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Introduction-Fluid-Mechanics-Solution-Chapter-06

Introduction-Fluid-Mechanics-Solution-Chapter-06 - Problem...

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Incompressible flow du dx dv dy + A sin 2 π ⋅ω t () A sin 2 π t = 0 = Hence du dx dv dy + 0 = Check for incompressible flow vA y sin 2 π t = uA x sin 2 π t = ρ 2 kg m 3 = ω 1 1 s = A2 1 s = The given data is Solution Find: Expressions for local, convective and total acceleration; evaluate at several points; evaluat pressure gradient Given: Velocity field Problem 6.6
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The governing equation for acceleration is The local acceleration is then x - component t u 2 π A ⋅ω x cos 2 π t () = y - component t v 2 −π A y cos 2 π t = For the present steady, 2D flow, the convective acceleration is x - component u du dx v du dy + Ax sin 2 π t A sin 2 π t A y sin 2 π t 0 + ... = u du dx v du dy + A 2 x sin 2 π t 2 = y - component u dv dx v dv dy + sin 2 π t 0 A y sin 2 π t A sin 2 π ( ( + = u dv dx v dv dy + A 2 y sin 2 π t 2 =
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12.6 m s 2 and 12.6 m s 2 Local t 0.5 s = 12.6 m s 2 12.6 m s 2 Total 0 m s 2 and 0 m s 2 Convective 12.6 m s 2 and 12.6 m s 2 Local t0 s = Evaluating at point (1,1) at t v u dv dx + v dv dy + 2 −π A ⋅ω y cos 2 π t () A 2 y sin 2 π t 2 + = y - component t u u du dx + v du dy + 2 π A x cos 2 π t A 2 x sin 2 π t 2 + = x - component The total acceleration is then
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x p ρ 2 π A ⋅ω x cos 2 π t () A 2 x sin 2 π t 2 + = x p ρ Du dt = Hence, the components of pressure gradient (neglecting gravity) are (6.1) The governing equation (assuming inviscid flow) for computing the pressure gradient is 12.6 m s 2 12.6 m s 2 Total 0 m s 2 and 0 m s 2 Convective 12.6 m s 2 and 12.6 m s 2 Local t1 s = 12.6 m s 2 12.6 m s 2 Total 0 m s 2 and 0 m s 2 Convective
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25.1 Pa m y comp. 25.1 Pa m x comp. t1 s = 25.1 Pa m y comp. 25.1 Pa m x comp. t 0.5 s = 25.1 Pa m y comp. 25.1 Pa m x comp. t0 s = Evaluated at (1,1) and time x p ρ 2 −π A ⋅ω y cos 2 π t () A 2 y sin 2 π t 2 + = y p ρ Dv dt =
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Problem 6.7 Given: Velocity field Find: Expressions for velocity and acceleration along wall; plot; verify vertical components are zero; plot pressure gradient Solution The given data is q2 m 3 s m = h1 m = ρ 1000 kg m 3 = u qx 2 π x 2 yh () 2 + 2 π x 2 + 2 + + = v qy h 2 π x 2 2 + + 2 π x 2 + 2 + + = The governing equation for acceleration is
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x - component u du dx v du dy + q 2 x x 2 y 2 + () 2 h 2 h 2 4y 2 x 2 yh + 2 + 2 x 2 2 + 2 ⋅π 2 = a x q 2 x x 2 y 2 + 2 h 2 h 2 2 π 2 x 2 + 2 + 2 x 2 2 +
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Introduction-Fluid-Mechanics-Solution-Chapter-06 - Problem...

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