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Unformatted text preview: 2 2 '( ) '( ) 0.127 0.167 '( ) 0.2 '( ) d R t dR t R t P t dt dt + + = These are deviation variables, and we are interested in the original variables R ( t ) and P ( t ). Recalling the definition of the deviation variables: '( ) ( ) SS R t R t R =-'( ) ( ) SS P t P t P =-And consequently: 5.14 – 2 '( ) ( ) dR t dR t dt dt = And: 2 2 2 2 '( ) ( ) d R t d R t dt dt = Substituting in the differential equation we obtain: ( ) 2 2 ( ) ( ) 0.127 0.167 ( ) 0.2 ( ) SS SS d R t dR t R t R P t P dt dt + +-=-Assuming that we start from the steady state in Figure E5.14, i.e. 15 SS P psi = and 8 SS R mm = , we get: ( ) 2 2 ( ) ( ) 0.127 0.167 ( ) 8 0.2 ( ) 15 d R t dR t R t P t dt dt + +- =-Which is equivalent to: 2 2 ( ) ( ) 1.31 7.88 ( ) 1.57 ( ) 39.5 d R t dR t R t P t dt dt + + = +...
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- Spring '09
- Derivative, dt dt, 8mm, Seborg