1. Derive the transfer functions H2(s)/Q(s) and H3(s)/Q(s) for the liquid-level shown below. The resistances are linear. Note that two streams are flowing from Tank 1 to Tanks 1 and 2. You are expected to show clearly how you derived the transfer function. a. For the unit step change in q, determine the deviation for H 3(3). b. For impulse change in q, determine the deviation for H 2(3)
TANK 1: Balance:
− − − −
q q
TANK 2: Balance:
qa
qb = A1
h 1
h1
Ra
Rb
= A1
dh 1
qa
dt dh 1
h1
Deviation: Q
− −
H 1
Ra
Rb
−
= A1
dH 1
H1
dt
Ra
− = A sH s
H 1 s
H 1 s
Ra
Rb
1
Transfer function:
= = H1 s
R a Rb
Q s
R a +R b +R a R b A 1 s
1
R2
= A2
dh 2
dt dh 2 dt
Deviation:
H 1
Laplace:
Q s
q2 = A2
h 2
Ra
dt
− −
3
3 +4
=
0.75
0.75 +1
=
1
1 3
1+3+ 1 3 1
−
H 2 R2
= A2
dH 2 dt
Laplace:
− = A sH s
H1 s Ra
H 2 s
2
R2
2
Transfer function:
= = =
H2 s
R2
H1 s
2
H1
1+ 2 2
3
6 +1
3
1 1+ 2 3
TANK 3: Balance: qb h1
Solving a)
− −
Ra
R3
= A3
3
dh 3
q3 = A3
h 3
∙ − ∙ − − ∙ −
dt dh 3 dt
Deviation:
−
H1 Ra
H 3 R3
= A3
dH 3
Ra
3
R3
3
R3
2
1+ 3 3
3
2 3
H1 s
6 +1
3
=
3
= 0.5 0.5 +
3
3 = 0.1547
2
=
2
=
2
= = =
H1 s
0.75 +1 6 +1 0 .5 3 56
2
Transfer function: H3 s
0.5
+
24 7
0.75 +1 4 1 3
14
6 +1 4
− − − −
− = A sH s H 3 s
1
7
1 6
Solving b)
dt
Laplace:
H1 s
=
=
2 2.25
0.75 +1 6 +1 36 9
7 6 +1
6
7
1 6
14 0.75 +1 4 3
3 = 0.504
3 1+ 3 2
Transfer Function:
= × = = = × = = 2
1
2
Q s
2
Q s
3
Q s 3
Q s
H1 2.25
0.75
×
0.75 +1
3
6 +1
0.75 +1 6 +1
1
3
H1 0.5
0.75
×
0.75 +1
2 3 6 +1
0.75 +1 6 +1
2. A thermocouple has the following characteristics when it is immersed in a stirred bath. Mass of thermocouple = 1 g Heat Capacity of thermocouple = 0.25 cal/g oC Heat transfer coefficient = 20 cal/cm 2hoC (for thermocouple and bath) Surface area of thermocouple = 3 cm 2 (a) Derive a transfer function model for the thermocouple relating the change in its indicated output T to the change in the temperature of its surroundings Ts. Assuming uniform temperature (no gradient in the thermocouple bead), no conduction in the leads, and constant physical properties. (b) If the thermocouple is initially out of the bath and at room temperature (23 oC), what is the maximum temperature that it will register if it is suddenly plunged into the bath (80oC) and held there for 20secs?
Solving a) Balance: mC
dT
= hA(Ts
dt
−
Solving b) T)
′′ − ′ ′ ′ −− ′ ℃ = 80
=
Deviation: mC
T s =
dT′ = hATs′ −T′
15s+1
Ts s =
= 57(1
Laplace:
=
′ = ′ −′()
57 s(15s+1)
15
)
− − −−
+ = 23 + 57(1
= 20 = 23 + 57(1 20 = 64.97
Transfer Function:
′
1
Laplace:
dt
T (s)
23 = 57
57
15
20 15
)
)
1
= Ts ′ (s) +1 ′T (s) 1 = Ts ′ (s) 15s+1
3.
An overdamped system consists of two first-order processes operating in series, where τ1=4 and τ2=1. Find the equivalent values of τ and ζ for this system.
τ ∙τ ζ τ∙ τ τ∙τ ∙ ζ ∙ ∙
Equations:
=
=
1
1+ 2
2
1 2
Solution: = 4 1 = 2 =
4+1
2
41
2
ζ−ζ − 2 = ζ ζ −
1=
2
+
1
2
1
(ζ≥1) (ζ≥1)
= 1.25
4. Determine the values of K and k of the closed-loop system, shown in the figure below, so that the maximum overshoot in unit-step response is 25% and the peak time is 2sec. Assume that J=1 kgm 2.
The closed-loop system transfer function is
Substituting the value of J,
The undamped natural frequency, ω n
−
=
=
=
2
2
+
+
+
+
, 2
=
The maximum overshoot, M p is
− − =
1
2
= 0.25
Solving for δ,
= 0.404
= 1.57
=
1
2
= 1.72
Therefore,
∙ ∙ 2
=
= 1.722
= 2.958
=
2
=
2(0.404)(1.72) 2.958
k = 0.469 sec
Given the peak time, t p which is 2,
=
=2
5. A water tank with vertical sides and a cross-sectional of 2m2 (shown in the figure below) is fed from a constant displacement pump, which may be molded as a flow source Qin(t). A valve, represented by a linear fluid resistance (Rf), at the base of the tank is always open and allows water to flow out. In normal operation the tank is filled to a depth of 10m. When time is equal to zero the power to the pump is removed and the flow into the tank is disrupted. If the flow through the valve is 10-6m3/s when the pressure across it is 1N/m2, determine the transfer function of pressure at the bottom of the tank.
The tank is represented as a fluid capacitance C f
Standard first-order form:
with a value:
+
=
where:
Assume input flow is zero, therefore,
A – Area of the tank
+
ρ
– Density of fluid g – Acceleration due to gravity
=
∙
=
−
1000 9.81
−
4
=
− ∙
5
1 10
6
Equation in terms of pressure across the fluid
0
where:
∙ ∙ 0 =
5
=0
The homogenous pressure response,
2
= 2.038 × 10
=
= 1000 9.81 1
0 = 9810
2
= 204
capacitance:
− =
1
+
1
Answer:
−
= 9810
204
2
6. A homogenous system consists of 3 tanks. Tank 1 is nested inside the Tank 2 which is connected to Tank 3. Heat is transferred by convection through the wall of inner tank. The data are enumerated below: a. The volume of Tank 1 is 1ft 2 while 2ft 2 for Tank 2 and Tank 3. b. The cross-sectional area for all tanks is 1ft 2. c. The overall heat transfer coefficient is 8BTU/(h ft2 F) d. The heat capacity of the fluid in each tank is 1BTU/(lb F). The density of each fluid is 30lb/ft3.
∙∙
∙
The flow of heat is changed according to unit step from 0 to 300BTU/h. Determine T2’(s)/T1’(s) and T’3(s)/T1’(s).
− − − − ρ
wC To
Tref + hA T1
T2
wC T2
Tref = V2 C
dT2 dt
Transfer Function:
′ ′
T2 s
Deviation:
′ ′− ′ − ′ ρ
wCTo + hA T1
′ ′
0 + 8 T1 s = T2 s 13 + 60s
Tank 2: Balance:
T2
wCT2 = V2 C
dT2
′
T1 s
8
60 + 13
dt
Tank 1: Balance:
Laplace:
′ ′ − ′ − ′ ρ
wCTo s + hA T1 s
=
T2 s
′
wCT2 s = V2 CsT2 s
′ ′
5To s + 8 T1 s T2 s 5T2 s = 60sT2 s 5To s + 8T1 s 8T2 s 5T2 s = 60sT2 s 5To s + 8T1 s = T2 s 13 + 60s
Since initial temperature is constant, the deviation is 0.
− − ρ hA T1
T2 = V1 C
dT 1 dt
Deviation:
Substitute the given values:
′ ′ − ′ − ′ ′ ′ − ′ − ′ ′ ′ ′
Q
Q′− T′ −T′ ρ hA
Laplace: hA
1
2
= V1 C
d
T′
1
dt
Q′− T′ −T′ ρ T′ 1
2
= V1 C
1
Substitute the given values:
− ′ T′ T′ T′ T′ ∙ T′ − ′ ′
Tank 3
T′ T′
Q (s) 8T 1 + 2 = 30 1 Q (s) + 2 = 1 30 + 8 300 8 + 1 = 1 30 + 8 60 + 13 300 8 = 1 30 + 8 60 + 13
Transfer Function:
T′ 1
=
6
60 +13 2 +2.9
+0.36
Stirred heater transfer function: 1 = 3 12 + 1 2 1 8 = 3 1 12 + 1 60 + 13 1 8 3 = 12 + 1 60 + 13 1 8 3 = 720 2 + 216 + 13 1
′ ′ ′ ∙T′ ∙ ′ ∙ T′′ T′
7. A temperature sensor having time constant characteristics is subjected to a sudden change in temperature of 25oC-200oC. If it has a time constant of 5 secs, what temperature will be indicated after 8 secs? Solution: Let: Ti = initial temperature, 25 oC Tf = final temperature, 200 oC t1 = 5 secs t2 = 8 secs
− = + − 2
2 1
− − ℃
8 = 200 + 25 8 = 164.67
200
8 5
8. A liquid storage system show below has normal operating conditions: q1= 10 ft 3/min q2 = 5 ft3/min h = 4 ft
The tank is 6ft in diameter, and the density of each stream is 60 lb/ft 3. Suppose that a pulse change in q 1 occurs as shown below: a) What is the transfer function relating H to Q 1? b) Derive an expression for h(t) for this input change. Solution: a) Assume that q is constant Material Balance over the tank, dh A q1 q 2 q dt Writing in deviation variable
(Q1 Q 2 ) Q A
dH dt
Laplace transform ,
Q1 (s) Q 2 (s) AsH(s) Transfer function: H(s) Q1 (s)
b)
1 As
q1 ' (t ) 5s(t ) 5s(t 12) Q1 ' (s)
5 s
5
e 12s s
5
5 12s A H' (s) Q1 ' (s) 2 A e 2 As s s 1
h' (t )
5 A
h(t ) 4
ts( t ) 5
5 A
( t 12)s( t 12)
4 0.177t
0 t 12
5 (12) 6.122 A
12 t
A
h(t ) 4
9. Determine Y (4) for the system response expressed by
=
2
(2 +4) (4 2 +0.8 +1)
Solution:
2 = 4(1 + )
1
(4
2
+ 0.8 + 1) 8 4 = + (4 2 + 0.8 + 1) (4 2 + 0.8 + 1)
Step response + impulse response
ϛ
Now, = 4 = 2 ; 2 = 0.8
ϛ
4
= 0.2and = = 2 2
Impulse response
∗ ∗
= 4 0.63 = 2.52
Step response = 8 1.15 = 9.2
4 = 1.26 + 9.2 = 10.46
10. When the system as shown in Figure (a) is subjected to a unit step input, the system output responds as shown in Figure (b). Determine the values of ‘K’ and ‘T’ from the response curve.
The maximum overshoot of 25.4% corresponds to ϛ= 0.4. From the response curve we have,
tp= 3
ω − − =
=
2
1
=
1
0.4
ωn=1.14 From the block diagram, the closed loop transfer function is
( )
( )
Hence ωn=
, 2ϛωn=
=
2
+ +
1
Therefore the values of K and T can be determined using
ϛω ∗ ∗ ω ∗ =
=
1
2
n
2
=
1
2 0.4 1.14
T 1.09 = 1.42
= 1.09
2
=3
11.
Problem: Obtain transfer functions C(s)/R(s) and C(s)/D(s) of the system shown above. Solution:
U(s) = G fR(s) + GcE(s) C(s) = G p[D(s) + GlU(s)] E(s) = R(s) - HC(s) C(s) = G pD(s) + GlGp[Gf R(s) + G c E(s)] C(s) = G pD(s) + GlGp{GfR(s) + G c[R(s) – HC(s)]} C(s) + GlGpGcHC(s) = GpD(s) + G lGp(Gf + G c)R(s) C(s) =
Gp D s + Gl Gp Gf + Gc R(s) 1 + Gl Gp Gc H
.1
To find transfer function C(s)/R(s), we let D(s) = 0 in eq. 1 C(s) R(s)
=
Gl Gp Gf + Gc
1 + Gl Gp Gc H
To obtain transfer function C(s)/D(s), we let R(s) = 0 in eq. 1 C(s) D(s)
=
Gp 1 + Gl Gp Gc H
12. A step change from 15 to 31 psi in actual pressure results in the measured response from a pressure-indicating element shown in the figure below. a) Assuming second-order dynamics, calculate all important parameters and write an approximate transfer function in the form
′′ ( )
( )
Where (psi).
′
=
2 2
+2
+ 1
is the instrument output deviation (mm),
′
is the actual pressure deviation
SOLUTION:
− − − − −− − − ′′
=
11.2
8
31
=
12.7
=
11.2
11.2
×
( )
( )
8
2
1
0.2342
2
=
= 0.47
= 0.234
= 2.3
2
1
1
/
= 0.47,
2
=
= 2.3
= 0.29
15
= 0.356
0.2
0.127
2
+ 0.167 + 1
13. The caustic concentration of the mixing tank shown below is measured using a conductivity cell. The total volume of solution in the tank is constant at 7 ft3 and the density (ρ = 70 lb/ft3) can be considered to be independent of concentration. Let Cm denote the caustic concentration measured by the conductivity cell. The dynamic response of the conductivity cell to a step change (at t=0) of 3 lb/ft3 in the actual concentration (passing through the cell) is also shown i n the figure. a) Determine the transfer function
′′ assuming the flow rates are equal and constant ( )
1( )
(w1 = w2 = 5 lb/min) b) Find the response for a step change in C 1 from 14 to 17 lb/ft 3 c) If the transfer function
′′ were approximately by 1(unity), what would be the step ( )
1( )
response of the system for the same input change?
SOLUTION: a) Caustic balance for the tank,
− ′ ′ ′ −′ ′′ ′ ′ − =
Since V is constant, For constant flows,
=
1 +
2 =
1
1
1
+
+
1
2
70 7 + 10
( )
=
2
5
=
( )
= 6
2 2
10 /
=
=
1 1 +
=
3
= 0.1 min (
0.5
49 +
+ 1
0
3
=
=1
)
′′ =
1
1
0.5
0.1 + 1
49 + 1
=
0.5
0.1 + 1 49 + 1
b)
′
3
=
1
′ ′ − − − − 1.5
=
= 1.5 1 +
0.1 + 1 49 + 1 1
49
(0.1
0.1
0.1
49
49
c)
′ ′ −− =
0.5
3
49 + 1
=
= 1.5(1
1.5
49 + 1 49 )
14. A heated process is used to heat a semiconductor wafer operated with first-order dynamics, that is, the transfer function relating changes in temperature T to changes in the heater input power level P is: ( ) = ( ) + 1
′′
where K has units [°C/kW] and has units [minutes]. The process is at steady state when an engineer changes the power input stepwise from 1.0 to 1.5 kW. She notes the following: (i) The process temperature initially is 80°C. (ii) Four minutes after changing the power input, the temperature is 230°C. (iii) Thirty minutes later the temperature is 280°C.
a. What are K and in the process transfer function? b. If at another time the engineer changes the power input linearly at a rate of 0.5 kW/min, what can you say about the maximum rate of change of process temperature: When will it occur? How large will it be?
SOLUTION:
15. A surge tank is designed with a slotted weir so that the outflow rate, w, is proportional to the liquid level to the 1.5 power; that is, w = Rh1.5 where R is a constant. If a single stream enters the tank with flow rate w i, find the transfer function H’(s)/W’(s). Identify the gain and all time constant. Verify units. The cross-sectional area of the tank is A. Density ρ is constant.
SOLUTION:
16.
Problem: The figure above shows a system with two inputs and two outputs. Derive C 1(s)/R 1(s), C1(s)/R2(s), C2(s)/R1(s), and C 2(s)/R 2(s). (In deriving outputs for R 1(s), assume that R 2(s) is zero and vice versa.) Solution:
C1 = G1(R1 – G3C2) C2 = G4(R2 – G2C1) C1 = G1[R1 – G3G4(R2 – G2C1)] C2 = G4[R2 – G2G1(R1 – G3C2)] C1 =
C2 =
G1 R1
−
− − − 1
G1 G3 G4 R2
G1 G2 G3 G4
G1 G2 G4 R1 + G4 R 2 1
G1 G2 G3 G4
Combined in the form of a transfer matrix
– −− C1 = C2
1
1
G1
G1 G2 G3 G4 G1 G2 G4 G1 G2 G3 G4
G1 G3 G4
1 1
− –
G1 G2 G3 G4 G4 G1 G2 G3 G4
R1 R2
When R2(s) = 0, the original block diagram can be simplified. Similarly, when R 1(s) = 0, the original block diagram can be also simplified
= –
= − −
− − =
– ( )
( )
=
17. A tank with constant outlet flow is connected to a stirred tank as shown below. The constant flow is 2ft 3/min. The cross-sectional area and volume of each tank is 2ft 2 and 1ft2 respectively. The feed with temperature T 1 entered the first tank and maintain the temperature as it enter the stirred tank. a. Determine the transfer function H 1(s)/Q(s) and T’ 2(s)/T’ 1(s). b. For a unit step change in the flow rate, determine the height of the fluid in the tank at t = 3min. c. For an impulse change in inlet temperature of magnitude 2, determine the deviation for T’2(2).
DEVIATION:
TANK 1: BALANCE:
− =
1
′ − ′ ′
1
DEVIATION:
=
′ − ′ ′ ′ − ′ ′ 1
2
1
LAPLACE:
=
1
1
′ = = ′ 1
+
1
Solving b)
2
1
− − − 2
2
2
TANK 2: BALANCE:
1
2
TRANSFER FUNCTION
1
= = 1
=
=
2
TRANSFER FUNCTION 1
2
LAPLACE:
1
1
2 =
1
2
=
1 1
=
∙ 1
2 1 1 1 = = 2 2 2
=
2
2+
2
′ ′ ′ ′ − ′
3 3 = = 1.5 1 2 = + 1.5 = 4 + 1.5 = 5.5
2
=
2
=
2 2
Solving c)
2
2+ 4
1
2+ = 4 2 2 = 0.0733
18. Problem: Find the time response of a standard second order system
=
1
2 2
+2
+ 1
to a step input when = 0 (underdamped). Solution:
θ θ
o
=
o
=
θ
i
T 2 s2
+ 2(0)Ts + 1
θ
T 2 s2
i
+1
θ
For a unit step input
θ
o
=
H/s T 2 s2 + 1
=
i =
H and
H s(T2 s 2 + 1)
=
θ
s = H/s
i
H T2 s(s2 +
1 ) T2
θ ωω Let
o
=
= 1/ H
s s2 +
2
Taking the Laplace Inverse Transform
− =
(
)
H = amplitude = frequency of oscillation in rad/s
