Modern Control Engineering

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

Taking the Laplace transform of Equation (4–44), assuming zero initial condition, we obtain

(4–47)

By using Equation (4–47) to eliminate W(s)from Equation (4–46), we obtain

from which we obtain the transfer function Z(s)/Y(s)to be

MultiplyingB/Ak 1 k 2 Bto both the numerator and denominator of this last equation, we get

Define Then the transfer function Z(s)/Y(s)becomes as follows:

A–4–10. Considering small deviations from steady-state operation, draw a block diagram of the air heat-

ing system shown in Figure 4–38. Assume that the heat loss to the surroundings and the heat

capacitance of the metal parts of the heater are negligible.

Solution.Let us define

steady-state temperature of inlet air, °C

steady-state temperature of outlet air, °C

G=mass flow rate of air through the heating chamber, kgsec

M=mass of air contained in the heating chamber, kg

c=specific heat of air, kcalkg °C

R=thermal resistance, °C seckcal

C=thermal capacitance of air contained in the heating chamber=Mc, kcal°C

steady-state heat input, kcalsec

Let us assume that the heat input is suddenly changed from to and the inlet air

temperature is suddenly changed from to Then the outlet air temperature will be

changed from to

The equation describing the system behavior is

Cduo= Ch+GcAui-uoBDdt

Q–o Q–o+uo.

Q–i Q–i+ui.

H– H–+h

H– =

Q–o=

Q–i=

Z(s)

Y(s)

=

T 1 s

T 1 T 2 s^2 +AT 1 +2T 2 Bs+ 1

Bk 1 =T 1 ,Bk 2 =T 2.

Z(s)

Y(s)

=

B

k 1

s

B^2

k 1 k 2

s^2 + a

2B

k 2

+

B

k 1

bs+ 1

Z(s)

Y(s)

=

k 2 s

Bs^2 +A2k 1 +k 2 Bs+

k 1 k 2

B

k 2 Y(s)=Ak 2 +BsB

k 1 +Bs

Bs

Z(s)+k 1 Z(s)

W(s)=

k 1 +Bs

Bs

Z(s)

H+h

Qi+ui Heater

Qo+uo

Figure 4–38

Air heating system.

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