Practice Problems with Solutions

Practice Problems with Solutions - Problem 2.41[Difficulty...

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Problem 2.41 [Difficulty: 2] Given: Velocity distribution between parallel plates Find: Force on lower plate Solution: Basic equations F τ yx A = τ yx μ du dy = du dy d dy u max 1 2 y h 2 = u max 4 h 2 2 y = 8 u max y h 2 = so τ yx 8 μ u max y h 2 = and F 8 A μ u max y h 2 = At the lower surface y h 2 = and h 0.1 mm = A 1 m 2 = u max 0.05 m s = μ 1.14 10 3 × N s m 2 = (Table A.8) Hence F 8 1 × m 2 1.14 × 10 3 × N s m 2 0.05 × m s 0.1 2 × mm 1 m 1000 mm × 1 0.1 1 mm 1000 mm 1 m × 2 × = F 2.28 N = (to the right)

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Problem 2.57 [Difficulty: 2] Given: Flow of three fluids between two plates Find: Upper plate velocity; Interface velocities; plot velocity distribution Solution: The shear stress is the same throughout (the velocity gradients are linear, and the stresses in the fluids at the interfaces must be equal and opposite). Given data F 100 N = h 1 0.5 mm = h 2 0.25 mm = h 3 0.2 mm = A 1 m 2 = μ 1 0.15 N s m 2 = μ 2 0.5 N s m 2 = μ 3 0.2 N s m 2 = The (constant) stress is τ F A = τ 100 Pa = For each fluid τ μ V y = or V τ ∆ y μ = where V is the overall change in velocity over distance y Hence V 12 τ h 1 μ 1 = V 12 0.333 m s = where V 12 is the velocity at the 1 - 2 interface Hence V 23 τ h 2 μ 2 V 12 + = V 23 0.383 m s = where V 23 is the velocity at the 2 - 3 interface Hence V τ h 3 μ 3 V 23 + = V 0.483 m s = where V is the velocity at the upper plate 0 0.1 0.2 0.3 0.4 0.5 0.25 0.5 0.75 1 Velocity (m/s) Position (mm)