Integrating both sides and using the initial

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Unformatted text preview: parating the Vo & t variables: . € Integrating both sides and using the initial condition V0(0)=K, we get: or € V0 ( t ) / K = 1 /(1 − Kt ) l Solution to MT2: Problem#2 (a) Mass conservation: => (1) (b) Bernoulli’s equation is applicable for an incompressible and inviscid fluid, in a steady flow, along a streamline. We assume that it can be used even when the flow is unsteady. We take z1=0. . In the present case: p0=p1=patm => (2) At t~0, we can assume that v0~0 and (z0-z1) ~ h0 => (c) Assuming a constant exiting velocity v1, then it is straight forward to esimate: TA = A0 A1 h0 10 m 2 = 2 g 0.2 m 2 3m ≈ 19.6 sec 2 × 9.81m.s−2 (d) Equation (2), with h(t) now being variable, becomes: . € (e) Combining this equation with equation (1), we obtain: . 2 2 v 0 = 2 gh ( t ) /[ A0 A12 − 1] Noting that : 2 v 0 = 2 gh ( t ) /[ A0 A12 − 1] or dh 2 € = −h1 / 2 ( t ) 2 g /[ A0 A12 − 1] dt This is a 1st order ordinary differential equation in h(t), with the radical term being just a constant. € € ________________ (f) Extra credit. Solving the above ODE by separating the h and t variables yields: dh 2 = − 2 g /[ A0 A12 − 1]dt h( t ) => 2 h ( t ) h( t ) h0 2 = − 2 g /[ A0 A12 − 1]...
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This note was uploaded on 01/24/2014 for the course ME 106 taught by Professor Morris during the Spring '08 term at University of California, Berkeley.

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