% confidence level (

2
 0.025 ),
t  2.052
P  t S
 2.052 * 0.029
 0.060
with 95% confidence level
Systematic uncertaint y of the pump efficiency :
 B
  Q
  Q

B
2
2
2

 BW  
 B P 
  
 
 
 W  
 P 

1/ 2
(Apply Eq. 7.6 and 7.24)
There are two uncertainties associated with the pressure transducer. Accuracy
is always treated as systematic. It is hard to predict how thermal stability will
affect the uncertainty. It may vary over a long time period and appear constant
over a shorter period of time. It is conservative (high estimate of uncertainty) to
consider it systematic.
Following Eq. 7.15 :
B P
2
2
 0.2
  0.2
 
* 1200   
 
* 1200  
  100
 
 100
1/ 2
 3.39 kPa
7.43
1.5
* 340)  5.10 rpm
100
 0.07 kW
BQ  (
BW
 5.10  2  339  2  0.07  2 
 
 
 
 
  340   700   5  
B
1/ 2
 0.021
B  0.793 * 0.021  0.017
Total uncertainty of the pump efficiency:
(d)

2
w   B  P


2 1/ 2

1/ 2
 0.017 2  0.060 2
 0.062 With 95% confidence level
  0.793  0.062 With 95% confidence level
7.44
7.51
(a)   Q  P/
in2
1 hp
ft 3
1min
lbf
*
* 100 2 * 144 2 *
550 ft.lbf/sec
min 60 sec
in
ft

6.5 hp
 0.806 or 80.6%
12
b)
Parameter
Mean Value
Stand. Deviation
Q
12.0 cubic
feet/min
100.0 psi
6.50 hp
0.2
Deg. Freedom
( )
15-1
1.5
0.20
15-1
15-1
P

 SQ P  2  SP  2  S  2 
  
 
  W  


Q
P



 W  

S
 0.2  2  1.5  2  0.20  2 
 
 
 
 
 100 
 6.50  
 12 
 0.038
1/ 2
Refer to Eqs. 7.6 and 7.15
1/ 2
S  0.031
c) Random uncertaint y : P  tS
7.45
To find t, we need to calculate degrees of freedom for  (Eq. 7.28)
S 
2 2
 

2
2
2
2
  
1   
 
 

S
S


P  
W  
   W
 
 
 P
2
1  
1
 
S

Q   
 Q  Q
 P
 
2
S 
4


4
1  SQ

 Q  Q




4

1
 
 P

4
 S P  4 
1  SW  
  
 


   W  
 P  
 0.031 


 0.806 
 
4
4
4
1  0.2 
1  1.5 
1  0.2 

 

 


14  12 
14  100 
14  6.5 
 30
4
Using table 6.6 for 95% confidence level (

2
 0.025)
t 2
P  t S
 2 * 0.031
 0.062
with 95% confidence level
Systematic uncertainty of the pump efficiency :
 B
  Q
  Q

B
2
2
2
  BP   BW  
  
 
 
  P   W  
1/ 2
(Refer to Eq. 7.6 and 7.15)
There are two uncertainties associated with the pressure transducer. Accuracy
is always treated as systematic. It is hard to predict how thermal stability will
affect the uncertainty. It may vary over a long time period and appear constant
over a shorter period of time. It is conservative (high estimate of uncertainty) to
consider it systematic.
Following Eq. 7.15 :
B P
2
2
 0.2
  0.2
 
 
* 170   
* 170  
  100
 
 100
 0.48 psi
1/ 2
7.46
1.5
* 12  0.18 ft 3 / min
100
B   0.10 kW
BQ 
 0.18  2  0.48  2  0.10  2 
 
 
 
 
  12   100   6.50  
 0.022
B
1/ 2
B  0.018
d)

Total uncertainty of the pump efficiency:
2
w   B  P


2 1/ 2
 0.018 2  0.062 2
 0.064

1/ 2
With 95% confidence level
  0.806  0.064 With 95% confidence level
7.47
7.52
(a)   Q  P/W
3
 340 
3  m 
 * 700 * 10 3 Pa 

*
10



 60 
 sec 

5  10 3 Watts 
 0.793 or 79.3%
(b)
Parameter
Q
P

 S
  Q

 Q
S
Mean Value
Stand. Deviation
340 lit/min
700 Kpa
5 KW
5.6
10
0.15
2
2
2

 S P 
 SW  
  
 
 
 P 
 W  

Deg. Freedom
( )
>30
>30
>30
1/ 2
 5.6  2  10  2  0.15  2 
 
 
 
 
 700 
 5  
 340 
 0.037
Refer to Eq. 7.6 and 7.15
1/ 2
S  0.029
(c)
For large number of measurements (   30 ),
P  2 S
 2 * 0.029
 0.058
With 95% confidence level
Using Table 6.6, for 95% confidence level (

2
 0.025 ),
t  2.052
P  t S
 2.052 * 0.029
 0.060
with 95% confidence level
Systematic uncertaint y of the pump efficiency :
1/ 2
 BQ  2  B  2  B  2 
   P      
 
(Apply Eq. 7.6 and 7.24)
  Q   P     


There are two uncertainties associated with the pressure transducer. Accuracy
is always treated as systematic. It is hard to predict how thermal stability will
B
7.48
affect the uncertainty. It may vary over a long time period and appear constant
over a shorter period of time. It is conservative (high estimate of uncertainty) to
consider it systematic.
Following Eq. 7.15 :
B P
2
2
 0.2

 0.2
 
* 1200  
 
* 1200   

 100
 
 100
1/ 2
 3.39 kPa
1.5
BQ  (
* 340)  5.10 lpm
100
BW  0.07 kW
 5.10  2  339  2  0.07  2 
 
 
 
 
  340   700   5  
B
1/ 2
 0.021
B  0.793 * 0.021  0.017
(d)

Total uncertainty of the pump efficiency:
2
w   B  P


2 1/ 2

1/ 2
 0.017 2  0.058 2
 0.060 With 95% confidence level
  0.793  0.060 With 95% confidence level
7.49
7.53
(a)
1/ 2
 BQ  2  B  2  B  2 
   P      
 
(Apply Eq. 7.6 and 7.24)
  Q   P     


There are two uncertainties associated with the pressure transducer. Accuracy
is always treated as systematic. It is hard to predict how thermal stability will
affect the uncertainty. It may vary over a long time period and appear constant
over a shorter period of time. It is conservative (high estimate of uncertainty) to
consider it systematic.
Following Eq. 7.15 :
B
B P
2
2
 0.2

 0.2
 
 
* 1200   
* 1200  

 100
 
 100
1/ 2
 3.39 kPa
1.5
BQ  (
* 340)  5.10 lpm
100
BW  0.07 kW
 5.10  2  339  2  0.07  2 
 
 
 
 
  340   700   5  
B
1/ 2
 0.021
B  0.82 * 0.021  0.017
Random uncertainty based on direct measurement of the pump:
P  2 S
 2 * 0.01
 0.02
with 95% confidence level
The total uncertainty of the pump efficiency:
(b)

2
w   B  P


2 1/ 2

1/ 2
 0.017 2  0.02 2
 0.026 With 95% confidence level
  0.82  0.03 With 95% confidence level
7.50
7.54 10 engines
  30.8%
S  0.25
DF    10  1  9
t  2.262 (From Table 6.6 @ 95% confidence.)
Random Uncertainty
P  tS / n 1/ 2  (2.262)(0.25) / (101/ 2 )  0.1788
Systematic Uncertainty
Using Systematic Uncertainty from example 7.8 B = 1.04
Uncertainty of Mean
wR  [B 2  P 2 ]1/ 2
 (104
. 2  0.18 2 )1/ 2
 106
.
 efficiency will be 30.8 11%
.
7.51
7.55
17
. * 1000 N
F

 10.67
2
kg
.5 V A .5 * 118
2 m 2
2
.
* 30 (
) *.3 m
sec
m3
C
C
C
C
 [( D . BF )2  ( D B ) 2  ( D BV )2  ( D BA ) 2 ]1/ 2
F

V
A
CD 
BCD
And so for SCD
CD
F
CD

CD
V
CD
A

1
1

 0.00628 N 1
2
2
.5 V A .5 * 118
. * 30 *.3

1
1700
F

 9.04( kg / m 3 ) 1
.
2
.5V A  2 .5 * 30 2 *.3 * 118
. 2

2F 1
2 * 1700
 .711 ( m / sec)1/ 2
. 3 
3
.5 A V
.5 * 118
. *.3 * (30)

F
1700

 35.566 ( m 2 ) 1
2 2
.5 V A
.5 * 118
. * (30 )2 *.3 2
Parameter
Mean Value Systematic Uncertainty
Standard deviation DF()
A(cm2)
3000
50
F(N)
1700
50
50
19
V(m/sec)
30
0.2
0.5
19
(kg/m3)
1.18
0.01
2
2
2
2 1/ 2
BCD  [(.00628 * 50)  (9.04 * .01)  (.711 * 0.2)  (35,566 * .005) ]  0.398
SCD  [(
C D
C
C
C
SDF ) 2  ( D SV ) 2  ( D S  ) 2  ( D S A ) 2 ]1/ 2
F
V

A
 [(.00628 * 50) 2  (.711 * .5) 2  0  0]1/ 2  0.474
 CD 
4
SCD
1 C D
1 C D
1 C D
1 C D
(
SF ) 4 
(
S  )4 
(
SV ) 4 
(
SA )4
 F F
 
  V
 A A
(0.474) 4
 37,for 95%confidence,t  2
1
1
4
4
(.00628 * 50)  0 
(.711 * 0.5)  0
19
19
2
w CD  [BCD
 (tSCD ) 2 ]1/ 2  [(.398) 2  (2 * .475) 2 ]1/ 2

 1.02,C D  10.671.03
Note:The computed value of CD is large for normal definitions of CD. The area
given is probably not the correct frontal area. This does not affect the problem
solution.
7.52
7.56
Parameter
P (kW)
mf (g/min)
HV (kJ/kg)
Mean Value Systematic Uncertainty
48.5
0.2
200
3.0
47,500
1,500
nominal value:  
Standard deviation
0.2
--300
DF()
20-1
--15-1
48.5
P

 0.306
mf HV (47,000)200 / (60 * 1000)
Systematic Uncertainty
B  [( / P  BP )  ( / mf  Bmf )2  ( / HV  BHV )2 ]1/ 2
 / P 
1
 1 / ( 200)(47,500) / (60)(1000) .006316
mf  HV
(kJ / sec) 1
 / mf   P / m 2 f HV  48.5 / (200 2 )(47,500) / 60 2  1000 2  9189
. (kg / sec) 1
 / HV   P / mf HV 2  48.5 / 200(47,500 2 ) / 60  1000  6.45 * 10 6 ( kJ / kg ) 1
B  [(.006316*.2)2  (9189
. * 3 / (1000  60))2  (6.45 * 10 6 (kJ / kg ) 1
B  1078
.
* 10 2
Standard deviation
S  [( /  * S )2  ( / mf * Smf )2  ( /  HV * SHV )2 ]1/ 2
S  [(.006316 *.2)2  (0)  (6.45 * 10 6 * 300)2 ]1/ 2
S  .00231
Degrees of Freedom:
   S4 / [( / . S )4 /    ( / mf Smf )4 /  mf  ( / HV . SHV )4 / HV
  .002314 / [(.006316*.2)4 / 19  0  (6.45 * 10 6 * 300)4 / 14]
   25, Table 6.6 for 95% confidence, t  2.060
Overall Uncertainty
w  [(B2  P2 )1/ 2  (B2  (t S )2 ]1/ 2
 [(1078
.
* 10 2 )2  [(2.060)(.00231)]2 ]1/ 2
w  12
. * 10 2 , 12
. percentage po int s.
7.53
7.57
The output range of the pressure transducer is 0 - 5 volt. The 10 volts input
range of DAS will be required.
Following the same procedure as Example 7.9
Calibration Errors
Systematic errors: Linearity and hysteresis of the transducer
.25 * 2000
B1  .25%FS  
 5 kPa
100
Random errors: The values given are random uncertaintys, assuming that they
have been obtained from a large (>30) number of samples.
.03
* 2000  0.6 kPa
Repeatability 0.03%FS  
100
 0.003%FS / C
Thermal span uncertainty
0.003

* 2000 * 10  0.6 kPa
100
0.6
 0.3 (t  2 from Table 6.6)
S1  Srepeatability 
2
0.6
 0.3
S2  Sthermal 
2
Data Acquisition Errors
The number of possible output state for this bipolar system is: 212  4096, half of
which will be for the positive outputs, or 2048 states. At maximum transducer
output, the A/D output will be 5/10 of the maximum or 1024 bits.
Systematic errors
Linearity B2  
Gain error B3  
Random errors:
2
* 2000  3.91 kPa
1024
2
* 2000  3.91 kPa
1024
Quantization error S3  [
0.5
* 2000] / 2  0.49 kPa
1024
No noise will be considered here .
Combined Errors
n
Bp  [  Bi2 ]1/ 2  [5 2  3.912  3.912 ]1/ 2  7.45 kPa
i 1
n
Sp   [  Si2 ]1/ 2  [0.3 2  0.3 2  0.49 2 ]1/ 2  0.65 kPa
i 1
W p  [Bp2  (tSp )2 ]1/ 2  [7.95 2  (2 * 0.65)2 ]1/ 2  7.6 kPa
So the uncertainty in pressure measurements with this system will be 7.6 kPa.
7.54
7.58
Zeroth Order Analysis
B0  0.5 C
p 0  0.4 C
W 0  ( 0.5 2  0.4 2 )1/ 2  0.64 C
First Order Analysis
P1  tS  2 * 2.0  4.0 C
Nth Order Analysis
B0  0.5
BL  2.5 C (Loading error )
P1  4.0 C
BN  (B02  BL2 )1/ 2  ( 0.5 2  2.5 2 )1/ 2  2.55 C
Wn  (BN2  P12 )1/ 2  (2.55 2  4.0 2 )1/ 2  4.7 C
7.59
A single-measurement test consists of the designing and applying an experiment
with a single measurand whereas the multiple-measurement test will measure
numerous variables. The single-measurement test will have a simpler task for
data collection since only a single measurand requires to be quantified, and in
many non-automated data collection system would prove to be a simple for the
experimenter. However, to be both precise and accurate, the other variables
involved needs to be controlled. The multiple-measurement test would allow for
an expanded view of an experiment, as in multiple data of different variables are
collected to ensure reliability.
An example of a single-measurement test would be to measure the height the
level of fluid in a Venturi tube. An example of the multiple-measurement test
would to measure rpm of an air compressor shaft and the energy consumption of
the air compressor determine its efficiency.
7.55
CHAPTER 8
8.1 We first find the axial stress:
  F / A  1000 / (1.0 1 / 16 )  16 ,000 psi
Rearranging Eq. 8.1:
E   / a  16000 / (1143  10 6 )  1.40  107 psi
Rearranging Eq. 8.6:
   t / a  ( 286 ) / 1143  0 .25
8.2 We assume that the bridge is initially balanced.
Vs = 2 V
Vo = 12.5 mV
S = 2.09
R2 = 120 
R3i = 120 
From Eq. 8.22, we find that the strain is:
  Vo ( R2  R3 i )2 / (VsSR2R3 i )
 (12 .5  10 3 )(120  120 )2 / ( 2  2 .09  120  120 )
 11960 strain
8.3 We assume that the bridge is initially balanced.
Vs = 3 V
a = 2500 strain
S = 2.00
R2 = 120 
R3i = 120 
From Eq. 8.22, we find that the strain is:

 Vo (R2  R3i )2 / ( VsSR2R3i )
2500  10 6
 V0 (120  120 )2 / ( 3  2. 00  120  120 )
Vo  3. 75 mV
8.1
8.4 Eq. 8.23 can be used to calculate  R1=  R3 and then micro strain can be
calculated:
Vs R 2 R3
R
R1 R 2 R 4
Vo 
( 3 
)
Eq 8.23


2
R2  R3  R3 R1 R2 R 4
R1  R 3 and R1  R 3
Vo 
(Assumed)
2R 2 R 3V3
(R 2  R 3 ) 2
V R  R3 
R 3 1
 10 6  0 2
R3 S
Vs 2R2 R3 .S
2
 microstrain  10 6 *
20  10 3 (120  120) 2
*
5
2 * 120 2 * 2.10
 3,809 microstrain
Alternatively, we could use Eq. 8.22 and divide the result by 2 since the output is twice
what it would be for a given strain with only one gage.
 10 6 *
8.5 Error due to temperature change:
0.5
* R * ΔT
ΔR 
100
0.5

* 120 * ( 40)
100
 -2.4 Ω
Output due to temperatur e change in R1 and R 3 :
VsR 2R 3  ΔR 3 ΔR 1 ΔR 2 ΔR 4




R 2  R 3 2  R 3 R1 R 2 R 4
R1  R 3 and R1  ΔR 3 (Assumed)
Vo 
V0 




Eq 8.23
2R 2 ΔR 3 Vs
R 2  R 3 2
2 * 120 * (- 2.4) * 5
120  120 2
 0.050 V  50 mV
The error can be eliminated by exposing unstrained resistances to an equivalent
(40 0C ) temperature
8.2
8.6 Use of the lead wires will initially cause the bridge to be unbalanced when there is
no strain. This is not the error we are considering since the initial output is subtracted
from the final, strained value. See the discussion at the top of page 229. So what we
want is the error in the change in voltage output due to the strain.
The apparent initial gage resistance will increase from 120  to 124  when the lead
wires are included (2 for each wire. This will affect gages R1 and R3.
For this problem, it is best to use Eq. 8.19 since this equation does not have the
assumption of initial balance:
Vo  Vs
R 3 R1  R 4 R
(R1  R 3 )(R1  R 4 )
In our case, R1 = R3, R1 = R3 = R and are the active gages. R is the change in
resistance due to the strain, not due to the lead wires. R2 = R4 = 120 are the fixed
resistors. For the case of no lead wires, the output, called Vo1 is:
R 3 R1  R 4 R 2
(120  R1 ) 2  120 2
120 2  2R1R1  R12  120 2
 Vs
Vo1  Vs
 Vs
(R1  R3  2R )(R1  R 4 )
(240  2R )(240 )
(240  2R )(240 )
Since R is small, R2 can be neglected compared to R. Furthermore, in the
denominator R is small compared to R, and the result is:
Vo1  Vs
2R1R1
(240)
2
For the case with the lead wires, the initial resistance of gages 1 and 3 is 124. Since
we are going to subtract out the initial strain, we need to compute the output twice –
once with gages strained and once without and compute the difference.
Vo 2  Vs
(124  R1 ) 2  120 2
124 2  120 2
2R  124
 Vs
 Vs
(120  124)(120  124)
(244)(244)
244 2
Taking the ratio:
124
Vo 2 244 2

 0.99973
120
Vo 1
240 2
If the voltage change without the lead wires is 20 mV, with the lead wires, the output will
be 19.99 mV, a negligible change.
Since the strain is proportional to the output voltage, the strain will appear to be only
0.99973 times as large as it would if there were no lead wire effect – again a negligible
change.
8.3
8.7 We can use Eq. 8.21 to calculate the output voltage due to resistance change of
one of the legs
V s R 1 ΔR 3
V0 
R 2  R 3 R 1  R 4 

3 * 120 * 120 - 119.11 
120  119.11 120  120
 5.58  10
-3

V  5.58 mV
This is with no output loading effect. For the Wheatstone bridge
with a 50 k across the output, the circuit looks like the sketch
and we need to find the current through the 50 k resistor (R5).
This topic is covered in introductory circuits courses and is best
done with a computer program. For this case, the current
through the 50 k can be computed to be 1.114x10-7 A and the voltage drop across the
resistor is IR = 1.114x10-7 x 50000 = 5.57 mV, only slightly less than without the loading
resistor. It should be noted that to the power supply, the bridge presents a resistance of
120  which is very small compared to 50 k and so we would not expect a large
loading effect. In practical bridge circuits, the output device will have an input
impedance of 1 M or more.
8.8 We can use Eq. 8.21 to calculate the output voltage due to resistance change of
one of the legs
V s R 1 ΔR 3
V0 
R 2  R 3 R 1  R 4 

5 * 120 * 120 - 120.20 
120  1 20 . 20 120  120 
 -2.08  10
-3
V  - 2.08 mV
This is with no output loading effect. For the Wheatstone bridge
with a 25 k across the output, the circuit looks like the sketch
and we need to find the current through the 50 k resistor (R5).
This topic is covered in introductory circuits courses and is best
done with a computer program. For this case, the current
through the 50 k can be computed to be 4.1532 x10-8 A and the voltage drop across
the resistor is IR = 4.1532 x10-8 x 50000 = -2.0766 mV, only slightly less than without
the loading resistor. It should be noted that to the power supply, the bridge presents a
resistance of 120  which is very small compared to 50 k and so we would not expect
a large loading effect. In practical bridge circuits, the output device will have an input
impedance of 1 M or more.
8.4
8.9 Considering the definition of gage factor (Eq. 8.10)
1 ΔR
 R
As a result,
S

S2
R1

S1
R2
120


 S2  2 * 

 120  10 
 1.85
8.10 Considering the definition of gage factor (Eq. 8.10)
1 ΔR
 R
As a result,
S

S2
R1

S1
R2
 120

 S2  2 * 

120

5


 1.92
8.11 L = 0.3 m
A = 1 mm20 mm = 0.00002 m2
E = 200 GPa
 = 0.27
Fa = 1500 N
R1 = R3 = R2 = R4 =120 
Vs = 2.5 V
S = 2.10
The axial stress is:
a  Fa / A  15000 / 0 .00002  7 .5  10 8 Pa
Rearranging Eq. 8.1:
a  a / E  7 .5  108 / 200  10 9  0 .00375
Rearranging Eq. 8.8:
dR / R  aS  0 .00375  2 .10  0 .007875
Using Eq. 8.23:
V s R 2 R 3  R 3  R 1  R 2  R 4 
V0 





R1
R2
R4 
(R 2  R 3 ) 2  R 3
2.5  120 2
0.007875  0.007875 
(120  120 ) 2
 0.00984V  9.84mV

8.5
8.12
(a) Using Eq. 8.23, we can obtain V0 and then Eq. 8.23 for strain calculation. The
strain of gages will have an opposite sign in upper and lower and upper parts
V0 

Vs R 2 R 3
ΔR 3 ΔR 2 ΔR 1 ΔR 4



(
)
2
R2
R1
R4
R 3  R 2  R 3
Vs R 2 4ΔΔ
V ΔR
 s
.
2
4R
R
R
Considerin g that R 1  R 2  R 3  R 4  120 Ω
ΔR 1  ΔR 3   ΔR 2   ΔR 4
Using Eq. 8.10,
ΔR/R  S 
Combining the above two equations :

V0
S. VS
60  10 -3
 10 6
2.00 * 10
 3,000 μstrain (both upper and lower parts, but with opposite signs)

(b)
Temperature changes will not affect the results, because they will cancel each other out (Eq. 8.23)
8.6
8.13
(a) Using Eq. 8.23, we can obtain V0 and then Eq. 8.23 for strain calculation. The
strain of gages will have an opposite sign in upper and lower and upper parts
V0 

Vs R 2 R 3
R 3  R 2 
2
(
ΔR 3 ΔR 2 ΔR 1 ΔR 4



)
R3
R2
R1
R4
V s R 2 4ΔΔ
V ΔR
 s
.
2
R
R
4R
Considerin g that R 1  R 2  R 3  R 4  120 Ω
ΔR 1  ΔR
3
  ΔR
2
  ΔR
4
Using Eq. 8.10,
ΔR/R  S 
Combining the above two equations :

V0
S. VS
40  10 -3
 10 6
2.05 * 5
 3,902 μstrain (both upper and lower parts, but with opposite signs)

(b)
Temperature changes will not affect the results, because they will cancel each other out (Eq. 8.23)
(c)
The same configuration cannot be used for measurement of strain due to axial forces, because, all
gases will have the same output and cancel each other out
As sketch of the gage is shown below :
8.7
8.14 The bridge should be arranged as shown.
Ra
RC
Rd
Rb
Using Eq. 8.23, the operation can easily be
shown.
Vo 
Vs R 2 R 3  R 3 R1 R 2 R 4




R 2  R 3 2  R 3 R1 R 2 R 4



If only one active strain gage is used, (assume R 3 )
Vo 
Vs R 2 R 3 R 3
R 2  R 3 2 R 3
If all four gage are used as shown, assuming :
R 1  R 2  R 3  R 4 and R1  R 2  R 3  R 4
Vo 
4R 2 R 3
R 2  R 3 
2
4
R 3
R3
Consequently, the output will be four times as the case of a single gage usage.
8.8
8.15 L = 0.3 m
A = 1 mm20 mm = 0.00002 m2
E = 200 GPa
 = 0.27
Fa = 1500 N
R1 = R3 = R2 = R4 =120 
Vs = 2.5 V
S = 2.10
The axial stress is:
a  Fa / A  15000 / 0 .00002  7 .5  10 8 Pa
Rearranging Eq. 8.1:
a  a / E  7 .5  108 / 200  10 9  0 .00375
From Eq. 8.6:
t   a  0 .27  0 .00375  0 .001013
Rearranging Eq. 8.8 for gages 1 and 3:
dR / R  aS  0 .00375  2 .10  0 .007875
For gages 2 and 4:
dR / R  tS  0 .001013  2 .10  0 .002126
Using Eq. 8.23:
  R 3  R 1 R 2 R 4 





R1
R2
R4 
(R 2  R 3 )  R 3
2.5  120 2
0.007875  0.007875  0.002126  0.002126 

(120  120 ) 2
 0.00125V  12.5mV neglects transverse effects on gages 
V0 
Vs R 2 R 3
2
8.9
8.16 Vs = 2V 0.2%
V0 = 12.5 mV 0.3%
S = 2.09 0.5%
R2 = 120  0.1%
R3i = 120  0.15%
We can use Eq. 8.22 to estimate the strain:
a  Vo ( R2  R3 i )2 / (VsSR2R3 i )
a
a
 12 .5  10 3 (120  120 )2 / ( 2  2 .09  120  120 )
 11,960 strain
To obtain the uncertainty we will use Eq. 7.4. To use this, we need the partial
derivatives with respect to each of the independent variables:
R 2  R 3i 2

120  120 ) 2


 0.957V 1
V 0
Vs SR 2 R 3 i
2  2.09  120  120
V R  R 3 i 

12.5  10 3 (120  120 ) 2
 0.00598V 1
 0 22
 2
V s
2  2.09  120  120
Vs SR 2 R 3 i
2
V R  R 3 i 

12.5  10 3 (120  120 ) 2
  0 22

 0.00572
S
Vs S R 2 R 3 i
2  2.09 2  120  120
2
2
V  2R 2  R 3 i  R 2  R 3 i  

 0 


R 2 VS S  R 2 R 3 i
R 22 R 3 i 

2
12.5  10 3  2(120  120 ) 120  120  



2  2.09  120  120
120 2  120 

 0sincebridgeinitiallybalanced 
R 2



0
R3 i R2
wR3i = 1200.0015 = 0.18
wR2 = 1200.001 = 0.12
wS = 2.090.005 = 0.0105
wVs = 20.002 = 0.004 V
wV0 = 12.510-30.003 = 0.0000375 V
8.10
 
w   
w R3 i
 R3 i
2
2
2
2
  

  
  
  
  
w R2    w S   
w Vs   
w V0 
  Vs
  S
  V0

  R 2
2
 [0  0.18   0  0.12   0.00572  0.0105    0.00598  0.004 
2
2
2



1/ 2
2
 0.957  0.0000375    0.00598  0.004   0.957  0.0000375  ]1/ 2
2

2
 0  0  3.6  10 9  0.57  10 9  1.29  10 9

1/ 2
 0.000074  74strain
So:
 = 11,962  74 strain.
note: the uncertainty in the gage factor is the dominant error source.
8.11
2
8.17 Start with Eq. 8.19 and substitute for R2 and R3:
V R  R 3t  R 3s R1  R 4 R 2i  R 2t 
V0  s 3 i
R 2i  R 2t  R 3i  R 3t  R 3s R1  R 4 
Noting that all the  terms are small, they can be neglected in the denominator.
V R R  R 1 R 3t  R1 R 3s  R 4 R 2i  R 4 R 2t 
V0  s 1 3 i
R 2i  R 3i R1  R 4 
Using Eq. 8.20, this becomes:
V0 
Vs R1R 3t  R1R 3s  R 4 R 2t 
R 2i  R 3i R1  R 4 
Since R2i = R3i (same gage type) R2i = R3i (same temperature) then , Eq. 8.20
indicates R1 = R4 and our expression for V0 becomes:
Vs R1R 3s 
V0 
R 2i  R 3i R1  R 4 
which is the same as Eq. 8.21 (with the  term in the denominator already eliminated).
Hence Eq. 8.22 follows.
8.18 S =2.05
R2 = 360 
R3i = 360 
Vs = 3 V
(a) From Figure 8.6 at 300C, apparent = 75 strain and the S variation is +1.25%.
Reaaranging Eq. 8.22:
 V SR 2 R 3 i 
V0   a  s
2 
 (R 2  R 3 i ) 
 3  2.05  1.0125 360  360 
 75  10 6 

360  360 2


V0  0.117mV
(b) If the structure is strained to a value of 500 strain at 300 C, the total strain is 75 +
500 = 575 strain and the gage factor is the same as in part a. Using the above
equation, the voltage output is:
 V SR 2 R 3 i 
V0   a  s
2 
 (R 2  R 3 i ) 
 3  2.05  1.0125 360  360 
 575  10 6 

360  360 2


V0  0.895mV
8.12
8.19 (a) Since the strain gage is connected to the bridge with a pair of 15 meter leads
with a resistance of 0.10 /m, the change in R3 is:
R3 = 2150.10 = 3 
The following information is given:
R2 = R3i = 120 
S = 2.05
Vs = 3.00 V
Also, it is assumed that the bridge is initially balanced so R1 = R4. We will use Eq. 8.21:
V s R1 R3
V0 
R2  R3i  R3 R1  R4 

3  R1  3
120  120  3R1  R1 
V0  0.0185V
The apparent strain, εapparent, on the bridge would then be:
V o ( R 2  R 3i ) 2
 apparent 
V s SR 2 R3i
(0.0185)(120  120) 2
3.00  2.05  120  120
 12,033strain

(b) Using Eq. 8.8, the change in resistance due to 800 strain is:
R = 2.0580010-6120 = 0.20 
Rtotal = 0.20 + 3 = 3.2 
Using the above equation again:
V0 

V s R1 R3
R2  R3i  R3 R1  R4 
3  R1  3.2
120  120  3.2R1  R1 
V0  0.0197V
The change in output due to the strain is then 0.0197 - 0.0185 = 0.0012 V. Better
resolution would have been obtained if the bridge had been initially balanced by
adjusting R2.and changing the scale on the voltmeter.
8.13
8.20 (a) Since the strain gage is connected to the bridge with a pair of 40 foot leads with
a resistance of 0.026 /ft, the change in R3 is:
R3 = 2400.026 = 2.08 
The following information is given:
R2 = R3i = 120 
S = 2.00
Vs = 3.00 V
Also, it is assumed that the bridge is initially balanced so R1 = R4. We will use Eq. 8.21:
V s R 1 R 3
V0 
R 2  R 3i  R 3 R1  R 4 

3  R1  2.08
120  120  2.08 R1  R1 
V0  0.0129V
The apparent strain, εapparent, on the bridge would then be:
V o ( R 2  R 3i ) 2
 apparent 
V s SR 2 R3i
(0.0129)(120  120) 2
3.00  2.00  120  120
 8,600 strain

(b) Using Eq. 8.8, the change in resistance due to 1000 strain is:
R = 2.00100010-6120 = 0.24 
Rtotal = 0.24 + 2.08 = 2.32 
Using the above equation again:
V0 

V s R 1 R 3
R 2  R 3i  R 3 R1  R 4 
3  R1  2.32
120  120  2.32R1  R1 
V0  0.0144V
The change in output due to the strain is then 0.0144 - 0.0129 = 0.0015 V. Better
resolution would have been obtained if the bridge had been initially balanced by
adjusting R2.and changing the scale on the voltmeter.
8.14
8.21 We can evaluate this effect using Equation 8.21. The non-linearity results from the
R3 term in the denominator. For a linear assumption, Equation 8.21 becomes:
Vout,lin 
VsR1R3
(R2  R3 )(R1  R4 )
Dividing this equation into Equation 8.21, we obtain:
Vout,nonlin
=
Vout,lin
( R1 + R 4 )
( R1 + R 4 + R 3 )

120  120
120  120  R 3
 0.95
Solving for R3, the result is 12.6 , or 10% of R3. For a gage factor of 2, this
corresponds to a strain of approximately 50,000 strain (5% elongation). This strain
would beyond the yield point of most metals.
8.22 Potential sources of error:




Temperature variation
Lead wires
Value of strain gage resistance
Value of gage factor
These errors can be minimized by careful temperature compensation, selection of lead
wires with low resistances and careful determination of gage resistances and gage
factors. Normally, gage manufacturers supply accurate values of gage factor and initial
resistance.
Considering Eq. 8.22:
2
V R  R 3 
  0 2
V sR 2R 3 S
8.23
Assuming that the uncertaint ies are in V 0 and V S ,
ε K
V0
Vs
[K 
R 2
 R3 
]
R 2R 3 S
2
By using Eq. 7.6
2
 w
 w V0 
wε
    VS
 [ 
 V

ε
S

 V0 
 [ 0.003
2
 1/2
 ]


2  0.001 2 ]1/2
 3.2  10 - 3
0.032%
8.15
8.24
The linear potentiometer circuit is diagrammed below:
Solving for Vo we get:




1


90
 1
1 



 R 2 5000 
Vo 




1


1000  R 2 
 1
1 



 R 2 5000 
Substituting results in the following table:
xR2
x/L=R2
Vo
0
100
200
300
400
500
600
700
800
900
1000
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0
8.841V
17.442V
25.912V
34.351V
42.857V
51.527V
60.461V
69.767V
79.568V
90.000V
The plot shows the output to be quite non-linear due to loading.
8.16
8.25
When centered the current through R1 is:
I
90
A
R1
 0.094 A
90 V




1

500  
1 
 1



 500 5000 
VA=905000.094 = 43
IR1= (9043)/500 = 0.094A; PWR = I2R = (0.094)2500 = 4.42W
IR2= 43/500 = 0.086A ; PWR = (0.086)2500 = 3.70W
Total PWR = 3.70+4.42 = 8.12W
500 
C
500 
R2
B
Can reduce power by increasing potentiometer resistance. However, output will be
more non-linear.
8.17
5000 
8.26 (a) Let R be the output resistance of the potentiometer, Rmax be the maximum
resistance of the potentiometer (500 ), Ri be the input resistance of the measuring
device, and Vs be the supply voltage. The ideal voltage output (with no loading) will be:
Vo  Vs R / Rmax
With the measuring device, the resistances R and Ri will have to be combined.
Rcomb  1/(1/ R  1/ R i )
The current flow through the loop is then:
I  Vs /(Rmax  R )  Rcomb 
The output voltage into the measuring device is then:
Vo  Vs Rcomb /(Rmax  R )  Rcomb   Vs 1/(1/ R  1/ Ri ) /(Rmax  R )  1/(1/ R  1/ Ri )
R is proportional to , the angle of the potentiometer, R  500 / 300
The systematic error is then:
B  Vs R / Rmax  1/(1/ R  1/ Ri ) /(Rmax  R )  1/(1/ R  1/ Ri )
(b) To determine when the error is a maximum, one could differentiate the above
formula for B. This is rather tedious and it is simpler to simply compute B as a function
of R and examine the results.
B (V)
B vs R
0.04
0.03
0.02
0.01
0
0
200
400
600
R (ohm)
The peak error occurs when R is about 340  which corresponds to  of 204 degrees.
8.18
8.27
Displacement (cm) = x
0.00
1.25
2.50
3.75
5.00
Output (V) = y
0.10
0.65
1.32
1.95
2.70
Plotting the above data gives:
Since we have only five points of data, we can simply use the formula:
N  xi yi   xi yi
where m is the slope of the line
m
2
N  xi2   xi 
c   y i  m  x i  N
where c is the y-intercept.
Substituting,
N 5
 x  12.5
 y  6.72
 x  46.875
 x y  24.925
i
i
2
i
i
m
i
5(24.925)  (12.5)(6.72)
 0.52
5(46.875)  (12.5) 2
c  6.72  0.52(12.5) 5  0.044
The calibration constant for the curve is thus the slope which is 0.52 V/cm.
For the least squares fit straight line, the data points are:
8.19
x
0.00
1.25
2.50
3.75
5.00
y
0.044
0.694
1.344
1.994
2.644
Deviation
0.056
0.044
0.024
0.006
0.056
Therefore, the maximum error is:
0.100.044 = 0.056 V
The mean deviation assuming it to be the average of the deviation values is 0.056 V.
8.20
8.28
Displacement (in.) = x
0.00
0.50
1.00
1.50
2.00
Output (V) = y
0.15
0.63
1.30
1.92
2.65
Plotting the above data gives:
Since we have only five points of data, we can simply use the formula:
N xi y i   xi y i
m
where mistheslope.
2
N  x i2   x i 
c
 y
i

 m x i N
where c is the y-intercept.
Substituting,
N 5
 x  5.00
 y  6.65
 x  7.50
 x y  9.795
i
i
2
i
i
m
i
5(9.795)  (5.00)(6.65)
 1.258
5(7.50)  (5.00) 2
c  6.65  1.258(5.00) 5  0.072
The calibration constant for the curve is thus the slope which is 1.258 V/in.
For the least squares fit straight line, the data points are:
8.21
x
0.00
0.50
1.00
1.50
2.00
y
0.072
0.701
1.330
1.959
2.588
Deviation
0.078
0.071
0.03
0.039
0.062
Therefore, the maximum error is:
0.150.072 = 0.078 V
8.29 A = 1 cm2 = 0.0001 m2
d = 0.2 mm = 0.0002 m
K = 1 for air
0 = 8.8510-12 C/N-m2
(a) Using Eq. 8.26:
C = K0A/d = 1(8.8510-12)0.0001/0.0002 = 4.4310-12 F = 4.43 pF
(b) Differentiating Eq. 8.26 wrt d, we obtain:
dC/dd = -K0A/d2 = -1(8.8510-12)0.0001/0.00022 = -22.1 pF/mm
(c) Consider the area to be LW so C = K0LW/d. Consider that the capacitance is
proportional to the overlapping area. If the plates are moved in the W direction, then W
will decrease. The derivative of C wrt to W is then:
dC/dW = K0L/d = -1(8.8510-12)0.01/0.0002 = 0.443 pF/mm. Since movement will
result in a decrease in W, the actual sensitivity is -0.443 pF/mm.
8.22
8.30
(a)
For a capacitive displacement sensor,
A
C  K 0
d
Where : K  Dielectric coefficient
 0  Permitivit y of Vacuum
A  Area of capacitor plates
d  distance between capacitor plates
For the transducer with constant d and variable A, C is a linear function of the variable A
(b)
Sensitivity of C with respect to d:
 A
C
 - K 02
d
d
So sensitivity of C with respect to d can be increased by reducing d and increasing
K and A
(c)

C
 K 0
d
A
Sensitivity of C with respect to A can be increased by increasing K and reducing d
8.31 A = .005.005 cm2 = 2.510-5 m2
d = 0.1 mm = 0.0001 m
K = 1 for air
0 = 8.8510-12 C/N-m2
(a) Using Eq. 8.26:
C = K0A/d = 1(8.8510-12)2.510-5/0.0001 = 2.2110-12 F = 2.21 pF
(b) Differentiating Eq. 8.26 wrt d, we obtain:
dC/dd = -K0A/d2 = -1(8.8510-12)2.510-5/0.00012 = -22.1 pF/mm
(c) Consider the area to be LW so C = K0LW/d. Consider that the capacitance is
proportional to the overlapping area. If the plates are moved in the W direction, then W
will decrease. The derivative of C wrt to W is then:
dC/dW = K0L/d = -1(8.8510-12)0.005/0.0001 = 0.443 pF/mm.
Since movement will result in a decrease in the overlapping W, the actual sensitivity is 0.443 pF/mm.
8.32
An angular digital encoder having 8 bits in each sector will have 256 sectors (28).
Thus, the angular- resolution will be 2/256 = 0.0245 rad. or  0.0123 rad.
8.23
8.33 Using Eq. 8.29
30
V (30 )  0   a(t )dt
0
using the trapezoidal rule:
30
30  0
0 a(t )dt  2  7 5  25  5  6  1  6  7  7  5  8  5  9  6
 178.3m / s
To get the displacement, we need to use the trapezoidal rule to obtain a(t), then
integrate that again.
t
r(t)
0
0
5
26.25
10
36.8
15
65.4
20
98.2
25
135.6
30
178.3
30
Now, x (t )  x o   V (t )dt
0
using the trapezoidal rule again
30
0  226.25  36.8  65.4  98.2  135.6  178.3
x (30) 
27
x (30)  1934m
8.34
Using Eq. 8.27 :
fD 
2Vcos

For straight line measuremen t (Vsl ) :   0
For actual measuremen t (Vm ) :   10 0
Error in apparent speed, B v  Vmt  Vsl
For the straight line measurement: fD 
For the actual measurement: fD' 
2  120 cos 0

2  120 cos10

'
D
f
 cos10  0.985
fD
Since he assumes that the angle is zero and the speed is proportional to the doppler shift, he
concludes that the speed is 120x0.985 = 118.18 and the error is 1.82 km/hr.
so
8.24
8.35
A standard hexagonal nut will have 6 pulses per revolution ( 6pulses/rev.). If the shaft
on which the nut is attached is turning at 50,000 rev/min, the number of positive pulses
can be calculated below:
6 pulses 50,000rev
1min
Positive pulses 


1rev
1min
60 sec
 5000 pulses / sec
Rearranging the above equation for positive pulses, we find 10 to be the number by
which the number of counts in a second is multiplied in order to produce a result in rpm.
8.36
A standard hexagonal nut will have 6 pulses per revolution ( 6pulses/rev.). If the shaft
on which the nut is attached is turning at 100 rev/sec or 6,000 rev/min, the number of
positive pulses can be calculated below:
6 pulses 6,000rev 1min
Positive pulses 


1rev
1min
60 sec
 600 pulses / sec
Rearranging the above equation for positive pulses, we find 10 to be the number by
which the number of counts in a second is multiplied in order to produce a result in rpm.
8.37 We will get 20 pulses/(4div0.5sec/div) or 10 pulses per second.
TheRPM 
10 pulses
1rev
60 sec


sec
6 pulses
min
RPM  100rpm
8.38 Counts/sec = ( Pulses/ revolution) * (Revolution / sec)
3000
 12 
60
 600 pulses/sec
8.25
8.39
RPM 
1
N*

60
Where N  360/ number of teeth
  period of individual cycles (sec)
60  sec/min
  angle between two consecutiv e teeth
60
Then : RPM 
 360
Assuming that  is measured accurately, using Eq. 7.6
w RPM w 
0.5


360
RPM

12
 0.017 or 1.7 %
8.40
RPM 
1
N*

60
Where N  360/ number of teeth
  period of individual cycles (sec)
60  sec/min
  angle between two consecutive teeth
60
Then : RPM 
 360
Assuming that  is measured accurately, using Eq. 7.6
wRPM w
0.2


360
RPM

6
 0.0033 or 0.33 %
8.41 (a) With a flashing rate of 3000 per minute, one mark will be visible.
(b) For flashing rates of 1000 and 1500 per minute, one mark will be visible. However,
for flashing rates of 6000 and 9000 per minute, we will see 2 and 3 marks , respectively.
8.42 There should only be a single mark on the rotating disk. Make the strobe flash at a
rate higher than the expected rpm. Reduce the flashing rate until only a single stationary
mark is visible. This should be the correct rpm. If the flashing rate is now doubled, there
should be two diametrically opposite marks.
8.26
8.43
cycles (reflections)
1
sec

 60
reflection
sec
min
rev
1
 600 *
 60
24
 1500 RPM
RPM 
8.44 F=100,000 lbf
max = 20,000 psi
E = 9.5E6 psi
 = 0.33
D0 = 6 in
max = F/A; 20000 = 100000/((62 - Di2)/4)
Di = 5.44 in
From Eq. 8.1,

20,000
a  
 2105 .0 strain
E 9.5  10 6
From Eq. 8.4,
T  0.33( 2105 .0 )  694 .7 strain
From Eq. 8.22, for each longitudinal gauge
Vout   aVs SR1R 4 /(R1  R 4 ) 2
 (2105.0  10  6 )(10)(2.1)(350  350) /(350  350) 2
 11.05mV
Similarly for the transverse gages, Vout = -3.647mV
Noting that the transverse gages will change the bridge output in the direction opposite
from the longitudinal gages, the total output is :
2(11.05)+2(3.647) = 29.4mV
8.27
8.45 F=400,000 N
max = 100,000 kPa
E = 6.55E7 kPa
 = 0.33
D0 = 15 cm
max = F/A; 100,000 = 400,000/((152 - Di2)/4)
Di = 14.8 cm
From Eq. 8.1,
a 

E

400,000
 6106.9 
strain
6.55  10 7
From Eq. 8.4,
 T  0.33(6106 .9)  2015 .3 
strain
From Eq. 8.22, for each longitudinal gauge
Vout   aV s SR1 R4 /( R1  R 4 ) 2
 (6106.9  10  6 )(10)(2.1)(400  400) /(400  400) 2
 32.06mV
Similarly for the transverse gages, Vout = -10.58mV
Noting that the transverse gages will change the bridge output in the direction opposite
from the longitudinal gages, the total output is :
2(32.06)+2(10.58) = 85.3mV
8.28
8.46 The solution to 8.23 is similar to 8.22 except that each gage will be altered by a
fixed amount of strain dependent on the temperature. Figure 8.6 indicates that at 200C
(a 180 C increase) the apparent microstrain is about -30 microstrain. For Problem 8.22,
the output was found to be 29.4 mV.
F=100,000 lbf
max = 20,000 psi
E = 9.5E6 psi
 = 0.33
D0 = 6 in
2
2
max = F/A; 20000 = 100000/((6 - Di )/4)
Di = 5.44 in
From Eq. 8.1,

20,000
a  
 2105 .0 strain
E 9.5  10 6
to this we add -30 to get 2075 strain
From Eq. 8.4,
T  0.33( 2105 .0 )  694 .7 strain
to this we add -30 to get -724.7 strain
From Eq. 8.22, for each longitudinal gauge
Vout   aVs SR1R 4 /(R1  R 4 ) 2
 (2075.0  10 6 )(10)(2.1)(350  350) /(350  350) 2
 10.89mV
Similarly for the transverse gages, Vout = -3.804mV
Noting that the transverse gages will change the bridge output in the direction opposite
from the longitudinal gages, the total output is :
2(10.89)+2(3.804) = 29.4mV
The output is unchanged from problem 8.22 because the full bridge is temperature
compensating.
8.29
8.47 The calibration constant can be determined by finding the slope (mV/N) and then
dividing by the supply voltage.
Using least squares method:
N 5
 x i  185.00
 y i  2.02
2
 x i  13,225.00
 x i y i  144.45
m
N  x i y i   x i y i
N  x i   x i 
2
2

5(144.45)  (185.00)(2.02)
 0.0077
5(13,225.00)  (144.45) 2
Thus, the calibration constant is:
0.0077mV / N
 0.00128mV / N  V
6V
Vout  F  (calibrationcons tan t )  V s
1.5  10 3  F  0.00128  10 3  12
F  97.7 N
8.30
8.48 The calibration constant can be determined by finding the slope (mV/lb) and then
dividing by the supply voltage.
Using least squares method:
N 5
 x i  37.00
 y i  1.96
2
 x i  529
 x i y i  27.8
m
N  x i y i   x i y i
N  x i   x i 
2
2

5( 27.8)  (37)(1.96)
 0.0521
5(529)  (37) 2
Thus, the calibration constant is:
0.0521mV / lb
 0.00868mV / lb  V
6V
Vout  F  (calibrationcons tan t )  V s
1.5  10  3  F  0.00868  10  3  12
F  14.4lb
8.49 R = 4 inch
I = 0.0045 in4
E = 29106 psi
F = 300 lbs
We substitute into and solve Eq.8.34:
F
EI
 2 
R3  
4  
300 
 (29  10 6 )(0.0045 )
 2 
( 4) 3   
4  
   0.02189inch
8.31
8.50 L = 0.5m
F = 200N
3490rev 1 min 2rad



 365.472rad / sec
min
60 sec 1rev
By definition:
T  FL  200(0.5)  100 N  m
P  T  100(365.472)  36547.2Watts
 36.55kW
8.51 L = 0.6m
F = 240N
3620rev 1 min 2rad



 379.086rad / sec
min
60 sec 1rev
By definition:
T  FL  240(0.6)  144 N  m
P  T  144(379.086)  54588.4Watts
 54.59kW
8.52 L = 15 in = 1.25 ft
F = 200 lbf
Speed = 1760 rpm
1760rev 1 min 2rad



 184.307 rad / sec
min
60 sec 1rev
By definition:
T  FL  200(1.25)  250 ft  lbf
P  T  250(184.307 )  46076.75 ft  lbf / sec
 83.78hp (or 62.47 kW )
8.32
8.53

W (mechanical energy)   (Torque)   (Radians/sec)
2 . RPM
60
2 . 2000 1
 20 (N.m) 
60
sec
 4,189 Watts
 4.189 kW
Energy dissipated in 1 hour:
E  Power  time  4.189  3600  15.08 MJ
 
8.54
Engine Power   (Torque)   (Radians/s ec)
2 . RPM
 
60
2 . 5,500 1
 600 (N.m) 
60
sec
 345,575 Watts
 345.6 kW
Considering some safety factors, one suitable dynamometer will be:


 7500 
 1 * 100  36%
7500 RPM Safety factor  
 5500 


750 N. m


 750 
 Safety factor   600  1 * 100  25%




450 KW


 450 
Safety factor   345  1 * 100  30%




Note: The specifications for the dynamometer have been chosen quite arbitrarily, to show that
we always need to allow for some safety factor, beyond the measurement need. Additionally,
what is chosen is just a guide to the selection of equipment that eventually will be chosen from
the available equipment in the market.
8.33
8.55
Engine Power   (Torque)   (Radians/sec)
2 . RPM
60
2 . 5,000 1
1
 450 (ft.lbf) 

60
sec 550 ft  lbf / hp
 428.399 hp (or 319.586 kW)
 
Considering some safety factors, one suitable dynamometer will be:


 7500 
7500 RPM Safety factor  
 1 * 100  50% 
 5000 


750 ft.lbf


 750 
 Safety factor   450  1 * 100  67% 
450 KW


 450 
Safety factor   320  1 * 100  41% 
Note: The specifications for the dynamometer have been chosen quite arbitrarily, to show that
we always need to allow for some safety factor, beyond the measurement need. Additionally,
what is chosen is just a guide to the selection of equipment that eventually will be chosen from
the available equipment in the market that will also fit the testing conditions and constraints.
8.56 Because in part of the cycle, the engine absorbs power (coasting down a hill for
example). The eddy current dynamometer (and the hydraulic dynamometer) cannot
power the engine. The electric dynamometer is normally used.
8.34
CHAPTER 9
9.1 P = 150,000 Pa (gage)
m = 2000 kg/m3
g = 9.8 m/s2
From Eq. 9.2 for s <<m,
P
150000

 7.65m
 m g 2000  9.8
This is an impractical height in most cases - a manometer fluid with greater
density is required.
R
9.2 h = 28.6 cm fluid (or 11.3 in fluid)
S = 1.5
From Eq. 9.2 for s <<m,
P  h m g  hS w g

(28.6cmfluid )1.5(1000kg / m 3 )(9.81m / s 2 )
(100cm / m)
 0.421kPa
9.3 h = 45.3 in Hg
S = 13.6
From Eq. 9.2 for s <<m,
P  h m g  hS w g

(45.3 in Hg )13.6(62.43 lbm / ft 3 )(3217
. ft / sec 2 )
(32.17 lbm  ft / lbf  sec 2 )(144 in 2 / ft 2 )(12 in / ft )
 22.3 psid
9.4 P=5.1 psi
Eq. 9.2, P = Rg; R = P/g
144  5.1  32.17
(144in 2 / ft 2 )(Plbf / in 2 )(32.17 ft  lbm / lbf  sec 2 )
R in Hg 

 10.4 in Hg
3
2
13.6  62.43  32.17  12
(  Hg lbm / ft )(32.17 ft / sec )(12in / ft)
R ft H2O 
(144in 2 / ft 2 )(Plbf / in 2 )(32.17 ft  lbm / lbf  sec 2 ) 144  5.1  32.17
. ft H 2O

 118
62.43  32.17
(  w lbm / ft 3 )(32.17 ft / sec 2 )
9.5 P = 65 kPa
Pmm Hg 
Pm H2O 
P
 Hg g

65000  1000
 487.7 mm Hg
13.6  1000  9.8
P
65000

 6.63 m H2O
 w g 1000  9.8
9.1
9.6 (a)
To convert column of liquid to pressure, we use the relationship:
[Where P is pressure,  is the density   H2O  S ]
P   gh


and g is gravitatio nal accelerati on
P (Pa )  h ( mmHg ) *
1m
m
kg
* 9.81
* 13.6 * 1000 3
2
1000 mm
sec
m
 133.42 h
So, to convert mm Hg to Pa we need multiply it by 133.42 (Pa/mmHg)
(b)
P ( psi )  h ( in H 2O )
1 ft
ft
lbm
* 32.17
* 62.42 3 *
2
12 in
sec
ft
1
* (1 ft 2 / 144 inches 2 )
ft. lb. m
32.17
lb f. sec 2
 0.03612 h (in H 2O )
So, to convert in H 2O to psi, we need to multiply it by 0.036 (Psi/in H 2O ).
p
N
 Pa , h 
2
g
m
Pa
h

kg
m 

1000 3 * 9.81

m
sec 2 

h(m)  1.019 * 10 - 4 * Pa
(c)
So, to convert Pa to m H 2O, we need to multiply it by 1.019 * 10 - 4 (m H 2O / Pa).
9.7 R = 130 mm
S = 13.6
From Eq. 9.2 for s <<m,
P  h m g  hS w g
 0.13  13.6  1000  9.81  17.34kPa
Since the sensing line is connected to the column port, this pressure will be
negative (below atmospheric) and the pressure should be stated as -17.34 kPa.
9.8 R = 25 in
S = 13.6
From Eq. 9.2 for s <<m,
P  h m g  hS w g
 25  13.6  62.43  32.17 /(32.17  1728)  12.28 psi
Since the sensing line is connected to the column port, this pressure will be
negative (below atmospheric) and the pressure should be stated as -12.28 psi.
9.2
9.9 The applied pressure is (Eq. 9.2):
 P   h  g  7 . 5 cm 
kg
m
1 kPa
1 m
 1000
 9 . 81

 0 . 736 kPa
3
2
1000 Pa
100 cm
m
sec
This pressure is applied to the inclined manometer. Using Eq.9.3:
P  R sin g
0.736kPa  R  sin7.5o  800
kg
m
 9.81 2
3
m
sec
R  0.72m
9.10 The applied pressure is (Eq. 9.2):
P  hg  3in /(12in / ft )(62.4lbm / ft 3 ) /(32.17lbm  ft / lbf  sec 2 )32.17 ft / sec 2  15.6lbf / ft 2
This pressure is applied to the inclined manometer. Using Eq.9.3:
P  R sin g
15.6(lbf / ft 2 )  Rsin7 o  50.0(lbm / ft 3 )32.17(ft / sec 2 ) / 32.17(lbm  ft / lbf  sec 2 )
R  2.56 ft
9.11 From Eq. 9.3: P  R sing
For this linear device, the sensitivity is constant:
R
1
1


 0. 0255 cm / Pa
o
P sin g sin 30  0. 8  1000  9. 81
0.5 mm in the reading R corresponds to 0.05cm/.0255 cm/Pa = 1.96 Pa
9.12 First convert the pressure value from cm water to Pa:
P  Rg  0.1  1000  9.8  980 Pa
Using Eq. 9.3:
P  R sin g
980  R sin 30  0.8  1000  9.8
R  250 mm
9.13 Using Eq. 9.3
ΔP  R.sinθ. ρg
Where R is the reading on the inclined column
ΔP  0.102(m) * (sin 15) * (0.7 * 1000
kg
m
) * 9.81 (
)
3
m
sec 2
 181.28 Pa
9.3
9.14
Considering Eq. 9.3, systematic error due to error in the angle of inclined
manometer:
P  Rg sin  , where R is the indicated length on the inclined tube.
For the actual and indicated cases we have:
 Pact  R  g sin  15 .5 and  Pind  R  g sin  15 .
 Pact   Pind
sin  15 .5  sin  15 .

Systematic error as a percentage of reading :
 Pact
sin  15 .5
 0 .0325 or 3 .25 %
An alternative to the above solution:
  P  R  g  sin   R  gcos   for small changes in 
P
 cot  
P
0 .5 o 
P
 cot  15 *
 0 . 0326 or 3.26 % of reading
P
180
9.15 The expansion of the aluminum will decrease the reading while the
expansion of the mercury will increase the reading. The temperature change is
15K.
The effect on R of the aluminum scale is:
Ral = -(2310-6 m/m-K)(15K)(760 mm) = -0.2622 mm
The increased temperature will decrease the density of the mercury. The
increase in volume for the mercury is:
V/V = 1.8210-415 = 2.7310-3.
Considering the condition at the the hotter temperature to be state 2 and the
condition at the initial temperature to be state 1:
V2  V1 (1  V / V )  V1 (1  0.00273)
V2 / V1  100273
.
The length of the mercury column is inversely proportional to the density (=m/V)
and hence
R2Hg/R1Hg = 1/2 =mV2/(mV1) = V2/V1 = 1.00273
Consequently, the new length of the mercury column is 760(1.00273) = 762.07
and the increase in length is 2.07 mm. The net change in reading is thus 2.07 0.2622 = 1.81 mm and the new reading is 761.81 mm.
9.4
9.16 Both aluminum and mercury will contract due to reduction in temperature.
The contraction of the aluminum will increase the reading while the contraction of
the mercury will decrease the reading. The temperature change is 5K.
The effect on R of the aluminum scale is:
Ral = -(2310-6 m/m-K)(-5K)(760 mm) = +0.0874 mm
The decreased temperature will increase the density of the mercury. The
decrease in volume for the mercury is:
V/V = -1.8210-45 = -0.9110-3.
Considering the condition at the the colder temperature to be state 2 and the
condition at the initial temperature to be state 1:
V2  V1 (1  V / V )  V1 (1  0.000910)
V2 / V1  0.99909
The length of the mercury column is inversely proportional to the density (=m/V)
and hence
R2Hg/R1Hg = 1/2 =mV2/(mV1) = V2/V1 = 0.99909
Consequently, the new length of the mercury column is 760(0.99909) = 759.308
and the decrease in length is 0.692 mm. The net change in reading is thus -0.692
+ 0.087 = -0.605 mm and the new reading is 759.395 mm.
9.17 V = 2.5 mV
Type R, 0oC reference junction
Interpolating from Table 9.2:
2. 5  2. 400
T  300 
( 350  300 )  310.1 deg. C
2. 896  2. 400
9.18 V = 3.7 mV
Type R, 32o F reference junction
Interpolating from Table 9.2:
3.7  3.407
(450  400)  427.85deg .C (or802.13F )
T  400 
3.933  3.407
9.19 V = 4.005 mV
Type S with ice reference
4. 005  3. 743
T  450 
( 500  450 )  476. 7 deg. C
4. 234  3. 743
9.20 Using Table 9.2 for K-type thermo-couple,
a)
b)
At T1  492 0 C
At 17.51 mV
V Output  20.30 mV
T  426.3
0
C
9.5
(By interpolat ion)
(By interpolat ion)
9.21 For the Fig. P9.16,
VT1  V100  VT1  Vo  - (V100  V0 ), where VT1  V0  and (V100  V0 ) can be obtained
from Table 9.2, V100 - V0  5.268
T1 (  C )
VT1  V0 (mV )
VT1  V100  DVM (mV )
0
20
40
60
80
100
120
140
160
180
200
0.000
1.019
2.058
3.115
4.186
5.268
6.359
7.457
8.560
9.667
10.777
-5.268
-4.249
-3.210
-2.153
-1.082
0.000
1.091
2.189
3.292
4.399
5.509
9.22 V = 30C
Type E with 30 C reference junction.
To use 0C ref junction tables, first find the voltage corresponding to 30C.
Interpolate in Table 9.2:
30  20
( 2. 419  1.192)  1. 806 mV
40  20
If we add this to the output for the 30C reference, we can use the 0C reference
tables. The 0C output is then 37.0 + 1.806 = 38.806 mV. Interpolating in Table
9.2, the temperature is:
38. 806  36. 999
T  500 
( 600  500 )  522. 3 deg. C
45. 085  36. 999
V30  1.192 
9.6
9.23 V = 20C
Type R with 20 C reference junction.
To use 0C ref junction tables, first find the voltage corresponding to 20C, which
corresponds to 0.111 mV in Table 9.2.
If we add this to the output for the 20C reference, we can use the 0C reference
tables. The 0C output is then 4.20 + 0.1111 = 4.211 mV. Interpolating in Table
9.2, the temperature is:
T  450 
4.211  3.933
(500  450)  462.2deg .C
4.471  3.933
9.7
9.24 The differential output is 22 mV. Using Table 9.2, the output for type J at 4C
is:
1.0194/20 = 0.204 mV
The output for the hot junction with a 0C reference would then be:
22.000 + 0.204 = 22.204 mV.
Interpolating in Table 9.2, the temperature is:
T = 400+50(22.204-21.846)/(24.607-21.846)
T = 406.5 C
9.25 The differential output is 24 mV. Using Table 9.2, the output for type K at 5C
is:
0.7985/20 = 0.200 mV
The output for the hot junction with a 0C reference would then be:
24.000 + 0.200 = 24.200 mV.
Interpolating in Table 9.2, the temperature is:
T = 500+100(24.200-20.640)/(24.902-20.640)
T = 583.5 C
9.26 Interpolating in Table 9.2, the voltage output for 1C is 0.0399 mV. This is the
voltage error which will have to be added to all readings. The systematic error on
the temperature depends on the temperature since the sensitivity of the
thermocouple depends on the temperature. The sensitivity of the type K
thermocouple can be obtained by differentiating the data in Table 9.2 to obtain
mV/deg.C. The error is then the voltage error divided by the sensitivity. The
following results are obtained:
T
sensitivity (mV/deg.C)
Error, deg. C
-250
0.010
-4.0
0
0.040
-1.0
500
0.043
-0.9
1300
0.036
-1.1
The uncorrected temperature readings will always be low relative to the correct
values. Type K thermocouples are fairly linear except at low temperatures.
9.8
9.27 Vs = 3 V, V0 = - 0.5V, R2 = 100 
(a) Neglecting the lead resistance, we use Eq. 9.11:
V  2V0
3  2( 0.5)
RRTD  R2 s
 100
 200 
Vs  2V0
3  2( 0.5)
(b) Including the effect of Rlead, we use Eq. 9.12:
4V0
3  2( 0.5)
4V0
V  2V0
RRTD  R2 s
 Rlead
 100
1
 201
3  2( 0.5)
3  2( 0.5)
Vs  2V0
Vs  2V0
(c)The error in resistance due to neglecting the lead resistance is (201-200)/201
= 0.5%.
(d)Interpolating for the temperature values corresponding to 200 and 201 ohms
in Table 9.3, we obtain 261.8 C and 264.5 C respectively. The temperature error
is thus -2.7 C if we neglect lead resistance..
9.28 R = 100 at 0C
Interpolating from Table 9.3, the resistance at 375C is 240.8 .
9.29 R2 = 100
Vs = 2 V
RRTD = 246.1 at 390 C from Table 9.3
(a) Neglecting the lead wire effect, we use Eq. 9.11:
V  2V0
RRTD  R2 s
Vs  2V0
246.1  100
2  2V0
2  2V0
V0  0.422V
(b) Considering the lead wires, we use Eq. 9.12:
4V0
V  2V0
RRTD  R2 s
 Rlead
Vs  2V0
Vs  2V0
2  2 V0
4 V0
246.1  100
2
2  2 V0
2  2 V0
V0  0. 417 V
9.9
9.30 Rlead = 2 
Vs = 2 V
RRTD = 123.6 from Table 9.3
To obtain V0, we use Eq. 9.12:
RRTD  R2
4V0
Vs  2V0
 Rlead
Vs  2V0
Vs  2V0
123.6  100
2  2V0
4V0
2
2  2V0
2  2V0
V0  0.104V
Assuming that the lead resistance is zero, the output is -0.105 V.
9.31 T1 = 20C = 293 K, P1 = 110 kPa (abs)
Assuming argon to be an ideal gas, the ideal gas law gives the following relation:
P1V1 P2 V2

T1
T2
For a constant volume, this becomes:
P1 P2

T1 T2
P2 
PT
1 2
T1
Differentiating wrt. T2:
dP2 P1 110,000


 375.4 Pa / K
dT2 T1
293
9.32 P1 = 110 kPa
T1 = 20 C = 293K
T2 = 120 C =393K
Assuming argon to be an ideal gas:
PV
PV
1 1
 2 2
T1
T2
With V1 = V2:
P2 
PT
110000  393
1 2

 147.5 kPa
T1
293
9.10
9.33 The volume of the bulb will change as the diameter cubed. For a given T,
the diameter is given by:
D2 = D1(1 + T)
So:
V  D23 = D13 (1 + T)3 and V/V0 =(1 + T)3
Then the ideal gas law gives:
P1V1 P2 V2

T1
T2
Consider T1 = 293K, T2 = 393 K, P1 = 110kPa (abs).
Neglecting the bulb expansion, V1 = V2:
PT
110000  393
P2  1 2 
 147.5 kPa
T1
293
Considering the volume change:
V1/V2 = 1/(1+16.510-6100)3 = 0.995
and then:
V PT
110000  393
P2  1 1 2  0.995
 146.8 kPa
V2T1
293
The bulb expansion changes the pressure 0.5% for a 100 C temperature rise.
9.34 Gas pressure thermometers work on the basis of perfect gas law:
PV= mRT
Where, P is absolute pressure, V is volume, m is mass, R is gas constant and T
is absolute temperature.
Assuming m, R and V to remain constant (for a gas pressure thermometer m and
R actually remain constant, but V may slightly vary that can be corrected for),
T2 P2
T

 P2  P1 * 2
T1 P1
T1
 110 *
120  273
 147.5 kpa
20  273
9.11
9.35 Gas pressure thermometers work on the basis of perfect gas law:
PV= mRT
Where, P is absolute pressure, V is volume, m is mass, R is gas constant and T
is absolute temperature.
Assuming m, R and V to remain constant (for a gas pressure thermometer m and
R actually remain constant, but V may slightly vary that can be corrected for),
The temperature scale given in oF, must be converted to absolute value, oR
T2 P2
T

 P2  P1 * 2
T1 P1
T1
 15psia *
459.67  250
 20.48 psia
459.67  60
9.12
9.36 Gas pressure thermometers work on the basis of perfect gas law:
PV
PV=mRT  T 
mR
Where, P is absolute pressure, V is volume, m is mass, R is gas constant and T
is absolute temperature.
Assuming that P is measured accurately, and m and R remain constant,
T 
P
P0 0
V and T 0 
V
mR
mR
where “0” refers to the condition filled. The error in measurement of T due to
T V
P
expansion in V is: T 
, showing that error in
 V or

T
V
mR
temperature, T, is directly proportional to error in volume, V.
Variation in volume is related to variation in length through V/V = 3d/d. Yhis
can easily be shown through a simple differentiation of V = d3/6 for sphere.
Now we need to calculate the error in volume in terms of thermal expansion
coefficient, as it relates length to temperature through, dL/L = dT , where L is
length and dL is differential of L. For the case of sphere, L is equivalent to
diameter, d. So for change in diameter due to increase in temperature,
 (d – diameter)/ (d – diameter)
 (T-T0)
= 12*10-6 (150-20)
= 1.56*10-3
As a result, the error in temperature will be: T/T
= V/V = 3d/d
= 4.68*10-3
So the error in temperature will be:
T = (150+273)* 4.68*10-3
= 2.0 K
9.37 Ts = 800 C = 1073K
Tw = 700 C = 973K
 = 0.3
h = 200 W/m2-K
Using Eq. 9.26:
Tgas - Ts =
=

h
 (Ts 4 - Tw 4 )
0.3
4
4
5.669  10 8 1073   973   36.5deg .C
200
9.13
9.38 Ts = 1000F = 1459.7 R
Tw = 900F = 1359.7R
 = 0.75
h = 150 Btu/Hft2F
Stefan Boltzmann constant = 5.669E-8 W/m2K4 = 0.1712E-8 Btu/hrft2R4
Using Eq. 9.26:

4
4
Tgas - Ts =  (Ts - Tw )
h
0.75
4
4
0.1712  10 8 1459.67   1359.67   9.60deg .R
=
150


9.39 Ts = 500C = 773K
Tw = 350C = 623K
Using Eq. 9.26:

4
4
 (Ts - Tw )
h
0.7
4
4
Tgas = 500 
5.669  10 8 773   623   541deg.C
200
Tgas - Ts =


9.40 Ts = 900 oF = 1,359.67 oR
Tw = 800 oF = 1,259.67 oR
Using Eq. 9.26:

4
4
Tgas - Ts =  (Ts - Tw )
h
0.75
4
4
Tgas = 900 
0.1712  10 8 1359.67   1259.67   908deg .F
150


9.41 Ts = 500 C = 773K
Tw = 460 C = 733K
 = 0.7
h = 200 W/m2-K
Using Eq. 9.26:
T gas - Ts =
=

h
 (Ts 4 - Tw 4 )
0.7
4
4
5.669  10 8  773   733   13.6deg .C
200
9.14
9.42
There are two errors associated with the gas temperature measurement,
conduction error (Eq. 9.22), and radiation error (Eq. 9.26)
Conduction error :
Ts  T
1

T0  T
cosh mL
(Eq. 9.22)
W
hP
W
, P   d  Π(0.001)
h  100 2 , K  20
mK
KA
m K
d 2 Π
A 
 (0.001) 2  0.785  10  6 m 2
4
4
Where m 
100 * 3.14  10 - 3
 141 m -1 , mL  1.41
20 * 0.785  10  6
coshmL  2.17
m
500 - T
1

 0.46
350  T
2.17
Δ
Conduction
 (T gas - Ts )  127.8
Radiation Error : Tgas - T s 
ΔT rad  (Tgas  Ts ) 
 T   627.8
0
o
C
C
ε
σ(T s4 - T w4 ) - - - - - - - - - - Eq. 9.26
h
0.2
4
4
* 5.66  10  5 [ 273  500   273  350  ]
100
 23.4 0 C
Total error in tempera ture measuremen t :
ΔT  ΔT Conduction  ΔT radiation  151.2
o
C
Note: The conduction error is very high due to the low length to diameter ratio of
the thermometer and the assumption that it is solid stainless steel.
9.15
9.43 Using Eq. 9.26, we can calculate the temperature correction due to
radiation:
T  Tgas  Ts 

 Ts4  Tw4 
h
0.7
4
4

* 5.669 * 10 8 [500  273   350  273  ]
200
 41K
Based on Eq. 7.5 and 7.6,
2
2
w T  w  
wh  
 
 
 
T   
 h  
 0.1 2

2
 
  0.15 
 0.1

 0.21
1/ 2
1/ 2
So the uncertainty in temperature correction is :
w T   0.21 * 41
  8.5 K
or T  41  8.5 K
9.44 V = 250 m/s
R = 0.8
Cp = 1200 J/kg-K
Rearranging Eq. 9.27:
Taw  Tstatic  R
2502
V2
 0. 8
 20. 8 deg. C
2cp
2  1200
9.16
9.45
Using Eq. 9.27, we can calculate the air velocity
V2
Tmeasured  Tgas  R
2c p
or
T R V 2

T 2c p
T
V  2c p
,
T
 1%
T
T T

R T
 2 * 1005
298K
J
* 0.01
*
kg. K 0.75
V  89.4 m/sec
9.46 Tdb = 80 F
Twb = 80 F
From Figure B.2, the relative humidity is 100%.
9.47 Tdb = 80 F
Twb = 80 F
From Figure B.2, the humidity ratio is 0.0223 lbwater/lbair. The mass of 1 lb of air
and the moisture would be 1.0223 lbm. Therefore, the moisture represents
0.0223/1.0223 = 0.0218 lbmmoisture/lbmmixture.
9.48 Tdb = 30C
Twb = 15C
From Figure B.1, relative humidity = 17% and the humidity ratio is 4.6 g/kg dry
air.
9.49
Tdb  25 o C, Twb  22 0 C
Reffering to Fig. 3.1, relative humidity, φ  78% and humidity ratio,
  15.6 g/kg dry air
9.17
9.50
From a psychrometric chart, at 75oF dry-bulb and 55oF wet-bulb temperature,
Relative humidity, Ф,= 25%
Humidity ratio, ω,= 0.004686 lbs moisture/lbs of dry air
9.18
CHAPTER 10
10.1
from appendix B-1 for T=25 C,
o
ρ = 996.95kg / m 3
µ = 0.903 × 10 −3 Ns / m 2
Assuming a machined inlet cone , C = 0.995 (Table 10.1)
Q=
CA2
2∆ P / ρ
1 − ( A2 / A1 ) 2
3
0.016m / s =
0.016 =
(0.995)π ( D2 ) 2 / 4
1 − ( D2 / 0.15) 4
2(30,000) / (996.95)
6.063D2 2
1 − (1,975.31D24 )
D2 = 0.051m
To check if the correct value of discharge coefficient (C) was used,
D
0.051
β = 2=
= 0.34
D1 0.15
ρ1 D1
V
Q
0.016
=
= 0.91m / sec
µ
A1 π
2
(0.15)
4
996.95 * 0.91* 0.15
Re =
= 149,882.1
0.903 × 10 −3
Re =
,V =
This value of Re is below the given value in Table 10.1. so the uncertianty in the value
.
of C is greater than ± 10%.
10.1
10.2
ρ = 62.17lbm / ft 3
from appendix B-2 for T=80F,
µ = 2.08lbm / hr .ft
Assuming a machined inlet cone , C = 0.995 (Table 10.1)
Q=
CA2
1 − ( A2 / A1 ) 2
0.6cfs =
0.6 =
2∆ P / ρ
(0.995)π ( D2 / 12) 2 / 4
1 − ( D2 / 6.065) 4
0.14794 D2
2(5 * 144) / (62.34 / 32.17)
2
1 − (7.391 * 10 − 4 D24
D2 = 2.01in
To check if the correct value of discharge coefficient (C) was used,
D
2.02
β = 2=
= 0.33
D1 6.065
ρ1 D1
V
Q
0. 6
=
= 3 ft / sec
π
µ
A1
2
(6.065 / 12)
4
62.17 * 3 * 6.065 / 12
= 162,722
Re =
2.08 / 3600
Re =
,V =
This value of Re is slightly below the given value in Table 10.1. so the uncertainty in
.
the value of C may be greater than ± 10%.
10.2
10.3
From Table 10.1, C=0.995
assuming T=20C,
ρ = 998.2 kg / m 3
β=
D2
= 0.57
D1
* 10 −3 N. s / m 2
.
µ = 1005
Q=
Q=
CA2
1 − ( A2 / A1 )2
2∆ P / ρ
(0.995)π (0.03)2 / 4
1 − (3 / 5.25)4
2(14,000) / 998.2
Q = 0.00394 m 3 / s
To check if we chose the correct value of discharge ceofficient (C),
0.00394
Q
. m / sec,
= 182
=
V=
A1  π 5.25 2 


 4 100 2 
Re =
ρVD 998.2 * 182
. * 5.25 / 100
= 94885
=
µ
.
* 10 −3
1005
The value Re is below the range of values in Table 10.1 so the value of C is more
uncertain than ±1%. This venturi should be calibrated for this application.
10.4
We will follow the same procedure as Example 10.2,
w a = [(
∂Q
∂Q
wC ) 2 + (
w ) 2 ]1/ 2
∂∆P ∆P
∂C
or according to Eq. 7.6
wa
w
w
= [( C ) 2 + ( ∆P ) 2 ]1/ 2
Q
C
∆P
= [(.01) 2 +.25(.015) 2 ]1/ 2 = 0.0125
.
The uncertainty in the flow rate will be 1.25% ≈ 12%
10.3
10.5
We need to calculate the mass flow rate and then volumetric flow rate in SCMM. Since
the flowing fluid is compressible, we need to compute the expansion factor, Y using Eq.
10.9:
1/ 2
 2 / γ  γ 1 − r ( γ −1) / γ  1 − β 4 



Y = r 
 γ − 1  1 − r 1 − β 4 r 2 / γ 

We have air so γ = 1.4 and ρ = 1.2. Assuming the upstream pressure is absolute, r =
690/700 = 0.99 and β = 7.5/10 = 0.75. The resulting Y is 0.991.
P
CYA2
700,000
2 ρ1 ∆P ρ1 = 1 =
m =
= 8.179kg / m 3
2 1/ 2
RT1 287.058 * 298.15
[1 − ( A2 / A1 ) ]
0.995 * 0.991 * 0.004418m 2
7.5
[1 − ( ) 4 ]1 / 2
10

m = 2.131kg / sec
= 127.86kg / min
m
Q( SCMM ) =
m =
2 * 8.179 * 10,000
ρ s tan d
127.86
= 106.55m 3 / min( s tan dard )
1.2
Calculating Re:
From Table B.3, at 25oC~300K, µ = 1.846*10-5 Ns / m 2
m
D
ρ
VD
Re =
= A
=
µ
=
µ
(2.131kg/sec)(0.1m)
6
2
2 = 1.47 × 10
−5
(π * 0.1 / 4m )(1.846 *10 Ns / m )
2
This is within the suggested range of 2 * 10 5 < Re <2 * 10 6 . in Table 10.1.
10.4
10.6
We need to calculate the mass flow rate and then volumetric flow rate in SCFM. Since
the flowing fluid is compressible, we need to compute the expansion factor, Y using Eq.
10.9:
1/ 2

 γ   1 − r (γ −1)/γ   1 − β 4  
Y = r 2 / γ 


4 2/γ  
γ
r
−
−
1
1


r 
−
1
β




We have air so γ = 1.4 and ρ = 0.075. Assuming the upstream pressure is absolute, r =
99/100 = 0.99 and β = 3/4 = 0.75. The resulting Y is 0.991.
CYA2
P
100 * 144
= 0.509lbm / ft 3
m =
2 ρ1 ∆P ρ = 1 =
2 1/ 2
RT1 53.34 * 530.30
[1 − ( A2 / A1 ) ]
m lbm / sec
0.995 * 0.991 * 0.049 ft 2
=
3
32.17lbm − ft / lbf − sec 2
[1 − ( ) 4 ]1 / 2
4

m = 4.020lbm / sec
= 241.2lbm / min
m
Q( SCFM ) =
2 * 0.509lbm / ft 3 * (1 * 144)lbf / ft 2
32.17 lbm − ft / lbf − sec 2
ρ s tan d
241.2
= 3216ft 3 / min( s tan dard )
0.075
Calculating Re:
From Table B.4, at 70F, µ = 0.04391 lbm/hr-ft

m
D
ρVD A
Re =
=
=
µ
µ
4.05( lbm / sec) * [4 / 12](ft)
=
[π / 4][
4 2 2
] (ft ) * 0.04391( lbm / hr − ft ) / 3600 (sec/ hr )
12
. × 10 6
= 127
This is within the suggested range of 2 * 10 5 < Re <2 * 10 6 . in Table 10.1.
10.5
10.7
Critical Flow Nozzle
d t = 5mm⇒ A = 1.9635 * 10 −5 m 2
 , and Q
Find m
P = 7atm( gage)
T = 20C
m =
A2 P0 γ
2 (γ +1) /(γ −1) 1 / 2
[
(
)
]
T01 / 2 R γ + 1
(1.9635 * 10 −5 )(1 + 7)(`101.325) * 10 3 1.4
2 (1.4+1) /(1.4−1) 1 / 2
[
(
)
]
1/ 2
287 1.4 + 1
(20 + 273)
m = 0.038kg / sec
m * 60
Q=
= 1.9SCMM
m =
ρ std
Max pressure downstream is Pcrit which for γ = 1.4 can be evaluated at 0.528 from Eq.
10.11.
Pcrit Pcrit
=
= .528,
P0
8
Pcrit = 4.2atm
10.8
Critical Flow Nozzle
 , and Q
Find m
d t = 0.3in⇒ A = 4.911 * 10 −4 ft 2
P = 7atm( gage) = 8atm(abs ) = 16,929.7lb / ft 2
T = 70 0 F = 529.67 o R
m =
A2 P0 γ
2 (γ +1) /(γ −1) 1 / 2
[
(
)
]
T01 / 2 R γ + 1
(4.911 * 10 − 4 )(1 + 7)(2,116.210)(32.17)
2 (1.4+1) /(1.4−1) 1 / 2
1.4
(
)
]
[
1/ 2
(53.3431)(32.17) 1.4 + 1
(70 + 459.67)
m = 0.1921lb / sec
m * 60
= 153.7 SCFM
Q=
m =
ρ
std
Max pressure downstream is Pcrit which for γ = 1.4 can be evaluated at 0.528 from Eq.
10.11.
Pcrit Pcrit
=
= .528,
P0
8
Pcrit = 4.2atm
10.6
10.9
Critical Flow Nozzle
D = 2mm
A = 3.1416 * 10 −6 m 2
Pcrit = 1atm
T = 20C
C = 1 (smalldiameter maymake C actuallymuchlower)
a) - We should calculate the line pressure (stagnation) for Pcrit=1atm
−γ
−1.405
2.405 .405
γ + 1 γ −1
Pcrit / P0 = [
] =[
]
= .527, For Pcrit = 1atm, P0 = 1.9atm
2
2
AP γ
2 2.405 / .405 1 / 2
2 γ +1 / γ −1 1 / 2 (3.1416 * 10 −6 )(3 * 101.325) 1.405
b) m = 21 / 20 [ (
]
)
(
[
)
] =
1/ 2
R γ +1
4157 2.405
T0
(20 + 273)
R 8314
= 4157 J / kgK
=
M
2
m = 0.593 * 10 −3 kg / s
P
1 * 101,325
= 0.083kg / m 3
ρs = s =
RT 4157 * 293
m
Q=
= 0.00713SCMS = 0.00713 * 60 = 0.428SCMM
R=
ρs
10.7
10.10 (a) For a critical flow nozzle to be accurate,
1
Pcrit
, Pdownstream ≤ Pcrit
=
[(γ + 1) / 2]γ /(γ −1)
P0
In this case,
P0 = 8 atm
T0 = 20 C
γ = 14
.
1
Pcrit
=
= 0.528, Pcrit = 4.23atm
1.4
P0
. + 1 .4
14
[
]
2
For Pchamber >4.23 the critical flow nozzle can not be used for accurate measurment of
the flow rate.
 = Q(SCMM ) * ρ s tan d / 60
(b) m
= 100 * 1.20 / 60 = 2.00kg / sec ρ s tan d =
 =
m
A2 =
=
A2 P0
T0
 T0
m
P0
P
101.325 * 1000
=
= 1.20kg / m 3
RT
287 * 293
γ +1
γ
2
(
) γ −1
R γ + 1)
γ
γ +1
2 γ −1
/
(
)
R γ +1
2.4
2 293
1 .4
2
/
(
) .4
8 * 101,325 287 1.4 + 1
= 0.00104m 2
= 10.4cm 2 ,D2 = 3.65cm
 = Q(SCMM ) * ρ s tan d / 60 = 2.00kg / sec
c −m
10.8
10.11
D1 = 3cm
D2 = 1cm
Pdownstream = 100kPa
P0 = 500kPA
(The static pressure is very close to this value; refer to any compressible flow text.)
Pdownstream Pcrit
<
= .528
P0
P0
m =
A2 P0
T0
R = 287
γ
γ +1
2 γ −1
(
) For air γ = 1.4
R γ +1
J
kgk
(π .012 / 4)500,000 1.4 2 2.4.4
( )
287 2.4
400
= 0.079kg / sec
=
10.9
10.12
D1 = 3cm = 1.18in
D2 = 1cm = 0.3in
Pdownstream = 14.7 psia
P0 = 70 psia
(The static pressure is very close to this value; refer to any compressible flow text.)
Pdownstream Pcrit
<
= .528
P0
P0
m =
2 γ −1
(
) For airγ = 1.4
R γ +1
R = 53.34
π × 0.32
=
4
= 0.099
γ +1
γ
A2 P0
T0
ft.lbf
lbm.R
ft.lbm
lbf
× 32.17
2.4
2
1.4
2
lbf . sec 2
in
×
× ( ) .4
ft.lbf
ft.lbm
2.4
(250 + 459.67) R
× 32.17
53.34
lbm.R
lbf . sec 2
in 2 × 70
lbm
sec
10.10
10.13
Orifice meter
ID = 25.5cm ⇒ A1 = πD 2 / 4 = 5.107 * 10 −2 m 2
Q = 0.090m 3 / s
∆P = 65kPa
V = Q / A = 0.090 / 5.107 * 10 − 2 = 1.7623m / s
Re = ρVD / µ =
Q=
(998.2kg / m 3 )(1.7623m / s )(.255m )
= 446,338
(1.005 * 10 −5 Ns / m 2
CA2 [2(P1 − P2 ) / ρ ]1 / 2
[1 − ( A2 / A1 ) 2 ]1 / 2
Assu min gC = 0.6
0.090 =
0.6 A2 [2(65,000) / 998.2]1 / 2
[1 − ( A2 / 5.107 * 10 − 2 ) 2 ]1 / 2
76.080 A2 = [1 − 383.41A22 ]1 / 2
A2 = 1.273 * 10 − 2 m 2
1.273
= .249
5.107
Eq.(10.13)C = .5959 + 0.312 β 2.1 − 0.184 β 8 + 91.71β 2.5 / Re .D75
⇒β =
C = 0.598 ≈ 0.6
A similar procedure is followed for calculation of
∆P for thevolumetric flow rateof 0.007m 3 / s,∆P = 393.2Pa . Compare it with 65 kPa.
10.11
10.14
D1 = 4.026 in
D2 = 2 in
T1 = 100 F
P1 = 150 psia
 =
m
∆P = 6 psi
D
β = 2 = 0.5
D1
ρ1 =
Y = 1 − (.410 +.35 β 4 )
= 1 − (.410 +.35*.5 4 )
Assu min g C = 0.6
CYA2
A
1− ( 2 )2
A1
2ρ 1∆P
P1
150 * 144
lbm
=
= 0.723 3
RT1 53.34 * 560
ft
∆P
P1γ
6
= 0.987
.
150 * 14
 lbm / sec
m
=
32.17 lbm − ft / lbf − sec 2
 = 2.673 lbm / sec
m
0.6 * 0.987 * (
π 22
4 * 144
2 4
)
1− (
4.026
)ft 2
2 * 0.723 lbm / ft 3 * (6 * 144) lbf / ft 2
32.17 lbm − ft / lbf − sec 2
Now we can check the value of C,

ρVD mD
=
Re =
µ
µA
lbm
hr . ft
2.673 lbm / sec* 3600 sec/ hr * (4.026 / 12)ft
= 794,900
Re =
.04594 lbm / hr − ft * (π * 4.026 2 / 4 / 144)ft 2
@100F, for air (Table B4)
µ = 0.04594
91.71β 2
Re .D75
 =2.69
 assuming C=0.6 is .5%. Using this C, m
C=0.603, the error in calculation of m
lbm/sec
Using Eq.10.13, C=0.5959+0.0312 β 2.1 − 0184
. β8 +
10.12
10.15
Orifice meter
Find Q
ID = 25.5cm = 0.255m
D 2 = 15cm = .15m
∆P = 14kPa
T1 = 10C
ρ = 999.7kg / m 3
µ = 1.308 * 10 3 N.S / m 2
Solution: C is a f (Re) and Re is unknown so take C=0.6,
Q=
Q=
CA2 [2(P1 − P2 ) / ρ ]1 / 2
[1 − ( A2 / A1 ) 2 ]1 / 2
0.6(π (.15 2 ) / 4)[2(14,000) / 999.7]1 / 2
[1 − (.15 / .255) 4 ]1 / 2
Q = 0.0598m 3 / s
V = Q / A = 0.060 / π (.255 2 ) / 4
V = 1.175m / s
Re = ρVD / µ = (999.7)(1.175)(.255) / 1.308 * 10 −3
Re = 228,972
β = D 2 / D1 = 0.15 / 0.255 = 0.58824
⇒ C = 0.5959 + 0.0312 β 2.1 − 0.184 β 8 +
91.71β 2.5
Re .075
C = 0.606
⇒Q 2 = 0.0604m 3 / sec(@1%difference )
10.13
10.16
Problem 11 with flow nozzle. Using Re from problem 10.11 (228,000),
⇒ C = 0.9975 − 0.00653(10 6 β / Re0 )0.5
C = 0.9975 − 0.00653[(10 6 (.15 /.255)) / 264,800].5
C = 0.967
If Re is not available, use the dominant term, 0.9975 for first estimate of C.
Q=
CA2 [2(P1 − P2 ) / ρ ]1/ 2
[1 − ( A2 / A1 ) 2 ]1/ 2
0.9975(π (.15 2 ) / 4)[2(14,000) / 999.7]1/ 2
[1 − (.15 /.255) 4 ]1/ 2
= 0.0964
Q=
⇒ Q = 0.0964 m 3 / s
V = Q / A = 0.1118 / π (.255)2 / 4
. m3 / s
V = 189
Re = ρVD / µ = (999.7)(189
. )(.255) / 1308
.
* 10 −3
Re = 368,000
C = 0.9975 −.00653[(10 6 (.15 /.255)) / 368,000].5
C = 0.989
⇒ Q = 0.0986 m 3 / s
(For the orifice, Q = 0.0606 m3/s.)
10.14
10.17
Square edged orifice
4.026 2
D1 = 4.026 A = π (
) / 4 = 8.8 * 10 − 2 ft 2
12
D2 = ?
Q = 15cfm
∆P = 2psi
Assu min gT = 60F ⇒ ρ = 62.34lbm / ft 3
µ = 2.71lbm / hr .ft
V = Q / A = (15 / 60) / 0.088
V = 2.83ft / s
Re = ρVd / µ = (62.34 / 32.17)(2.83)( 4 / 12) /(2.71 / 32.17 * 3600 ))
Re = 79,081
Assu min gC = 0.6at first
Q = CA2 [2(P1 − P2 ) / ρ ]1 / 2 /[1 − ( A2 / A1 ) 2 ]1 / 2
15 / 60 = 0.6 A2 [2(2 * 144 ) /(62.34 / 32.17)]1 / 2 /[1 − ( A2 / 8.727 * 10 − 2 ) 2 ]1 / 2
0.25 = 0.6 A2 [17.241] /[1 − 131.30 A22 ]1 / 2
0.25 2 [1 − 131.3 A22 ] = 107.01A22
A2 = 2.329 * 10 − 2 ,D2 = 0.1722ft (2.07in )
⇒ β = D2 / D1 = 2.07 / 4 = 0.5175
⇒ C = 0.5959 + 0.0312 β
2 .1
91.71β 2.5
− 0.184 β +
Re .075
8
C = 0.607
⇒ A2 = 2.304ft ,D2 = 0.17ft (2.06in )
10.15
10.18
D1 = 4in
∆P = 1psi
D2 = 3in A2 = 4.909 * 10 −2 ft 2
 =?
m
Q = ?(SCFM )
T = 70F ρ s tan d = 0.075lbm / ft 3 (at1atm,70F )
P = 8atm = 117.6psi µ = .04391lbm / hr .ft
P
8 * 14.7 * 144
ρ=
=
= 0.598
RT
53.35 * 530
without compressibility effect (Y = 1)
Assume C = 0.6
 = CYA2 [2(P 1− P2 ) / ρ ]1/ 2 / [1 − ( A2 / A1 ) 2 ]1/ 2
m
 lbm / sec
0.6 * 1 * (.0491) ft 2 [2(1 * 144) lbf / ft 2 × (0.598 lbm / ft 3 / 32.17 lbm − ft / lbf − se
m
=
[1 − (3 2 / 4 2 ) 2 ]1/ 2
32.17 lbm − ft / lbf − sec 2
 = 2.65 lbm / s
m
We must now compute a better value for C. Calculate Re: From Table B.4, at 70F, µ =
0.04391 lbm/hr-ft

m
D
ρVD A
2.65( lbm / sec) * [4 / 12](ft)
Re =
=
=
= 0.837 × 10 6
4 2 2
µ
µ
[π / 4][ ] (ft ) * 0.04391( lbm / hr − ft ) / 3600 (sec/ hr )
12
9171
. β 2.5
=0.63 (β=0.75)
⇒ C = 0.5959 + 0.0312 β 2.1 − 0.184 β 8 +
Re.075
 = 2.63 lbm/s.
with this C, m

m
2.63 * 60
Q=
60 =
= 2104 scfm
0.075
ρs
Compressibility effect
Y = 1 − (0.410 + 0.35 β 4 )
P1 − P2
(1)(144)
= 1 − (0.410 + 0.35(3 / 4) 4 )
117.6(144)(14
P1γ
. )
Y = 0.997
with compressibility effect
 = 2.63 * 0.997 = 2.62 lbm / sec .
m
The difference, 0.3%, is small in this case.
10.16
10.19
D1 = 10cm
ρ = 998.2 kg / m3
A1 = 7.8540 * 10 −3 m 2
β = 0.7
A2 = 3.8485 * 10 −3 m 2
Water at 20C
1 < Q < 3 (m 3 / min)
β = D2 / D1 ⇒ D2 = 7cm
C = 0.995
Q=
(Table 10.1)
CA2 [2∆P / ρ ]1/ 2
[1 − ( A2 / A1 )2 ]1/ 2
ρQ 2 [1 − ( A2 / A1 )2 ]
∆P =
2C 2 A22
(998.2)(1 / 60)2 [1 − (3.8485 / 7.8584)2 ]
2( 0.995)2 ( 3.8485 * 10 −3 )2
∆PQ =1 = 7.183kPa
∆P =
∆PQ = 3 = 64.65kPa
64 − 7 = 57
⇒ Range > 65kPa
Span
10.20
For air we should base our calculations on air mass flow rate. Assume pressure in pipe
is given as absolute.
 air = ρ s tan d * SCMM 60SCMM :72kg / min
m
= 1.20 * SCMM 20SCMM :24kg / min
CYA2
P
7 * 101,325
 =
2 ρ 1∆P ρ 1 = 1 =
m
RT1
287 * 293
1 − ( A2 / A1 ) 2

m
∆P = [
CYA2
1 − ( A2 / A1 ) 2
2 ρ1
] 2 = 8.435kg / m 3
72 / 60
1 − .7 4 2
] C = .995,assu min gY = 1
*
π * .07 2
2 * 8.435
.995 * 1 *
4
= 4423.7Pa = 4.4kPa
We can choose the upper end of the range range of the differential pressure transducer
to be approximately 5kPa. Using the same methof, at the low end, the pressure
difference is about 500 Pa.
=[
10.17
10.21 Designation: Nozzle (N), Orifice (O), Venturi (V), length of Insertion (L), Pressure
Drop (∆P), Pressure loss (PL), accuracy (w) and cost ($)
Length of Insertion:
LV > LN > LO
Pressure Drop:
∆PO > ∆PV> ∆PN
The venturi discharge coefficient is slightly larger than for a nozzle so for the
same flow, the pressure drop will be less.
Pressure Loss:
PLO > PLN > PLV
The venturi has a diffuser so the pressure loss is less. Both the nozzle and the
orifice plate have a pressure loss associated with the highest velocity but due to
the vena contracta, the orifice maximum velocity is higher.
Cost:
$V > $N > $0
It has been assumed that other factors including the smoothness and the tolerances in
manufacturing of the three meters are the same. The accuracy will depend on the
accuracy of the discharge coefficient. For the venturi, ASME gives a 1% tolerance, for
the nozzle 2% and for the orifice 0.6% so the orifice is the most accurate, the venturi
next and the nozzle last. However, with calibration, this order may not apply.
10.18
10.22 Qmax = Vmax · A
πD 2
= Vmax ·
4
π (0.05) 2
= 5.89 * 10-3 m3/sec
=3·
4
Considering that the elevation is constant, Eq. 10.6 becomes:
Q=
CA2
1/ 2
 2∆P 
 ρ 


1/ 2
  A 2 
1 −  2  
  A1  
For maximum flow, ∆P will become maximum.
 A2 
d
 which is unknown, so d will need to be determined through trial
C = f   = f 

A
D
1


and error. We will assume that C = 0.60 and determine d, then check for C value. Using
the equation for Q:
C 2 A22  2∆P 
2
Eq. 10.6
Q =

2 
ρ

A  
1 −  2 
 A1 
0.6 2 A22  2 * 40,000 
Assuming ρ = 1000 kg/m3, (5.89*10-3)2 =
2 

 A2   1000 
1 −  
 A1 
πd 2
⇒ d = 0.0349 m = 3.49 cm
A2 = 0.9566 * 10 −3 m 2 =
4
Now we can check if C (discharge coefficient) was chosen correctly, using equation
91.71β 2.5
Eq. 10.13 : C = 0.5959 + 0.0312 β2.1 – 0.184 β8 +
Re 0D.75
d 3.49
β=
=
= 0.698
5
D
ρVD 1000 * 3 * 0.05
Re =
=
= 166,113 (µ has been take @ 25 °C)
µ
0.903 * 10 −3
91.71 * 0.698 2.5
2.1
8
C = 0.5959 + 0.0312 (0.698) – 0.184 (0.698) +
(166,113)0.75
C = 0.604
which is a about 0.66% higher than the assumed value.
The answer to the specific questions:
(a) The maximum flow will correspond to the maximum pressure drop allowed, in this
case 40 kPa. So Qmax = 5.89*10-3 m3/sec as calculated above.
10.19
(b) Expected pressure drop at ½ of the maximum flow: Q = Qmax/2 = 2.94*103 m3/sec.
The change in flow will affect the ReD, but the effect on C (discharge coefficient) will be
minimum. To calculate pressure drop, Eq. 10.6 will be used
Q=
CA2
 A
1 −  2
  A1



 ∆P
Q
= 
Qmax  ∆Pmax
2






1/ 2
1/ 2
 2∆P 
 ρ 


1/ 2
or
 Q
⇒ ∆P = ∆Pmax 
 Qmax
 ∆P
Q
(c) Turn-down ratio: max =  max
Qmin  ∆Pmin
Qmax  40 
= 
Qmin  5 
1/ 2



2

∆Pmax 40
 = 4 = 4 = 10kPa

1/ 2
, assuming C value will remain constant.
= 2.83 or the turn-down ratio of flow meter will be: 2.8
10.20
10.23
Qmax = Vmax · A
πD 2
= Vmax ·
4
π (0.05) 2
= 5.89 * 10-3 m3/sec
=3·
4
We need to use Eq. 10.10 to estimate C
1/ 2
 10 6 β 

C = 0.9975 − 0.00653
 Re D 
Assuming C = 0.99, we can find the nozzle area, and diameter (A2 and d)
C 2 A22  2∆P 
2
Eq. 10.6
Q =

2 
ρ

A  
1 −  2 
 A1 
0.99 2 A22
2
 2 * 40000 
5.89 * 10 −3 =
2
 1000 
A2
1−
2
π
2
 * 0.05 
4

-3
2
A2 = 0.630*10 m
πd 2
=
or
d = 0.0283m
4
= 2.83 cm
To check C:
d 2.83
ρVD 1000 * 3 * 0.05
β= =
= 0.566 , Re =
=
= 166,113 (µ has been taken @ 25 °C)
D
5
µ
0.903 * 10 −3
(
)
 10 6 * 0.566 

C = 0.9975 − 0.00653
 166,113 
assumed C
1/ 2
= 0.9854 which is about 0.45% less than the
The answer to the specific questions are:
(a) The maximum flow will correspond to the maximum pressure drop allowed, in this
case 40 kPa. So Qmax = 5.89*10-3 m3/sec as calculated above.
(b) Expected pressure drop at ½ of the maximum flow: Q = Qmax/2 = 2.94*103 m3/sec
The change in flow will affect the ReD, but the effect on C (discharge coefficient) will be
minimum. To calculate pressure drop, Eq. 10.6 will be used
10.21
Q=
CA2
 A
1 −  2
  A1



 ∆P
Q
= 
Qmax  ∆Pmax
2






1/ 2
1/ 2
 2∆P 
 ρ 


1/ 2
or
 Q
⇒ ∆P = ∆Pmax 
 Qmax
2

∆Pmax 40
 =
=
= 10kPa
4
4

1/ 2
 ∆P 
Q
(c) Turn-down ratio: max =  max  , assuming C value will remain constant.
Qmin  ∆Pmin 
1/ 2
Qmax  40 
=   = 2.83 or the turn-down ratio of flow meter will be: 2.8
Qmin  5 
10.22
10.24
Qmax = Vmax · A
= Vmax ·
πD 2
4
π (0.05) 2
=3·
= 5.89 * 10-3 m3/sec
4
We will use the data from Table 10.1 for C value of machined entrance cone,
C = 0.995. Checking for the Reynolds number,
ρVD 1000 * 3 * 0.05
Re =
=
= 166,113 (µ has been taken @ 25 °C)
µ
0.903 * 10 −3
It turns out that this value is about 17% below those of Table 10.1, and will be prudent
to calibrate any chosen venturi with the assumption for C based on data of Table 10.1..
Q2 =
C 2 A22
 2∆P 

2 
 A2   ρ 

1 − 
 A1 
(5.89 * 10 )
−3 2
Eq. 10.6
(0.995 )2 A22
=
1−
 2 * 3000 


 1000 
A22
2
π
2 
 0.05 
4

2
−3
A2 = 0.627 * 10 m ⇒ d = 0.0283m = 2.83cm
The value of β = d/D = 0.566, which is within the range of data in Table 10.1.
The answer to the specific questions are:
(a) The maximum flow will correspond to the maximum pressure drop allowed, in this
case 40 kPa. So Qmax = 5.89*10-3 m3/sec as calculated above.
(b) Expected pressure drop at ½ of the maximum flow: Q = Qmax/2 = 2.94*103 m3/sec
The change in flow will affect the ReD, and it may affect C (discharge coefficient), but it
is not expected to be very significant.
To calculate pressure drop, Eq. 10.6 will be used.
1/ 2
Q=
CA2
 A
1 −  2
  A1



2



1/ 2
 2∆P 
 ρ 


1/ 2
 ∆P
Q
(c) Turn-down ratio: max =  max
Qmin  ∆Pmin
Qmax  40 
=

Qmin  5 
1/ 2
2
 Q   C max 
Q
C  ∆P 
 
 ⇒ ∆P = ∆Pmax 

=

Q
C
P
Q
C
∆


max
max
max
max




or
C
∆Pmax
40
Assu min g max ≈ 1, ∆P =
=
= 10kPa
4
4
C



1/ 2
, assuming C value will remain constant.
= 2.83 or the turn-down ratio of flow meter will be: 2.8
10.23
2
10.25
In this problem ∆Pmin corresponds to 100 lit/min, ∆Pmax corresponds to 300(1+0.5) = 450
lit/min
Range of the differential pressure transducer (∆Pmin to ∆Pmx) using Eq. 10.6:
CA2
Q=
1/ 2
 2∆P 
 ρ 


1/ 2
  A 2 
1 −  2  
  A1  
Since the issue is determing the range, it is not usually needed to determine the
parameters very accurately, Thus we will assume that value of C = 0.6.
  A 2 
2
ρQ 1 −  2  
  A1  
1
To determine ∆P: ∆P =
2
C 2 A22
∆Pmin =
 100 * 10 −3
0.5 * 1000 * 
60

∆Pmax = ∆Pmin



2
π * 0.02 2 
0 .6 2 

4


Q
*  max
 Qmin



  4 2 
1 −   
  25  
2
= 38,078Pa = 38Kpa
2
 450 
∆Pmax = 38 * 

 100 
∆Pmax= 770 kPa
2
Note: This pressure is unreasonably high for pressure drop across an orifice. The pipe
and orifice are too small for this level of flow.
10.24
10.26
In this problem ∆Pmin corresponds to 10 cfm, ∆Pmax corresponds to 30(1+0.5) = 45 cfm
Range of the differential pressure transducer (∆Pmin to ∆Pmx) using Eq. 10.6:
1/ 2
 2 ∆P 
CA2
Q=

1/ 2 
  A 2   ρ 
2
1 −   
  A1  
Since the issue is determing the range, it is not usually needed to determine the
parameters very accurately, Thus we will assume that value of C = 0.6.
  A 2 
2
ρQ 1 −  2  
A
1
  1  
To determine ∆P: ∆P =
2
C 2 A22
∆Pmin =
2
2
 ft 3  
1

 × 1 − 
1
 sec   81 ×
= 2,484.56 psf = 17.25 psi
2
2
ft.lbm


1
32.17
 π ×   
lbf . sec 2
2

12  

2
2
.6 × 
 ft
4






lbm  10 
.5 × 62.4 3 ×  
ft
 60 
( )
∆Pmax
Q 
= ∆Pmin *  max 
 Qmin 
2
 30 
∆Pmax = 17.25 *  
 10 
∆Pmax = 155.28 psi
2
Note: This pressure is unreasonably high for pressure drop across an orifice. The pipe
and orifice are too small for this level of flow.
10.25
10.27
Note: It is assumed that all pressures given in the problem statement are in absolute
value.
Mass flow of critical flow nozzle:
γ +1
γ  2  γ −1
 =


Eq. 10.12
m
T0 R  γ + 1 
For the conditions given:
m max = 0.1 kg/sec
To = 25 – 300 °C
Po = 100 kPa
D = 5 cm
R = 287 J/kg·K
for air
γ = 1.4,
γ has a slight variation with T that will be neglected here.
4
The nozzle should be sized for the highest temperature, so d =
A2 (nozzle size)
π
should be determined based on T° = 300 °C.
γ +1 −1 / 2


0

γ  2  γ −1 
m T



*
A2 =
 R  γ + 1 
P0


A2 P0
−1 / 2
1.4 +1


0.1 (273 + 300 )  1.4  2  1.4 −1 
A2 =



 287  1.4 + 1
1000 * 10 3


−6
−3
2
A2 = 59.22 * 10 m ⇒ d = 8.68 * 10 m = 0.868cm
Note: We have assumed that the upstream static and stagnation conditions are the
same, which is often a reasonable assumption. This is a good assumption for the cases
that Mach Number (M) is less than 0.2. In this case, the upstream mach number is
about 0.1.
The answer to the specific questions:
(a) Pressure and temperature instruments need to be connected one diameter
upstream of the nozzle. To insure critical flow, a pressure gage needs to be installed
downstream too.
(b) For the flow to remain critical, downstream pressure should satisfy:
P0
Eq. 10.11
Pdownstream ≤ Pcrit =
γ
 γ + 1 γ −1
 2 


For lower pressure of 400 kPa,
400
Pcrit =
= 211kPa
1 .4
1.4 + 1 0.4
 2 


10.26
(c)
γ +1


A2 P0  γ  2  γ −1 
 =


m
0  R  γ + 1
 
T  


1/ 2
Eq. 10.12
1.4 +1


59
.
22
*
10
*
500
*
10
1
.
4
2

 1.4−1 

 =
m


 287  1.4 + 1

(273 + 150 )


 = 0.058 kg/sec
m
−6
1/ 2
3
 = ρ st * Q(SCMS )
To obtain the airflow in standard cubic meters per second, m
ρst is usually calculated at 1 atm, 20 °C,
P
101,325
=
= 1.205kg / m 3
ρ st =
RT 287 * 293
0.058
Q(SCMS ) =
= 0.048 SCMS
1.205
= 2.88 SCMM
10.27
10.28 Note: It is assumed that all pressures given in the problem statement are in
absolute value.
Mass flow rate of critical flow nozzle:
γ +1
γ −1
γ  2 


T0 R  γ + 1 
For the conditions given:
m max = 0.1 kg/sec
T° = 25 – 300 K
P0 = 100 kPa
D = 5 cm
R = 188.9 J/kg·K
γ = 1.289
 =
m
A2 P0
Eq. 10.12
The nozzle should be sized for the highest temperature, so d =
should be determined based on T° = 300 °C.
A2 =
 T
m
ρ0
0
γ +1


γ −1


2
γ




*
 R  γ + 1 


4
π
A2 (nozzle size)
−1 / 2
−1 / 2
1.289 +1


0.1 (273 + 300 ) 1.289 
2
 1.289 −1 
A2 =



1000 * 10 3 Pa 188.9  1.289 + 1


-6
2
-3
= 49.45*10 m ⇒
d = 7.94*10 m
= 0.794 cm
Note: We have assumed that the upstream static and stagnation conditions are the
same, which is often a reasonable assumption. This is a good assumption for the cases
that Mach Number (M) is less than 0.2. In this case, the upstream Mach number is
about 0.1
(a) Pressure and temperature instruments need to be connected one diameter
upstream of the nozzle. To insure critical flow, a pressure gage needs to be installed
downstream too.
(b) For the flow to remain critical, downstream pressure should satisfy:
P0
Pdownstream ≤ Pcrit =
γ
 γ + 1 γ −1
 2 


For lower pressure of 400 kPa,
400
Pcrit =
= 219kPa
1.289
0
.
289
1.289 + 1


2


10.28
Eq. 10.11
(c)
 =
m
 =
m
A2 P0
T0
γ +1
γ  2  γ −1


R  γ + 1
49.45 * 10 −6 * 500 * 10 3
(273 + 150 )
1.289 +1
2
1.289 
 1.289 −1


188.9  1.289 + 1
 = 0.058 kg/sec
m
 = ρ st * Q(SCMS )
To obtain the airflow in standard cubic meters per second, m
ρst is usually calculated at 1 atm, 20 °C,
P
101,325
ρ st =
=
= 1.831kg / m 3
RT 188.9 * 293
0.0058
= 0.032 SCMS
Q=
1.831
= 1.91 SCMM
10.29
10.29 Note: It is assumed that all pressures given in the problem statement are in
absolute
We need to use Eq. 10.8 because the flow is compressible
CYA2
 =
Eq. 10.8
m
2 ρ 1 (P1 − P2 )
2
A 
1 −  2 
 A1 
P1
RT1
The nozzle area needs to be determined when T1 is maximum, and consequently ρ1 is
minimum. To determine A2, we need to substitute for the other parameters in the above
equation. We will assume the value of 0.99 for both C and Y, and then check them. If
needed we will use a trial and error process to correct for C and Y.
P
1000 * 10 3
ρ1 = 1 =
= 6.081 kg/m3
RT1 287 * (273 + 300 )
⋅
0.99 * 0.99 A2
m = 0 .1 =
2 * 6.081 * 40 * 10 3
2
A 
1 −  2 
 A1 
ρ1 =
where:
A2 = 0.1462 * 10 −3 m 2 =
πd 2
d = 0.0136m = 1.36 cm
Checking the value of C and Y:
C = .9975 – 0.00653
4
 10 6 β 


 Re D 
1.36
= 0.272
5 .0
ρVD (ρVA1 )D m D
Re D =
=
=
=
µ
µA1
µA1
Eq. 10.10
β=
0.1 * 0.05
(1.846 * 10 )π (0.405)
−5
2
= 1.38 * 10 5
µ has been taken from Table B.3.
 10 6 * 0.272 
 = 0.988
C = 0.9975 − 0.00653 
5 
 1.38 * 10 
The value of C is about 0.2% lower than the assumed value.
γ −1

 2
 γ γ 1− r γ 1− β 4 
Y = r

2

 γ − 1 1− r
1− β 4r γ 

P
1000 − 40
where r = 2 =
= 0.96, β = 0.4 and γair = 1.4
1000
P1
10.30
Eq. 10.9
1.4 −1


2
1 .4


1
.
4
1
0
.
96
1 − 0.272 4
−
1 .4
Y =  0.96
2 
1.4 − 1 1 − 0.96

1 − 0.272 4 * 0.96 1.4 

Y = 0.956
Now we can replace these values for C and Y to find new values for A2 and d.
A2 = 0.1514*10-3 m2 and d = 0.0139 m = 1.39 cm
Rechecking for C and Y might produce slightly different values and improve the
accuracy of the results.
The pressure taps need to be one diameter upstream and ½ diameter downstream.
Upstream temperature should also be one diameter upstream,.
For the case of P1 = 500 kPa
and T1 = 150 °C
We will take C and Y to be same as above. They can be corrected as shown above, if
needed.
P1
500 * 10 3
ρ1 =
=
= 4.119 kg/m3
RT1 287 * 423
 =
m
0.988 * 0.956 *
π (0.0139 )
 1.39 2
1 −  2
 5
= 0.058 kg/sec
2
4



2
2 * 4.119 * 20 * 10 3
10.31
10.30
Rotameter calibrated for air @
T = 20C
P = 1 atm
Used @
T = 20C
P = 5 atm
Find flow correction factor
Using Eq. 10.21, and 10.23
Q = Va Aa
2∀gρ b 1/ 2
]
ρ
Q 5atm V5
CD Abf ρ f 5
=
=
= ( f 1 )1/ 2
ρf 5
Q 1atm V1 [ 2∀gρ b ]1/ 2
CD Abf ρ f 1
[
PM
ρ=
RT
So,
P1M
ρ
1
P
, ( f 1 )1/ 2 = ( RF )1/ 2 = ( 1 )1/ 2 = ( )1/ 2 =.45
P5M
ρf 5
5
P5
Q5atm
= 0.45
Q1atm
RT
10.32
10.31
1. air @ T = 20C
P = 1 atm
2. air @ T = 200C
P = 5 atm
Following Example 10.10,
ρf1
ρf2
P1M
RT1
P T
=
= 1 2
P2 M T1 P2
RT2
Q2
ρ
= ( f 1 )1/ 2
Q1
ρf2
(For a perfect gas, ρ =
PM
)
RT
1(200 + 273)
5(20 + 273)
= 0.32
U sin gEqs. 10.21 and 10.23
=
Q5atm
ρ
= ( 1atm )1/ 2 =.321/ 2
Q1atm
ρ 5atm
= 0.57
10.32
1. air, M=29
2. propane, C 3 H 8
M = 44.1
Using Eqs. 10.22 and 10.23
Qp
Qa
(
=(
ρ a 1/ 2
)
ρp
PaMa
M
RTa 1/ 2
=[
] = ( a )1/ 2 = 0.81
Pp M p
Mp
RTp
Ma 1/ 2
29 1/ 2
) =(
) = 0.811
Mp
44.1
Propane has been assumed to be a perfect gas.
10.33
10.33
1. air
2.CO2
M=29,
M = 44,
ρ
QCO 2
= ( a )1/ 2
ρ CO 2
Qa
=[
T=20 C,
T = 20C,
P = 1 atm
P = 5 atm (gage )
Pa M a 1/ 2
)
RTa
Pa M a 1/ 2
)
=
=(
PCO 2 M CO 2 1/ 2
PCO 2 M CO 2
(
)
RTCO 2
(
(1)(29) 1/ 2
] = 0.33
(1 + 5)(44)
10.34
S b = 7 .8
S oil = 0.84
Q oil
Qwater
2∀g ( ρ b − ρ f ) 1 / 2
]
C D Abf ρ f
=
2∀g ( ρ b − ρ w ) 1 / 2
]
[
C D Abf ρ w
[
ρ − ρ oil 1 / 2 ρ w 1 / 2
Q oil
) (
)
=( b
ρ b − ρ water
ρ oil
Qwater
7.8 − 0.84 1 / 2 1 1 / 2
)
) (
0.84
7 .8 − 1
= 1.10
=(
Major source of error is the assumption of equality of viscosity for both fluids. Because
of the difference in density and viscostiy, the Reynolds number of the flow will be
different for the two fluids, resulting in uncertianty in the above calculations.
10.34
10.35
QNEW
QSS
=(
2∀g ( ρ NEW − ρ f ) 1/ 2
]
CD Abf ρ f
=
2∀g ( ρ SS − ρ f 1/ 2
[
]
CAbf ρ f
[
ρ NEW − ρ f 1/ 2
)
ρ SS − ρ f
4.0 − 1 1/ 2
)
7.8 − 1
= 0.66
The range of the rotameter when using the new float will be 66% of the flow meter with
the stianless steel float.
=(
10.36
The thermal mass flow meters:
⋅
q
m =
Cp (T − T )
Eq.10.25
2 1
Under similar conditions (same q and ∆T), for different fluids :
 CO
m
c
2
= p, Air
 air
m
c p, CO
2
 CO2 = m
 Air *
m
c p,air
c p, CO
2
= 10 *
1.004
= 11.92 g/sec
0.842
10.35
10.37
We can use Eq. 10.27 to estimate the velocity of water:
fD
(Eq. 10.27)
St
This equation is valid for Re of 500 to 100,000, and the value of St is 0.2 - 0.21
65 * 0.02
V =
= 6.19 m/sec
0.21
To check for the Reynolds number, using the water properties from Table B.1
ρVD 999 * 619 * 0.02
Re D =
=
µ
1.156 * 10 −3
= 106,906 which is slightly above the range.
No significant error is expected.
V
=
10.38
First we use Eq. 10.27 to estimate the velocity, then calculate the flow rate.
fD
V=
Eq. 10.27
St
1400 * 0.003
=
0.21
= 20 m/sec
To check the velocity of the above calculation, we need to check the Reynolds number. Air
properties are from Table B.3
ρVD
Re D =
µ
0.8826 * 20 * 0.003
=
= 2,316 which is in the range of 500 - 100,000 for St = 0.21
2.286 * 10 −5
Assuming the V values corresponds to the pipe average velocity (this is a major assumption )
(πD )
Q = VA = V
2
Pipe
4
= 20
(π * 0.1 )
2
4
= 0.157 m 3 /sec
10.36
10.39
r 02 − r 2
V
=
VCL
r 02
V =
VCL
r
Vav =
2
0
(r 02 − r 2 )
1
πr 02
∫
r0
0
2πrVdr
r0
Vav . = [ ∫ (VCL − r 2
0
[VCL r 2 −
VCL
r
2
0
)2πrdr ] / πr 02
2 VCL 4 r0 2
r ] 0 / r0
4 r 02
1
= VCL
2
r0
AverageFlow = Q = ∫ (2πrV )dr
0
r0
Q = ∫ 2π
0
Q=
VCL
r 02
(r 02 − r 2 )rdr =
2πVCL
r 02
[r 02
r 2 r 4 r0
− ]0
2
4
πr 02VCL
2
10.40
V = 10m / s
∆P = ?
ρ = 1.1kg / m 3
C =1
V2 = C 2 × ( P1 + gzρ1 − P2 − gzρ2 ) / ρ
V2 = C 2 ∆P / ρ
V22 ρ = 2 ∆P
∆P = 0.5 * (10m / s ) 2 (1.1kg / m 3 )
∆P = 55Pa
10.37
10.41
V = 100 ft / s
∆P = ?
ρ = 0.075lbm / ft 3
C =1
V2 = C 2 × ( P1 + gρz1 − P2 − gρz 2 ) / ρ
V2 = C 2∆P / ρ
V22 ρ = 2∆P
∆P = [0.5 * (100 ft / s ) 2 (0.075lbm / ft 3 )] /[(32.17lbm. ft / lbf .s 2 )(144in 2 / ft 2 )]
∆P = (0.0810 psi ) * (27.7in.water / 1 psi )
∆P = 2.24in.water
10.42
Using Eq. 10.31, and considering that there is no gravitational effect in this problem,
V=C
2 ∆P
ρ
,C =1
2 * 5 × 103
=1
1 .0
= 100 m/sec
10.43
Using Eq. 10.31, and considering that there is no gravitational effect in this problem,
V=C
=1
2∆P
ρ
, C =1
ft.lbm
in 2
lbf
2 * 0.5 2 *144 2 * 32.17
lbf.sec 2
ft
in
lbm
0.075
ft 3
= 249
ft
sec
10.38
10.44
Using Eq. 10.31, and assuming C=1,
V=C
2 ∆P
ρ
∆P = ρw g∆hw = 1000 * 9.81 * 0.20
= 1,962 kPa
2 * 1962
1000
= 1.98 m/sec
V =1
10.45
Using Eq. 10.31, and assuming C=1,
V=C
2∆P
ρ
∆P = ρ w g∆hw = (62.34 * 32.17 * 0.83) /(32.17 *144)
= 0.36 psi
V =1
2 * 0.36
= 7.31
lbf
in 2
ft.lbm
*
144
* 32.17
2
2
in
ft
lbf.sec2
lbm
62.4 3
ft
ft
sec
10.39
10.46
Considering Fig. 10.27
ρ = 990 kg / m 3
P1 = PV + ρ 1gh1 = PV + (990)(9.81)(h1 )
P1 = PV + 97119
. h1
P2 = PV + ρ 2 gh2
∆Pi = P2 − P1 = g ( ρ 2 h2 − ρ 1h1 ) = −58.8 kPa
(transducer reading )
gρ 1h1 = 58.8 kPa
This remains the same between the two neasurements.
∆Pf = g ( ρ 2h2 − ρ 1h1 ) = −10 kPa
solving for h2 ,
ρ 2gh2 = −10 + 58.8 = 48.8 kPa
h2 =
48.8 * 1000
= 5.09 m
978 * 9.81
10.47
C = 1500 ft/s
d = 5 ft
t = 2 L/C
2 * 5ft
t=
1500 ft / s
t = 6.67 * 10 −3 s
10.40
CHAPTER 11
11.1 (a) The time after which the thermocouple will follow the gas temperature
steadily can be obtained from Eq. 11.12a;
t
y  KA e   t   


 Ka  (e

t

 1)  t 

To satisfy this condition: e
t

 1 for  = 0.286 sec.
t
Theoretically lim e t 0  0 , but if approximately 99% approach to the
steady state is satisfactory, then:
e
t

 0.01  t  132
. sec
(b) The continuing error will be:
y  y S.S.  y e  KAt     KAt   KA
Where KA  10C / sec
y  10  0.286  2.86C
11.2 These devices are usually first order, and we can use Eq. 11.9a (response
of a first order system to a step input) for this purpose:
t
t
y
y  KX o  1  e   
  1  e    0.98



KX o 
e
t

 0.02
(sec)
t98%(sec
)
0.1
0.391
1
3.91
10
39.1
11.3 Response of a first order system to a sine input is (Eq. 11.14 and 11.15):
KX o
y (t ) 
sint   
1   2 2
   tan 1  
 = 2f = 20.1 = 0.2 rad/sec
(sec) =
y (t )
1


kXo
1   2 2
(rad) =
0.1
1
10
0.998
0.84
7
-0.56
0.15
7
-1.41
-0.063
Note: 2 rad  360 deg.
11.1
11.4 For a thermocouple (first order system) to closely follow the temperature
variation, time constant of the thermocouple () should be much smaller than the
time scale (period) of temperature variation, thermo<<variation. All thermocouples
of problem 11.2 satisfy this condition.
Note: the smaller the thermocouple, the more closely it will follow the
temperature variation of the building.
11.5 For a thermocouple (first order system) to closely follow the temperature
variation, time constant of the thermocouple () should be much smaller than the
time scale (period) of temperature variation, thermo<<variation. All thermocouples
of problem 11.2 satisfy this condition.
Note: the smaller the thermocouple, the more closely it will follow the
temperature variation of the oven.
11.6 Time constant of a thermocouple is calculated from Eq. 11.16,

d3 


m V

mc
where:

 6 
hA
A
d2
For copper from Table B.5, =8930 kg/m3 and c=385 J/(kgꞏK)
Therefore:
   0.002 3 
5
m  8930
  3.741  10 kg
6


A   ( 0.002) 2  12.57  10 6 m 2
3.741  10 385

5
 gas
 liq . 
100  12.57  10 6
3.741  10 5 385


3000  12.57  10 6
. sec
 114
 0.38 sec
11.2
11.7
The thermocouple spherical junction diameter, d, of 1/16 is equivalent to:
1
1 ft
in 
16
12in
 0.005208 ft
d
Time constant of a thermocouple is calculated from Eq. 11.16,
 d 3 

m  V   
mc
where:

6 

hA
A  d 2
From Table B.5, copper has a density, , of 8,930 kg/m3 and a specific
heat, c, of 385 J/kg.K. These values are equivalent to:
kg
1 lbm
m3
lbm
  8930 3 
 0.02832 3  554 3
m 0.4563 kg
ft
ft
J
kg 5 K
Btu
1 Btu
c  385

 0.4563
 o  0.0925
kgK 1055.0 J
lbm 9 F
lbm o F
Therefore:
   0.0052083 
  4.098 10 5 lbm
m  554
6


A  d 2   (0.005208) 2  8.52110 5 ft 2
 gas 
4.098 10  0.0925  3600  3.2 sec
 liq. 
4.098 10  0.0925  3600  0.32 sec
5
50  (8.52110 5 )
5
500  (8.521 10 5 )
11.3
11.8
a) Time constant of the thermocouple:
mc

hA
The junction of the thermocouple is made of a combination of Chromel
and Alumel. For properties of the junction, we take the average properties
of the two (for lack of better available data). From data of Table B.5,
   2 8600  8730
 1

 8665 kg / m 3
2
2
c  c 2 526  450

 488 J / kg  K
c 1
2
2
 d 3 
   0.0013 
6
m  
  8665
  4.54  10 kg
6
 6 


A  d 2   0.001  3.14  10 6 m 2
2
(4.54  10 6 )488
 141
. sec
500(3.14  10 6 )
b) Response of first order system (such as a thermocouple) to a step
change in input is represented by Eq. 11.9, where K=1 and y= T  Ti and
x= TfinalTi (refer to Ex. 11.1). So the response of the thermocouple will be:
t
T  Ti
T

 1 e 
Tfinal Tfinal  Ti

For
t
t
T
 0.99  1  e   e   0.01 or t   ln 0.01  6.49 sec
Tfinal
c) If the diameter of the thermocouple is doubled, then:
 d 3 
 2  c
 6 
m c
 2 1 (because d2 = 2d1)
2  2 
hA2
hd 22
So,
2 = 21 = 21.41 = 2.82 sec
t2 = -2ln0.01 = -2.82ln0.01
=13.0 sec
11.4
11.9
a) The thermocouple spherical junction diameter, d, of 1/16 is equivalent
to:
1
1 ft
in 
16
12in
 0.005208 ft
d
Time constant of the thermocouple:
mc

hA
The junction of the thermocouple is made of a combination of Chromel
and Alumel. For properties of the junction, we take the average
properties of the two (for lack of better available data). From Table B.5,
   2 8600  8730

 8665kg / m 3
 1
2
2
c1  c2 526  450
c

 488J / kg  K
2
2
These material properties are equivalent to:
kg
1 lbm
m3
lbm
  8665 3 
 0.02832 3  538 3
m 0.4563 kg
ft
ft
J
kg 5 K
Btu
1 Btu

 0.4563
 o  0.1173
c  488
kgK 1055.0 J
lbm 9 F
lbm o F
Therefore:
   0.0052083 
 d 3 
  3.979  10 5 lbm


m  
 538

6


 6 
A  d 2   0.005208  8.52110 5 ft 2
2
(3.979 10 5 )  0.1173  3600

 3.94sec
50  (8.521 10 5 )
b) Response of first order system (such as a thermocouple) to a step
change in input is represented by Eq. 11.9, where K=1 and y= T  Ti and
x= TfinalTi (refer to Ex. 11.1). So the response of the thermocouple will be:
t
T  Ti
T

1 e 
T final T final  Ti
For
t
t
T
 0.99 1  e   e   0.01ort   ln 0.01  18.16 sec
T final
c) If the diameter of the thermocouple is doubled, then:
11.5
 d 23 
c
 6 
 
m2 c

 2 1 (because d2 = 2d1)
hA2
hd 22
So,
2 = 21 = 23.94= 7.88 sec
t2 = -2ln0.01 = -7.88ln0.01
=36.29 sec
2 
11.6
11.10 Thermocouple is a first order system. As derived in Ex. 11.1, the response
of the thermocouple is represented by:
t
t
T T
T  Ti
T

 1  e   final
e 
Tfinal  Ti
Tfinal Tfinal  Ti
t
T
 e   ln T / Ti    t 
Ti
By drawing lnT / Ti  vs. t, the slope of the line be the negative inverse
of the time constant, . In this case:
Ti = TfinalTi = 020 = -20 C
T = TfinalT = 0T = T C
or
ln(TTi
0
t
T(c)
0.1
0.5
1
2
3
16.7
8.1
3.3
0.6
0.1
-1
-2
-3
-4
-5
-6
0
0.5
1
1.5
2
2.5
3
3.5
4
T
Ti
0.835
0.405
0.165
0.03
0.005
T
Ti
-0.180
-0.903
-1.802
-3.507
-5.298
ln
Time - sec
A quick method to find the slope is to pick two points. Selecting t = 0.1 and
t = 1 sec, we have:
1802
 ( .180)
.
then   1 / a  0.555 sec
a
 1801
.
1  0.1
Alternatively, we could use linear regression on the data for ln T / Ti 
vs. t. The resulting time constant is 0.569 sec., 3% higher than the above
value.
11.7
11.11
Thermocouples are first order systems with the following responses to a
step function (TfTi),
t
T  Ti
mc
 1  e  where  
Tfinal  Ti
hA
For the two thermocouples, assuming spherical junctions,
 d 13 
   0.0013 
6
  8930
m1   1 
  4.676  10 kg
6


 6 
 d 2 3 
   0.002 3 
5
  8930
m2   2 
  3.741  10 kg
6


 6 
A1  d 1   0.001  3.14  10 6 m 2
2
2
A2  d 2   0.002  12.56  10 6 m 2
2
1 
2
(4.676  10 6 )385
 0.191 sec
3000(3.14  10 6 )
(3.741  10 5 )385
 0.382 sec
3000(12.56  10 6 )
The response of thermocouple of Ti = 20 C and Tf = 100 C are given in the
following table:
t
T1
T2
1
99.59
94.16
2
100.00 99.57
5
100.00 100.00
10 100.00 100.00
They will read the effectively the same temperature after 2 seconds.
2 
11.8
11.12
Thermocouples are first order systems with the following responses to a
step function (TfTi),
t
T  Ti
mc
 1  e  where  
Tfinal  Ti
hA
From Table B.5, copper has a density, , of 8,930 kg/m3 and a specific
heat, c, of 385 J/kg.K. These values are equivalent to:
1 lbm
kg
m3
lbm
  8930 3 
 0.02832 3  554 3
m 0.4563 kg
ft
ft
J
kg 5 K
Btu
1 Btu
c  385

 0.4563
 o  0.0925
kgK 1055.0 J
lbm 9 F
lbm o F
For the two thermocouples, assuming spherical junctions,
1
1 in
 0.002604 ft
d1  in 
32
12 ft
1
1 in
 0.005208 ft
d 2  in 
16
12 ft
 d13 
   0.002604 3 


  5.122  10 6 lbm
 554
m1  1 

6


 6 
 d 3 
   0.0052083 
  4.098 10 5 lbm
m2   2  2   554
6


 6 
A1  d1   0.002604   2.130 10 5 ft 2
2
2
A2  d 2   0.005208  8.521 10 5 ft 2
2
1 
2
(5.122  10 6 )  0.0925  3600
 0.053 sec
1500  (2.130  10 5 )
(4.098  10 5 )  0.0925  3600
2 
 0.107sec
1500  (8.52110 5 )
The response of thermocouple of Ti = 75oF and Tf = 32 oF are given in the
following table:
t
T1
T2
1
32.00
32.00
2
32.00
32.00
5
32.00
32.00
10
32.00
32.00
They will read the effectively the same temperature at 1 second.
11.9
11.13 Equation 11.12 represents the response of a first order system (such as a
thermocouple) to a ramp input:
t
y  KA e   t   


where yTTi (i is initial); and KAramp slope; and K =1 (input has same
units as output).
The output is then given by:
t
t
T  Ti  A e   t     10 / 60 e   t   




Time delay in indicated measurement is  = 10 sec. This estimate is for
the times larger than the initial transient period.
The ideal output (actual temp. variation) would be TTi = At
Instantaneous error in temperature reading is then the difference between
the indicated temperature and the ideal temperature
t
t
(Tindic  Ti )  (Tiact  Ti )  TindicTact  10 / 60 e   t     10 / 60t  10 / 60  e   1




After the initial transitory period:
Tindicated  Ti  10 / 60t   
10  C 
. C

  10 sec  167
60  sec 
That is, the output of the thermometer will read 1.67 degrees low.
Tact  Tindicated   A 
11.14 (a) Average, maximum and minimum of the measured temperature:
1 
Tave   Tdt  200; Tmax  200  10  1  210C; Tmin  200  10   1  190C
 0
 = 0.5 rad/sec
(b) The measured temperature (T=200+10cos0ꞏ5t) is the response of a
first order system to a sinusoidal input of the form:
Tin  Tave  Tamp sint 
and the output form:
Tamp
Tout  Tave 
sin(t   )
1   2 2
Tamp
2
 10  Tamp  10 1  0.5  5  26.9C
2 2
1  
The phase lag of the output:
=tan-1() = tan-1(0.55) = 1.19 rad
The input will have an advance of 1.19 rad (68.2 deg) with respect to the
output. Consequently, the input time variation will be:
T = 200 + 26.9cos(0.5t+1.19) C
11.10
11.15 (a) Average, maximum and minimum of the measured temperature:
1 
Tave  0Tdt  300;Tmax  300  15  1  315 o F ;Tmin  300  15    1  275 o F  =

1.5 rad/sec
(b) The measured temperature (T=300+15cos1ꞏ5t) is the response of a
first order system to a sinusoidal input of the form:
Tin  Tave  Tamp sin  
t
and the output form:
Tamp
Tout  Tave 
sin(t   )
1   2 2
Tamp
 15  Tamp  15 1  1.5  5 2  113.5 o F
2 2
1  
The phase lag of the output:
=tan-1() = tan-1(1.55) = 1.44 rad
The input will have an advance of 1.44 rad (82.4 deg) with respect to the
output. Consequently, the input time variation will be:
T = 300 + 113.5cos(1.5t+1.44) oF
11.16 We will convert this equation to the standard form of:
1 d 2 y 2 dy
Eq. 11.22: 2 2 
 y  kx (t )
 n dt
 n dt
5 d 2y
1 dy
x (t )

y 
2
1000 dt
1000 dt
1000
1000
2
 200   n  14.14 rad / sec
n 
5
By comparison:
2
1

   0.007
 n 1000
Equilibrium response of the system:
x (t )
In the absence of dynamic effects, y eq (t  ) 
 0.025
1000
11.17 Following the same process as problem 11.11,
1 d 2 y 2 dy
Eq. 11.22:

 y  kx (t )
2
2
 n dt
 n dt
1 d 2y
2 dy
50

y 
sin( 50t )
2
100 dt
100 dt
100
2
 n  100   n  10rad / sec
By comparison: 2
2

   0.1
 n 100
In the absence of dynamic effects: yeq = 1/2sin(50t)
11.11
11.18 Amplitude and phase response of the equation in Prob. 12 can be
calculated from Eq. 11.30 and 11.31:
y amp
y eamp

y amp
kX o
1


n
   tan 1
2
1 2
n
2
1
2
;

2 

 
 1   2    2   
 n  
  n 


In Prob. 11.12: n = 10 rad/sec,  = 50 rad/sec  = 0.1
y amp
1

 0.042
1
kX o
2
2 2
 2500 
50  

 1 
   2(0.1)  

100 
10  

50
2(0.1)
10  2.4 
   tan 1
2
 50 
1  
 10 
2
2
11.19
The equation for a mass, spring, damping system is given as Eq. 11.23,
and the general form of it is given by Eq. 11.22:
d2y
dy
m 2 
 ky  F( t ) ; where m=mass, =damping ratio, k=spring
dt
dt
constant
Substituting for m, k, and :
d2y
dy
0.020 2  5  1000y  F( t ) ; all units in MKS
dt
dt
d2y
dy
F( t )
standard form: 0.02  10 3 2  5  10 3
y
1000
dt
dt
1
1
 k  2 1000  2
n     
  223.6rad / sec
 m
 0.02 

5


 0.56 The system is under damped.
1
1/ 2
(
.
)

2
1000
0
02
2km  2
11.12
11.20
In this problem, we can assume that effectively we have a mass, spring,
damping system governed by Eq. 11.22. For the given transducer, k = 106
N/m and m = 2 kg.
a- n = (k/m)1/2 = (106/2)1/2 = 707.1 rad/sec
b- In this case  = 0.1 and  = 25 Hz
Amplitude ratio is given by:
y amp
1

1
kX o
2
2 2
2


  

 1  2    2
 
 n  
  n 


y amp
1

.
 1051
 5.1% error in amplitude
1
kX o
2
2






2

( 50 ) 2 
50  

  2( 0.1)
 1 
 
2

707.1 
  10 6  
 
 

2

 




n
   tan 1
2
1 2
n
2
   tan 1
50
707.1  2.68 0
(50 ) 2
20.1
1
 10 6 


 2 
2
11.13
11.21
In this problem, k = 1 N/m = 1106 N/m
m = 250 N = 25.5 kg; and  = 0.07. For this underdamped system, the
actual amplitude will be greater than the ideal amplitude.
Governing equations for amplitude and phase response of the system are
Eq. 11.30 and 11.31.
y amp
1

1 =1.005
kX o
2 2
  2 2 


 
1  2    2 
 n   n  
1
2
  2  2 
 2

1  2     0.14    0.995
n  
  n  
solving:
n 

=.071
n
k
110 6 N / m

 63.2rad / sec
m
250kg
 = .071n = 4.5 rad/sec
11.22
In this problem, k = 5 lbf/in = 5106lbf/in=60106 lbf/ft
m = 50lbm; and  = 0.05. For this underdamped system, the actual
amplitude will be greater than the ideal amplitude.
Governing equations for amplitude and phase response of the system are
Eq. 11.30 and 11.31.
y amp
1
=1.005

1
kX o
2
2
2

2  
 
 1  2    2
 
 n  
  n 


1
2
2 2

2 

 


 1 
   0.1
   0.995
n  
  n 2 




solving:
=.071
n
n 
k
ft  lbm
60 10 6 lbf / ft

 32.17
 6,213rad / sec
m
50lbm
lbf  sec 2
 = 0.071n = 441 rad/sec
11.14
11.23
2500 Nx1000mm / m
 10  10 6 N / m
0.250mm
k
N
1
n 
 10  106 
 1000rad / sec
m
m 10kg
Under a step input (sudden application of load) for  = 0.1, Eq. 11.28 can
be used. The maximum deviation from equilibrium is:
e  n t
1
 0.995
1 2
k
for  = 0.1, e  nt  0.005  0.995  0.005
Solving for t, t = 53msec
11.24
k
500 lbf
 50000 lbf / in  600,000lbf / ft
0.01in
k
lbf
ft  lbm
1
 600,000

 32.17
 982.47rad / sec
m
ft 20lbm
lbf  sec 2
Under a step input (sudden application of load) for  = 0.1, Eq. 11.28 can
be used. The maximum deviation from equilibrium is:
e  nt
1
 0.995
1  2
n 
for  = 0.1, e   t  0.005  0.995  0.005
Solving for t, t = 54msec
n
11.15
11.25 We should calculate natural frequency and damping ratio of the system
using
Eq. 11.35 through 11.37:
8314
R
.
 298  448.7m / sec
T  13
m
16
 d 2 
  0.005 2 
3
6
Vs  L
  0.2
  4  10 m
4
4




C  RT  
Vt  2  10 6 m 3
Then:
C
n 
L 0.5  Vt Vs

448.7
 2243.4 rad / sec
0.2 0.5  0.5
5
.  10
115
32
 32
 14.7 kg 3
Rl 
2
2
m sec
ds
0.005

.  10 16
P
PM 10001013

 6.54 kg / m 3 (M=16 for Methane)

8314298
RT R T
14.7 kg 3
0.2m
Rl L
m sec
0.5  0.5  0.5  10 3
0.5  Vt Vline 

3
2C
2 6.54kg / m 448.7m / sec


Because  << 1 Eq. 11.28 can be used to calculate the time lag to detect
the pressure loss to 90%,
e  nt
e  nt
1
 0.9;
 0.1
1  2
1  2
t
 ln 0.1
 n
 2 sec
Damping might be increased by putting a restriction in the sensing line.
This should reduce the settling time.
11.16
11.26 If instead of natural gas the line carries compressed air (under similar
condition), the delay time will be as follows:
For air C, , and  have different values then natural gas.  = 1.810-5 N-s/m2 , k
= 1.4 and M=29 from Table B.2.
8314
R
.
 298  345.8m / sec
T  14
M
29
 d 2 
  0.005 2 
3
6
Vs  L
  0.2
  4  10 m
4

 4 

C  RT  
Vt  2  10 6 m 3
18
.  10 5
32

32
 23 kg 3
Rl 
2
2
m sec
ds
0.005


.  10 3 29
P
PM 1000 10  1013


. kg / m 3

 1186
RT R T
8314298
n 

345.8
C

 1729 rad / sec
0.2 0.5  0.5
L 0.5  Vt Vs
230.2
Rl L
0.5  Vt Vline 
0.5  0.5  5.6  10 4
2C
21186
. 345.8
Because  << 1 Eq. 11.30 can be used to calculate the time lag to detect
the pressure loss to 90%,
e  nt
e  nt
1
 0.9;
 0.1
1  2
1  2
t
 ln 0.1
 n

ln 0.1
5.6  10 1729
4
 2.4 sec
Damping might be increased by putting a restriction in the sensing line.
This should reduce the settling time.
11.17
11.27 If we use Table 11.1 for very low damping ratios, for an amplitude error of
1%, /n=0.1, or  = 0.1n = 1kHz.
To find a more exact value for  to result in at most 1% amplitude error, we
should use Eq. 11.30:
y amp
1
=1.01

1
kX o
2
2
2

2  
 
 1  2    2
 
 n  
  n 


1
2
2 2

2 

 


 1 
   2
   0.99
  n 2 
 n  


Solving for (/n)
 /  n = 0.0101
 = 1.01 kHz
  = 1.01 kHz
Which is essentially the same as the value obtained from Table 11.1.
11.28 This is similar to problem 11.20.
According to Table 11.1, for amplitude error of less than 1% for a second
order system (such as the pressure transducer under consideration), /n
= 0.1 where  is the frequency of input signal.
The solution for natural frequency and damping ratio for Problem 11.18 is:
Using Eq. 11.35 through 11.37:
8314
R
.
 298  448.7m / sec
T  13
16
m
 d 2 
  0.005 2 
6
3
3
6
Vs  L
  0.2
  4  10 m ; Vt  2  10 m
4
 4 


C  RT  
Then:  n 
Rl 

C
L 0.5  Vt Vs

448.7
 2243.4 rad / sec
0.2 0.5  0.5
115
.  10 5
32

32
 14.7 kg 3
2
m sec
ds 2
0.005
.  10 16
P
PM 10001013

 6.54 kg / m 3 (M=16 for Methane)

RT R T
8314298
14.7 kg 3
0.2m 
Rl L
m sec
0.5  0.5  0.5  10 3

0.5  Vt Vline 
3
2C
2 6.54kg / m 448.7m / sec


so,  = 0.1 n = 224 rad/sec = 36 Hz
11.18
11.29 We will use Eq. 11.44 and 11.45 to calculate the natural frequency and
damping ratio,
Natural frequency:
Vse
C
fn 
4 L BCvt  Vt  Vse
From Table B.1 for water at 10C, B = 2.11109 N/m2,  = 999.7 kg/m3,  =
1.30810-3 N-s/m2.
1
1
 B  2  211
.  10 9  2
C  
  1453 m / sec
 
 999.7 
Vt = 210-6 m3
4
4  ds 2 
Vse  2 Vs  2 
 L =(1/)(0.005)20.15=1.210-6 m3

  4

BCv   BCvt  Vt  Vse 
CVt  3  10 5 (cm 3 / psi )10 6 (m 3 / cm 3 )[14.7 / 101  10 3 ]psi / Pa
 4.35  10 15 m 3 / Pa
BCv = 2.111094.3510-15+210-6+1.210-6
= 12.310-6
Cv = 5.810-15 m3/Pa
1453
12
.  10 6
fn 
 757Hz
4  0.15 12.3  10 6
Damping ratio:
32 

3
d s

 10 3
.
LCv 32 1308

3

 0.005
  0.155.8  10   17.  10
15
4
999.7
11.30 From the solution for Problem 11.22,
n = 757 Hz
 = 1.710-4
If  = 10 Hz, /n = 10/757 = 0.013 and substituting these values into Eq.
11.32 results in:
output amp
1

1
input amp
2
2 2
2


  

 1  2    2
 
 n  
  n 


1
=
 10002
.
1
2
2
2
 1  0.013 2  2  17
.  10 4  0.013 


In practice the amplitude ratio will be one.

 

11.19
11.31 Sudden change in pressure line corresponds to a step change in the input
of the transducer ( in this case second order system). The behavior of the
system is represented be Eq. 11.28,
 1

y
 1  e  nt 
sin  n t 1   2   
ye
 1   2

From the Problem 11.22 solution:  = 1.710-4, n = 757 rad/sec
The maximum deviation of the amplitude will be:
y max  y e
e  nt

 0.05
ye
1  2


1  17
.  10 7

2

1
 ln 0.05
 23 sec
 n
17
.  10 4  757
This is rather long, because there is little damping involved! The method
used to estimate damping may predict a low value.
t
ln 0.05

11.20
11.32 Eq. 11.44 is used to calculate the natural frequency,
Vse
C
fn 
4 L BCVt  Vt  Vse
Uncertainty in calculation of f n in term of Cvt can be calculated through RSS
relation (Refer to Chapter 7, Eq. 7.4.)
1/ 2
 f
2 
w fn  ( n ) 2 w Cvt


 Cvt
1/ 2
fn
 CVse

[
(BCvt  Vt  Vse ) 1/ 2 ]
Cvt Cvt 4L
CVse1/ 2
1
* (  B)(BCvt  Vt  Vse ) 3 / 2
4L
2
w fn
w Cvt
1 f
1
 ( n )w Cvt   B
Eq.A
fn
f n Cvt
2 BCvt  Vt  Vse
From Problem 11.22:
From Table B.1 at 10C, B = 2.11109 N/m2,  = 999.7 g/m3

1
2
1
 211
 B
.  10 9  2
C  
  1453 m / sec
 
 999.7 
Vt = 210-6 m3
4
4  ds 2 
Vse  2 Vs  2 
 L =(1/)(0.005)20.15=1.210-6 m3

  4

BCv   BCvt  Vt  Vse 
CVt  3  10 5 (cm 3 / psi )10 6 (m 3 / cm 3 )[14.7 / 101  10 3 ]psi / Pa
 4.35  10 15 m 3 / Pa
BCv = 2.111094.3510-15+210-6+1.210-6
Cv = 5.810-15 m3/Pa
= 12.310-6
1453
12
.  10 6
 757Hz
4  0.15 12.3  10 6
Substituting into Eq. A above:
w fn
0.5  ( 2.11  10 9 )( 0.25  4.35  10 15

 0.093 = 9.3%
fn
12.3  10 6
This is the uncertainty in natural frequency due to a 25% uncertainty in
determination of compliance.
fn 
11.21
11.33 We will follow the same procedure as Example 11.7.
From Table B.1 for water at 10C, B = 2.2109 N/m2,  = 998.2 kg/m3,  =
1.00510-3 N-s/m2.
3
0.003 * 10 6
V
15 m
2
9
10
.
*
Compliance Cvt 


Pmax
Pa
1013
. * 10 3
150 *
14.7
B
C  ( ).5  (2.20 * 10 9 / 998.2).5  1484 m / sec

Vs  Ld s2 / 4  0.25
Vse 
4

2
0(.005) 2
 4.91 * 10 6 m 3
4
. * 10 6 m 3
Vs  199
Vt  2.8cm 3  2.8 * 10 6 m 3
Cv  Cvt  Vt / B  Vse / B
 2.9 * 10 15 
fn 
2.8 * 10 6 199
. * 10 6

 5.08 * 10 15 m 3 / Pa
2.2 * 10 9
2.2 * 10 9
Vse / B
1484
C

4L Cvt  Vt / B  Vse / B 4 * 0.25
.  10 6
199
2.2  10 9  626 Hz
5.08 * 10 5
.
* 10 3  * 0.25 * 5.08 * 10 15
32 LCv
32 * 1005
. * 10 4

 16
3
3

998.2
 (0.005)
d s
These values are comparable to those of Example 11.7. Lower value of f n is
mostly due to greater sensing line length (L).

11.22
11.34 Note: In this problem, we have to use the fluid properties of water at 100C.
In the problem statement, 200 oC is cited which is not included in Table B-1, so
100C is used in this solution instead. Data for temperatures greater than 100 oC
are available from other sources. Viscosity data is available from heat transfer
texts and bulk modulus data can be obtained by differentiating specific volume
data from saturated and compressed liquid tables [B=P/(/)].
N sec
At 100C from Table B  1,   958.4 kg / m 3 ,  .284 * 10 3
m2
and B  207 * 10 7 N / m 2 . Following Example 117
. , Speed of sound,
2.07 * 10 9 .5
B
)  1469.4 m / sec
C  ( ) .5  (
958.4

Compliance:
.003 * 10 6
V
Cvt 

 2.90 * 10 15 m 5 / N
P 150 / 14.7101325
volume:
. * 10 6 m 3
Vs  0.1 *  (.005) 2 / 4  196
Vse 
4

2
Vs  (
4

2
)196
. * 10 6  7.96 * 10 7 m 3
2.8 * 10 6 7.96 * 10 7

 4.64 * 10 15
2.07 * 10 9 2.07 * 10 9
The natural frequency is obtained from Eq. 11.48,
Vse
C
fn 
4L BCvt  Vse  Vse
Cv  Cvt  Vt / B  Vse / B  2.90 * 10 15 
1469.4
7.96 * 10 7

4 * 0.1 2.07 * 10 9 * (4 * 10 15 )
 1057.5 Hz
The Damping Ratiois obtainedfromEq.11.45:

32 LCv

d s3

32 * (.284 * 10 3 )  * 0.1 * 4.64 * 10 15
958.4
 (.005) 3
 3.52 * 10 5
11.23
11.35 The amplitude distortion of an accelerometer can be calculated from Eq.
11.30,
1
 1005
.
2
 2
2 1/ 2
[(1  2 )  (2 /  n ) ]
n
2 2
 2
(1  2 )  (12
.
) .99
n
n
2
y
 n2
y 2 .56 y  0.01  0
y .0185


.136 or   109 Hz
n
The other value of the solution is y  0.542,

 0.736 which corresponds to
n
w=589 Hz. Examining Figure 11.8, for a damping ratio of 0.5 (close to the 0.6
here) it can be seen that a given value of the amplitude ratio will occur at two
values of . The lower frequency is limiting.
11.36 Amplitude distortion in measurement of vibration with a frequency of 1500
Hz,

1500

.15
 n 10,000
U sin g Eq. 1130
.
1
amplitude ratio 
[(1 
 2
 2 1/ 2
)  (2
) ]
2
n
n
2
1
[(1.15 )  (2*.02*.15) 2 ]1/ 2
 1023
.
So the amplitude distortion will be 2.3%

2 2
11.24