Biomedical Engineering Reference
In-Depth Information
TABLE 4-4
Comparison Between Electrical and Hydraulic Elements
Block
Describing Equation
Energy/Power
1
L
1
2
Li
2
Inductor
i
=
Vdt
E
ind
=
1
I
1
2
Iq
2
Hydraulic inertiance
q
=
(
P
1
−
P
2
)
dt
E
HI
=
i
=
C
dV
dt
1
2
CV
2
Capacitor
E
cap
=
q
=
C
d
(
P
1
−
P
2
)
dt
1
2
C
(
P
1
−
P
2
)
2
E
mass
=
Hydraulic capacitance
V
R
1
R
V
2
Resistor
i
=
P
res
=
1
R
(
P
1
−
P
2
)
1
R
(
P
1
−
P
2
)
2
Hydraulic resistance
q
=
P
HR
=
The mass of the fluid cylinder is equal to the product of the volume of the fluid and the
density
m
=
ρ
V
=
ρ
AL
(4.57)
Therefore,
(
P
1
−
P
2
)
A
=
ρ
AL
d
dt
(4.58)
But the volume flow rate is
q
=
A
v
; therefore,
(
P
1
−
P
2
)
A
=
ρ
L
dq
dt
(4.59)
The pressure difference can be written as
P
2
)
=
ρ
L
A
dq
dt
=
I
dq
dt
(
P
1
−
(4.60)
where the hydraulic inertiance,
I
is defined as
=
ρ
L
A
I
(4.61)
As with the mechanical elements discussed earlier, hydraulic elements can be thought of
as equivalent to their electrical counterparts, which allows SPICE models to be used to
simulate fluid flow. The equivalence is shown in Table 4-4.
4.6
SYSTEM RESPONSE
If the input to a system changes, the output will change in two different stages. The first will
a transient which will settle into the steady-state response. The solution to the differential
equations discussed earlier is used to determine analytically what these responses will be.
