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Solutions_F11_ECH157_Midterm

Solutions_F11_ECH157_Midterm - Problem 1 Part 1 For tank 1...

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Problem 1: Part 1 For tank 1, from the materials balance and the head-flow 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 re-written 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 = 0 . 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 )= 0 . 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
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Therefore, the overall steady state gain is 4 ft/ ( ft 3 /min ) and the time constant is 16 min .
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