A Model Answer for. Problem Set #7


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1 A Model Answer for Problem Set #7 Pipe Flow and Applications Problem.1 A pipeline 70 m long connects two reservoirs having a difference in water level of 6.0 m. The pipe rises to a height of 3.0 m above the water level in the upper reservoir at a distance of 40 m from the entrance before falling to the lower reservoir. If the pipe is 1. m diameter and the friction coefficient f = 0.015, find the discharge and the pressure at highest point in the pipeline 1
2 Problem.1  sol Apply B.E bet. 1 & (Neglect minor losses) = h f1 h f1 = 6 = ( 8fL/g π D 5 ) Q = (8*0.015*70 / g π (1.) 5 ) Q Q = 4.09 m 3 /sec v =Q/A = 3.6 m/sec Apply B.E bet. 1 & 3 (Neglect minor losses) = 9 + v / g + P 3 / γ + h f13 h f13 = (8fL/g π D 5 ) Q = (8*0.015*40 / g π (1.) 5 ) Q P 3 / γ = 3  (3.6) / g  (8*0.015*40 / g π (1.) 5 ) (4.09) = Problem. Water discharges from a reservoir into the atmosphere through a pipeline 39 m long. The pipeline, which has a sharp entrance, is 50 mm diameter (f = 0.0) for the first 15 m then suddenly enlarges to 75 mm diameter (f = 0.05) for the rest of its length. If the discharge is maintained at.8x103 m 3 /sec, calculate the water head in the reservoir taking into account all losses.
3 Problem.  sol v 1 = Q/A 1 = 1.46 m/sec & v = Q/A = m/sec Apply B.E bet. 1 & H = v / g + h l1 H = v / g v 1 / g + (8f 1 L 1 /g π D 15 ) Q + ( v 1 v ) /g + (8f L /g π D 5 ) Q H = 0.89 m Problem.3 Oil of kinematic viscosity 0.1 stoke flows through a smooth pipe of 30 cm diameter. If the virtual slope is 1/800, obtain the discharge 3
4 Problem.3  sol h f = (8fL/g π D 5 ) Q The virtual slope = slope of total energy line = h f /L =(8f/g π D 5 ) Q = 1/800 f Q = * (1) assume f 0 = sub. In (1)  Q0 = m 3 /sec v 0 = Q 0 / A = m/sec R n0 = v0 D / ν = Moody chart  Smooth pipe  f 1 = f 1 = sub. In (1)  Q1 = m 3 /sec v 1 = Q 1 / A = 0.57 m/sec R n1 = v 1 D / ν = Moody chart  Smooth pipe   f = 0.06 f = sub. In (1)  Q = m 3 /se Problem.4 Water from a large reservoir discharges into the atmosphere through a 100 mm diameter pipe 450 m long ending with a nozzle of diameter.45 cm. If the pipe, which discharges at a level 1 m below that of the water in the reservoir, has a sharpedged entrance and roughness of 0.1 mm, calculate the discharge knowing that the coefficient of the nozzle is 0.98 and the viscosity of water is 0.01 poise. 4
5 Problem4  sol Apply B.E bet. 1 & = v / g + h l1 H = v / g vp / g + (8fL p /g π D p5 ) Q v / g v = Q / A & v p = Q / A p H = Q / ga Q / ga p + (8fLp/g π D p5 ) Q Q / ga Q fq = (1) ЄS / D = 0.1/10 = assume f 0 = sub. In (1)  Q 0 = m 3 /sec R n0 = ρ v 0 D / µ = ρ Q 0 D / Aµ = Moody chart  ЄS / D = f 1 = sub. In (1)  Q 1 = m 3 /sec R n1 = ρ Q 1 D / µ = Moody chart  ЄS / D = f = sub. In (1)  Q = m 3 /sec Problem5 Two reservoirs are connected by a pipeline, which is 150 mm diameter for the first 6 m and 55 mm diameter for the remaining 15 m. The difference in water levels is 6 m. If the entrance and exit are sharpedged and the change in diameter is sudden, determine the losses and calculate the discharge. Sketch the hydraulic gradient and the total energy lines. (f = 0.04). 5
6 Problem5  sol Apply B.E bet. 1 & = h l16 = 0.5 v 1 / g + (8f 1 L 1 /g π D 15 ) Q + ( v 1  v ) /g + (8f L /g π D 5 ) Q + v / g v 1 = Q/A 1 & v = Q/A 6 = 0.5 Q / g A 1 + (8f 1 L 1 /g π D 15 ) Q + (Q/A 1 Q/A ) /g + (8f L /g π D 5 ) Q + Q / g A Q = m 3 /sec Problem.6 A tank delivers water through a pipeline to a lower tank with a rectangular sharpedged weir of 5 cm width and (0.00) crest level; refer to figure (1). If the weir coefficient of discharge is 0.6, determine the steady discharge and the water head (h) above the weir crest taking into account all losses. 6
7 Problem.6  sol Apply B.E bet. 1 & (the crest level is the datum ) = h h l1.5 = h + [ 0.5 v / g + (8fL/g π D 5 ) Q + v / g ].5 = h + [ 1.5 Q / ga + (8fL/g π D 5 ) Q ].5 = h Q (1) Q = /3 C d B g h 1.5 Q = /3 * 0.6 * 0.5 g h 1.5 Q = h () From 1 &.5 = h h 3 h = 0.8 m & Q = 0.1 m 3 /sec H =.5 h = 1.7 m Problem.7 A, B, C and D are four points on a pipeline. Sections AB, BC and CD are straight with lengths 00 m, 300 m and 00 m, respectively. Elevations above datum of points A, B, C and D are 78 m, 764 m, 60 m and 614 m respectively. If the pressure at point A is 1. kg/cm and at point D is 19.3 kg/cm, find the pressure at a point 700 m above datum. (Neglect minor losses ) 7
8 Problem.7  sol T.E A = Z A + v / g + P A / γ = * 10 4 / v / g = v / g T.E D = Z D + v / g + P D / γ = * 10 4 / v / g = v / g T.E D > T.E A The flow direction from D to A hl = T.E D T.E A = 13 m hl / m= 13 / 700 m E C /300 = 80/ E C = m Apply B.E bet. D & E v / g * 10 4 /1000 = v / g + PE / γ + h lde Where h lde = hl/m * DE = 13/700 * ( ) = 6.81 m PE / γ = m PE = t/m Problem.8 A compound pipeline 8 km long is made up of a pipe 10 cm diameter for length of 1 km, 0 cm for km, 5 cm for 1.5 km and 30 cm for 3.5 km. It is required to replace the compound pipe by an equivalent pipe for the same total length and discharge. Find the diameter of the new pipe assuming all pipes have the same friction coefficient. 8
9 Problem.8  sol Neglect the minor losses h f (pipe1) = h f (pipe ) (8fL 1 /g π D 15 ) Q + (8fL /g π D 5 ) Q + (8fL 3 /g π D 35 ) Q + (8fL 4 /g π D 45 ) Q = (8fL/g π D 5 ) Q 1000/(0.10) /(0.0) /(0.5) /(0.30) 5 = 8000/D 5 D = cm Problem.9 Two reservoirs are connected by a 600 m long, 30 cm diameter pipe ( f = 0.03). The flow produced by the difference in the water levels is 170 lit/sec. If a new pipe 00 m long, 30 cm diameter (f = 0.0) is laid parallel to an equal length of the old pipe, determine the new discharge. If it is required to double the original discharge, calculate the required length of the new pipe 9
10 Problem.9  sol Apply B.E bet. 1 & H = h f = (8fL/g π D 5 ) Q H = m h f = 8f L/gπ D5 Q = k Q k 1 = 8f L 1 /gπ D 15 = k = 8f L /gπ D 5 = k 3 = 8f L 3 /gπ D 35 = Q 1 = Q + Q (1) Apply B.E bet. 1 & = k 1 Q 1 + k Q = k 1 Q 1 + k Q () Problem.9  sol Apply B.E bet. 1 & = k 1 Q 1 + k 3 Q = k 1 Q 1 + k 3 Q (3) from & 3 k Q = k 3 Q 3 Q = Q (4) Sub. from 4 in 1 Q 1 = Q (5) Q 3 = m 3 /sec Q = m 3 /sec 10
11 Problem.9  sol Sub. from 5 in 3 Q 1 = m 3 /sec k 1 = 8f (600 L)/gπ D 5 1 = L k = 8f L/gπ D 5 = 1.0 L k 3 = 8f L/gπ D 5 3 = 0.68 L 0.34 = Q + Q (1) Apply B.E bet. 1 & = k 1 Q 1 + k Q = k 1 Q 1 + k Q () Apply B.E bet. 1 & = k 1 Q 1 + k 3 Q = k 1 Q 1 + k 3 Q (3) from & 3 k Q = k 3 Q 3 Q = Q (4) Sub. from 4 in = Q 3 Q 3 = m 3 /sec Q = m 3 /sec Sub. in = ( L) * (0.3) L * (0.158) L = m Problem.10 A water main is 100 mm diameter and 4.8 km long. Supplies, that are arranged along the water main, uniformly draw a discharge q of 7.5 lit/hr per meter of its length. Calculate the difference in head between the pipe entrance and the last point of supply. Take f =
12 Problem.10  sol Q= (Q in Q out ) / L If Q out = 0 Q in = q L = 7.5 / (1000 * 60 * 60 ) * 4.8 * 1000 = 0.01 m 3 /sec h f = 1/3 [8f L/gπ D 5 Q ] h f = 1/3 [8*0.0*4.8*1000 /gπ (0.10) 5 (0.01) ] = 6.44 m Problem.11 Water flows between two reservoirs with water level difference of 1.5 m through a pipe 4 km long, 50 cm diameter. It is required to feed a third reservoir, which is 15 m lower than the highest reservoir. The new pipe is 1.5 km long and is connected to the main pipe at a point 1.0 km from its entrance. Find the diameter of this new pipe such that water discharges equally into both reservoirs. Take f = 0.03 for all pipes and plot the Hydraulic Gradient line. 1
13 Problem.11  sol Apply B.E bet. 1 & = h f1 + h f =.5 + (8fL 1 /g π D 15 ) Q +(8fL /g π D 5 ) (Q/) Q = m 3 /sec Apply B.E bet. 1 & = h f1 + h f3 = (8f 1 L 1 /g π D 15 ) Q +((8f L /g π D 5 ) Q D 3 = m Problem.1 A pump delivers water through two pipes laid parallel and connected to each other at the pump outlet. One pipe is 100mm diameter and 45 m long and the other pipe is 150 mm diameter and 60 m long. They both discharge to the atmosphere at 6 m and 8 m, respectively, above pump outlet. Determine the total head at the pump outlet if the flow rate through it is m 3 /sec. Take the datum at the pump outlet and assume f =
14 Problem.1  sol = Q + Q3. Q = Q (1) Apply B.E bet. 1 & H P = v /g + 8f L /gπd 5 Q H P = 6 + Q /ga + 8f L /gπ D 5 Q H P = Q () Apply B.E bet. 1 & 3 H P = v3/g + 8f L 3 /gπ D 35 Q 3 H P = 8 + Q3/gA3 + 8f L 3 /gπ D 35 Q 3 H P = Q (3) From & Q = Q (4) sub. From 1 in ( Q 3 ) = Q 3 Q 3 = > (rejected) OR. Q 3 = m 3 /sec. Q = m3/sec Sub in 3 The total head at the pump outlet = 9.0 m 14
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