Problem 4.189 - ω R 2 ⋅ V ⋅ α R 3 3 ⋅ ⎛ ⎜ ⎝...

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Problem 4.189 [Difficulty: 3] Given: Data on rotating spray system Find: Differential equation for motion; steady speed Solution: Basic equation: Rotating CV Assumptions: 1) No surface force; 2) Body torques cancel; 3) Steady flow; 5) Uniform flow; 6) L<<r The given data is Q1 5 L min = R 225 mm = d5 m m = ρ 999 kg m 3 = For each branch V 1 2 Q π 4 d 2 = V 6.37 m s = A π 4 d 2 = A 19.6 mm 2 = The basic equation reduces to a single scalar equation (FOR EACH BRANCH) V r 2 ω V × r × α r × + () × ρ d A r V xyz ⎯⎯ × ρ V xyz ⎯⎯ d = where α is the angular acceleration But r 2 ω V × r × α r × + () × 2 ω r V α r 2 + = (r and α perpendicular) The volume integral is then V r 2 ω V × r × α r × + () × ρ d
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Unformatted text preview: ω R 2 ⋅ V ⋅ α R 3 3 ⋅ + ⎛ ⎜ ⎝ ⎞ ⎠ − ρ ⋅ A ⋅ = For the surface integral (FOR EACH BRANCH) A → r V xyz → ⎯⎯ × ρ ⋅ V xyz → ⎯⎯ ⋅ ⌠ ⎮ ⎮ ⌡ d R V ⋅ ρ ⋅ Q 2 ⋅ = Combining ω R 2 ⋅ V ⋅ α R 3 3 ⋅ + ⎛ ⎜ ⎝ ⎞ ⎠ − ρ ⋅ A ⋅ R V ⋅ ρ ⋅ Q 2 ⋅ = or α 3 A R 2 ⋅ ω − V ⋅ A ⋅ R ⋅ Q V ⋅ 2 − ⎛ ⎝ ⎞ ⎠ ⋅ = (1) The steady state speed ( α = 0 in Eq 1) is then when ω max − V ⋅ A ⋅ R ⋅ Q V ⋅ 2 − = or ω max Q 2 A ⋅ R ⋅ − = ω max 28.3 − 1 s = ω max 270 − rpm =...
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