MCE 366 System Dynamics, Spring 2011
Set 8
Reading: Chapter 7, Sections 7.3, and 7.4. Skip from page 362 to the end of Section 7.4.
Problems: 7.15, 7.16, 7.20, 7.22, 7.23, 7.24, 7.25, 7.28, 7.29
Important Examples: 7.3.1, 7.3.2, 7.4.1, 7.4.3, 7.4.4, 7.4.5
Solutions
7.15 a) The model is
dh
g
=− h
dt
R
where A = 20. The time constant is τ = RA/g. Taking the time to empty to be 4τ , we
obtain τ = 200/4 = 50 s. Thus
A
R=
τg
= 24.525 m−1 s−1
A
b) The model is
A
Thus
hss =
dh
g
=3− h
dt
R
3R
3(24.525)
=
= 7.5 m
g
9.81
8-1
7.16 a) The model is
A
The time constant is
τ=
g
dh
=− h
dt
R
RA
150(2)
=
= 93.168 sec
g
32.2
The time to decay by 98% (that is, approximately the time to empty) is approximately 4τ ,
regardless of the initial height, and is 4τ = 372.671 sec.
b) The model is
g
dh
= 0.1 − h
A
dt
R
The time constant is τ = 93.168 and the steady-state height is hss = 0.1R/g = 0.466 ft.
The response is
h(t) = 0.466 1 − e−t/93.168
(1)
Setting h(t) = hss /3 = 0.466/3 in (1) gives
1
= 1 − e−t/93.168
3
or
t = −93.168 ln
2
= 37.776 sec
3
7.20 If h < D,
ρA
1
1
dh
=
(ps + pa − ρgh − pa ) + qmi =
(ps − ρgh) + qmi
dt
R1
R1
If h ≥ D,
ρA
dh
1
1
1
1
=
(ps +pa −ρgh−pa )+qmi − (pa +ρg(h−D)−pa ) =
(ps −ρgh)+qmi − ρg(h−D)
dt
R1
R2
R1
R2
7.22 Applying conservation of mass to each tank gives
ρA1
ρA2
ρg
dh1
= qmi −
h1
dt
R1
dh2
ρg
ρg
=
h1 −
h2
dt
R1
R2
Note that ρ cancels out in the second equation.
8-2
7.23 a) Applying conservation of mass to each tank gives
ρA1
ρA2
ρg
dh1
= qmi −
(h1 − h2 )
dt
R1
dh2
ρg
ρg
=
(h1 − h2 ) −
h2
dt
R1
R2
Note that ρ cancels out in the second equation.
b) Substituting the given values we obtain
ρA
dh1
ρg
= qmi − (h1 − h2 )
dt
R
dh2
g
g
= (h1 − h2 ) −
h2
dt
R
3R
Applying the Laplace transform with zero initial conditions, we obtain
4A
g
Qmi (s)
g
H1 (s) − H2 (s) =
R
R
ρ
As +
g
4g
− H1 (s) + 4As +
H2 (s) = 0
R
3R
Let b = g/RA. After dividing both equations by A, they can be expressed as
(s + b) H1 (s) − bH2 (s) =
Qmi (s)
ρ
4
−bH1 (s) + 4s + b H2 (s) = 0
3
Using Cramer’s method to solve these equations, we obtain
s + b Q (s)/ρ
mi
H2 (s) = −b
0
where
Thus
s+b
D=
−b
−b
4s + 4b/3
bQmi (s)
/D =
ρD
4
16
1
bs + b2
= (s + b)(4s + b) − b2 = 4s2 +
3
3
3
H2 (s)
b/ρ
= 2
Qmi (s)
4s + (16/3)bs + (1/3)b2
8-3
7.24 Applying conservation of mass to each tank gives
ρA
dh1
ρg
= − (h1 − h2 )
dt
R
ρg
ρg
dh2
= qmi + (h1 − h2 ) −
h2
dt
R
3R
If we divide both equations by ρA and let b = g/RA, these equations can be expressed as
2ρA
dh1
= −b(h1 − h2 )
dt
2
dh2
qmi
b
=
+ b(h1 − h2 ) −
h2
dt
ρ
3R
Applying the Laplace transform with zero initial conditions, we obtain
(s + b) H1 (s) − bH2 (s) = 0
4
Qmi (s)
−bH1 (s) + 2s + b H2 (s) =
3
ρ
Using Cramer’s method to solve these equations, we obtain
0
H1 (s) = Qmi (s)/ρ
−b
2s + 4b/3
bQmi (s)
/D =
ρD
where
s+b
D=
−b
−b
2s + 4b/3
Thus
10
1
4
bs + b2
= (s + b) 2s + b − b2 = 2s2 +
3
3
3
H2 (s)
b/ρ
= 2
Qmi (s)
2s + (10/3)bs + (1/3)b2
The characteristic roots are
s=
−10b/3 ±
p
5 1√
100b2 /9 − 8b2 /3
= − ±
19 b = −1.56b, −0.1069b
4
6 6
8-4
7.25 From Example 7.4.4 the damping constant is given by
c=
128µLA2
πD4
Substituting the given and desired values, we obtain
2000 =
128(0.9)LA2
πD4
which gives
LA2
2000π
=
= 54.5415
4
D
0.9(128)
(1)
So we have three parameters to select: L, A, and D.
Let n be the ratio of the piston area A to the area Ao of the hole through the piston.
A
Ao
n=
(2)
But
D
Ao = π
2
and thus
A=
2
nπD2
4
From (1),
LA2
n2 π 2
=
L
= 54.5416
D4
16
so
16(54.5415)
88.419
=
2
2
n π
n2
Now we try various values for n to see if we obtain a reasonable value for the piston
length L. Using n = 50 gives L = 0.035 m, which is 1.38 in., which is a reasonable length.
Now we pick the piston area A. A cylinder diameter of 0.05 m (1.96 in.) gives
L=
A=π
0.05
2
2
= 1.963 × 10−3 m2
So from (2),
A
1.963 × 10−3
D
Ao =
=
= 3.9 × 10−5 = π
n
50
2
Thus the hole diameter is
D = 7.1 × 10−3 m
which is about 0.3 in.
8-5
2
So one of many possible designs is
piston diameter = 0.05 m
piston length, L = 0.035 m
piston hole diameter, D = 7.1 × 10−3 m
7.28 The equivalent mass of the cylinder is
me = m +
I
R2
Use this instead of m in equation (7) of Example 7.4.5 to obtain
I
m + 2 ẍ = (R1 + R2 )ρA2 ẋ = A(p1 − p2 )
R
8-6
7.29 This is like Example 7.4.5. Define the pressures p3 and p4 to be the pressures on the
left and right-hand sides of the piston. The mass flow rates through the resistances are
qm1 =
1
(p1 + pa − p3 )
R1
(1)
qm2 =
1
(p4 − p2 − pa )
R2
(2)
From conservation of mass,
qm1 = qm2
(3)
qm1 = ρAẋ
(4)
and
Combining equations (1) through (4), we obtain
p1 + pa − p3 = R1 ρAẋ
(5)
p4 − p2 − pa = R2 ρAẋ
(6)
Adding equations (5) and (6) gives
p4 − p3 + p1 − p2 = (R1 + R2 )ρAẋ
or
p3 − p4 = p1 − p2 − (R1 + R2 )ρAẋ
(7)
From Newton’s law, for small angles,
mL21 θ̈ = − (mg sin θ) L1 + (f cos θ) L2 = −mgL1 θ + f L2
(8)
where, from equation (7),
f = A(p3 − p4 ) = A(p1 − p2 ) − (R1 + R2 )ρA2 ẋ
(9)
Substitute equation (9) into (8), using the small-angle relation ẋ = L2 θ̇, and collect
terms to obtain the desired model:
mL21 θ̈ + L22 (R1 + R2 )ρA2 θ̇ + mgL1 θ = L2 A(p1 − p2 )
(10)
This is simply a linear, second-order equation of the form:
θ̈ + aθ̇ + bθ = c(p1 − p2 )
(11)
If there is no resistance in the hydraulic system, then a = 0 and the system is neutrally
stable.
8-7