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Control ENG HW_Part_39

# Control ENG HW_Part_39 - -Also for tank 2 dt dh A R h R h h...

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8.12 sketch the response Y(t) if ) 1 2 . 1 ( ) ( 2 2 + + = - s s e s Y S Determine Y(t) for t = 0,1,5, 2 2 2 2 2 2 2 2 ) 8 . 0 ( ) 6 . 0 ( ) 8 . 0 ( 8 . 0 1 ) 8 . 0 ( ) 6 . 0 ( ) 1 2 . 1 ( ) ( + + = + + = + + = - - - s e s e s s e s Y S S S 0 ) ( , 14 . 0 ) 5 ( , 5 0 ) 1 ( , 1 0 ) 0 ( 0 2 )) 2 ( 8 . 0 sin( 25 . 1 ) ( ) 2 ( 6 . = = = = = = = = - = - - Y t Y t Y t Y t for t t e t Y t Problem 8.13 The system shown is at steady state at t = 0, with q = 10 cfm A 1 = 1ft 2 ,A 2 =1.25ft 2 , R 1 = 1 ft/cfm, R 2 = 0.8 ft/cfm. a) If flow changes fro 10 to 11 cfm, find H 2 (s). b) Determine H 2 (1),H 2 (4),H 2 ( ) c) Determine the initial levels h 1 (0),h 2 (0) in the tanks. d) obtain an expression for H 1 (s) for unit step change.

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Writing mass balances, ( ) ) 1 tan ( 1 1 1 2 1 k for dt dh A R h h q = - - At steady state ( ) S S s s h h R h h q 2 1 1 2
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Unformatted text preview: --Also for tank 2 ( ) dt dh A R h R h h 2 2 2 2 1 2 1 =--At steady state ( ) 8 10 * 8 . 8 . 1 2 2 2 1 = = = =-S S s s h h h h 18 1 = S h C) 18 1 = S h ft ft h 8 ) ( 2 = The equations in terms of deviation variables dt dH A Q Q 1 1 1 =-where 1 2 1 1 R H H Q-= dt dH A Q Q 2 2 2 1 =-2 2 2 R H Q = 1 8 . 2 8 . 1 ) ( ) ( ) ( 2 2 1 2 1 2 2 1 2 2 + + = + + + + = s s s R A s R s Q s H τ...
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Control ENG HW_Part_39 - -Also for tank 2 dt dh A R h R h h...

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