Unformatted text preview: 1234567898 9.1 a) Flowrate pneumatic transmitter: 15 psig  3 psig qm(psig)= (q gpm  0 gpm) + 3 psig 400 gpm0 gpm psig = 0.03 q (gpm) + 3 psig
gpm Pressure current transmitter: 20 mA  4 mA Pm(mA)= ( p in.Hg − 10 in.Hg) + 4 mA 30 in.Hg  10 in.Hg mA = 0.8 p (in.Hg) − 4 mA
in.Hg Level voltage transmitter: 5 VDC  1 VDC hm(VDC)= (h(m)  0.5m) + 1 VDC 20 m  0.5 m VDC = 0.205 h(m) + 0.897 VDC
m Concentration transmitter: 10 VDC  1 VDC Cm(VDC)= (C (g/L)2 g/L)+1 VDC 20 g/L  2 g/L VDC = 0.5 C (g/L)
g/L b) The gains, zeros and spans are: Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 91 GAIN
ZERO
SPAN PNEUMATIC
0.03psig/gpm
0gal/min
400gal/min VOLTAGE
CURRENT
VOLTAGE
0.8mA/in.Hg 0.205 VDC/m 0.5VDC/g/L
10 in.Hg
0.5m
2g/L
20 in.Hg
19.5m
18g/L *The gain is a constant quantity
9.2 a) The safest conditions are achieved by the lowest temperatures and
pressures in the flash vessel.
VALVE 1. Fail close
VALVE 2. Fail open
VALVE 3. Fail open
VALVE 4. Fail open
VALVE 5. Fail close
Setting valve 1 as fail close prevents more heat from going to flash drum
and setting valve 3 as fail open to allow the steam chest to drain. Setting
valve 3 as fail open prevents pressure build up in the vessel. Valve 4
should be failopen to evacuate the system and help keep pressure low.
Valve 5 should be failclose to prevent any additional pressure buildup. b) Vapor flow to downstream equipment can cause a hazardous situation
VALVE 1. Fail close
VALVE 2. Fail open
VALVE 3. Fail close
VALVE 4. Fail open
VALVE 5. Fail close
Setting valve 1 as fail close prevents more heat from entering flash drum
and minimizes future vapor production. Setting valve 2 as fail open will
allow the steam chest to be evacuated, setting valve 3 as fail close prevents
vapor from escaping the vessel. Setting valve 4 as fail open allows liquid
to leave, preventing vapor build up. Setting valve 4 as failclose prevents
pressure buildup. c) Liquid flow to downstream equipment can cause a hazardous situation
VALVE 1. Fail close
VALVE 2. Fail open
VALVE 3. Fail open
VALVE 4. Fail close
VALVE 5. Fail close 92 Set valve 1 as fail close to prevent all the liquid from being vaporized
(This would cause the flash drum to overheat). Setting valve 2 as fail open
will allow the steam chest to be evacuated. Setting valve 3 as fail open
prevents pressure buildup in drum. Setting valve 4 as fail close prevents
liquid from escaping. Setting valve 5 as fail close prevents liquid buildup
in drum 9.3 a) Assume that the differentialpressure transmitter has the standard range of
3 psig to 15 psig for flow rates of 0 gpm to qm(gpm). Then, the pressure
signal of the transmitter is 12 PT = 3 + 2 q 2
q m dP 24 KT = T = 2 q
dq qm 2.4/qm , q = 10% of qm 12/qm , q = 50% of qm 18/qm , q = 75% of qm 21.6/qm , q = 90% of qm KT = b) Eq. 92 gives
1/ 2 ∆P q = Cv f (1) v gs = qm f ( 1 ) For a linear valve,
f (1) = 1 = αP , where α is a constant. KV = dq
= qm α
dP Hence, linear valve gain is same for all flowrates 93 For a squareroot valve,
f ( 1 ) = 1 = αP
q α 1
q αq
dq
1
KV =
= qm α
= m
= m m
dP
2 1
2 q
2 p 5qmα , q = 10% of qm qmα , q = 50% of qm 0.67qmα , q = 75% of qm 0.56qmα , q = 90% of qm KV = For an equalpercentage valve,
f (1) = R 1 −1 = R αP −1
KV = q dq
= qm αR 1 −1 ln R = qm α ln R dP qm 0.1qmαlnR , q = 10% of qm 0.5qmαlnR , q = 50% of qm 0.75qmαlnR , q = 75% of qm 0.9qmαlnR , q = 90% of qm KV = c) The overall gain is
KTV = KTKV
Using results in parts a) and b)
For a linear valve
2.4α , q = 10% of qm 12α , q = 50% of qm 18α , q = 75% of qm 21.6α , q = 90% of qm KTV = 94 For a squareroot valve
KTV = 12α for all values of q For an equalpercentage valve 0.24αlnR , q = 10% of qm 6.0αlnR , q = 50% of qm 13.5αlnR , q = 75% of qm 19.4αlnR , q = 90% of qm KTV = The combination with a squareroot valve gives linear characteristics over
the full range of flow rate. For R = 50 and α = 0.067 values, a graphical
comparison is shown in Fig. S9.3 7
Linear valve
Square valve
% valve
6 5 K
TV 4 3 2 1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 q/qm Fig. S9.3. Graphical comparison of the gains for the three valves d) In a real situation, the squareroot valve combination will not give an
exactly linear form of the overall characteristics, but it will still be the
combination that gives the most linear characteristics. 95 9.4 Nominal pressure drop over the condenser is 30 psi ∆Pc = Kq2
30 = K (200)2 ∆Pc = , K= 3
psi
4000 gpm 2 3
q2
4000 Let ∆Pv be the pressure drop across the valve and ∆P v , ∆P c be the
nominal values of ∆Pv , ∆Pc, respectively. Then, ( ) ( ) ∆Pv = ∆ P v + ∆ Pc −∆Pc = 30 + ∆ Pv − 3
q2
4000 (1) Using Eq. 92 ∆P
q = C v f (1) v g s 1/ 2 (2) and q ∆ Pv Cv = f (l ) g s −1/ 2 200 ∆ P v = 0.5 1.11 −1/ 2 (3) Substituting for ∆Pv from(1) and Cv from(3) into (2) , ∆ Pv q = 400 1.11 a) −1/ 2 3 2 30 + ∆ P v − 4000 q f (1) 1.11 ∆ Pv = 5
Linear valve: f (1) = 1 , and Eq. 4 becomes
q 35 − 0.00075q 2 l=
188.5 1.11 −1 / 2 96 −1/ 2 (4) Equal % valve: f (1) = R 1 −1 = 20 1 −1 assuming R=20 q 35 − 0.00075q 2 −1 / 2 ln 1.11
188.5 l = 1+
ln 20
250 200 q(gpm) 150 Linear valve
Equal % valve 100 50 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 l (valve lift) Figure S9.4a. Control valve characteristics for ∆ P v = 5 b) ∆ P v = 30
Linear valve: f (1) = 1 , and Eq. 4 becomes
q 60 − 0.00075q 2 l=
76.94 1.11 −1 / 2 Equal % valve: f (1) = 20 1 −1 ; Eq. 4 gives q 60 − 0.00075q 2 −1 / 2 ln 1.11 76.94 l = 1+
ln 20 97 300
Linear valve
Equal % valve
250 q (gpm) 200 150 100 50 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 l (valve lift) Figure S9.4b. Control valve characteristics for ∆ P v = 30 c) ∆ P v = 90
Linear valve: f (1) = 1 , and Eq. 4 becomes
q 120 − 0.00075q 2 l=
44.42 1.11 −1 / 2 Equal % valve: f (1) = 20 1 −1 ; Eq. 4 gives q 120 − 0.00075q 2 −1 / 2 ln 1.11 44.42 l = 1+
ln 20 98 300 250 Equal % valve 200 150 100
Linear valve
Equal % valve
50 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 l (valve lift) Figure S9.4c. Control valve characteristics for ∆ P v = 90 Conclusions from the above plots:
1) Linearity of the valve
For ∆ P v = 5, the linear valve is not linear and the equal % valve is
linear over a narrow range.
For ∆ P v = 30, the linear valve is linear for very low 1 and equal
% valve is linear over a wider range of 1 .
For ∆ P v = 90, the linear valve is linear for 1 <0.5 approx., equal %
valve is linear for 1 >0.5 approx.
2) Ability to handle flowrates greater than nominal increases as ∆ P v
increases, and is higher for the equal % valve compared to that for the
linear valve for each ∆ P v .
3) The pumping costs are higher for larger ∆ P v . This offsets the
advantage of large ∆ P v in part 1) and 2) 99 9.5
Let ∆Pv/∆Ps = 0.33 at the nominal q = 320 gpm ∆Ps = ∆PB + ∆Po = 40 + 1.953 × 104 q2
∆Pv= PD  ∆Ps = (1 –2.44 × 106 q2)PDE – (40 + 1.953 × 104 q2)
(1  2.44 × 10 6 × 320 2 )PDE  (40 + 1.953 × 10 4 × 320 2 )
= 0.33
(40 + 1.953 × 10 4 × 320 2 )
PDE = 106.4 psi
Let qdes = q = 320 gpm
For rated Cv, valve is completely open at 110% qdes i.e., at 352 gpm or the
upper limit of 350 gpm ∆p
C v = q v q s − 1
2 (1 − 2.44 × 10 × 350 )106.4 − (40 + 1.953 × 10 × 350 ) = 350 0.9 −6 2 Then using Eq. 911 q 66.4 − 4.55 × 10 − 4 q 2 −1 / 2 ln 0.9
101.6 l = 1+
ln 50 910 −4 2 − 1
2 400 350 300 q (gpm) 250 200 150
Cv = 101.6
Cv = 133.5
100 50 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 l (valve lift) Figure S9.5. Control valve characteristics From the plot of valve characteristic for the rated Cv of 101.6, it is evident
that the characteristic is reasonably linear in the operating region 250 ≤ q
≤ 350.
The pumping cost could be further reduced by lowering the PDE to a value
that would make ∆Pv/∆Ps = 0.25 at q = 320 gpm. Then PDE = 100.0 and
for qdes = 320 gpm, the rated Cv = 133.5. However, as the plot shows, the
valve characteristic for this design is more nonlinear in the operating
region. Hence the selected valve is Cv = 101.6 911 9.6
a)
1
0.9
0.8
0.7 f 0.6
0.5
0.4
0.3 Linear 0.2 "Square root" 0.1 "Square" 0
0 0.2 0.4 0.6 0.8 1 l The "square" valve appears similar to the equal percentage valve in Fig. 9.8 b)
/ d1 ) 1 =0 1 =0.5 1 =1 1/ 2 1 ∞ 0.707 0.5 1 1 1 1 21 0 1 2 Gain ( df Valve
Quick open
Linear
Slow open The largest gain for quick opening is at 1 =0 (gain = ∞), while largest for
slow opening is at 1 =1 (gain = 2). A linear valve has constant gain.
c) q = C v f (1)
For ∆Pv
gs gs = 1 , ∆Pv = 64 , q = 1024 Cv is found when f (1) =1 (maximum flow): 912 Cv = d) ∆Pv g s = 1024 gal/min 1024
gal.in
=
=128
8
min.(lb)1/2
64 lb/in 2 1 in terms of applied pressure
1 =0
1 =1 when p = 3 psig
when p = 15 psig Then 1 =
e) q (1 − 0)
1
( p − 3) =
p − 0.25
(15 − 3)
12 q = 128 1 2 ∆Pv for slow opening ("square") valve
2 1 = 128 ∆Pv p − 0.25 12 128
2
=
∆Pv ( p − 3) = 0.8889 ∆Pv ( p − 3) 2
144 p=3 , q = 0 for all ∆Pv p =15 , q = 128 ∆Pv
= 0 for ∆Pv = 0
= 1024 for ∆Pv = 64 looks O.K 9.7 Because the system dynamic behavior would be described using deviation
variables, all that is important are the terms involving x, dx/dt and d2x/dt2.
Using the values for M, K and R and solving the homogeneous o.d.e:
0.3 d 2x
dx
+ 15,000 + 3600 x = 0
2
dt
dt This yields a strongly overdamped solution, with ζ=228, which can be
approximated by a first order model by ignoring the d2x/dt2 ter
913 9.8 A control system can incorporate valve sequencing for wide range along
with compensation for the nonlinear curve (Shinskey, 1996). It features a
small equalpercentage valve driven by a proportional pH controller. The
output of the pH controller also operates a large linear valve through a
proportionalplusreset controller with a dead zone. The system is shown
in Fig. E9.8 Reagent Linear pHC
Percent Influent Figure S9.8. Schematic diagram for pH control Equalpercentage valves have an exponential characteristic, similar to the
pH curve. As pH deviates from neutrality, the gain of the curve decreases;
but increasing deviation will open the valve farther, increasing its gain in a
compensating manner. As the output of the proportional controller drives
the small valve to either of its limits, the dead zone of the twomode
controller is exceeded. The large valve is moved at a rate determined by
the departure of the control signal from the dead zone and by the values of
proportional and reset. When the control signal reenters the dead zone, the
large valve is held in its last position. The large valve is of linear
characteristic, because the process gain does not vary with flow, as some
gains do. 914 Note: in the book’s second printing, the transient response in this problem will be
modified by adding 5 minutes to the time at which each temperature reading was taken. We wish to find the model:
′
Tm ( s )
Km
=
T ′( s ) τm s + 1 where Tm is the measurement
T is liquid temperature
From Eq. 91,
Km = range of instrument output 20 mA  4 mA 16 mA
mA
=
=
=0.04 o
o
o
o
range of instrument input
400 C  0 C
400 C
C From Fig. 5.5, τ can be found by plotting the thermometer reading vs.
time and the transmitter reading vs. time and drawing a horizontal line
between the two ramps to find the time constant. This is shown in Fig.
S9.9.
Hence, ∆τ = 1.33 min = 80 sec
To get τ, add the time constant of the thermometer (20 sec) to ∆τ to get
τ = 100 sec.
122
120
118
T (deg F) 9.9 <Time constant> 116
114
112
110 Thermometer
Transmitter 108
106
2 2.5 3 3.5
time (min) 4 4.5 5 Figure S9.9. Data test from the Thermometer and the Transmitter 915 9.10 precision = 0.1 psig
= 0.5% of full scale
20 psig accuracy is unknown since the "true" pressure in the tank is unknown
resolution = 0.1 psig
= 0.5% of full scale
20 psig repeatability = ±0.1 psig
=±0.5% of full scale
20 psig 9.11 Assume that the gain of the sensor/transmitter is unity. Then,
′
Tm ( s )
1
=
T ′( s ) ( s + 1)(0.1s + 1) where T is the quantity being measured
Tm is the measured value
T ′ (t) = 0.1 t °C/s , T ′ (s) = ′
Tm ( s ) = 0 .1
s2 1
0.1
× 2
( s + 1)(0.1s + 1) s ′
Tm (t ) = −0.0011e −10t + 0.111e − t + 0.1t − 0.11 Maximum error occurs as t→∞ and equals 0.1t − (0.1t − 0.11) = 0.11 °C
If the smaller time constant is neglected, the time domain response is a bit
different for small values of time, although the maximum error (t→∞)
doesn't change. 916 2 1.8 1.6 1.4 T', Tm' (C) 1.2 1 0.8 0.6
Tm'(t)
T'(t) 0.4 0.2 0 0 2 4 6 8 10 12 14 16 18 20 t(s) Figure S9.11. Response for process temperature sensor/transmitter 917 ...
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 Spring '08
 Cummings
 Amplifier, Valve, flow control valve

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