Modern Control Engineering

142 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systems

In the form of the standard vector-matrix representation, we have

which is the state equation, and

which is the output equation.

A–4–3. The value of the gas constant for any gas may be determined from accurate experimental obser-

vations of simultaneous values of p, v, and T.

Obtain the gas constant Rairfor air. Note that at 32°F and 14.7 psia the specific volume of air

is 12.39 ft^3 lb. Then obtain the capacitance of a 20-ft^3 pressure vessel that contains air at 160°F. As-

sume that the expansion process is isothermal.

Solution.

Referring to Equation (4–12), the capacitance of a 20-ft^3 pressure vessel is

Note that in terms of SI units,Rairis given by

Rair=287 N-mkg K

A–4–4. In the pneumatic pressure system of Figure 4–29(a), assume that, for t<0, the system is at steady

state and that the pressure of the entire system is Also, assume that the two bellows are identi-

cal. At t=0, the input pressure is changed from to Then the pressures in bellows 1 and

2 will change from to and from to respectively. The capacity (volume) of each

bellows is 5*10–4m^3 , and the operating-pressure difference (difference between piandp 1 or

difference between piandp 2 ) is between –0.5*10^5 Nm^2 and0.5*10^5 Nm^2. The corresponding

mass flow rates (kgsec) through the valves are shown in Figure 4–29(b). Assume that the bellows

expand or contract linearly with the air pressures applied to them, that the equivalent spring con-

stant of the bellows system is k=1*10^5 Nm, and that each bellows has area A=15*10–4m^2.

¢p

P

–

P +p 2 ,

–

P

–

P +p 1

– P

–

P +pi.

– P

–

.

C=

V

nRair T

=

20

1 53.3 620

=6.05* 10 -^4

lb

lbfft^2

Rair=

pv

T

=

14.7 144 12.39

460 + 32

=53.3 ft-lbflb°R

B

y 1

y 2

R = B

1

0

0

1

RB

x 1

x 2

R

B

x# 1

x# 2

R= D

-

1

R 1 C 1

1

R 1 C 2

1

R 1 C 1

• a

1

R 1 C 2

+

1

R 2 C 2

b

TB

x 1

x 2

R +D

1

C 1

0

0

1

C 2

TB

u 1

u 2

R

Bellows 1 Bellows 2

Valve 1 Valve 2

(a) (b)

x

Area

CCA

q 1 q 2

R 1 R 2

P+p 1 P+p 2

P+pi

Valve 2

0.5 105 Valve 1

–3 10 –5

1.5 10 –5

• 0.5  105

Dp(N/m^2 )

q(kg/sec)

Figure 4–29

(a) Pneumatic

pressure system;

(b) pressure-

difference-versus-

mass-flow-rate

curves.

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