Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
Chapter 10: Control Systems Design and Mechatronics
The answers to these problems are not necessarily unique since there is some latitude in equivalent
ways they can be formulated. In these problems, as well as G for gain, the letter C is used meaning
‘Command’. In some instances, such as on-off control, C & G may only have a value of either 0 or 1.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
306
10-1) The circuit diagram below is an open loop control for tuning on or off a light. Draw a block
diagram of the control. Why is it considered open loop?
Need: Block diagram equivalent to circuit diagram.
Know: Circuit diagram for a simple light circuit. Closing the switch will enable
current I to pass through the bulb’s filament given by Ohm’s law, I = V/R. The power
from the bulb is P = I × V.
Switch
Light Bulb
R = 100. Ω
V = 10. V
How: Input on closing switch is a gain G of 100%.
Solve: Block 1 is switch for which G = I = V/R = 10./100. = 0.10 A. when switch is
turned on and C = 1; block 2 is bulb for which P = I 2 × R = 0.102 × 100. = 1.0 W.
C
I=G×C
Switch, G
= V/R
I
I2 V
P
P P==I R
Bulb
The bulb is on and stays on provided the switch is closed. Day or night, needed or not
until the switch is manually turned off. That is why this is considered open loop
control.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
307
10-2) The circuit diagram below represents a closed loop system for lighting control in an otherwise
dark room. Assume you want a light bulb to provide constant illumination of strength LD even if the
strength of the light bulb changes with age. Light control is achieved by a suitable light sensitive
sensor that compares bulb’s measured output LB to the desired light level. The sensor returns a
measured light signal LM. The command signal is the difference between LD and LM and is
communicated to a controller that sets a variable circuit voltage from 0 – 100.V. The bulb’s intensity
is set by this voltage. Draw a block diagram of the control, indicating on the diagram which parts of
the conceptual sketch correspond to the blocks.
Bulb output LB
Switch
Sensor,
sets LD
V = 0 - 100. V
Light bulb, RB
Controller circuit
Need: Block diagram equivalent to circuit diagram.
Know - How: Circuit similar to Exercise 1 except that incoming light is detected by a
sensor that operates an actuator that turns the light up or down.
Solve:
C ∝ LD - LM
+
Desired
Light, LD
-
V =G × C
LB
V
P = V2/R
Δ(light)
Control
Bulb
Sensor
LM
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
308
10-3) Part of the mechanism of a toilet is the system that refills the toilet tank after a
flush. A conceptual sketch is shown below. Water enters the tank through an annular
opening around the plug; eventually the water level, WL, in the tank rises and the float
also rises pushing down the plug and turning off the supply of fresh water. Draw a block
diagram of the system, indicating on the diagram which parts of the conceptual sketch
correspond to the blocks.
Plug
Pipe
Lever
Float
level
Water
Need: Block diagram coresponding to physical diagram.
Know – How: This is a feedback device with the incoming water rate regulated by the
lever/float mechanism. Need blocks for plug, water rate and water level plus
comparator for water level position.
Solve:
ΔWL = WL - WLAct
WL
+
-
Plug position
Water Flow
WLAct
Plug position
= GP x ΔWL
Water Flow=
GW x
Plug Position
Actual
Water Level =
GF x Water Flow
Plug, GP
Water Flow, GW
Float, GF
WLAct
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
309
10-4) The conceptual sketch for an open loop control for a toaster is shown. Draw a block diagram of
the system, indicating on the diagram which parts of the conceptual sketch correspond to the blocks.
Toast
Resistor
Switch
Plug
Radiant heat
Need: Block diagram corresponding to open-loop toaster.
Know: Need blocks for switch, heater and toast color.
How: Switch block shows the line voltage when the switch is turned on; the power
block converts the voltage to power as it dissipates in the resistor representing the
heater and finally the last block represents the browning of the toast proportional to the
energy absorbed by the toast.
Solve: C is either 0 (unplugged) or 1 (plugged in to a power socket).
Voltage, V
Plug, C
V=G×C
Switch, G
Toast
color
Power, P
P = V 2/R
Heater
Copyright ©2010, Elsevier, Inc
Brown rate
Toast
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
310
10-5. The conceptual sketch for a closed loop control for a toaster is shown below. The controller is a
bimetallic switch. It works by combining two metals of different thermal expansion so that, when
heated, it bends away from the more expansive metal and can then open a pair of contacts. The heat is
interrupted when the contacts are opened and is re-established when the bimetallic strip cools. The
heat source is “waste” heat generated in the resistor that also heats the toast. The stationary contact is
adjustable so that the clearance between the two contacts determines the set point as to when the
heater is on or off.
Resistor
Toast
Bimetallic
Switch
Plug
Radiant heat
Bimetallic strip
V
Contacts
Current I or 0
Bimetallic switch
Draw a block diagram of the system, indicating on the diagram which parts of the conceptual sketch
correspond to the blocks.
Need: Block diagram corresponding to closed-loop toaster.
Know: Have blocks for open loop toaster
How: Just add block for bimetallic switch as a feedback.
Solve: As in the previous exercise, C is 0 or 1.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
Voltage, V
C
Contact
set point
+-
Angle of
Strip
Power, P
Toast
temperature
V=G×C
P = V2/R
Browning
rate
Contacts
Heater
Toast
Bending fn
Bimetallic
The feedback loop determines whether the switch is closed. It operates as an on/off
controller and achieves the proper toast color by the physical design of the bimetallic
controller.
Copyright ©2010, Elsevier, Inc
311
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
312
10-6. The conceptual diagram of an open loop control for the heat of a room is shown.
Furnace gain
GSw (0, 1)
Switch
Room
Q
Q
Radiator
The room achieves a given temperature as a function of the added heat from the radiator and the heat
loss QL from the room. But, for simplicity, ignore QL as being too slow to influence the immediate
response. Draw a block diagram of the system that determines TRoom, indicating on the diagram
which parts of the conceptual sketch correspond to the blocks. Assume the controller switch is a
simple off/on (0,1) and the furnace responds with a gain G to produce heat at a rate Q 1.
Need: Block diagram corresponding to physical diagram.
Know – How: Need blocks for switch, furnace, radiator, and room.
Solve:
Q
Input I
I = GSw
Switch
GSw= 0,1
Q= GF x I
Furnace,
GF
QRad
QRad = Grad x Q
Radiator,
GRad= 1
TRoom
TRoom=
GR x QRad
Room, GR
1
Controllers may actuate their controlled output using either current or voltage. The advantage of current as a control
signal is that it is undiminished over long distances (conservation of electric charge), something useful perhaps for
geometrically large installations.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
313
10-7. The conceptual diagram and block diagram of an open loop control for a lawnmower engine are
shown below. Assume the control is turned on at time t = 0 and left on thereafter. Use the
proportional gains for the throttle = 0.001 kg/s and for the engine = 1000. kW/kg/s as shown in the
block diagram. Prepare a spreadsheet and graph showing the power of the engine each second for a
total of 20 seconds.
Throttle control
Fuel
Engine
AIR
AF
AF
C = 0 or 1
AF=.001C kg/s
AF
AF kg/
Input
I = 0, 1
Throttle
P = 1000. × AF
kW
P, kW
Power, kW
Engine
AF = GTh x I
Power = G E x AF
Throttle GTh
= 0.001 kg/s
Engine GE = 1000 kW/kg/s
Need: Graph of power for given AF and with stated gains.
Know – How: Use spreadsheet to calculate power as a function of time; note there
will be a discontinuity at t = 0 since there is no inertia terms in this model to limit the
initial surge of power.
Solve: I = 1 means the lawn motor is on, 0 means it is off.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
B
C
D
Throttle
=0.001
gain
Engine gain
=1000
t s
CI (0
(0or
or1)1)
0
1
1
1
1
1
17
18
19
20
21
22
23
24
25
26
0
0
=1+B22
=1+B23
=1+B24
=1+B25
A
B
17
18
19
20
21
22
23
24
25
26
t s
Power, kW
0
0
1
2
3
4
1.200
1.000
0.800
0.600
0.400
0.200
0.000
E
AF kg/s
=$D$17*C21
=$D$17*C22
=$D$17*C23
=$D$17*C24
=$D$17*C25
=$D$17*C26
C
Throttle
gain
Engine
gain
D
0.001
P kW
=$D$18*D21
=$D$18*D22
=$D$18*D23
=$D$18*D24
=$D$18*D25
=$D$18*D26
E
1000
CI (0 or 1) AF kg/s P kW
0
0
0.000
1
0.001
1.000
1
0.001
1.000
1
0.001
1.000
1
0.001
1.000
1
0.001
1.000
Power response
0
10
Time, s
Copyright ©2010, Elsevier, Inc
314
20
F
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
315
10-8. The conceptual diagram and block diagram for open loop control of a 1000. W microwave
oven are shown below (The “magnetron” is the device that delivers the microwaves into the oven’s
cavity.) Assume the control is turned on to value C = 1 at time t = 0 and left on thereafter. Prepare a
spreadsheet and graph showing the power delivered to the food each second for a total of 6 seconds.
MagnetronPower Device
C
P = 1000. × C
Control
adjustment
Microwave
MicrowavePower
Power
Magnetron
Food
Need: Graph and spreadsheet of power produced in microwave oven.
Know – How: Use block information with gain given there.
Solve:
A
B
16 Magnetron gain
1000
17
18
19
20
21
22
23
A
16 Magnetron gain
17 1000
18
19
20
21
22
23
C
t, s
D
C (0,1)
0
1
2
3
4
5
6
B
C
t, s
0
=C17+1
=C18+1
=C19+1
=C20+1
=C21+1
=C22+1
E
P, W
0
1000
1000
1000
1000
1000
1000
D
E
C (0,1) P , W
0
=D17*$A$17
1
=D18*$A$17
1
=D19*$A$17
1
=D20*$A$17
1
=D21*$A$17
1
=D22*$A$17
1
=D23*$A$17
Copyright ©2010, Elsevier, Inc
0
1
1
1
1
1
1
P, W
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
316
1200
Power, W
1000
800
600
400
200
0
0
2
4
6
Time, s
Note: The rate of change of power in the first second is not realistic – its rise time is not defined in
this simple model.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
317
10-9. Suppose a simplified thermostatic room controller operates a furnace that can deliver heat to a
room at the rate of 1000. watts. The initial temperature of the room is TR = 20.0°C, and that the
outside temperature is T0 = 10.0°C.
For this example assume the room will lose some heat over a relatively long period when you are
tracking the temperature. To simplify, assume the furnace delivers the net heat being the difference
between what the furnace produces and the heat lost from the room, QNet = Q - QL.
Draw a single block of a block diagram modeling the surroundings of the heating system under the
assumptions that the room gains heat at the rate of QNet = Q - QL = (1000. - QL) where QL = 20. × (TR
–To) watts, and that the temperature of the room at time t +10 minutes is the temperature at time t
minutes plus 0.0005 × QNet. Give the room temperature at intervals of 10. minutes up to 100. minutes.
Need: Block diagram + temperature response of room (use a spreadsheet)
Know - How: Block diagram will need specifics of transfer function;
TR, t + 10 = TR, t + 0.0005 [1000. - 20. × (TR, t - T0 )]
and QNet = (1000. - QL) where QL = 20. × (TR –To)
Solve:
TR, t + 10 = TR, t + 0.0005 × QNet
TRoom, t+1
Room
transfer fn
QNet = [1000 – 20. (TR, t – 10.0)]
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
A
Q , in W
Q loss cft
T 0, C
TR, 0 C
t , minutes
Copyright ©2010, Elsevier, Inc
0
10
20
30
40
50
60
70
80
90
100
B
1000
20.0
10.0
20.0
T R,t C
20.0
20.4
20.8
21.2
21.6
22.0
22.3
22.7
23.1
23.5
23.8
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
A
B
Q , in W
Q loss cft
T 0, C
TR, 0 C
1000
=20
=10
20
t , minutes
T R,t C
0
=10+A18
=10+A19
=10+A20
=B15
=B18 + 0.0005*($B$12 - $B$13 * (B18-$B$14))
=B19 + 0.0005*($B$12 - $B$13 * (B19-$B$14))
=B20 + 0.0005*($B$12 - $B$13 * (B20-$B$14))
24.0
Temperature, C
12
13
14
15
16
17
18
19
20
21
318
23.0
22.0
21.0
20.0
0
20
40
60
Time, minutes
Copyright ©2010, Elsevier, Inc
80
100
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
319
10-10. The conceptual sketch below shows a model for a closed loop control of a lawnmower engine.
Unlike the lawn mower motor of exercise 7, there is an additional feedback device. It takes advantage
of a cooling fan that is directly connected to the output shaft of the lawn mower. This fan blows on a
flap indicated in the diagram, which is preset to a desired position, and which is kept in that position
against a spring tension if the fan is rotating at a desired speed. The desired speed is set by initializing
the spring with a tension SSP corresponding to a desired operating condition such as 3,000 RPM.
If the fan rotates too fast (i.e., the engine speed is too high) the flap is blown clockwise causing a
mechanical linkage to close the throttle and reduce engine speed. If the speed is too low, the flap is
pulled counterclockwise by the spring.
Draw a block diagram for the this control system, assuming: 1) that the control unit sends an error
signal in terms of the spring tension, (SSP (set point) – S) to throttle a proportional amount of air/fuel
GTh(SSP – S) kg/s, 2) resulting in engine power P = GE × AF, and 3) the speed response function for
the motor speed is N = GSR × P. Finally the controller converts engine speed to spring tension by the
relationship S = GSTN.
Need: Block diagram of feedback control for lawnmower motor
Know - How: The difference in spring tension between what is set and the actual
(mediated by the engine speed) produces a change in the amount of fuel and air
mixture fed to the engine. In turn, this produces a change in the power output of the
engine, which in turn blips the engine speed. Finally the loop is closed as the engine
speed moves the controller flap accordingly in a feedback mode.
Solve:
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
AF
P
320
N
SSP-S
+
-
GTh(SSP-S)
P= GE × AF
N = GSR × AF
S
Copyright ©2010, Elsevier, Inc
S = GST × N
S
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
321
10-11. Suppose that for the open loop toaster control you sketched in exercise 5, the variable that you
really want to control is d (for darkness), where d is the energy absorbed by the toast, in units of
joules. Assume the initial energy of the toast (in units of joules) is d = 0, and d t+1 = dt + K × P × Δt,
where P is the power in watts delivered by the heating coil at time t and K is the proportional gain
relating d to P. Fill in the world block to correspond to this model.
If the line voltage is 115 V and the resistor is 13.2Ω, and the darkness gain is 2.5, plot d in joules for
10. s.
Need: Toaster block model based on Exercise 5. Want d = fn(t) graphed for 10
seconds.
Know: V =115 V, R = 13.2Ω, K = 2.5
How: Enter the equation for darkness of the toast: d t+1 = dt + K × P × Δt into relevant
block.
Solve:
Power, W
Current, A
I = G × V/R
Switch, G = 1
B
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
P = I2R
d t+1 = dt +
2.5. × P × t
Heater
Toast
C
V
R
I
d0
K
D
115
13.2
8.71
0.00
2.5
V
Ohms
A
J
3.0E+04
2.5E+04
Toast color, joules
Plug,
V
Color, J
2.0E+04
1.5E+04
1.0E+04
5.0E+03
t, s
0
1
2
3
4
5
6
7
8
9
10
d, J
0.0E+00
0.00E+00
0
2.50E+03
5.01E+03
7.51E+03
1.00E+04
1.25E+04
1.50E+04
1.75E+04
Copyright ©2010, Elsevier, Inc
2.00E+04
2.25E+04
2.50E+04
5
Time, s
10
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
B
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
V
R
I
d0
K
C
115
13.2
=C13/C14
0
2.5
t, s
0
=B21+1
=B22+1
=B23+1
=B24+1
=B25+1
=B26+1
=B27+1
=B28+1
=B29+1
=B30+1
322
d, J
=$C$16+$C$17*($C$15^2)*$C$14*B21
=C21+$C$17*($C$15^2)*$C$14*(B22-B21)
=C22+$C$17*($C$15^2)*$C$14*(B23-B22)
=C23+$C$17*($C$15^2)*$C$14*(B24-B23)
=C24+$C$17*($C$15^2)*$C$14*(B25-B24)
=C25+$C$17*($C$15^2)*$C$14*(B26-B25)
=C26+$C$17*($C$15^2)*$C$14*(B27-B26)
=C27+$C$17*($C$15^2)*$C$14*(B28-B27)
=C28+$C$17*($C$15^2)*$C$14*(B29-B28)
=C29+$C$17*($C$15^2)*$C$14*(B30-B29)
=C30+$C$17*($C$15^2)*$C$14*(B31-B30)
Copyright ©2010, Elsevier, Inc
D
V
Ohms
A
J
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
323
10-12. The conceptual sketch below shows a model of a closed loop control for heat in the room.
Thermostat
Room
Net heat, QNet
Prepare a block diagram for this closed loop control. Label each block on the block diagram with the
name of the corresponding component or components on the sketch. To simplify, assume the furnace
delivers the net heat being the difference between what the furnace produces and the heat lost from
the room, QNet = Q - QL.
The thermostat operates by comparing TSP (the thermostat’s set point to TM (the measured room
temperature) and sending that signal to the furnace. The furnace is ‘on’ if that difference is positive
and vice versa.
Need: Block diagram corresponding to heating system.
Know - How: Start with the TSP (the thermostat’s set point) and compared to TM (the
measured room temperature). Add blocks for the furnace heat Q, the heat losses QL,
and a transfer function converting their difference to the room temperature.
Solve:
QNet
C =1
Tsp-Tm
+
-
C(0,1) = 1
QNet = QNet
×C
Thermostat
Furnace
TM = GRm
× QNet
Room
Copyright ©2010, Elsevier, Inc
TM
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
324
10-13. Use the block diagram developed in exercise 12. Assume the room is initially at T0 = 15°C and
the set point for the furnace is 22°C. The furnace heat rate is Q = 1000. W and the room losses are QL
= 20. × (TM – T0).
The response function for the room is 0.003 × QNet°C per minute when the furnace is on and 0.003 ×
QL °C per minute when it is off.
Plot the temperature response of the room for 10 minutes. (Hint: Use ‘IF’ statement, IF(test, value if
true, value if false) to indicate when the furnace is on and when it is off.)
Need: Room temperature as a function of time, TM(t) = _____?
Know: TSetPoint = 22°C, TInitial = 15°, response functions for furnace on and off; block
diagram describing feedback control in exercise 12.
How: Use spreadsheet analysis. For clarity, first program the criterion to switch on the
furnace (TM ≤ TSet point); then compute the incremented temperatures when the furnace
is on (or decremented temperature when it is off).
Solve: Use the ‘IF’ function, IF(test, value if true, value if false)
B
1000
20.0
10.0
15
22.0
11
12
13
14
15 Tset point,
16
17
t, minutes Furnace
18
on/off
0
19
1
20
2
21
3
22
4
23
5
24
6
25
7
26
8
27
9
28
10
29
C
Set point = 22 C
30.0
Room temperature
A
Q , in W
Q loss cft
T 0, C
T initial, C
T M(t ), C
1
1
1
0
1
0
0
0
1
0
0
25.0
20.0
15.0
10.0
5.0
0.0
15.0
17.7
20.2
22.6
21.9
24.2
23.3
22.5
21.8
24.1
23.2
0
5
Time, minutes
Copyright ©2010, Elsevier, Inc
10
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
11
12
13
14
15
16
17
A
Q , in W 1000
Q loss cft =20
T 0, C
=10
T initial, C 15
Tset poin =22
B
t,
Furnace on/off
minutes
18
=IF(C19 >= $B$15, 0, 1)
19 0
20 =A19+1 =IF(C20 >= $B$15, 0, 1)
21 =A20+1 =IF(C21 >= $B$15, 0, 1)
22 =A21+1 =IF(C22 >= $B$15, 0, 1)
23 =A22+1 =IF(C23 >= $B$15, 0, 1)
C
T M(t ), C
=B14
= C19 + 0.003 * ($B$11 * B19 - $B$12 *(C19-$B$13))
= C20 + 0.003 * ($B$11 * B20 - $B$12 *(C20-$B$13))
= C21 + 0.003 * ($B$11 * B21 - $B$12 *(C21-$B$13))
= C22 + 0.003 * ($B$11 * B22 - $B$12 *(C22-$B$13))
Copyright ©2010, Elsevier, Inc
325
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
326
10-14. Suppose a car employs a digital speed sensor as shown in Figure 10.15, and the car’s wheels
are each 0.60 meters in diameter. If the speed sensor consists of the magnet and coil and the magnet
induces one pulse in the coil for every revolution of the axle, how many pulses per second will the
coil send to the pulse counter if the car is traveling at 65. miles per hour?
Need: Number of pulses per second
Know: Diameter of wheel is 0.60 meters; speed of car is 65. mph or 29. meters per
second.
How: This is a basically a problem based on laws of geometry (e.g., distance traveled
by wheel in one revolution is π × diameter) in which one has to be consistent with
units conversion.
Solve: Translational speed, v = ωR = 2π (N/60)× D/2 if N is in RPM units:
[revolutions/minute][minute/seconds][m/revolution] = 29.1 m/s; ∴N (RPS) = 29. × 60.
/(π × 0.5 × 0.60) = 15. pulses/s if one pulse per revolution
(Often there are multiple pulses per revolution set by slotting say 10 cuts uniformly
around the diameter of the rotating shaft. There would be 154 pulses/s if 10 cuts.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
10-15.
327
Do you really need a cruise control?
Suppose you are driving in the Midwest on a
flat road; you are bored with the monotony of
the terrain and you set a brick against the
accelerator pedal that maintains the vehicle’s
speed to 60. mph. Unexpectedly you drive
onto a construction road that is 10. m deep. Ignoring friction and wind losses, will your vehicle still
be at 60. mph at the bottom of the construction road and, if not, what will its speed S be?
Need: Speed of vehicle at bottom of construction road.
Know - How: In the absence of frictional or wind losses, conservation of energy
applies and the GPE of the vehicle will be recovered as TKE at the bottom of the hill.
Initial speed, v = 27 m/s.
Solve: At the bottom of the hill, the final energy is ½ mvinit2 + mgh = ½ mvfinal2 where v
is the final speed. In SI units: v2(final) = v2(initial) + 2gh = 272 + 2 × 9.81 × 10.0 = 925
(m/s)2. ∴v(final) = 30.4 m/s or 68. mph.
Even in this modest change of the surroundings (a 10. m dip), your speed has
increased by 13%. Cruise control is therefore a popular option for today’s cars.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
328
10-16. In the chapter, we have discussed but not derived, an equation for the transient inertial limited
response to engine power and to windage losses. The equation is
St +1 ≅ St2 +
2 ( Δt ) ( P(engine) − P ( losses ) )
m
in which P(engine) and P(losses), respectively, are the engine’s power output and the air resistance
losses. The vehicle has mass m. Starting with Newton’s Law of Motion, derive this equation in which
the vehicle has sped up to St+1 from St in a time interval of Δt.
. (You will need elementary calculus to solve this problem.)
Need: Derive St +1 ≅ St2 +
2 ( Δt ) ( P(engine) − P ( losses ) )
m
Know - How: Start with Newton’s Law of Motion.
F(“engine”)
Solve: First a sketch:
F(“losses”)
If horizontal, there are two forces: one propelling the car, one opposing its forward
motion. Newton’s Law deals with the net effect of these on the car’s motion. Their
combined effect is to accelerate or decelerate the vehicle depending on which term is
the larger.
dv
For v in the direction of motion: m = { F (" engine ") − F (" losses ")} .
dt
Power is F × v; ∴multiply each side of this equation by v;
Then mv
dv
= P(engine) − P(losses ) .
dt
v
Simple calculus shows that
dv 1 dv 2
=
;
dt 2 dt
2Δt
[ P(engine) − P(losses)] in which Δ means the difference in time Δt.
m
Note that Δv 2 ≡ ( St2+Δt − St2 ) .
Hence Δv 2 =
Finally substitute and St +1 ≅ St2 +
2 ( Δt ) ( P(engine) − P ( losses ) )
m
as required.
Again, the assumption has been that P(losses) are constant over the integration step.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
329
10-17) Referring to Figure 10.12, when a throttle plate rotates in a circular plenum, its effective
blocking area is πab where a, b are the principal axes of an ellipse. Show that:
ΔA
= 1 − cos θ
A
where A is the fully open area of the plenum and ΔA is the open area when the plate has been turned
through an angle θ.
ΔA
= 1 − cos θ
A
Know: That the circular plate appears elliptical and is thus of area πab.
Need: Proof that
How: Check Figure 10.12 for geometry: at the partial open position a = R (major or
vertical axis) and b = Rcosθ (minor or horizontal axis).
Solve: The open area is
πR 2 − πR × R cosθ = πR 2 (1 − cosθ )
Hence the fractional open area is π R 2 (1 − cos θ ) π R 2 = 1- cos θ .
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
330
10-18 Some skeptics have suggested that we are not entering a man-made period of global warming
but instead are entering a mini ice age (as Earth has done many times before in its 4+ billion year
history). Consider this simple block model of the conflicting possibilities:
TAtm
TAtm + TCalc
+
TCalc
TCalc = Gp1(TCalc + TAtm)
+
TCalc
TCalc
TCalc + TGlobal warming
+
+
TGlobal warming
TGlobal warming =
Gp2(TCalc+ TGlobal warming )
TGlobal warming
In this model the calculated warm temperatures are subject to a cooling feedback, the result of which
G p1G p 2T Atm
is the predicted temperature of Earth. Show that TGlobalwar min g =
(1 − G p1 )(1 − G p 2 ) in which Gp1 is the
gain for global warming and Gp2 is the gain for global cooling. If K is defined as Gp2/Gp1, plot the
Earth’s predicted temperature from the years 2000 to 2050 for a) K = 1.001, b) K = 1.000 and c) K =
0.999. Assume Gp1 = 0.50011 as in Example 7.
Need: TGlobal warming = ________ °C between 2000 and 2050 for three cases, K = Gp2/Gp1 =
1.01, 1.00 and 0.99.
Know: First loop is positive feedback in which TCalc =
G p T Atm
(1 − G )
(Example 10.6) and
p
TGlobalwar min g =
G p 2TCalc
(1 − G ) since it’s also described by a simple positive feedback loop. (It’s a
p2
positive feedback since more cold = more ice and snow = more reflected sunlight = more
cold)
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
331
How: Eliminate TCalc between the two equations.
Solve: TGlobalwar min g =
G p 2TCalc
(1 − G ) = (1 − G )(1 − G ) = (1 − KG )(1 − G )
p2
N
1
2
3
4
5
6
KG p21T Atm
G p1G p 2T Atm
p2
p
O
Year
2000
=1+N3
=1+N4
=1+N5
p1
P
Tatm C
=K4
=P3
=P4
=P5
K = 1.001
Tglobal warming C
=$K$6*O3*($K$5^2)/( (1-$K$5)*(1- $K$6*$K$5) )
=$K$6*O4*($K$5^2)/( (1-$K$5)*(1- $K$6*$K$5) )
=$K$6*O5*($K$5^2)/( (1-$K$5)*(1- $K$6*$K$5) )
=$K$6*O6*($K$5^2)/( (1-$K$5)*(1- $K$6*$K$5) )
Q
R
K = 1.000
Tamb C Tglobal warming C
=K4
=$K$7 * Q3 * ($K$5^2)/((1-$K$5) * (1 - $K$7 * $K$5))
=R3
=$K$7 * Q4 * ($K$5^2)/((1-$K$5) * (1 - $K$7 * $K$5))
=R4
=$K$7 * Q5 * ($K$5^2)/((1-$K$5) * (1 - $K$7 * $K$5))
=R5
=$K$7 * Q6 * ($K$5^2)/((1-$K$5) * (1 - $K$7 * $K$5))
S
Tatm C
=K4
=T3
=T4
=T5
T
K =0.999
Tglobal warming C
=$K$8 * S3 * ($K$5^2)/((1-$K$5) * (1 - $K$8 * $K$5))
=$K$8 * S4 * ($K$5^2)/((1-$K$5) * (1 - $K$8 * $K$5))
=$K$8 * S5 * ($K$5^2)/((1-$K$5) * (1 - $K$8 * $K$5))
=$K$8 * S6 * ($K$5^2)/((1-$K$5) * (1 - $K$8 * $K$5))
18
17.5
G lobal Tem perature, C
1
2
3
4
5
6
p1
17
16.5
16
K = 1.001
15.5
K = 1.000
K = 0.999
15
14.5
14
13.5
13
2000
2010
2020
2030
2040
2050
Year
J
1
2
3
4
5
6
7
8
9
10
K
L
M
K = Gp2/GP1
or K =
Cooling gain/heating gain
Start T, C
15
Gp1
0.50012
a) K
1.001
b) K
1.000
c) K
0.999
N
Year
O
Tatm C
2000
2001
2002
2003
2004
2005
2006
2007
15
15.044
15.089
15.134
15.179
15.224
15.269
15.314
P
Q
R
S
T
K = 1.001
K = 1.000
K =0.999
Tglobal warming CTamb C Tglobal warming C Tatm C Tglobal warming C
15.044
15
15.014
15
14.984
15.089
15.014
15.029
14.984
14.969
15.134
15.029
15.043
14.969
14.953
15.179
15.043
15.058
14.953
14.938
15.224
15.058
15.072
14.938
14.922
15.269
15.072
15.087
14.922
14.907
15.314
15.087
15.101
14.907
14.891
15.359
15.101
15.116
14.891
14.876
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
332
Note how much the solution is sensitive to the ratio of the gains describing the cooling and
heating with the case of K = 1.000 being the solution already found that ignored the imputed
cooling. This type of phenomenon is known as the ‘Tipping Point’; if the ratio is only off by 1
part in a 1,000 the solution is entirely different not only in magnitude but in sign too. No
wonder each side has its adherents.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
333
10-19. You work for the control and guidance systems division of a major aircraft manufacturer with
a major government contract. You observe employees who regularly leave work early while being
paid for time not worked 2. What do you do?
a)
b)
c)
d)
Ignore it since there is nothing you can do anyway.
Talk to your supervisor about it.
Report it to the government representative on the contract.
Blow the whistle and talk to a newspaper reporter.
Use an Engineering Ethics Matrix to reach your conclusions.
1) Apply the Fundamental Canons: Engineers, in the fulfillment of their professional duties, shall:
1) Hold paramount the safety, health and welfare of the public. Since government contracts
are involved, so is the welfare of the public. This weighs in favor of choices b), c), and
d), and against choice a).
2) Perform services only in areas of their competence. Does not apply.
3) Issue public statements only in an objective and truthful manner. Blowing the whistle
tempts one to make subjective or even untruthful statements for effect. So option d)
should be done with care, and only after other avenues have been explored.
4) Act for each employer or client as faithful agents or trustees. This weighs in favor of
options b) and c).
5) Avoid deceptive acts. Option a), ignoring what you know to exist, is deceptive.
6) Conduct themselves honorably, responsibly, ethically, and lawfully so as to enhance the
honor, reputation, and usefulness of the profession. Again, this weighs heavily against
option a), somewhat against option d) (since whistle blowing can have negative effects on
a profession unless done with absolute integrity and truthfulness) and in favor of options
b) and c).
3) Summarize in an Engineering Ethics Matrix
Options a) Ignore
Canons
Hold
paramount the
safety, health
and welfare of
the public.
Perform
services only
in the area of
your
b) Talk to
supervisor
c) Report
d) Blow
whistle
Does not
apply
Does not
apply
Does not
apply
Does not apply
Does not
apply
Does not
apply
Does not
apply
You may not
be a competent
public
spokesman
2. General Dynamics Corporation paid $1,800,000 to settle a suit for over billing the government for testing F-16
fighters. Company employees knew that General Dynamics billed the defense department for thousands of hours that
were never worked by using falsified time cards from about 50 employees who regularly left the plant early.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
competence
Issue public
statements
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful agents
or trustees
Avoid
deceptive acts
Conduct
themselves
honorably …
334
Silence here is Meets canon
untruthful
Meets canon
Reporter
interviews
endanger
slanting the
truth
Endangers
employer
Meets canon
Endangers
employer
Deceptive
Meets canon
Going to
government
before
informing
employer is
not being a
faithful agent
Meets canon
Dishonorable
Meets canon
Meets canon
Meets canon
Meets canon
Solution: The canons rule out option a), and suggest a phased response. Try b) first. If
it doesn’t work, escalate to c). Use d) only as a last resort if all else fails, remembering
that it is very difficult to be an objective and fully truthful whistleblower.
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
335
10-20. You are a new engineer working for a motorcycle manufacturer that produces a bike with a
known control system instability. This instability can cause the rider to lose control at high speed and
crash. You supervisor says that you should ignore it because everyone knows about it and it would
be too expensive to fix. Besides, a new control algorithm would cause more harm since the drivers
expect the bikes to behave in a certain way. What do you do?
a) Attempt to convince your supervisor that it will be cheaper to fix the flaw than pay the
subsequent law suits.
b) Suggest that a warning label be put on the bikes about riding at high speeds.
c) Talk to your corporation’s legal office about your professional obligations.
d) Blow the whistle and talk to a newspaper reporter.
Use an Engineering Ethics Matrix to reach your conclusions.
1) Apply the Fundamental Canons: Engineers, in the fulfillment of their professional
duties, shall:
2) Hold paramount the safety, health and welfare of the public. Safety is involved. This
weighs in favor of all the choices
3) Perform services only in areas of their competence. Does not apply
4) Issue public statements only in an objective and truthful manner. Blowing the whistle
tempts one to make subjective or even untruthful statements for effect. So option d)
should be done with care, and only after other avenues have been explored.
5) Act for each employer or client as faithful agents or trustees. This weighs in favor of
options a), b) and c).
6) Avoid deceptive acts. Does not apply.
7) Conduct themselves honorably, responsibly, ethically, and lawfully so as to enhance the
honor, reputation, and usefulness of the profession. This weighs somewhat against
option d) (since whistle blowing can have negative effects on a profession unless done
with absolute integrity and truthfulness) and in favor of options a), b) and c).
2) Summarize in an Engineering Ethics Matrix:
Options a) Attempt to
convince
Canons
Addresses
Hold
paramount the canon, but
safety, health does not put
and welfare of safety
paramount
the public.
You probably
Perform
are not an
services only
expert in law
in the area of
or cost benefit
your
calculation
competence
b) Suggest
warning label
c) Talk to
legal office
d) Blow
whistle
Addresses
canon, but
does not put
safety
paramount
Meets canon
Meets canon
Meets canon
Meets canon
You may not
be a
competent
spokesman
Copyright ©2010, Elsevier, Inc
Kosky, Wise, Balmer, Keat: Exploring Engineering, Second Edition
Issue public
statements
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful agents
or trustees
Avoid
deceptive acts
Conduct
themselves
honorably …
336
Does not
apply
Warning must
be objective
and truthful
for this to
meet canon
Does not
apply
Difficult to
enforce
truthfulness
upon a
reporter
Addresses
canon, but not
as strongly as
insisting on a
fix
A slightly
deceptive way
of addressing
safety
Meets canon
Addresses
canon, but not
as strongly as
insisting on a
fix
Meets canon
unless
warning is
deceptive
Meets canon
Meets canon
Not being a
faithful agent
Meets canon
Meets canon
Meets canon
May meet
canon if this
appears only
option left
Solution: The canons suggest a phased response, escalating the response if necessary
in the order a), b), c), then d).
Copyright ©2010, Elsevier, Inc