Question

Question 3 (a) Water flows through a horizontal pipeline of constant 400 mm diameter in a water treatment plant. The pipe bends through a 70° angle. In order to design a thrust block for the bend, calculate the magnitude and line of action of the force exerted by the water on the pipe. The discharge through the pipe is 0.4 m/s. The water pressure at the inlet is equivalent to 22 m head of water. [10 marks] (b) Oil of relative density 0.85 is flowing through a pipe of 40 mm diameter. Determine the maximum flow rate possible if the flow is to remain laminar The oil has a viscosity of 92 x 10 kg/ms [3 marks] (c) Water is being stored in a cylindrical tank 3 m tall and 1 m in diameter Water can be removed through a tap which is 2 cm in diameter and is located on the side of the tank, 0.1 m above the base. If the tank is initially full, calculate how long it will take to lower the water level in the tank by 1 m. The tap can be assumed to act like a small orifice with Co = 0.7. [7 marks)

Answer 1

Page no- • Solution - (a) Given that - Diameter of Pipe, d = 400mm MY PI = 22m X700 LS o is = 0.4 m/s. PA ---- K u ... PLAN ① and . Applying Bernouli's equation b/w op +21+x - L: 42 the neglecting loves + 2t 29 " & There is no change in diameter of pipe = V, = V₂ As the pipe is laid horizontal = Z, = Z2. Now, The control volume is shown in the figure along . with various forces acting on it. Along -axis Net Force = Total Momentum flux Total Momentum Huxl oncr.in going out of evin going into CV in x-diren" + x-direction tx-division 81

Page no-2 =>P, A - Fx - P A LOS700 = SUV, cost0° - BUVO F = 880. (1-10370") + PIA (1-c0570°) [.:P, = = 1000x0.40.4(-40%70') + 22x9.81x10x5 x64)3(1-00570) A x(0.4² = 18682.670 ../fx = 18.683 KN (k ) Thus net force on Pipe in x direction - Net force on evinxdy" 18.683 kW] < >) Along Y-direction Net force on CV = Total Momentum Total momentum flux In tyodah (flux going out of going into CV in cvinty-don loul ty-don - Fy ~PA sinto° = 3 Svo sin 70° -0 . => Fy = (0,4 + $ ¢v) Sin 70' 22x9.81x10²x A x(4)² + 1000x0.480.4 lxsinzo 26681.614N 26.682 KN Y = 26.682.XN) (1) Thus net force on Pipe in yodr" – Net Force onCV in yodan [Fyp = 26.682KN (1) :. Net Force on Pipe a Fx 8 + Fu = 118,683) + (26.682)² => fp = 32.573 kN] Direct" affe To Jay estari / Frp) = tan (26,682) = 55°

Page no-3 force on Pipe bend, Ifo = 32.573 kN original direction (x-axia). art at 55° from (6) Given that Specific Granity of oil, b=0.85 Diameter of Pipe, D = 40mm = 0.04m Dynamic vincosity of Oil, Mo = 92x103 kglms To keep the flow Laminar in pipe, Reynolds No., Re €2000 =) S. VD < 2000 Mo => 0.85X1000* vx0.04 < 2000 92X1023 V5 5.412 m/s V< 5.4 n that - k Cylinderical Tank. -daun a k=3m Cd = 0.7 0.1 m Dim -Orifice

Page no-4 When the water level will be at a height i'm above the orifice, ' velocity of water coming out of orifice, varagh. velocity of wash In a small time 'dt, water level falls by 'dh, Thus- volume of water on fall = volume of water leaving In height by 'dh' the orifice in at time. Area oftank xdha ot - I XD2x dn =-(d" E x10.02) 3J2 x3.8lxh.lt ov. hoog maken d =- 806.292. Integrating both sides under the limits to tox=T and h=2,9 to h=1.8m 1.9 Say = -806.292 2.9 = 523.34 seconds, - T=-806,292 X ash I 12.9 T= 8 min 43.34 sec