Set 6

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_______________________
Last Name, First
CHE426: Problem set #61
1. A step change of magnitude 4 is introduced into a system having the transfer function
Y (s)
10
= 2
s  1.6 s  4
X (s)
Determine (a) Percent overshoot, (b) Rise time, (c) Maximum value of Y(t), (d) Ultimate
value of Y(t), and (e) Period of oscillation. Use the following formulas:
Step response for  < 1.
Y(t)/10 = 1 
1
1  2

t
t

exp    sin  1   2

 



Overshoot = exp  
 1  2

1
1
f=
=
2
T




1  2

Solution
(a) Percent overshoot = 25.4%
(b) Rise time = 1.0814 s
(d) Ultimate value of Y(t) = Y(∞) = 10
(c) Maximum value of Y(t) = 12.54
(e) Period of oscillation = 3.4278 s
Rise time
 1  2

1
  tan 

 




2. The two-tank system shown in Figure E-2 is operating at steady state. At time t = 0, 10 ft3
of water is quickly added to the first tank. Determine the maximum deviation in level (feet)
in both tanks from the ultimate steady-state value and the time at which each maximum
occurs. Data: A1 = A2 = 10 ft2, R1 = 0.1 ft/cfm, R2 = 0.35 ft/cfm.
Note: Q1 = h1/R1, if y(t) = (0) (unit impulse) then Y(s) = 1.
3
20 ft /min
10 ft
3
h1
R1
Q1
A1
h2
R2
Q2
A2
Figure E-2
Solution
t = 1.7539 min
H2d(t) = 0.6059 ft
3.1 Determine y(t = 0), y(t = 0.6), and y(t = ∞) if Y(s) =
1 25( s  1)
s s 2  2 s  25
Solution
y(t = 0) = 0, y(t = 0.6) = 2.05, and y(t = ∞) = 1.
4.1 Sketch the response y(t) if Y(s) = exp(2s)/[s2 + 1.2s + 1]. Determine y(t) for t = 0, 1, 5,
∞.
Solution
y(t) =0 for t = 0, 1, ∞
>> t=5;
yt = 0.1396
5.2 The two tanks shown in Fig. E-6 are connected in an interacting fashion. The system is
initially at steady state with q = 10 cfm. The following data apply to the tanks: A1 = 1 ft2, A2
= 1.25 ft2, R1 = 1 ft/cfm, and R2 = 0.8 ft/cfm.
(a) If the flow changes from 10 to 11 cfm according to a step change, determine
H2(s),i.e., the Laplace transform of H2 where is the deviation in h2.
(b) Determine H2(1), H2(4), and H2(∞).
Q
h1
h2
R2
R1
Q2
A2
A1
Figure E-6
Solution
(a)
H2(s) =
1
0.8
2
s s  2.8s  1
(b)
t = 1 min  H2(t) = 0.8+B*exp(-2.3798)+C*exp(-0.4202) = 0.1777 ft
t = 4 min  H2(t) = 0.8+B*exp(-2.3798*4)+C*exp(-0.4202*4) = 0.6191 ft
t = ∞  H2(t) = 0.8 ft
6.1 The overhead vapor from a depropanizer distillation column is totally condensed in a
water-cooled condenser at 120oF and 230 psig. The vapor is 98 mol % propane and 2 mol %
isobutene. The vapor design flow rate is 40,000 lb/h and average latent heat of vaporization
is 128 Btu/lb. Cooling water inlet and outlet temperatures are 75 and 100oF, respectively. The
condenser heat transfer area is 1000 ft2. The cooling water pressure drop through the
condenser at design rate is 50 psi. A linear-trim control valve (air-to-closed, when CO = 20
mA, PV = 15 psig) is installed in the cooling water line. The pressure drop over the valve is
25 psi at design with the valve half open.
The process pressure is measured by an
electronic (4-20 mA) pressure transmitter whose range is 150-400 psig. An analog electronic
proportional controller with a gain of 2 is used to control process pressure by manipulating
cooling water flow. The electronic signal from the controller (CO) is converted into a
pneumatic signal in the I/P transducer.
Vapor
Cooling
water
Control valve
Condenser
PV
I/P
Reflux drum
PT
PM
PC
CO
SP
a) Calculate the cooling water flow rate (gpm) at design conditions. Water density is 62.3
lb/ft3 and 1 ft3 = 7.48 gal
____________
409 gpm
b) If the cooling water flow rate is 250 gpm at design conditions, calculate the size
coefficient (Cv) of the control valve.
Cv = _________
Cv = 100 gpm/psi0.5
(c) Calculate the value of the signal PM at design condition ________
PM = 9.12 mA
(d) Calculate the value of the signal PV at design conditions
_________
CO = 9 psig
(e) Suppose the process pressure jumps 20 psi, determine value for CO
CO = 9.44 mA
_________
7.
Express the function given the graph in the t-domain
f(t) = (t - 1)*u(t -1) - 2*(t - 4)*u(t - 4) + (t - 5)*u(t - 5) -u(t - 6)
8. A thermometer having first-order dynamics with a time constant of 1 min is at 100oF. The
thermometer is suddenly placed in a bath at 110oF at t = 0 and left there for 0.167 min, after
which it is immediately returned to a bath at 100oF. Calculate the thermometer reading at t =
0.5 min.
_________
101.1oF
References
1. D.R. Coughanowr and S. LeBlanc, Process Systems Analysis and Control, McGraw-Hill,
3nd edition, 2008.
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