Control ENG HW_Part_46

Control ENG HW_Part_46 - For inner tank dT1 - (3) dt Q - hA...

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For inner tank Q - hA (T 1 – T 2 ) = ρ C V 1 dt dT 1 --------------------- (3) Qs - hA (T 1s – T 2s ) = 0 ------------------------------- (4) (3) – (4) gives Q’ - hA (T 1 ’– T 2 ) = ρ C V 1 dt dT ' 1 Taking L.T and putting numerical values Q(s) – 10 T 1 (s) + 10 T 2 (s) = 50 s T 1 (s) Q(s) = 500/s and T 2 (s) = T 1 (s) / (2+ 5s) ) ( 50 5 2 ) ( 10 ) ( 10 500 1 1 1 s sT s s T s T s = + + - + + - = 1 5 2 1 5 ) ( 50 1 s s s T s ) 1 15 25 ( ) 5 2 ( 50 ) ( 2 1 + + + = s s s s s T ( ) ( ) 50 18 . 26 50 82 . 3 ) 5 2 ( 2 ) ( 1 + + + = s s s s s T ( ) ( ) 50 18 . 26 29 . 5 50 82 . 3 71 . 94 100 ) ( 1 + - + - = s s s s T 50 18 . 26 50 82 . 3 ' 1 e 5.29 - e 94.71 - 100 (t) T t t - - = and 50 18 . 26 50 82 . 3 1 e 5.29 - e 94.71 - 200 (t) T t t - - = For t=0,5,10 and ∞ T(0) = 100 o F T(5) = 134.975 o F T(10) = 155.856 o F T(∞) = 200 o F
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Q – 10.1. A pneumatic PI controller has an output pressure of 10 psi, when the set point and pen point are together. The set point and pen point are suddenly changed by 0.5 in (i.e. a step change in error is introduced) and the following data are
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This note was uploaded on 11/13/2011 for the course COP 4355 taught by Professor Koslov during the Spring '10 term at University of Florida.

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Control ENG HW_Part_46 - For inner tank dT1 - (3) dt Q - hA...

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