Download Problems and Answers 1. Explain the basic control strategy

1
Problems and Answers
1.
Explain the basic control strategy employed in a room airconditioning system. What is the controlled variable?
What is the manipulated variable?
Soln:
The basic strategy of the room air-conditioner can be described
as follows:
• The room temperature is measured by a "thermostat", which
is nothing more than a sensor of temperature. Thus
temperature is a controlled variable.
• The measured temperature is compared to a set point in the
thermostat. Often this is simply a bimetal strip which closes a
contact when the temperature exceeds some limit.
• If the temperature is too low then the compressor and
distribution fan of the air-conditioner are turned on. This
causes room air to be circulated though the unit and thereby
cooled and exhausted back into the room. Thus, the
manipulated or controlling variable is the temperature of the
re-circulated air.
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2.
An Resistence-Temperature Detector (RTD) has
α 0 = 0.005/ oC , R = 500Ω and a dissipation constant
PD = 30mW / oC at 20oC . The RTD is used in a bridge
circuit as show in the figure with R1 = R2 = 500Ω and R3 a
variable resister used to null the bridge. If the supply is 10
volts and the RTD is placed in a bath of 0oC , find the
value of R3 to null the bridge.
R1
V
R2
c
a
b
D
R3
RTD
3
Soln:
First we find the value of the RTD resistance at 0 o C without
including the effects of dissipation.
From the linear
approximation relationship between resistance and temperature,
we have
R(T ) = R(T0 )[1 + α 0 (T − T0 )]Ω
= 500[1 + 0.005(0 − 20)]Ω
= 450Ω
Without the effects of self-heating, we would expect the bridge
to null with R3 equal to 450Ω as well.
However, self-heating exists. Assuming the RTD resistance is
still (an approximation) 450Ω . The current I to three
significant figures is found from
I=
10
V
=
= 0.011A
(
)
500
450
R2 + R T
+
The power is
P = I 2 R = 0.0112 * 450 = 0.054W
We then get the temperature rise
P 0.054
∆T =
=
= 1.8o C
PD 0.030
Thus, the RTD is not actually at the bath temperature of 0 o C
but at a temperature of 1.1o C .
As a consequence, the RTD resistance should be
R(T ) = R(T0 )[1 + α 0 (T − T0 )]Ω
= 500[1 + 0.005(1.8 − 20)]Ω
= 454.5Ω
Thus, the bridge will be null with R3 = 454.5Ω .
The indicated temperature is 1.8o C
temperature rise.
for the self-heating
4
Using an RTD with α 0 = 0.0034/ oC and R = 100Ω at
3.
20 oC , design a bridge and op amp system to provide a
0.0- to 10.0- volt output for a 20 oC to 100 oC temperature
variation. The RTD dissipation constant is 28 mW / oC .
Soln:
First, we find the resistance of the RTD at the two temperature
extremes:
R(20 o C ) = 100Ω
R(100o C ) = 100[1 + 0.0034(100 − 20)] = 127.2Ω
If we use the RTD in a bridge with the rest of the arms at 100Ω ,
then it is will null at 20 oC .
Assuming a 10 volt bridge excitation voltage, we find the offnull voltage at 100 oC as
100
127.2
)10 = −0.599V
∆V = (
−
100 + 100 100 + 127.2
To get the required output of 10 volt, we need a gain of
K = 10 / 0.599 = 16.7
The following circuit will provide the result:
5V
100Ω
167kΩ
100Ω
10kΩ
Vout
100Ω
RTD
10kΩ
167kΩ
5
4.
For the industrial process, vapor flows though a chamber
containing a liquid at 100 oC . A control system will regulate the
vapor temperature. A measurement must be provided to convert
50 oC - 80 oC into 0 - 2.0 volts. The error should not exceed
±1o C . If the liquid level rises to the tip of the transducer, its
temperature will rise suddenly to 100 oC . This event should
cause an alarm.
Temp. Transducer
Vapor Out
Vapor In
100 o C
Heater
6
Soln:
This is an example of a midrange temperature measurement. Let
us use an RTD because the output over the 30 o C range will be
substantially linear. The specifications are:
At 65o C , R = 150Ω , α = 0.004/ oC and PD = 30mW / oC
The three resistance of interest are at 50 o C , 80 o C and 100 o C .
From the linear RTD relationship of resistance and temperature,
we have
At 50 o C , R = 150[1 + 0.004(50 − 65)]Ω = 141Ω
At 80 o C , R = 150[1 + 0.004(80 − 65)]Ω = 159Ω
At 100 o C , R = 150[1 + 0.004(100 − 65)]Ω = 171Ω
For a 1o C error because of self-heating, we can find the
maximum current through the RTD.
The maximum power and maximum current are
P = PD ∆T = (30mW / oC )(1o C ) = 30mW
I = P / R = 30mW / 159Ω = 13.7 mA
Although an op-amp could be used, let us place the RTD in a
bridge circuit and use the offset voltage for measurement.
The small range of resistance will not cause any appreciable
nonlinear effects, and the bridge can be nulled at 50 o C , which
will simplify the signal conditioning.
7
The bridge is excited from a 5.0 volt source because this value is
common. Taking R4 be the RTD, and the value of R2 can be
determined by the requirement that the current be below
13.7 mA .
The voltage across the RTD at 80 o C will be
V = IR = (13.7mA)(159Ω) = 2.17V
Therefore, R2 = (5 − 2.17) / 13.7 = 206.5Ω .
Let us use 220Ω for R2 , because this is a standard value and
will ensure that the current is low and the error condition is
satisfied. To null the bridge at 50 o C , we will take R1 = 220Ω
and use a trimmer to set R3 = 141Ω .
(To check: The actual current flowing in the RTD:
5
At 80 o C , I =
= 13.19mA
220 + 159
5
At 50 o C , I =
= 13.85mA which is slightly above
220 + 141
the maximum and is regarded as acceptable).
Assuming the impedance to be very high, the bridge offset
voltages are:
141 * 5
141 * 5
At 50 o C , ∆V =
−
= 0V
220 + 141 220 + 141
159 * 5
141 * 5
At 80 o C , ∆V =
−
= 0.1447V
220 + 159 220 + 141
171 * 5
141 * 5
At 100 o C , ∆V =
−
= 0.2338V
220 + 171 220 + 141
8
To boost the 80 o C voltage to 2.0 volts, the required gain is
(2/0.1447)=13.8. Because the 5volt source used is ground
referenced, we must use a differential amplifier for the bridge
offset voltage.
The figure shows the required amplifier with gain. The
comparator reference voltage is
Vref = 13.8 * 0.2338 = 3.23V
5V
138 Ω
10 k Ω
R 2 = 220 Ω
R 1 = 220 Ω
10 k Ω
V out
R 3 = 141 Ω
10 k Ω
10 k Ω
R 4 = RTD
5V
548 Ω
V ref
= 3 . 23 V
1k Ω
A la r m
C o m p arato r
9
5.
As a water tank loses heat, the temperature drops by 2K
per minute. When a heater is on, the system gains
temperature at 4K per minute. A two position controller
has a 0.5 min control lag and a neutral zone of ± 4% of the
setpoint about a setpoint of 323K. Plot the heater
temperature versus time. Find the oscillation period.
Solu.:
Setpoint is 323 K, ± 4% neutral zone gives the range of
operation as [310, 336]
(=[323-323*4%, 323+323*4%] = [323-12.92, 323+12.92])
Let us assume we start at the setpoint value, then the
temperature will drop linearly at
T1 (t ) = T (t s ) − 2(t − t s )
where t s = time at which we start observation.
The heater will start at temperature of 310 K (4% below
setpoint), after which the temperature will rise according to
T2 (t ) = T (t h ) − 4(t − t h )
where t h = time at which heater goes on. When the temperature
reaches 336 K, the heater goes off and the system temperature
drops by 2 K/min until 310 K is reached.
Because of the control lag = 0.5 min, there is an undershoot of
1-K (2*control lag = 2*0.5= 1), and an overshoot of +2-K
(4*control lag = 4*0.5= 2).
The system response is then plotted as in the Figure using the
two equations.
Notice that the period is 21.5 minutes.
10
Temp (K)
Upper setpoint
t (min)
Lower setpoint
11
6. Analog Implementation of:
P Control: u = K p e
t
I Control: u = K I ∫ e(τ )dτ
0
D Control: u = K D
de
dt
Soln:
P Control: u = K p e
R2
R
R1
Ve
R
Vout
Inverter
Vout R2
=
Ve
R1
12
t
I Control: u = K I ∫ e(τ )dτ
0
C
R
R1
R
Ve
Vout
Inverter
D Control: u = K D
de
dt
R2
R1
Ve
R
C
R
Vout
Inverter
Resistance R1 is added for stability of the circuit against rapidly
changing signals. The value of R1 is determined by a
requirement that the derivative action stops or "breaks" at some
frequency f c that is higher that any expected signal frequency.
Thus, spontaneous oscillation is prevented, but no measurement
information is lost. Generally, speaking, we make f c 10 to 100
13
times the maximum expected signal frequency. If f s is the
maximum signal frequency, then the requirement can be written
1
R1C <<
, or 2πR1C << Ts
2πf s
How about composite modes?
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6.
What is integrator wind-up? Give at least three possible
solutions to the problem.