x

Disp.

(m)

x

Vel.

(m/s)

x

Acc.

(m/s2)

m

Mass

(kg)

k

Spring cons.

(N/m)

c

Damp. Cons.

(Ns/m)

f

Force

(N)

θ

Angular pos.

(rad)

θ

Angular vel.

(rad/s)

θ

Ang.

acc.

(rad/s2)

IG

mass m.

ineratia

(kg-m2)

Kr

Rot. Spring const.

(Nm/rad)

Cr

Rot. damp. const.

Nm/(rad/s)

M

Moment

(Nm)

q

Charge

(Coulomb)

i

Current

(Amper)

L

Induc.

(Henry)

1/C

C:Capac.

(Farad)

R

Resistance

(Ohm)

V

Voltage

(Volt)

Vf

Volume

(m3)

Qf

Flow rate

(m3/s)

If

Fluid inertia

(kg/m4)

1/Cf

Cf:Fluid capacitance

Rf

Fluid resistance

P

Pressure

(N/m2)

Ht

Heat

(Joule)

Qt

Heat flow rate

(J/s)

1/Ct

Ct:Thermal capacitance

Rt

Thermal resistor

T

Temp.

(oC)

. ..

. ..

Analogy in Mechanical (trans.-rot.), Electrical, Fluid, Thermal SystemsAnalogy in Mechanical (trans.-rot.), Electrical, Fluid, Thermal Systems

Fluid inertia in pipes:

Fluid reservoir capacitance:

g

AC d

f Cf: Fluid reservoir capacitance , Ad:Cross section area of

reservoir, g=9.81 m/s2

If:Fluid inertia, ρ: Density, L: Pipe length A: Cross section of pipe

Fluid resistance of pipes in laminar flow:

4f d

L128R

Rf:Resistance, µ:Viscosity, L:Pipe length, d:Pipe diameter

Reynolds number in laminar flow<2000 d

Q4 fReynolds Number=

D.Rowell & D.N.Wormley, System Dynamics:An Introduction, Prentice Hall, 1997

Vf

Volume

(m3)

Qf

Flow

Rate

(m3/s)

If

Fluid inertia

(kg/m4)

1/Cf

Cf:Fluid capacitance

Rf

Fluid resistance

P

Pressure

(N/m2)

A

LIf

1

21pk1

Qk1

34

Q3

52

Q5

pa

Equivalent Electrical Circuit of Fluidic Systems

In this lesson, we will learn and study the modeling fluidic systems by creating equivalent circuits. There is an analogy between the fluidic and electrical quantities.

Let’s consider Example 5.1. A fluidic system is given here. We will create the equivalent circuit for the system by using the analogy.

pn Vn

Fluid. Elec.

Rn Rn In Ln

Qn

nq

Cn Cn

Example 5.1

pn Vn

Fluid. Elec.

Rn Rn In Ln

Qn

nq

Cn Cn

Fluidic System

Equivalent Electrical Circuit

-

Va

+

Vk1

R1L1

1q

1pk1In the fluidic system the pumps is connected the pipe #1.

The pump’s output pressure is Pk. The atmospheric pressure is Pa. The top of the tank #1 is open to the atmosphere.

In the electrical circuit, a supply voltage operating at Vk-Va is connected to the resistor R1 and inductor L1.

R1 represents the resistance to a flow in the pipe #1.

L1 represents the inertia of a flow in the pipe #1.

pa

Fluidic System

Equivalent Electrical Circuit

-

Va

+

Vk1

R1L1

1q

1pk1

Qk1pa

1

2

In the fluidic system the tank #1 is connected to both of the pipe #1 and the pipe#2.

However, in the electrical system, the capacitor #1 is connected to both of the resistor R1, inductor L1 and the resistor R2, inductor L2.

The other end of the capacitor #1 is connected to the voltage Va because the tank #1 is open to the atmospehere.

The capacity of a tank in a fluidic system corresponds to a capacitor in an electrical circuit.

There is an external flow with the flow rate Qk1 in the fluidic system. The current supply qk1_dot is placed in the electrical circuit.

1kq 2q

R2 L2

C1

Fluidic System

Equivalent Electrical Circuit

-

Va

+

Vk1

R1L1

1q

1pk1

Qk1pa

1

2

1kq 2q

R2 L2

C1

34

Q3A

3q

R3 L3

4q

R4 L4

-+

V4

A

In the fluidic system, the pipe #2 is connected to both of the pipe #3 and the pipe#4 at the point A. On the other hand, in the electrical system, the line with the resistor R2, inductor L2 is connected to the lines the resistor R3, inductor L3 and the resistor R4, inductor L4 at the point A.

There is a pressure P4 in the pipe #4 due to the placement of pipe #4 in the vertical direction because the flow is opposite to the gravity in pipe #4. So, the voltage supply V4 is placed in the circuit due to the analogy. The positive end of the V4 is connected to the point A. The current produced from V4 flows in the opposite direction.

Fluidic System

Equivalent Electrical Circuit

-

Va

+

Vk1

R1L1

1q

1pk1

Qk1pa

1

2

1kq 2q

R2 L2

C1

34

Q3A

3q

R3 L3

4q

R4 L4

-+

V4

A

52

Q5

R5 L5

C2

5q

In the fluidic system the tank #2 is connected to both of the pipe #1 and the pipe#2. On the other hand, in the electrical system, the capacitor #2 is connected both of the resistor R4, inductor L4 and the resistor R5, inductor L5.

Electrical systems can be analyzed instead of fluidic systems

Nowadays computer aided engineering (CAD/CAE) is used.

Dynamic (Transient) behaviour

Steady-state behaviour

By analyzing the equivalent circuit, transient or steady-state dynamic behavior at a desired point of a circuit can be calculated.

Thus, the corresponding flow rates and pressures at a desired line of a fluidic system are found.

Nowadays, circuits and fluidic systems can be modeled and analyzed by the computers.

Before the developments in computer technology, in order to analyze fluidic systems engineers have used the equivalent circuits, which are produced easily and cheaply.

Complex fluidic systems have been easily analyzed with the experiments of circuits.