HWK3514_HWK_7_solution_Fall_2007

HWK3514_HWK_7_solution_Fall_2007 - ME 3514 System Dynamics...

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ME3514 - HWK - Burdisso - 68 ME 3514 – System Dynamics HOMEWORK #7-Solution DUE: 10/31/07 1. (10pts) In the liquid level system shown in the following figure, the head is kept at 1m for 0 t < . The inflow valve opening is changed at 0 = t , and the inflow rate is sec 05 . 0 3 m for 0 t . Determine the time needed to fill the tank to 2.5-m level. Assume that the outflow rate Q s m 3 and head H m are related by H Q 02 . 0 = The capacitance of the tank is 2 2 m . Remark: To solve this problem, please linearize the equation of motion around 1 Hm = . F i g u r e : Liquid-level system Solution: The equation of motion in terms of the absolute level () Ht is (conservation of mass) i dH CQ t Q t dt =− The valve equation is then ( ) 0.02 ( ) Qt = There are two options: a) Replace ( ) 0.02 ( ) = into EOM: 0.02 ( ) ( ) i dH t CH t Q t dt += ( 1 a ) b) Differentiate ( ) 0.02 ( ) = w.r.t. and then replace: ( ) 0.02 ( ) 2( ) dQ t dH t dt dt = ) 0.02 i dQ t t Q t dt ( 1 b ) NAME: Q qt =+ ii Qt Q qt H ht
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ME3514 - HWK - Burdisso - 69 The first option is better because it solves for h(t) directly. The steady-state condition is: 3 1 0.02 0.02 sec Hm QH m = == Thus, the increase in flow rate at t=0 can be modeled as a step input, i.e. 3 ( ) 0.03 1( ) sec i qt t m . Thus, the input flow rate is 3 () 0 .03 1 sec ii Qt Q qt Q =+ =+ × (2) Approach #1: Linearization of EOM Linearization of ( ) Qt about 1 = using Taylor’s series expansion: 0.02 ( ) 0.02 ( ) 2 () 0 .01() 0.01 H Ht H H Q qt ht dq dh dt dt +− −= ± ± ± ( 3 ) Note that H ht Replacing (2) and (3) into (1a,b), the new linear EOM becomes 2 0.01 ( ) ( ) 0.03 1( ) 200 ( ) 3 1( ) i dh t q t t dt dh t t dt += = × × ( 4 a ) with initial condition (0) 0 h = . Note that the time constant is 200 1/ 0.005 To r T = = The solution to (4a) is ( ) 0.005 () 31 t e =− The time needed to fill the tank to 2.5-m level is 1 0.005 11 1 1.5 ( ) 1.5 3 1 0.005 1 0693 3 0693 138.6sec 0.005 t H e t Ln t ⎛⎞ = = = ⎜⎟ ⎝⎠
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ME3514 - HWK - Burdisso - 70 REMARK: The same solution is obtained by solving for q ( t ). That is () 0.005 200 ( ) 0.03 1( ) ( ) 0.03 1 t dq t qt t e dt +=× = and ( ) 0.005 ( ) 0.01 ( ) ( ) 3 1 t ht e →= ± Approach #2: Exact solution Integrating (1) ( ) ( ) ( ) 0.02 ( ) 2 ( ) 0.05 0.02 ( ) i CdH t Q t H t dt dH t H t dt =− Then 1 2.5 1 01 2 0.05 0.02 ( ) 2 128.8sec 0.05 0.02 ( ) t dt dH Ht td t d H = == = ∫∫ Error in Approach #1 is: 138.6 128.8 1.06 6% 128.8 Error or
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ME3514 - HWK - Burdisso - 71 2. (10pts) At steady state the flow rate throughout the liquid-level system shown in the following figure is Q , and the heads of tanks 1 and 2 are 1 H and 2 H , respectively. At 0 = t , the inflow rate is changed from Q to q Q + , where q is a small change in the inflow rate. The resulting changes in the heads () 2 1 & h h
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This homework help was uploaded on 04/18/2008 for the course ME 3514 taught by Professor J.renno during the Fall '08 term at Virginia Tech.

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HWK3514_HWK_7_solution_Fall_2007 - ME 3514 System Dynamics...

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