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Unformatted text preview: Problem 1: Part 1 For tank 1, from the materials balance and the headflow relation we can obtain A 1 dh 1 dt = F i F 1 = F i h 1 R 1 (1) The deviation variables are defined as follows: h ′ 1 = h 1 h 1 (0) , F ′ i = F i F i (0) (2) And the level of tank 1 is initially at steady state, which implies that dh 1 dt vextendsingle vextendsingle vextendsingle vextendsingle t =0 = F i (0) h 1 (0) R 1 = 0 ⇒ F i (0) = h 1 (0) R 1 (3) Then the materials balance equation can be rewritten in terms of the deviation variables: A 1 dh ′ 1 dt = F ′ i h ′ 1 R 1 (4) Taking the Laplace transform of both sides and rearranging the equation yields ( A 1 s + 1 R 1 ) H ′ 1 ( s ) = F ′ i ( s ) (5) Therefore, the transfer function between F ′ i and h ′ 1 is G 1 ( s ) = H ′ 1 ( s ) F ′ i ( s ) = 1 A 1 s + 1 R 1 = 1 4 s + 0 . 5 (6) Similarly, the materials balance equation for tank 2 is A 2 dh 2 dt = F 1 F 2 = h 1 R 1 h 2 R 2 (7) We define the deviation variable h ′ 2 = h 2 h 2 (0) (8) and assume that tank 2 is also initially at steady state, then we obtain h 2 (0) = h 1 (0) R 2 R 1 (9) Thus, the materials balance equation for tank 2 in terms of the deviation variable is A 2 dh ′ 2 dt = h ′ 1 R 1 h ′ 2 R 2 (10) Again, by taking the Laplace transform and rearranging the resulting equation yields the transfer function between h ′ 1 and h ′ 2 : G 2 ( s ) = H ′ 2 ( s ) H ′ 1 ( s ) = 1 R 1 A 2 s + 1 R 2 = . 5 8 s + 0 . 25 (11) So the transfer function between F ′ i and h ′ 2 is G ( s ) = G 1 ( s ) G 2 ( s ) = . 5 (4 s + 0 . 5)(8 s + 0 . 25) = 4 256 s 2 + 40 s + 1 = 4 16 2 s 2 + 2 · 16 · 1 . 25 s + 1 (12) 1 Therefore, the overall steady state gain is...
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This note was uploaded on 01/28/2012 for the course ECH 157 taught by Professor Palagozu during the Fall '08 term at UC Davis.
 Fall '08
 PALAGOZU

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