Question

Q4: The pump is used to deliver of water from a river to a storage reservoir. The four pipes are connected in series of Inverse L shape which means the first pipe in vertical direction and others in horizontal direction. The first pipe in vertical position of 500m length of 150 mm diameter cast iron pipe, the second pipe is bend at 90' with a radius of 0.75m, the third pipe in horizontal direction of 100m length of 160mm diameter of which then transitions to a fourth pipe in same direction of 50m long of 100 mm diameter. If the difference between the water levels in the river and the reservoir is 20 m and the water discharges into the atmosphere at the reservoir through a fully open gate valve, Considering all minor losses of head which occur, determine the discharge? Given Co-efficient of friction factor is 0.01 as same for all sections. Take Kbend = 0.2. Kinematic viscosity of water = 0.2 stoke Draw a neat sketch. Find also (1) What is the Power lost due to the difference of water level between river and reservoir? (11) What would be the nature of flow based on Reynolds number for each pipe? (8 marks)

Answer 1

Given, Four Pipes are connected in serios, 1st pipe dia meter = 150mm I length = 500m 2" pipe bend at 90 of rading of 0.75m. Same de 3rd pipe diameter = 160mm = length = 1.178m (ro) A length = 100m. 4th pipe diameter = 100mm length = som the head of water levels between river and reservoir = 20m Cofficient of friction factor f = 0.01 khead = 0.2 Kinematic Viscosity of water, Uwater = 0.2 cm / See Let Vi, V₂, V and Vq are the Velocities in the 1st and ard and 4th pipe respectively. From continuity, we have AV = A, V = Aqva - Agle. Vis VERT - - Car DV 14 a viz Centro - 0.527 Lupa - 05.2V= 2:257

The head loss in the pipe , + + + ' zgdq 0.2%V/ rg 01011.178 XV, 20 - OSV + 0.01x 500 x V + 2x9181 2x9,81 XOS 2x 9.3) 2x9,818 015 + 0,-1) 0.01% 100x (6.87€ v.)? 05x62-25 V) ta 2x9.31 X0.100 2x9.81 + 0.01 X 50X (2-25V) + (2.25V,) 2x9.31 x 0110 249.81 20 = 0.0257+ 1.698 v," +0.01016" + 0.004 v1 +0.2.45 V . + 0,129 vr + 1.29 + 0.258 vr 20 = 3,6591 Y → = 2000 . 3.659 V = 2.33 m/s Rate of flow, Q = Areax Velocity - I (4,0x V - I (0-15)x 233 = 0.0411 m/s Power lost , p = egQH 1000x981 X 0.0411 x 20 = 8063.82 Joule

Size The equivalent to thang the pipe is ebtained from return terms [1344454] espero tempo com o d = 01138m evd Reynold's numbu, Re = 1000 X 2-33 X 0'138 = 321540 > 4000 is turbulent flow: It