From the given lt table 14 e b t 81 e 4 and s

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Unformatted text preview: e given LT Table #14: e B (t ) = 8(1 − e 4 ) and [s + (3 / 4)] 2 s from Equation (3-6), we have e A (t ) = 6 + 2(1 − e − 3t 4 ) qin [4] (33.33%) Consider the following fluid system. Recall that the volume of the fluid in the tank is given by 1 V = kh 3 , and the flow rate through a uniform pipe and 3 a valve can be approximated by q = ∆h / R and q = k t ∆h respectively, where ∆h is difference across the valve. the Ps=ρghs h pressure R qs (a) Determine the flow rates qin, qs and qo at the equilibrium operating point ( hs , h ). At equilibrium, h −h dV =0 ⇒ qin = q s − q o where qs = s and q o = k t h R dt (b) Obtain a linearized approximation for the volume of fluid, and for the resistance of the valve at the operating point. ∂V 1 ˆ ˆ V = (kh 2 )h V = kh 3 V =V + (h − h ) ∂h h= h 3 12 ∂h ˆ 2 ˆ ˆ q o = k t h or h = 2 q o , h = h + q = q (q − q ) or h = 2 q o q 0 = Ro q 0 , ∂q kt kt hence Ro = 2 q 2 q o or Ro = 2 q o o 2 q kt kt o 2h = q o ˆ ˆ ˆ (c) Derive H ( s ) in terms of H s ( s ) and Qin ( s ) about the operating point. dV = q s + q in − q o dt ˆ d (V + V ) ˆ ˆ ˆ = (q s + q s ) + (qin + q in ) − (q o + q o ) dt ˆ dV ˆ ˆ ˆ = (q s ) + (q in ) − (q o ) + (q s ) + (q in ) − (q o ) dt kh 2 ˆˆˆ ˆ h dh h s − h ˆ = + qin − Ro dt R or kh 2 See part (a) ˆ ˆ dh 1 2q ˆ hs ˆ + + 2 h = + qin R dt R k t ˆ ˆ H ( s ) + RQin ( s ) ˆ H ( s) = s R Rkh 2 s + 1 + R o qo...
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