ME_Fluid Mechanics_922 - 2017 WORKBOOK Detailed Explanations of Try Yourself Questions Mechanical Engineering Fluid Mechanics and Machines 1 Fluid

# ME_Fluid Mechanics_922 - 2017 WORKBOOK Detailed...

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Mechanical Engineering Fluid Mechanics and Machines WORKBOOK WORKBOOK WORKBOOK WORKBOOK WORKBOOK 2017 Detailed Explanations of Try Yourself Questions
© Copyright T1 : Solution V W θ W cos θ W sin θ W F viscous θ N Balancing forces along the inclined plane. F viscous = W sin θ μ = W sin θ V = θ μ = × × × × × = 0.5625 m/s Fluid Properties 1
© Copyright 3 Workbook T2 : Solution Power = T ω Calculating torque, Torque = F × radius F = μ μ = 2 × 10 –1 Ns/m 2 A = π D l = π × × = 0.1414 m 2 v = π × = × π × × = 1.131 m/s y = 2.5 mm = 2.5 × 10 –3 m F = × × × × = 12.79 N Torque = F × radius = × = 0.576 Nm ω = × π × = 25.12 rad/s P = 0.576 × 25.12 = 14.47 Watt 14.5 Watt
© Copyright T1 : Solution A ρ = 0.12 kg/m 3 0.4 m 0.5 m ρ w = 10 kg/m 3 3 ρ Hg = 13.6 × 10 kg/m 3 3 P A + (0.12 × 9.81 × 0.9) – {13.6 × 10 3 × 9.81 × 0.5} + {10 3 × 9.81 × 0.5} – {13.6 × 10 3 × 9.81 × 0.5} + {10 3 × 9.81 × 0.5} – {13.6 × 10 3 × 9.81 × 0.5} + {10 3 × 9.81 × 0.5} – {13.6 × 10 3 × 9.81 × 0.5} = 0 P A + 1.05948 – 66708 + 2 × {4905} – 2 × {66708} = 0 P A = 190312.94 Pascal Pressure in terms of height of water column = × = 19.39 m of H 2 O column T2 : Solution outside volume = 600 m 3 inside volume = (V– 600) ρ ice berg = 915 kg/m 3 ρ sea water = 1025 kg/m 3 Fluid Statics 2
© Copyright 5 Workbook Let the total volume of iceberg be “ v ”. Buoyancy force = Weight of iceberg ρ sea water × ( V – 600) × 9.81 = ρ iceberg × v × 9.81 1025 ( v – 600) = 915 V 1025 v – 915 v = 1025 × 600 v = × = 5590.9 m 3 Weight of the iceberg = ρ iceberg × v iceberg × 9.81 = 915 × 5590.9 × 9.81 = 50184757.04 N = 50.185 MN T3 : Solution h M G B r Specific gravity of the cylinder is “ S ”. Given, r = For floatation to be unstable GM < 0 We have GM = I Calculating height of cylinder inside water, let it be “ Y ”. Buoyancy force = Weight of cylinder ρ w × Volume of cylinder inside × g = ρ cylinder × Volume of cylinder × g ρ w × π r 2 × Y × 9.81 = s × ρ w × π r 2 × h × 9.81 Y = sh I = ( ) π π × = π × π × × ×
6 Mechanical Engineering Fluid Mechanics and Machines © Copyright = × = × × = BG = OG OB = = = ( ) I < 0 ( ) < 0 2 – 9 s (1 – s ) < 0 2 < 9s – 9s 2 or 9 s 2 – 9 s + 2 < 0 9 s 2 – 3 s – 6 s + 2 < 0 3 s (3 s – 1)– 2 (3 s – 1) < 0 (3 s –1)(3 s – 2) < 0 ⎞ ⎛ ⎟ ⎜ ⎠ ⎝ < 0 + + 0.333 0.667 + + < s < T4 : Solution F B T + W D T = 1.5 m = 5.5 kN F buoyancy = Tension + Weight ρ w × Volume 5 g = Tension + Weight, Weight = F buoyancy – Tension = [ ] ρ × × π × × × = [ ] × × π × × × = 17335.7 – 5500 = 11835.7 N = 11.8 kN
© Copyright 7 Workbook T5 : Solution 30° 5 sin30° 5 sin30° 30° F h = ρ × g × A x = 1000 × 10 × 2 × 5 × sin30° × 1 × 5 × sin30° = 125 kN/m F v = θ × π × × × 1000 × 10 = ( ) × π × × × ° × ° × 1000 × 10 = 22.65 kN/m F R = + = + = 127.03 kN/m
© Copyright T1 : Solution x = 0 x = 0.375 m v = 1.5 m/s v = 15 m/s Let the velocity by given by u = a + b x At x = 0, u = 1.5 a = 1.5 At x = 0.375, u = 15 b

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