Solution (Assignment-3)

Solution (Assignment-3) - Assignment -3 (Solution) Question...

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Assignment -3 (Solution) Question 1: Let Point 1: At inlet pipe, Point 2: At bathroom Speed (v 1 ) and pressure (p 1 ) at the inlet pipe Speed (v 2 ) and pressure (p 2 ) at the bathroom y 1 = Height at the inlet = 0 y 2 = Height at the bathroom = 5 m d 1 = 2.0 cm, r 1 = 1.0 cm d 2 = 1.0 cm, r 2 = 0.5 cm v 1 = 1.5 m/s Q 1 = Q 2 , Hence A 1 v 1 = A 2 v 2 v 2 = (A 1 /A 2 )v 1 = (r 1 /r 2 ) 2 * v 1 = (1/0.5) 2 * (1.5) = 6.0 m/s From Bernoulli’s equation ) ( ) ( 2 1 1 2 2 1 2 2 2 1 y y g v v p p - + - = - ρ ) ( ) ( 2 1 1 2 2 1 2 2 1 2 y y g v v p p - - - - = p 2 = (4*10 5 ) – ½ (1*10 3 ) (6 2 – 1.5 2 ) – (1.0*10 3 )(9.8)(5-0) = 3.3*10 5 Pa Volume flow rate is = Q = A 2 v 2 = π (0.5*10 -2 ) 2 (6.0) = 4.7 * 10 -4 m 3 /s Question 2: Pressure (p) = F/A = (Mg)/( πd 2 /4) = [(4)(2.1*10 3 )(9.8)]/ [(π)*(0.22) 2 ] = 5.4 * 10 5 Pa Mechanical advantage of the system is the ratio of areas = MA = A o /A t = (22) 2 /1 2 = 484 Required input force = F t = F o /MA = (2.1 * 10 3 ) (9.8)/484 = 43 N
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Solution (Assignment-3) - Assignment -3 (Solution) Question...

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