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ELECTROMECHANICAL SYSTEMS

43

where Va is the applied voltage, and L and R are the inductance and the resistance of the armature circuit, respectively. From (2.79),

i =

I

dω

(2.82)

kt

dt

Combining (2.80)–(2.82), we obtain

L I d2ω

+

R I dω

+ ke ω = Va .

(2.83)

kt dt 2

kt dt

Equation (2.83) is the model for a simple d.c. motor, describing the change of the angular velocity with the applied voltage. In many applications the motor inductance is small and can be neglected. The model then becomes

R I dω

kt dt

+ ke ω = Va .

Models of more complex d.c. motor circuits are given in the following examples.

Example 2.11

Figure 2.20 shows a d.c. motor circuit with a load connected to the motor shaft. Assume that the shaft is rigid, has negligible mass and has no torsional spring effect or rotational damping associated with it. Derive an expression for the mathematical model for the system.

Solution

Since the shaft is assumed to be massless, the moments of inertia of the rotor and the load can be combined into I , where

I = IM + IL

where IM is the moment of inertia of the motor and IL is the moment of inertia of the load. Using Kirchhoff’s voltage law, we can write the following equation for the motor circuit:

Va − vb = L

di

+ Ri,

(2.84)

dt

L

i

R

ω

b

Va

d.c. motor

Load

IM

IL

Figure 2.20 Direct current motor circuit for Example 2.11

44 SYSTEM MODELLING

where Va is the applied voltage and L and R are the inductance and the resistance of the armature circuit, respectively. Substituting (2.80), we obtain

Va = L

di

+ Ri + ke ω

dt

or

˙

˙

(2.85)

Va = Li + Ri + ke θ .

We can also write the torque equation as

dω T + TL − bω = I .

dt

Using (2.78),

Idω + bω − kt i = TL dt

or

¨

˙

− kt i = TL .

(2.86)

I θ

+ bθ

Equations (2.85) and (2.86) describe the model of the circuit. These two equations can be represented in matrix form as

0

0

¨

+

ke

L

˙

+

0

R

i

=

Va

i˙

i˙

I

0

θ

b 0

θ

0

−kt

θ

TL .

2.4 FLUID SYSTEMS

Gases and liquids are collectively referred to as fluids. Fluid systems are used in many industrial as well as commercial applications. For example, liquid level control is a well-known application of liquid systems. Similarly, gas systems are used in robotics and in industrial movement control applications.

In this section, we shall look at the models of simple liquid systems (or hydraulic systems).

2.4.1 Hydraulic Systems

The basic elements of hydraulic systems are resistance, capacitance and inertance (see Figure 2.21). These elements are similar to their electrical equivalents of resistance, capacitance and inductance. Similarly, electrical current is equivalent to volume flow rate, and the potential difference in electrical circuits is similar to pressure difference in hydraulic systems.

Hydraulic resistance

Hydraulic resistance occurs whenever there is a pressure difference, such as liquid flowing from a pipe of one diameter to to one of a different diameter. If the pressures at either side of a hydraulic resistance are p1 and p2, then the hydraulic resistance R is defined as

p1 − p2 = Rq

FLUID SYSTEMS

45

Figure 2.21 Hydraulic system elements

where q is the volumetric flow rate of the fluid.

Hydraulic capacitance

Hydraulic capacitance is a measure of the energy storage in a hydraulic system. An example of hydraulic capacitance is a tank which stores energy in the form of potential energy. Consider the tank shown in Figure 2.21(b). If q1 and q2 are the inflow and outflow, respectively, and V is the volume of the fluid inside the tank, we can write

q1 − q2 =

d V

d h

(2.87)

= A

.

dt

dt

Now, the pressure difference is given by

p1 − p2 = hρg = p

or

h =

p

.

(2.88)

ρg

Substituting in (2.87), we obtain

q1 − q2 =

A d p

(2.89)

.

ρg

dt

Writing (2.89) as

d p

(2.90)

q1 − q2 = C

,

dt

we then arrive at the definition of hydraulic capacitance:

C =

A

(2.91)

.

ρg

Note that (2.90) is similar to the expression for a capacitor and can be written as

1

(q1

− q2)dt .

(2.92)

p =

C

Hydraulic inertance

Hydraulic inertance is similar to the inductance in electrical systems and is derived from the inertia force required to accelerate fluid in a pipe.

46 SYSTEM MODELLING

Let p1 − p2 be the pressure drop that we want to accelerate in a cross-sectional area of A, where m is the fluid mass and v is the fluid velocity. Applying Newton’s second law, we can write

dv

m

= A( p1 − p2).

(2.93)

dt

If the pipe length is L, then the mass is given by

m = Lρ A.

We can now write (2.93) as

dv

Lρ A

= A( p1

− p2)

dt

or

dv

(2.94)

p1 − p2 = Lρ

,

dt

but the rate of flow is given by q = Av, so (2.94) can be written as

p1 − p2 =

Lρ dq

(2.95)

.

A

dt

The inertance I is then defined as

I = Lρ ,

A

and thus the relationship between the pressure difference and the flow rate is similar to the relationship between the potential difference and the current flow in an inductor, i.e.

dq

(2.96)

p1 − p2 = I dt .

Models of some hydraulic systems are given below.

Example 2.12

Figure 2.22 shows a liquid level system where liquid enters a tank at the rate of qi and leaves at the rate of qo through an orifice. Derive the mathematical model for the system, showing the relationship between the height h of the liquid and the input flow rate qi .

qi

h

R

qo

A, ρ, g

Figure 2.22 Liquid level system

FLUID SYSTEMS

47

Solution

From (2.89),

qi − qo =

A d p

ρg

dt

or

qi =

A d p

+ qo .

(2.97)

ρg

dt

Recalling that

(2.97) becomes

p = hρg,

d h

(2.98)

qi = A

+ qo .

dt

Since

so that

p1 − p2 = Rqo ,

q

o =

p1 − p2

=

hρg

,

R

R

substituting in (2.98) gives

d h

ρg

(2.99)

qi = A

+

h.

dt

R

Equation (2.99) shows the variation of the height of the water with the inflow rate. If we take the Laplace transform of both sides, we obtain

qi (s) = Ash(s) + ρg h(s) R

and the transfer function of the system can be written as

h(s)

=

1

;

qi (s)

As + ρg/ R

the block diagram is shown in Figure 2.23.

Example 2.13

Figure 2.24 shows a two-tank liquid level system where liquid enters the first tank at the rate of qi and then flows to the second tank at the rate of q1 through an orifice R1. Water then leaves

qi(s)

1

h(s)

As + ρg /R

Figure 2.23 Block diagram of the liquid level system

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