Unformatted text preview: =
¶F s s
V1 ¶T ss V1 ¶Q ss r pV1 c 1 b =
1 ¶f 2 (T  T ) ¶f F ¶f  Fs = 1s 2 s ; b = 2 = s ; b = 2 =
2
3
¶F ss V ¶T ss V ¶T ss V 2
1
2
2
2 By subtracting the steadystate equation and defining deviat ion variables (like F = F  F ), we obtain the fo llowing equat ions: s
dT 1
= a F + a 2 T + a 3Q 1
1 dt d 2 T = b F + b T + b T 1
21
32
dt Taking the Laplace transformat ion, sT ( s ) = a F ( s + a 2 T ( s ) + a 3Q ( s ) )
1
1
1 s 2 ( s ) = b1 F ( s + b2 T ( s + b3T ( s ) T ) 1 ) 2 and rearranging, a 3 a 1
F ( ) +
s
Q ( ) s
s  a 2 s  a 2 b b T ( s = 1 F ( s + 2 T ( s )
2)
1)
s  b s  b 3
3
T1 ( s = ) Replacing the first equation into the second equation and rearranging again, T2 ( s = ) ù
b b é a a 3 1
F ( s + 2 ê 1 F ( s +
)
)
Q ( s )ú
s  b s  b ë s  a 2 s  a 2 3
3
û T ( s =
2) b é
a ù
b a 3 1
1
2
)
)
ê1 + s  a ú F ( s + s  b s  a Q ( s s  b ë
3
2 û
3 2 The transfer funct ions can now be ident ified,
T ( s = g TF F ( s + g TQ Q ( s ) )
2)
T (...
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This note was uploaded on 01/12/2014 for the course CHE 4198 taught by Professor Hjortso,m during the Fall '08 term at LSU.
 Fall '08
 Hjortso,M
 pH

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