Modern Control Engineering

Example Problems and Solutions 143

Defining the displacement of the midpoint of the rod that connects two bellows as x, find the
transfer function Assume that the expansion process is isothermal and that the
temperature of the entire system stays at 30°C. Assume also that the polytropic exponent nis 1.

Solution.Referring to Section 4–3, transfer function can be obtained as

(4–38)

Similarly, transfer function is

(4–39)

The force acting on bellows 1 in the xdirection is and the force acting on bellows 2
in the negative xdirection is The resultant force balances with kx, the equivalent
spring force of the corrugated sides of the bellows.

or

(4–40)

Referring to Equations (4–38) and (4–39), we see that

By substituting this last equation into Equation (4–40) and rewriting, the transfer function
is obtained as

(4–41)

The numerical values of average resistances R 1 andR 2 are

The numerical value of capacitance Cof each bellows is

whereRair=287N-mkg K. (See Problem A–4–3.) Consequently,

R 1 C=0.167*10^10 *5.75*10–9=9.60sec

R 2 C=0.333*10^10 *5.75*10–9=19.2sec

By substituting the numerical values for A, k,R 1 C, and R 2 Cinto Equation (4–41), we obtain

X(s)

Pi(s)

=

1.44* 10 -^7 s

(9.6s+1)(19.2s+1)

C=

V

nRair T

=

5 * 10 -^4

1 287 (273+30)

=5.75* 10 -^9

kg

Nm^2

R 2 =

d¢p

dq 2

=

0.5* 105

1.5* 10 -^5

=0.333* 1010

Nm^2

kgsec

R 1 =

d¢p

dq 1

=

0.5* 105

3 * 10 -^5

=0.167* 1010

Nm^2

kgsec

X(s)

Pi(s)

=

A

k

AR 2 C-R 1 CBs

AR 1 Cs+ 1 BAR 2 Cs+ 1 B

X(s)Pi(s)

=

R 2 Cs-R 1 Cs

AR 1 Cs+ 1 BAR 2 Cs+ 1 B

Pi(s)

P 1 (s)-P 2 (s)= a

1

R 1 Cs+ 1

-

1

R 2 Cs+ 1

bPi(s)

ACP 1 (s)-P 2 (s)D=kX(s)

AAp 1 - p 2 B=kx

AAP

–

+p 2 B.

AAP

–

+p 1 B,

P 2 (s)

Pi(s)

=

1

R 2 Cs+ 1

P 2 (s)Pi(s)

P 1 (s)

Pi(s)

=

1

R 1 Cs+ 1

P 1 (s)Pi(s)

X(s)Pi(s).