hydraulics_analysis

hydraulics_analysis - : u t ( ) = h t ( ) U s ( ) = 1 / s q...

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Example 1 Input/output EOM For an input of pump supply pressure p s and output volumetric flow rate q 2 we have the following I/O EOM (as derived in lecture): R 1 CI P  q 2 + R 1 CR T + I P [ ] q 2 + R T + R 1 [ ] q 2 = p s = p 0 u t ( ) where C = A ρ g = tank capacitance R T = R P + R V + R N Note that the above is a second-order differential equation. Writing this I/O equation in the standard second-order form:  q 2 + R 1 CR T + I P R 1 CI P q 2 + R T + R 1 R 1 CI P q 2 = p 0 R 1 CI P u t ( )  q 2 + 2 ζω n q 2 + ω n 2 q 2 = K n 2 u t ( ) where n 2 = R T + R 1 R 1 CI P n = R T + R 1 R 1 CI P 2 n = R 1 CR T + I P R 1 CI P n = R 1 CR T + I P 2 R 1 CI P = 1 2 R T I P + 1 R 1 C ζ = R 1 CR T + I P 2 R T + R 1 R 1 CI P = 1 2 R T + R 1 R T R 1 C I P + I P R 1 C K n 2 = p 0 R 1 CI P K = 1 n 2 p 0 R 1 CI P = p 0 R T + R 1 A
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Transfer function Taking LT of above (w/ zero ICs) and solving for Q 2 s ( ) gives: Q 2 s ( ) = K ω n 2 s 2 + 2 ζω n s + n 2 U s ( ) = G s ( ) U s ( ) Response to step input
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Unformatted text preview: : u t ( ) = h t ( ) U s ( ) = 1 / s q 2, ss = ? t s = ?? % OS = ?? q 2 p a p a q 2 p 2 q 2 p 1 p a q 1 q 1 q 2 p a Textbook Problems For both problems: R 1 = R , R 2 = 3 R , A 1 = A , A 2 = 4 A and B = g / AR Derivation of transfer functions provided in the lecture notes. Problem 7.22 G s ( ) = H 2 s ( ) Q mi s ( ) = 3 B / A 12 s 2 + 13 Bs + B 2 h 2, ss = ?? t s = ?? Problem 7.23 G s ( ) = H 2 s ( ) Q mi s ( ) = 3 B / A 12 s 2 + 16 Bs + B 2 h 2, ss = ?? t s = ??...
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hydraulics_analysis - : u t ( ) = h t ( ) U s ( ) = 1 / s q...

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