Hydraulics. B.E. (Civil), Year/Part: II/II. Tutorial solutions: Pipe flow. Tutorial 1

Hydraulics B.E. (Civil), Year/Part: II/II Tutorial solutions: Pipe flow Tutorial 1 -by Dr. K.N. Dulal Laminar flow 1. A pipe 200mm in diameter and 20km long conveys oil of density 900 kg/m 3 and viscosity 0.08 NS/m 2 at 10 lps. a) Compute Reynolds number and state whether the flow is laminar or turbulent. b) Compute loss of head c) Compute the maximum velocity d) Compute the power required to maintain the flow. Diameter of pipe (D) = 200 mm = 0.2m Length of pipe (L) = 20 km = 20000m Density of oil (ρ) = 900 kg/m 3 Viscosity of oil (µ) = 0.08 NS/m 2 Rate of flow (Q) = 10 lps = 0.01 m 3 /s a) Reynolds no. (Re) =? Average velocity of flow = 0.318 m/s As Re<2000, the flow is laminar. b) Head loss (h f ) =? c) Maximum velocity (u max ) =? Maximum velocity (u max ) = 2V = 2x0.318 = 0.636 m/s d) Power required to maintain flow (P) =?

= 4070 W 2. A fluid of viscosity 5 poise and specific gravity 1.2 flows through a pipe of diameter 100mm. The maximum shear stress at the pipe wall is 150 N/m 2. Find (a) pressure gradient, (b) average velocity, and (c) Reynolds number. Diameter of pipe (D) = 100 mm = 0.1m Radius of pipe (R) = 0.1/2 = 0.05m Viscosity of oil (µ) = 5 poise = 5/10NS/m 2 = 0.5 NS/m 2 Specific gravity of fluid (S) = 1.2 Density of fluid (ρ) = 1.2x1000 = 1200 kg/m 3 Maximum shear stress (τ 0 ) = 150 N/m 2 (a) Pressure gradient (dp/dx) =? At r = R, shear stress, τ = τ 0 N/m 2 per m (b) Average velocity (V) =? =3.75 m/s C) Reynolds no. (Re) =? As Re<2000, the flow is laminar.

3. Oil of density 800 kg/m 3 and dynamic viscosity 0.2 poise flows through 50mm diameter pipe of length 500m at the rate of 0.2 lps. Determine (a) Reynolds number, (b) Centerline velocity, (c) pressure gradient, (d) loss of pressure in 500m length and (e) wall shear stress. Diameter of pipe (D) = 50 mm = 0.05m Radius of pipe (R) = 0.05/2 m = 0.025m Length of pipe (L) = 500m Density of oil (ρ) = 800 kg/m 3 Viscosity of oil (µ) = 0.2 poise = 0.2/10 NS/m 2 = 0.02 NS/m 2 Rate of flow (Q) = 0.2 lps = 0.0002 m 3 /s a) Reynolds no. (Re) =? Cross sectional area (A) = = 0.001963 m 2 Average velocity of flow = 0.102 m/s b) Centerline velocity (v max ) =? v max =2V =2x0.102 = 0.204 m/s c) Pressure gradient (dp/dx) =? d) Loss of pressure in 500m length =? N/m 2 per m N/m 2 per m Loss of pressure in 500m length = 26x500 = 13000 N/m 2 e) Wall shear stress (τ 0 ) =? = 0.325 N/m 2

4. Oil ofspecific gravity 0.6 and kinematic viscosity 0.0002 m 2 /s flows through an inclined pipe of diameter 60mm. The angle of inclination is 30 0. The pressures at 10 m apart are 360 KPa and 275 KPa respectively. Assuming steady laminar flow, find the direction and rate of flow. Justify that the flow is laminar. 2 Specific gravity of fluid (S) = 0.6 Density of fluid (ρ) = 0.6x1000 = 600 kg/m 3 Kinematic viscosity (υ) = 0.0002 m 2 /s Dynamic viscosity (µ) = υρ = 0.0002x600 = 0.12 NS/m 2 Diameter of pipe (D) = 60 mm = 0.06m Length of pipe (L) = 10m Pressure at section 1 (P 1 ) = 360 Kpa = 360000Pa Pressure at section 1 (P 2 ) = 275 Kpa = 275000Pa Cross sectional area (A) = = 0.002827 m 2 Z 1 = 0 Z 2 = 10Sin30 = 5m Total energyat 1 (E 1 ) = = Total energy at 2 (E 2 ) = = As E 1 >E 2, the flow takes place from section 1 to section 2. 1 30 0 Rate of flow (Q) =? Loss of head (h f ) =E 1 E 2 = 61.16 51.72 = 9.44 m Also Average velocity (V)= 5.2 m/s Q =A V = 0.002827x5.2 = 0.0147 m 3 /s Computing Reynolds no. (Re) As Re<2000, the flow is laminar

5. Crude oil of dynamic viscosity 0.2 NS/m 2 and specific gravity 0.75 flows through 20mm diameter vertical pipe. Two gauges have been fixed at 20m apart. The pressure gauge fixed at higher end reads 20 N/cm 2 and that at the lower end reads 60 N/cm 2. Assuming steady laminar flow, find the direction and rate of flow. Justify that the flow is laminar. Specific gravity of fluid (S) = 0.75 Density of fluid (ρ) = 0.75x1000 = 750 kg/m 3 Dynamic viscosity (µ) = 0.2 NS/m 2 20m Diameter of pipe (D) = 20 mm = 0.02m Length of pipe (L) = 20m Pressure at section A (P a ) = 20N/cm 2 = 20x10 4 N/m 2 Pressure at section B (P b ) = 60 N/cm 2 = 60x10 4 N/m 2 Cross sectional area (A) = = 0.000314 m 2 Z b = 0 Z a = 20 m V = Velocity of flow Total energy ata (E a ) = = A B Total energy head B (E b ) = = As E b >E a, the flow takes place from B to A (upwards) Rate of flow (Q) =? Loss of head (h f ) = E b E a = 81.55 47.2 = 34.35 m Also Average velocity (V) = 0.79 m/s Q = A V = 0.000314x0.79 = 0.000248 m 3 /s = 0.248 lps Computing Reynolds no. (Re) As Re<2000, the flow is laminar

6. Oil ofspecific gravity 0.9 and viscosity 10 poise flows through a pipe of diameter 100mm. The velocity at the center is 2m/s. Find (a) mean velocity and the radial distance at which this occurs, (b) pressure gradient in the direction of flow, (c) shear stress at the pipe wall, (d) Reynolds number, and (e) velocity at a distance of 30mm from the wall. Specific gravity of fluid (S) = 0.9 Density of fluid (ρ) = 0.9x1000 = 900 kg/m 3 Dynamic viscosity (µ) = 10 poise = 10/10 NS/m 2 = 1 NS/m 2 Diameter of pipe (D) = 100 mm = 0.1m Cross sectional area (A) = = 0.007854 m 2 Velocity at the center (u max ) = 2m/s a) Mean velocity (V) =? V = 1 m/s Radial distance at which mean velocity occurs = 0.707R = 0.707x0.1/2 =0.035m b) Pressure gradient (dp/dx) N/m 2 per m c) Wall shear stress (τ 0 ) =? d) Reynolds no. (Re) =? = 80 N/m 2 e) Velocity (u) at y = 0.03m =? r= R-y = 0.05-0.03 = 0.02m

= 1.68 m/s 7. Oil of viscosity 0.1 NS/m 2 and relative density 0.9 is flowing through a pipe of 50mm diameter. The pressure drop in a length of 300m is 680 Kpa. Assuming steady laminar flow, find the rate of flow and the shear stress at the pipe wall. Justify that the flow is laminar. Viscosity of oil (µ) = 0.1 NS/m 2 Relative density = 0.9 Density of fluid (ρ) = 0.9x1000 = 900 kg/m 3 Diameter of pipe (D) = 50mm = 0.05m Cross sectional area (A) = = 0.001963 m 2 Pressure drop (P1-P2) = 680 KPa = 680000 Pa Length (L) = 300m Rate of flow (Q) =? Shear stress at the pipe wall (τ 0 ) =? Computing average velocity (V) using Hagen-Poiseuille equation, V = 1.77 m/s Q = AV = 0.001963x1.77 = 0.003475 m 3 /s = 3.475 lps Computing Reynold no. (Re) As Re<2000, the flow is laminar. = 28.33 N/m 2

8. Glycerin of viscosity 0.9 NS/m 2 and density 1260 kg/m 3 is pumped along a horizontal pipe of diameter 0.01 m and length 100m at a flow rate of 1.8 lit/min. Determine Reynolds no. and verify the nature of flow. Calculate the pressure lost in the pipe due to frictional effects and calculate the maximum flow rate for laminar conditions to prevail. Viscosity of glycerin (µ) = 0.9 NS/m 2 Density (ρ) = 1260 kg/m 3 Diameter of pipe (D) = 0.01m Length of pipe (L) = 100m Cross sectional area (A) = = 0.000079 m 2 Flow rate (Q) = 1.8 lit/min = (1.8x10-3 /60) = 0.03x10-3 m 3 /s Reynolds no. (Re) =? Pressure loss (P 1 -P 2 ) =? Maximum flow rate for laminar condition (Qmax)=? Average velocity of flow = 0.379 m/s As Re<2000, the flow is laminar. = 10915200Pa = 10915.2Kpa Upper limit of laminar flow is reached when Re max = 2000 Qmax = 0.012 m 3 /s 9. Water flows through a 50mm diameter pipe sloping upwards at 45 0 to the horizontal. At a section some distance downstream of the inlet, the pressure is 700 Kpa and at a section 350m further along the pipe, the pressure is 450 Kpa. Assuming laminar flow, determine the average shear stress at the wall of the pipe and the loss of head. Take viscosity of water = 1.02x10-3 Pa-S.

Diameter of pipe (D) = 50mm Radius of pipe (R) = 25mm = 25x10-3 m Pressure at 1 (P 1 ) = 700 Kpa Pressure at 2 (P 2 ) = 450 Kpa Length (L) =350m Z 1 = 0 Z 2 = 35Sin45 0 = 24.75m Shear stress at pipe wall =? Head loss (h f ) =? 1 45 0 2 For inclined pipe, = 2.6 N/m 2 = 1.57 m/s = 0.73m Turbulent flow 10. A pipeline carrying water has average height of irregularities projecting from the surface of the boundary of the pipe as 0.2mm. What type of boundary is it? The shear stress developed is 8 N/m 2. Take the value of kinematic viscosity of water as 0.01 stokes. Roughness height (k) = 0.2 mm = 0.2x10-3 m Shear stress (τ 0 ) = 8 N/m 2 Kinematic viscosity (υ) = 0.01 stokes = 0.01x 10-4 m 2 /s Shear velocity (V * ) = = 0.0894 m/s Thickness of laminar sub layer = 1.3x10-4 m = 1.5 As 0.25< <6, the boundary is transition.

11. A liquid of specific gravity 0.9 and absolute viscosity 6.5x10-4 NS/m 2 flow through a pipe of diameter 150mm at the rate of 40 lps. If the loss of head in 100m length of pipe is 4.5m, determine whether the pipe is rough or smooth and find the average roughness height. Use Moody chart. Specific gravity of liquid (S) = 0.9 Density of liquid (ρ) = 0.9x1000 = 900 kg/m 3 Absolute viscosity (µ) = 6.5x10-4 NS/m 2 Diameter of pipe (D) = 150mm = 0.15m Cross-sectional area (A)= = 0.01767 m 2 Flow rate (Q) = 40 lps = 0.04 m 3 /s Head loss (h f ) =4.5m Length (L) = 100m Velocity (V) = = 2.26 m/s As Re>4000, the flow is turbulent. f = 0.026 For Re = 4.7x10 5 and f = 0.026, the Moody chart shows that the pipe behaves as rough. For Re = 4.7x10 5 and f = 0.026 k/d = 0.003 k/0.15 = 0.003 k = 0.00045m = 0.45mm 12. A 1m diameter pipe is to carry a water discharge of 1 m 3 /s at the minimum loss of energy (smooth pipe). What will be the permissible height of surface roughness? Take kinematic viscosity = 10-6 m 2 /s and use the resistance equation for smooth pipe. Diameter of pipe (d) = 1m Discharge (Q) =1 m 3 /s Cross-sectional area of pipe (A) = = 0.7854 m 2

Velocity of flow (V) = Q/A = 1/0.7854 = 1.27 m/s = 1.27x10 6 Permissible roughness height (k) =? For minimum loss of energy, pipe should act as smooth. For Re>10 5, friction factor for smooth pipe is ( ) Solving for f f = 0.0111 For smooth pipe, ( ) k = 0.0000635m = 0.0635mm 13. A 150mm diameter pipe with k = 0.01mm carries water over a length of 100m with a pressure loss of 26 Kpa. State whether the pipe will act as smooth or rough pipe. Also determine maximum velocity, average velocity and discharge. Take kinematic viscosity = 10-6 m 2 /s. Diameter of pipe (d) = 0.15m Cross-sectional area of pipe (A) = = 0.01767 m 2 k = 0.01x10-3 m Length (L) = 100m Pressure loss (P 1 -P 2 ) = 26000 Pa Maximum velocity (V max ) =? Average velocity (V) =? Discharge (Q) =? Shear stress at the wall = 9.75 Pa Shear velocity = 0.098 m/s Thickness of laminar sub layer

As = 0.000118m = 0.01x10-3 /0.000118 = 0.08, the pipe will act as hydrodynamically smooth pipe. At y = R = 0.075m, maximum velocity occurs V max = 2.7 m/s For average velocity (V), V = 2.35 m/s Q = AV = 0.01767x2.35 = 0.0415 m 3 /s 14. A smooth brass pipeline 100mm diameter and 1km long carries water at the rate of 12lps. The friction factor is expressed by where Re is Reynolds number. Take kinematic viscosity = 0.02 Stokes and density = 1000 kg/m 3. Calculate (a) Reynolds number and verify the nature of flow, (b) loss of head, (c) wall shearing stress, (d) center line velocity, (e) shear stress and velocity at 25mm from the center line, and (f) thickness of laminar sub-layer. Diameter of pipe (D) = 100mm = 0.1m Cross-sectional area (A) = = 0.007854 m 2 Length (L) = 1km = 1000m Discharge (Q) = 12lps = 0.012 m 3 /s Kinematic viscosity (υ) = 0.02 Stokes = 0.02X10-4 m 2 /s Density (ρ) = 1000 kg/m 3 Average velocity (V) = a) Reynolds no (Re) =? = 1.53m/s As Re>4000, the flow is turbulent. b) Loss of head (h f ) =?

= 23.86m c) Wall shear stress (τ 0 ) =? = 5.85 N/m 2 d) Centerline velocity (V max ) =? Shear velocity (V * ) = = 0.0765 m/s For turbulent flow in smooth pipes at y= D/2 = 0.1/2 = 0.05m, v= v max v max = 1.86 m/s e) r= 25mm = 0.025m Shear stress (τ) and velocity (v) at 25 mm from center line =? = 2.925 N/m 2 y= R-r = 0.05-0.025 = 0.025m v = 1.73 m/s (f) Thickness of laminar sub-layer (δ ) =?

= 0.0003m = 0.3mm 15. Mean velocities at the mid-point and quarter point of 0.2m diameter pipes are 1.7m/s and 1.45m/s respectively. If the flow in the pipe is turbulent, determine the discharge, friction factor and the average height of roughness projections. Diameter of pipe (D) = 0.2m Radius of pipe (R) =0.1m Cross-sectional area (A)= = 0.0314 m 2 Mean velocity at mid point = V max = 1.7 m/s r = 0.2/4 = 0.05 y = R-r = 0.1 0.05 = 0.05m Mean velocity at 0.05m (v) = 1.45 m/s Discharge (Q) =? Friction factor (f) =? Roughness height (k) =? From Prandtl s equation V * = 0.14 m/s Average velocity (V) = 1.225 m/s Q =AV = 0.0314x1.225 = 0.0384 m 3 /s Also f =0.1

K = 0.019 m 16. A 30cm diameter pipe conveys water in turbulent flow regime. The friction factor is 0.02. If the centerline velocity is 4m/s, estimate the discharge through the pipe. Diameter of pipe (D) = 0.3 m Cross-sectional area (A) = = 0.07 m 2 f = 0.02 Centerline velocity (v max ) = 4m/s Discharge (Q)=? (a) (b) Solving eq. (a) and (b), V = 3.37 m/s Q =AV = 0.07x3.37= 0.23 m 3 /s 17. In a 500mm diameter rough pipe carrying water, the velocity and velocity gradient at 50mm from the wall are 4.4m/s and 5.5x10-3 S -1 respectively. Determine the water discharge, roughness height, friction factor, wall shear and pressure gradient. Diameter of pipe (D) = 0.5m Cross-sectional area of pipe (A) = = 0.1963 m 2 y = 50mm = 0.05m At y = 0.05m, velocity (v) = 4.4m/s and velocity gradient (dv/dy) = 5.5x10-3 S -1 Discharge (Q) =? Roughness height (k) =? Friction factor (f) =? Wall shear =? Pressure gradient =? For rough pipe, (a)

Differentiating a w.r.t. y (b) Substituting the value of V * from b to a Solving for k k = 5x10-4 m = 0.5mm Substituting k into b V * = 0.22 m/s Average velocity (V) for rough pipe V =4.46 m/s Q = AV = 0.1963x4.46 = 0.875 m 3 /s f= 0.0195 Wall shear stress =48.5 Pa Pressure gradient = 388 Pa per m 18. A smooth pipe of 200mm diameter carries oil of density 850 kg/m 3 and kinematic viscosity 3x10-6 m 2 /s at a discharge of 30 lps. What will be the wall shear, thickness of laminar sublayer and velocity at the edge of laminar sublayer? Diameter of pipe (D) = 0.2m Cross-sectional area of pipe (A) = = 0.0314 m 2 Discharge (Q) = 30 lps = 0.03 m 3 /s Average Velocity (V) = Q/A = 0.03/0.0314 = 0.955 m/s

Wall shear =? Thickness of laminar sublayer =? Velocity at the edge of laminar sublayer =? = 63667 Re 4000 to 10 5, f for smooth pipe is given by = 0.0199 = 1.92 N/m 2 At = 0.047 m/s, velocity (v) = 0.00074m v = 0.54 m/s 19. A certain 100mm diameter pipe which acts as rough pipe gave 0.026 and 0.03 as its friction factor values after 5 and 10 years of operation. Estimate the friction factor after 15 years of service. Diameter of pipe = 100mm Friction factor after 5 year = 0.026 Friction factor after 10 year = 0.03 Friction factor after 15 year =? For rough pipe, Finding roughness values after 5 and 10 years of service K 5 = 0.294mm K 10 = 0. 481mm

(a) (b) Solving a and b and K 0 = 0.107mm After 15 years, = 0.107+15X0.0374 = 0.668mm f 15 = 0.0332 20. Water flows through a 500mm diameter commercial pipe of 100m length of pipe and average roughness height 0.465mm. What is the maximum discharge at which this pipe will act as smooth pipe? What is the minimum discharge at which this pipe will act as rough pipe? Diameter of pipe (D) = 0.5m Cross-sectional area of pipe (A) = = 0.1963 m 2 k = 0.465x10-3 m = 0.465mm (a) When the pipe just ceases to act as smooth pipe, Also f = 0.0197 ( )

Re = 64997 V = 0.13m/s Q = AV = 0.1963x0.13 = 0.025 m 3 /s (b) When the pipe acts just as rough pipe, f = 0.0193 Re = 1547991 V = 3.096m/s Q = AV = 0.1963x3.096 = 0.607 m 3 /s 21. A pressure drop of 11.5 Kpa is measured with gauges placed 7m apart on a pipe with 10cm diameter transporting water with kinematic viscosity of 0.6x10-6 m 2 /s. Compute wall shear stress, shear velocity, friction factor, mean velocity, maximum velocity, Reynold number and flow rate. Use exponential velocity distribution for turbulent flow. Pressure drop(p 1 -P 2 ) = 11500 Pa Length (L) = 7m Diameter of pipe (D) = 0.1m Kinematic viscosity = 0.6x10-6 m 2 /s Velocity profile Average velocity V for this type of velocity distribution is

a. Wall shear stress =? = 41 Pa b. shear velocity =? = 0.2m/s c. friction factor (f) =? Assuming n = 7 in the exponential equation = 0.2 d. Mean velocity (V) =? V = 1.26m/s e. Maximum velocity (V max ) =? V max = 1.54m/s f. Reynold no. (Re) =? = 2.1x10 5 g. Flow rate (Q) =? Q = AV = = 0.0098 m 3 /s

Head loss in pipes 22. At a sudden enlargement of a water main from 250mm to 500mm diameter, the hydraulic gradient rises by 10mm. Estimate the rate of flow. 1 2 1 2 Diameter of smaller pipe (D 1 ) = 250 mm = 0.25m Cross-sectional area (A 1 )= = 0.049 m 2 Diameter of larger pipe (D 2 ) = 500 mm = 0.5m Cross-sectional area (A 2 ) = 0.19635 m 2 Rise of hydraulic gradient, i.e. = 10mm = 0.01m Rate of flow (Q) =? Applying Bernoulli s equation from smaller pipe to larger pipe section where h e =head loss due to sudden enlargement (a) From continuity equation A 1 V 1 = A 2 V 2

From a and b, (b) V 2 = 0.181 m/s Q =A 2 V 2 = = 0.19635x0.181 = 0.0355 m 3 /s 23. The rate of flow of water through a horizontal pipe of diameter 200mm is 0.3 m 3 /s. The diameter of the pipe is suddenly enlarged to 400mm. The pressure intensity in the smaller pipe is 15 N/cm 2. Determine (a) loss of head due to sudden enlargement, (b) pressure intensity in the larger pipe, and (c) power lost due to enlargement. 1 2 1 2 Rate of flow (Q) = 0.3 m 3 /s Diameter of smaller pipe (D 1 ) = 200 mm = 0.2m Cross-sectional area (A 1 )= = 0.0314 m 2 Diameter of larger pipe (D 2 ) = 400 mm = 0.4m Cross-sectional area (A 2 )= = 0.1256 m 2 Pressure in smaller pipe (P 1 ) = 15 N/cm 2 = 15x10 4 N/m 2 Velocity in smaller pipe (V 1 ) = Q/A 1 = 0.3/0.0314 = 9.55 m/s Velocity in larger pipe (V 2 ) = Q/A 2 = 0.3/0.1256 = 2.38 m/s a) Loss of head due to sudden enlargement (h e ) =?

= 2.62 m b) Pressure intensity in the larger pipe (P2) =? Applying Bernoulli s equation from smaller pipe to larger pipe section where h e =head loss due to sudden enlargement Z 1 = Z 2 P 2 = 167067 N/m 2 = 16.7 N/cm 2 (c) Power lost due to enlargement (P) =? = 7711 W = 7.71 KW 24. A horizontal pipe of diameter 500mm is suddenly contracted to a diameter of 300mm. The pressure intensities in the large and smaller pipe are given as 140 Pa and 120 Pa respectively. Find the loss of head due to contraction if C C = 0.62. Also determine the rate of flow of water. 1 C 2 C 1 2 Diameter of larger pipe (D 1 ) = 500 mm = 0.5m Cross-sectional area (A 1 )= = 0.19635 m 2

Diameter of smaller pipe (D 2 ) = 300 mm = 0.3m Cross-sectional area (A 2 )= = 0.0707 m 2 Pressure in larger pipe (P 1 ) = 140Kpa = 140x10 3 Pa Pressure in smaller pipe (P 2 ) = 120Kpa = 120x10 3 Pa Coefficient of contraction (C c ) = 0.62 From continuity equation A 1 V 1 = A 2 V 2 Head loss due to contraction (h c ) =? Rate of flow (Q) =? (a) [ ] [ ] (b) Applying Bernoulli s equation from smaller pipe to larger pipe section Z 1 = Z 2 (c) From a and c V 2 = 5.67 m/s = 0.614 m Q = A 2 V 2 = 0.0707x5.67 = 0.4 m 3 /s

25. Two reservoirs whose water surface elevation differs by 12m are connected by the following horizontal compound pipe system starting from the high level reservoir as shown in the figure. L 1 =200m, D 1 = 0.2m, f 1 = 0.008 L 2 =500m, D 2 = 0.4m, f 2 = 0.006 Compute all the losses of head. Sketch EGL and HGL lines. 12m A C V 1 B V 2 D 3m D 1, L 1, f 1 D 2, L 2, f 2 Applying Bernoulli s eq. between the free surface of the first reservoir and the second reservoir Total loss = 12m From continuity equation A 1 V 1 = A 2 V 2 (a) Total loss = h i + h f1 + h e + h f2 + h o = 12 (b) h i = entry loss at A = = 0.0255 = 0.0255 2 = h f1 =head loss in pipe AB = = = 0.0407 2 = h e = Head loss due to sudden expansion = = = h f2 =head loss in pipe CD = = =

h 0 = Exit loss = = substituting above values in eq. (a) V 2 = 1.24 m/s V 1 = 4V 2 = 4.96 m/s h i = 0.408V 2 2 = 0.408 x 1.24 2 = 0.63m h f1 = 0.6512V 2 2 = 6.512 x 1.24 2 = 10m h e = 0.4587 V 2 2 =0.4587 x 1.24 2 = 0.71m h f2 = 0.3822V 2 2 = 0.3822 x 1.24 2 = 0.58m h 0 = 0.1274V 2 2 = 0.051x 1.24 2 = 0.08m Position of TEL and HGL At A Elevation of EGL =15 h i = 15-0.62 = 14.37m Elevation of HGL =Elevation of EGL V 1 2 /2g = = 13.11m at B Elevation of EGL = Elevation of EGL at A h f1 = 14.37-10 = 4.37m Elevation of HGL = Elevation of EGL V 1 2 /2g = = 3.11m at C Elevation of EGL = Elevation of EGL at B h e = 4.37-0.71 = 3.66m Elevation of HGL = Elevation of EGL V 2 2 /2g = = 3.58m at D Elevation of EGL = Elevation of EGL at C h f2 = 3.66-0.58= 3.08m Elevation of HGL = 3m EGL and HGL lines HGL TEL A C V 1 B V 2 D D 1, L 1, f 1 D 2, L 2, f 2

26. The following are the data for a piping system shown in the figure. Pipe l D f AP 250m 30cm 0.02 PB 150m 20cm 0.025 BC 100m 15cm 0.025 Take entry loss coefficient = 0.5 and bend loss coefficient = 0.5, and exit loss coefficient = 0.1 Discharge through the system = 0.2 m 3 /s Compute all losses and find the pressure head on the suction and discharge side of the pump. Plot EGL and HGL lines. If the efficiency of the pump is 75%, what is input power to the pump. B 4 C Nozzle dia = 7.5cm 1 30m Water A 3m 2 3 Datum P Discharge (Q) = 0.2 m 3 /s V = Q/A A AP = = 0.0707 m 2 A PB = = 0.0314 m 2 A BC = = 0.01767 m 2 C/s of nozzle = =0.0044 m 2 V AP = 0.2/0.0707 = 2.83m/s V PB = 0.2/0.0314 = 6.37m/s V BC = 0.2/0.01767 = 11.32m/s Velocity through nozzle (V j ) = 0.2/0.0044 = 45.45m/s Computation of head losses

Head loss at entrance (hl 1 ) = = 0.2m Head loss in pipe AP (h f1 ) = = = = 6.8m Head loss in pipe PB (h f2 ) = = = = 38.8m Head loss in pipe BC (h f3 ) = = = = 108.8m Head loss at bend (hl 2 ) = = = 1.03m Head loss at exit (hl 3 ) = = 52.6m Applying Bernoulli s equation between 1 and 2 P 2 = pressure at suction end of pump = -43244.5 Pa Applying Bernoulli s equation between 3 and 4 P 3 = pressure at discharge end of pump = 2900933 Pa Power = Q(P 3 -P 2 ) = 0.2x(290093+43244.5) = 588835.5W = 588.8KW Input power to pump = power/efficiency = 588.8/0.75 = 785KW Position of EGL and HGL lines At A Elevation of EGL = Water surface elevation h l1 = 3-0.2 = 2.8m Elevation of HGL = Elevation of EGL V 1 2 /2g = 2.8-0 =2.8m at 2 (inlet of pump) Elevation of EGL = Elevation of EGL at A h f1 = 2.8-6.8 = -4 Elevation of HGL = Elevation of EGL V AP 2 /2g = at 3 (outlet of pump) = -4.41m Elevation of EGL = Elevation of EGL at 2 + head supplied by pump = =296.1m Elevation of HGL = Elevation of EGL V PB 2 /2g = = 294m at bend Elevation of EGL = Elevation of EGL at 3-h f1 = 296.1-38.8 = 257.3m

Elevation of HGL = Elevation of EGL V PB 2 /2g = = 255.2m d/s of bend Elevation of EGL = Elevation of EGL at bend h l2 = 257.3-1.03 = 256.3 Elevation of HGL = Elevation of EGL V BC 2 /2g = = 249.8m At nozzle Elevation of EGL = Elevation of EGL at bend h lf3 = 256.3-108.8 = 147.5m Elevation of HGL = Elevation of EGL V BC 2 /2g = =141m Plot of EGL and HGL B 4 C 1 EGL HGL Water A 2 3 Datum P

Three types of pipe flow problems Tutorial 2 1. Determine the head (energy) loss for flow of 150 lps of oil flowing through 400m of 200mm diameter cast-iron pipe. Take υ = 0.00001 m 2 /s and K = 0.25mm. Use Moody s diagram. Rate of flow (Q) = 150 lps = 0.15 m 3 /s Length (L) = 400m Diameter of pipe (D) = 200 mm = 0.2m Cross-sectional area (A)= = 0.0314 m 2 υ = 0.00001 m 2 /s and K = 0.25mm Head loss (h f ) =? Average velocity (V) = Q/A = 0.15/0.0314 = 4.77 m/s Reynolds no. (Re) = = 95400 k/d = 0.25x10-3 /0.2 = 0.00125 For k/d = 0.00125 and Re = 95400, f = 0.024(from Moody s chart) = 55.6 m 2. Water flows through a 300mm diameter steel pipe with a head loss of 6m in 300m. Determine the rate of flow. Use (a) Moody s diagram and (b) Colebrook-White equation for obtaining f and compare the result. Take υ = 1.13x10-6 m 2 /s and K = 3mm. Head loss (h f ) = 6m Length (L) = 300m Diameter of pipe (D) = 300 mm = 0.3m Cross-sectional area (A)= = 0.0707 m 2 υ = 1.13x10-6 m 2 /s and K = 3mm k/d =3x10-3 /0.3 = 0.01 Rate of flow (Q) =?

a. Using Moody diagram Trial 1: For k/d = 0.01, take trial value of f = 0.038 from Moody s diagram = 1.76 m/s Reynolds no. (Re) = = 467257 Trial 2: For k/d = 0.01 and Re = 467257, f = 0.038 (from Moody s diagram) As f is same for trial 1 and 2, the trial and error procedure is stopped here. f = 0.038 and V = 1.76 m/s Q = AV = 0.0707x1.76 = 0.124 m 3 /s b. Using Colebrook-White equation Trial 1: For k/d = 0.01, take trial value of f = 0.038 = 1.76 m/s Reynolds no. (Re) = = 467257 ( ) ( ) f = 0.3803 As f is same as assumed f value, the iteration is stopped here. f = 0.03803 and V = 1.759 m/s Q = AV = 0.0707x1.759 = 0.124 m 3 /s 3. Determine the size of steel pipe required to convey 250 lps oil through 3000m with a head loss of 25m. Take υ = 1.0x10-5 m 2 /s and K = 0.046mm. Use Moody s diagram. Flow rate (Q) = 250 lps = 0.25 m3/s Length (L) = 3000m Head loss (h f ) = 25m υ = 1.0x10-5 m 2 /s and K = 0.046mm Diameter of pipe (D) =?

[ ] = 0.62f (a) [ ] = (b) Trial 1: Assume f = 0.02 D = (0.62x0.02) 1/5 = 0.416m Re =31830/0.416 = 7.7x10 4 k/d = 0.046x10-3 /0.416 = 0.00011 Trial 2: For K/D = 0.00011 and Re = 7.7x10 4, f = 0.0195 (from Moody s diagram) As the difference in f for trial 1 and 2 is very small, the trial and error procedure is stopped here. D = (0.62x0.0195) 1/5 = 0.413m Take 420mm diameter pipe. 4. A smooth pipe carries 0.3 m 3 /s of water discharge with a head loss of 3m per 100m length of pipe. Determine the diameter of the pipe. Use friction factor equation for smooth pipe as and take kinematic viscosity = 10-6 m 2 /s. Flow rate (Q) = 0.3 m 3 /s Length (L) = 100m Head loss (h f ) = 3m υ = 10-6 m 2 /s Diameter of pipe (D) =? = 0.248f [ ] (a) [ ] = (b) For first trial, assume f = 0.01 = 0.301m With this value, new f is

= 0.0111 Second trial: f = 0.0111 = 0.3075m With this value, new f is = 0.01114 As the difference in f is very small, the trial and error is stopped here. = 0.308m = 308 mm Adopt 310mm diameter pipe. 5. Determine the size of steel pipe required to carry water at 30 lps if the permissible energy gradient is 0.05. Will the boundary act as smooth or rough? Take dynamic viscosity of water = 10-3 Pa.S and k = 0.045mm. Flow rate (Q) = 0.03 m 3 /s Energy gradient (h f /L) = 0.05 = 10-3 Pa.S k = 0.045x10-3 m Diameter of pipe (D) =? ( ) = 0.001487f [ ] (a) [ ] = (b) For first trial, assume f = 0.02 = 0.124m =3.1x10 5 k/d = 0.045x10-3 /0.124 = 0.000363 For above K/D and Re, f from Moody diagram = 0.0175 Trial2:

= 0.121m =3.15x10 5 k/d = 0.045x10-3 /0.121 = 0.000372 For above K/D and Re, f from Moody diagram = 0.0175 As the value of f is same for trial 2 and 3, the iteration is stopped. Diameter of pipe = 0.121m. Take 130mm dia. pipe. = 31 As is in between 17 and 400, the boundary is in transition. 6. A 2cm diameter 20km long pipeline connects two reservoirs filled with water open to the atmosphere. What is the discharge in the pipeline if the surface elevation difference of the reservoirs level is 5m? m 2 /s. 5m Diameter of pipe (D) = 2cm = 0.02m Length of pipe (L) = 20km = 20000m Surface elevation difference between two reservoirs (H) = 5m Kinematic viscosity of water =1.02x10-6 m 2 /s Discharge (Q) =? Neglecting minor loss, H = h f where h f = head loss due to friction Trial and error for determination of friction factor, f Trial 1 Assume f = 0.05 = 0.044m/s Reynold s number = 863 As Re<2000, the flow is laminar, for which f is given by f = 64/Re = 64/863 = 0.07

Trial 2 Computing V and Re with f = 0.07 V = 0.037m/s, Re = 725 f = 64/725 = 0.09 Trial 3 Computing V and Re with f = 0.09 V = 0.033m/s, Re = 647 f = 64/647 = 0.1 Trial 4 Computing V and Re with f = 0.1 V = 0.0313m/s, Re = 614 f = 64/614 = 0.104 As the difference in f is small between trial 3 and 4, the trial is stopped here. f = 0.104 For f = 0.104, V = 0.0307m/s Q = AV = = 9.6x10-6 m 3 /s = 0.0096 lps 7. A pressure drop of 600 Kpa is measured over a 400m length of 15cm diameter horizontal wrought iron pipe with absolute roughness of 0.045mm. Estimate the flow rate, if the specific gravity of the liquid is 0.9 and kinematic viscosity is 10-5 m 2 /s. Use Moody chart. Pressure drop (P 1 -P 2 ) = 600 Kpa Length (L) = 400m Head loss (h f ) = = 67.95m Diameter of pipe (D) = 150 mm = 0.15m υ = 10-5 m 2 /s and K = 0.045mm Rate of flow (Q) =? k/d =0.045x10-3 /0.15 = 0.0003 Trial 1: Take trial value of f = 0.025 = 4.47 m/s Reynolds no. (Re) = = 6.7x10 4 Trial 2: For k/d = 0.0003 and Re = 6.7x10 4, f = 0.027 (from Moody s diagram)

= 4.3 m/s Reynolds no. (Re) = = 6.45x10 4 Trial 3: For k/d = 0.0003 and Re = 6.45x10 4, f = 0.027 (from Moody s diagram) As f is same for trial 2 and 3, the trial and error procedure is stopped here. f = 0.027 and V = 4.3 m/s Q = AV = = 0.076 m 3 /s

Siphon 8. A siphon of diameter 150mm connects two reservoirs having a difference in elevation of 15m. The length of the siphon is 400m and the summit is 4m above the water level in the upper reservoir. The length of the pipe from upper reservoir to the summit is 80m. Determine the discharge through the siphon and pressure at the summit. Consider all losses. Take f = 0.02. Diameter of siphon (D) = 150mm = 0.15m Cross-sectional area (A)= = 0.01767 m 2 Difference in elevation (H) = 15m Length of siphon (L) = 400m Height of summit above water level in the upper reservoir (h) = 4m Length of inlet leg (l) = 80m Atmospheric pressure head(p A / and P D / )) = 0 V A = V C = 0 V B = V= Velocity of flow f = 0.02 Discharge through the siphon (Q) =? Pressure at summit (P B ) =? B h A H C Z a Z b Datum

Applying Bernoulli's equation between two points A and C (expressing Pressure values in terms of absolute pressure and considering all losses) V = 2.316 m/s Q = AV = 0.01767x2.316 = 0.041 m 3 /s Applying Bernoulli's equation between the point A and summit B (expressing Pressure values in terms of absolute pressure) = 2.98m absolute (-7.32 m gauge) 9. A siphon of diameter 200mm connects two reservoirs having a difference in elevation of 20m. The length of the siphon is 800m and the summit is 5m above the water level in the upper reservoir. If the separation takes place at 2.8m of water absolute, find the maximum length of the siphon from upper reservoir to the summit. Take f = 0.016 and atmospheric pressure = 10.3m of water. Neglect minor losses. Diameter of siphon (D) = 200mm = 0.2m Cross-sectional area (A)= = 0.0314 m 2 Difference in elevation (H) = 20m Length of siphon (L) = 800m Height of summit above water level in the upper reservoir (h) = 5m Pressure head at summit (P B / ) = 2.7 m of water absolute Atmospheric pressure head= 10.3m of water absolute f = 0.016

V A = V C = 0 V B = V= Velocity of flow Length of inlet leg (l) =? Applying Bernoulli's equation between two points A and C (neglecting minor losses and working in terms of absolute pressure) V = 2.48 m/s B h A H C Z a Z b Datum Applying Bernoulli's equation between the point A and summit B (expressing Pressure values in terms of absolute pressure and neglecting minor losses)

l =87.2m 10. A siphon of diameter 200mm connects two reservoirs having a difference in elevation of 40m. The total length of the pipe is 6000m. The pipe crosses ridge. The summit of the ridge is 7m above the level of the water in the upper reservoir. Find the minimum depth of pipe below the summit of the ridge, if the absolute pressure head at the summit of the siphon is not to fall below 2.7m of water. Take f = 0.03 and atmospheric pressure head =10.3 m of water. The length of siphon from the upper reservoir to the summit is 500m. Find the discharge also.neglect minor losses. Diameter of siphon (D) = 200mm = 0.2m Cross-sectional area (A)= = 0.0314 m 2 Difference in elevation (H) = 40m Length of siphon (L) = 6000m Height of summit of the ridge above water level in the upper reservoir = 7m Pressure head at summit (P b / ) = 2.7 m of water absolute Atmospheric pressure head= 10.3m of water absolute f = 0.03 Length of inlet leg (l) = 500m V A = V C = 0 V B = V= Velocity of flow Depth of pipe below the summit of the ridge (x) =? Applying Bernoulli's equation between two points A and C(expressing Pressure values in terms of absolute pressure and neglecting minor losses) Applying Bernoulli's equation between two points A and C (neglecting minor losses and working in terms of absolute pressure)

V = 0.933m/s Q = AV = 0.0314x0.933 = 0.03 m 3 /s x B 7m A H C Z a Z b Datum Applying Bernoulli's equation between the point A and summit B (expressing Pressure values in terms of absolute pressure and neglecting minor losses)

x = 2.77m 11. Water from a main canal is siphoned to a branch canal over an embankment by means of a wrought iron pipe of 100mm diameter. The length of the pipeline up to the summit is 30m and the total length is 90m. Water surface elevation in the branch canal is 10m below that of main canal. Take f = 0.025 and consider all losses. a) If the total quantity of water required to be conveyed is 0.05m 3 /s, how many pipelines are needed? b) What is the maximum permissible height of the summit above the water level in the main canal so that the water pressure at the summit may not fall below 20 Kpa absolute, the barometer reading being 10m of water? Diameter of siphon (D) = 100mm = 0.1m Cross-sectional area (A)= = 0.00785 m 2 Difference in elevation (H) = 10m Length of siphon (L) = 90m Length of inlet leg (l) = 30m f = 0.025 V A = V C = 0 V B = V= Velocity of flow a) no. of pipelines for discharge of 0.05m 3 /s =? Applying Bernoulli's equation between two points A and C (working in terms of absolute pressure) V = 2.86 m/s Q = AV = 0.0785x2.86 = 0.02245 m 3 /s Discharge through a 100mm diameter pipe = 0.02245 m 3 /s no. of pipelines for discharge of 0.05m 3 /s = 0.05/0.002245 = 3

B h A H C Z a Z b Datum b) Atmospheric pressure = 10.0 m of water absolute Pressure at summit = 20 KPa of water absolute Pressure head at summit(p b / )= 20x10 3 /9810 =2.04m of water absolute Maximum permissible height of the summit above the water level(h) =? Applying Bernoulli's equation between the point A and summit B (expressing Pressure values in terms of absolute pressure) h = 4.2m

12. The siphon pipe of diameter 200mm shown in the fig. below discharges 120 l/s of water to the atmosphere. Find the total loss of head from point A to C in terms of the velocity head V 2 /2g. Find also the pressure head at B if the total loss of head from A to B is two-thirds of the total loss of head. B 2m A Water 1.5m C Diameter of siphon (D) = 200mm = 0.2m Cross-sectional area (A)= = 0.0314 m 2 Discharge (Q) = 120lps = 0.12 m 3 /s V a = 0 V b = V c = V= Velocity of flow Head loss between A to C (hl AC ) =? Pressure at B (P B )=? Head loss between A to B Velocity of water (V) = Q/A = 0.12/0.0314 = 3.82m/s Applying Bernoulli's equation between two points A and C (neglecting minor losses and working in terms of absolute pressure)

k = 1.017 = = 0.756m Head loss between A to B = = 0.504m Applying Bernoulli's equation between two points A and B (neglecting minor losses and working in terms of absolute pressure) =7.05m (abs) (-3.25m gauge) 13. A siphon pipe 75mm in diameter discharges oil of sp gr 0.84 from a reservoir into air as shown in the fig. The total loss of head from point 1 to 2 is 2V 2 /2g and from 2 to 3 is 3V 2 /2g where V is the velocity of oil in the pipe. Find the rate of flow in the siphon pipe and the absolute pressure at point 2. Take atmospheric pressure = 101Kpa. 2 1.5m 1 3.5m Water 3 Diameter of siphon (D) = 75mm = 0.075m

Cross-sectional area (A)= = 0.004418 m 2 sp gr of oil = 0.84 Sp wt of oil = 0.84x9810 = 8240.4N/m 3 Head loss between 1 to 2 Head loss between 2 to 3 Total head loss between 1 and 3 (hl) = Atmospheric pressure (P atom ) =101Kpa P 1 = P 3 = P atm Discharge (Q) =? Absolute pressure at 2 (P 2 ) =? V 1 = 0, V 2 = V 3 = V = Velocity of flow Applying Bernoulli's equation between two points 1 and 3 (neglecting minor losses and working in terms of absolute pressure) V = 2.55 m/s Q = AV = 0.004418x2.55 = 0.01126 m 3 /s = 0.663m Applying Bernoulli's equation between two points 1 and 2 (neglecting minor losses and working in terms of absolute pressure) P 2 = 80445Pa 14. Water is siphoned out of a tank by means of a bent pipe ACB 24m long and 25mm in diameter. A is below the water surface and 150mm above the base of the tank. AC is vertical and 9m long and CB is 15m long with discharge end B 1.5m below the base of the tank. If the atmospheric pressure head is 10.2m of water and the siphon action of C ceases when the absolute pressure is 1.8m of water, find the limiting velocity of water in the pipe and the depth of water in the tank when the siphon action ceases. Take f = 0.032.

C D Water h A 0.15m 1.5m B Diameter of siphon (D) = 25mm = 0.025m Cross-sectional area (A) = = 0.00491 m 2 Length ACB (L) = 24m length AC (l) = 9m Atmospheric pressure = 10.2m of water Absolute pressure at C = 1.8m of water Velocity (V) =? Depth of water in tank (h+0.15) =? Applying Bernoulli s equation between D and B (Taking datum through point B and working in terms of absolute pressure) (a)

Applying Bernoulli s equation between D and C (Taking datum through point B and working in terms of absolute pressure) (b) Solving a and b V = 1.5 m/s h = 1.78m Depth of water in the tank = 1.78+0.15 = 1.93m 15. An overhead tank of 5mx10m cross-section in a village is used to supply water to another village across a hill. For the configuration shown in the figure below, calculate the discharge conveyed. Is there an upper limit on the discharge conveyed by the pipe? Given length AT = 2km and TB = 1km. 1 EGL HGL T 20m A d = 10cm l = 3km f = 0.03 15m B Applying Bernoulli's equation between two points 1 (water surface of tank) and B (working in terms of absolute pressure)

= 0.0051 m 3 /s V = 0.66 m/s Theoretically the pressure at summit can be reduced to absolute vacuum. However, practically the pressure can be reduced up to 2.5m of water absolute in order to avoid blockage of flow. Computing the pressure at summit with the velocity computed above Applying Bernoulli's equation between two points 1 (water surface of tank) and T (working in terms of absolute pressure) As pressure at summit is less than 2m, the upper limit of discharge should correspond to 2.5m pressure absolute) at summit. With this pressure, the limiting velocity and corresponding discharge is computed. = 0.00507 m 3 /s V = 0.646 m/s

Tutorial 3 Pipes in series and parallel 1. Three pipes of lengths 800m, 600m and 300m and of diameters 400mm, 300mm and 200mm respectively are connected in series. The ends of the compound pipe are connected to two tanks, whose water surface levels are maintained at a difference of 15m. Take f = 0.02. Determine the rate of flow (a) neglecting minor losses and (b) considering all losses. c) What will be the diameter of a single pipe of length 1700m and f = 0.02, which replaces three pipes? Length of pipe 1 (L 1 ) = 800m, Diameter of pipe 1 (D 1 ) = 400 mm = 0.4m Length of pipe 2 (L 2 ) = 600m, Diameter of pipe 2 (D 2 ) = 300 mm = 0.3m Length of pipe 3 (L 3 ) = 300m, Diameter of pipe 3 (D 3 ) = 200 mm = 0.2m f= f 1 = f 2 = f 3 = 0.02 Difference of water level (H) = 15m Rate of flow (Q) =? 1 H 2 3 a) Neglecting minor losses A 1, A 2, A 3 = Cross-sectional area of pipe 1, 2 and 3 V 1, V 2, V 3 = Velocity in pipe 1, 2 and 3 From continuity, A 1 V 1 = A 2 V 2 = A 3 V 3 = 1.77V 1

= 4V 1 For pipes in series V 1 = 0.675 m/s Discharge (Q) = A 1 V 1 = = 0.0848 m 3 /s b) Considering all losses where = entry loss, = loss due to sudden contraction between 1 and 2, = loss due to sudden contraction between 2 and 3, and = exit loss V 1 = 0.662 m/s Discharge (Q) = A 1 V 1 = = 0.0832 m 3 /s c) Length of single pipe (L) =1700m Diameter of equivalent single pipe (D) =? D = 0.2665m = 266.5mm

2. Two pipes of lengths 2500m each and diameters 80cm and 60cm respectively, are connected in parallel. If the total flow is 250 lps, find the rate of flow in each pipe.take f = 0.024. Length of pipe 1 (L 1 ) = 2500m, Diameter of pipe 1 (D 1 ) = 80cm = 0.8m Length of pipe 2 (L 2 ) = 2500m, Diameter of pipe 2 (D 2 ) = 60cm = 0.6m Total flow (Q) = 250 lps = 0.25 m 3 /s f = 0.024 Rate of flow in pipe 1 and 2 (Q 1 and Q 2 ) =? Computing r using = 15.13 = 63.76 For parallel pipes h f1 = h f2 Q 1 = 2.05Q 2 (a) Q = Q 1 + Q 2 0.25 = Q 1 + Q 2 (b) From a and b Q1= 0.169 m 3 /s Q2 = 0.081 m 3 /s 3. A pipe of diameter 300mm and length 1000m connects two reservoirs, having difference of water levels as 15m. Determine the discharge through the pipe. If an additional pipe of diameter 300mm and length 600m is attached to the last 600m length of the existing pipe, find the increase in discharge. Take f = 0.08 and neglect minor losses. a) Length of pipe (L) = 1000m, Diameter of pipe (D) = 300mm = 0.3m Cross- sectional area of pipe (A) = = 0.0707 m 2 Difference in level (H) = 15m f= 0.08 Discharge (Q) =?

H Q = 0.0743 m 3 /s b) Length of pipe CD (l1) = 400m Length of pipe DE (l2) = 600m Length of pipe DF (l3) = 600m Diameter of all pipes = D= D1 = D2 = 300mm = 0.3m Increase in discharge =? A C Q 1 D Q 2 H B E Q 3 F Since diameters and lengths of pipes DE and DF are equal, Q 2 = Q 3 Q 1 = Q 2 + Q 3 Q 2 = Q 1 /2 V 1 = Q 1 /A 1 = Q 1 /0.0707 = 14.14Q 1 V 2 =Q 2 /A 2 = Q 2 /0.0707 = 14.14Q 2 =14.14Q 1 /2=7.07Q 2 Applying Bernoulli s equation between A and B, taking flow through CDF Q 1 = 0.1001 m 3 /s

Increase in discharge = Q 1 -Q = 0.1001-0.0743 = 0.0258 m 3 /s 4. Two sharp ended pipes of diameters 60mm and 100mm respectively, each of length 100m are connected in parallel between two reservoirs which have a difference of level of 20m. If f= 0.32 for each pipe, calculate the rate of flow for each pipe and also the diameter of single pipe 150m long which would give the same discharge if it were substituted for the original two pipes. Length of pipe 1 (L 1 ) = 100m, Diameter of pipe 1 (D 1 ) = 60mm = 0.06m Length of pipe 2 (L 2 ) = 100m, Diameter of pipe 2 (D 2 ) = 100mm = 0.1m Difference in reservoir level (H)= 20m f= 0.32 1 H 2 a) Rate of flow (Q 1 and Q 2 ) =? Computing r using = 3401013 = 244462.8 For parallel pipes H= h f1 =h f2 Q 1 = 0.00242 m 3 /s = 2.42 lps Q 1 = 0.009 m 3 /s = 9 lps b) Discharge through single pipe (Q) = Q 1 + Q 2 = 0.002424 + 0.009= 0.01142 m 3 /s Length of pipe (L) = 150m

H = h f =20m Diameter of single pipe (D) =? D = 0.120 m = 120 mm 5. Two pipes have a length L each. One of them has diameter D and the other a diameter d. If the pipes are arranged in parallel, the loss of head when a total quantity of water Q flows through them is h. If the pipes are arranged in series and the same quantity Q flows through them, the loss of head is H. If d = D/2, find the ratio of H to h, neglecting minor losses and assuming same f. Length of pipe 1 (L 1 ) = L, Diameter of pipe 1 (D 1 ) = D Length of pipe 2 (L 2 ) = L, Diameter of pipe 1 (D 2 ) = d Loss of head for pipes in parallel = h Loss of head for pipes in series= H d = D/2 a) For parallel pipes, Discharge through main pipe = Q Q 1 and Q 2 = Discharge through parallel pipes h f1 = h f2 = h Q 1 2 parallel (a) (b) Equating above expressions Q 1 = 5.657 Q 2 Q = Q 1 + Q 2 = 5.657 Q 2 + Q 2 Q 2 = 0.15Q Q 1 = 0.85Q b) For pipes in series Discharge = Q H = h f1 +h f2 From a and b, (c) and 1 2

[ ] Q 1 = 0.85Q and Q 2 = 0.15Q [ ( ] ) H/h = 45.8 6. Three pipes of same length L, diameter D and friction factor f are connected in parallel. Determine the diameter of the pipe of length L and friction factor f which will carry the same discharge for the same head loss. 1 Length of pipes 1, 2 and 3 = L 1 = L 2 = L 3 = L Diameter of pipes 1, 2 and 3 = D 1 =D 2 = D 3 = D Friction factor = f r 1 = r 2 = r 3 = r 2 3 For parallel pipes, h f = h f1 = h f2 = h f3 (a) Q = total discharge, Q 1, Q 2, Q 3 = Discharge through pipes 1, 2 and 3 Q = Q 1 + Q 2 + Q 3 h f1 = h f2 = h f3 rq 1 2 = rq 2 2 = rq 3 2 Q 1 = Q 2 = Q 3 Q = 3Q 1 = (b) where V = velocity through each pipe L, d, v For single pipe, discharge = Q, Length = L, diameter = d, velocity =v For single pipe, Q = (c) Equating b and c (d) Head loss in single pipe= (e) Equating a and e

(f) From d and f d =1.55D 7. For a town water supply, a main pipeline of diameter 0.3m is required. As pipes more than 0.25m diameter are not readily available, two parallel pipes of same diameter were used for water supply. If the total discharge in the parallel pipes is same as in the single main pipe, find the diameter of the parallel pipes. Assume same f for all pipes. Diameter of main pipe (D) = 0.3m Friction factor = f 1 2 D 1 = Diameter of parallel pipes 1 and 2, L 1 = Length of parallel pipes 1 and 2, V 1 = Velocity in parallel pipes 1 and 2 L= Length of single pipe, V = Velocity in single pipe Loss of head in single pipe = Loss of head in parallel pipes = Equating above expressions (a) For parallel pipes, Q = total discharge, Q 1, Q 2,= Discharge through pipes 1, 2 r 1 = r 2 = r h f1 = h f2 rq 1 2 = rq 2 2 Q 1 = Q 2 Q= Q 1 + Q 2 Q = 2Q 1 = Total flow in single pipe =sum of flow in parallel pipes From a and b (b)

D 1 = 0.227 m = 227mm, say 230 mm Use two pipes of 230 mm diameter. 8. A pipe of diameter 0.4m and of length 2500m is connected to a reservoir at one end. The other end of the pipe is connected to a junction from which two pipes of lengths 1500m and diameters 300mm run in parallel. These parallel pipes are connected to another reservoir, which is having level of water 10m below the water level of the above reservoir. Determine the total discharge if f = 0.06. Neglect minor losses. Length of pipe CD (l 1 ) = 2500m, Diameter of pipe CD (D 1 ) = 0.4m Length of pipe DE (l 2 ) = 1500m Length of pipe DF (l 3 ) = 1500m Diameter of pipes DE and DF = D 2 = D 2 = 300mm = 0.3m Cross- sectional area of pipe CD (A 1 ) = = 0.1256 m 2 Cross- sectional area of pipe DE and DF (A1 = A2 = = 0.0707 m 2 Difference of reservoir level (H) = 10m f= 0.06 Total discharge (Q) =? A C Q D Q 1 H B E Q 2 F Since diameters and lengths of pipes DE and DF are equal, Q 1 = Q 2 Q = Q 1 + Q 2 Q 1 = Q 2 = Q/2 V = Q/A 1 = Q/0.1256 = 7.961Q V 1 =Q 1 /A 1 = Q/(2x0.0707) = 7.07Q = V 2 Applying Bernoulli s equation between A and B, taking flow through CDF

Q = 0. 0711 m 3 /s 9. For a piezometric head difference (Z A -Z B )=7.5m Determine the discharge in each of the three pipes lines and the total discharge for the following data. Pipe Length (m) Diameter (mm) f 1 300 200 0.02 2 200 250 0.018 3 300 300 0.015 1 2 A B 3 D 1 = 0.2m, L 1 = 300m, f 1 = 0.02 D 2 = 0.25m, L 2 = 200m, f 2 = 0.018 D 3 = 0.3m, L 3 = 300m, f 3 = 0.015 Z A Z B = h f = 7.5m Q 1 =?, Q 2 =?, Q 3 =?, Total Q =? Computing r using = 1549.6 = 304.7 =153 For parallel pipes, h f1 = h f2 = h f3 = h f

Q 1 = 0.0695 m 3 /s Q 2 = 0.1569 m 3 /s Q 2 = 0.2214 m 3 /s Total discharge (Q) = Q 1 + Q 2 + Q 3 = 0.0695+0.1569+0.2214 = 0.4478 m 3 /s 10. Three pipes which are laid in parallel have the following details Pipe Length (m) Diameter (mm) f 1 1500 1000 0.02 2 1200 800 0.02 3 1600 1200 0.024 If the discharge through the system is 4.0m 3 /s, determine the discharge distribution in the three pipes and heads. 1 D 1 = 1m, L 1 = 1500m, f 1 = 0.02 D 2 = 0.8m, L 2 = 1200m, f 2 = 0.02 Q Q D 3 = 1.2m, L 3 = 1600m, f 3 = 0.024 2 Total discharge (Q) = 4.0m 3 /s Q 1 =?, Q 2 =?, Q 3 =? h f1 =?, h f2 =?, h f3 =?, 3 Computing r using = 2.48 = 6.05 =1.275 For pipes in parallel, Taking first and second

Taking second and third From continuity = 0.844 m 3 /s = 1.56x0.844 = 1.318 m 3 /s = 2.178x0.844 = 1.838 m 3 /s h f1 = h f2 = h f3 = 4.31m = 4.31m 11. A flow Q of 800 lps flows through the pipe system. What is the pressure drop between A and B if the elevation of A is 100m and that of B is 200m? Neglect minor losses. Take f = 0.02 for all pipes. The diameter of all pipes is 300mm. Q C 300m A 300m E I II III 500m 500m 500m D 300m B 300m F Q Head loss Discharge through branch I = Q 1 Discharge through branch II = Q 2 Discharge through branch III = Q 3 Head loss through branch I ( )

Head loss through branch II Head loss through branch III For parallel connections, ( ) Expressing Q 2 and Q 3 in terms of Q 1 Taking first and second Taking first and third From continuity, = 0.23 m 3 /s = 1.48x0.23 = 0.34 m 3 /s = Q 1 = 0.23 m 3 /s = = 39.3m Applying Bernoulli s equation between A and B through branch II (Neglecting velocity head) = 1366533 Pa = 1366.5 KPa 12. A pipeline 30m long connects two tanks which have a difference of water level of 12m. The first 10m of pipeline from the upper tank is 0f 40mm diameter and the next 20m is of 60mm diameter. At the change in section, a valve is fitted. Calculate the rate of flow when the valve is fully open assuming that its resistance is negligible and that f for both pipes is 0.0216. In order to restrict the flow, the valve is partially closed. If K for the valve is now 5.6, find the percentage reduction in flow.

1 L1 =10m, 40 mm dia. H=12m L2 = 20m, 60 mm dia. 2 f = 0.0216 for both pipes From continuity equation A 1 V 1 = A 2 V 2 Applying Bernoulli s equation between section 1 and 2 H = Total loss a. Neglecting loss due to valve Total loss = entry loss + friction loss in pipe 1+ loss due to sudden expansion + friction loss in pipe 2+exit loss V2 = 2.44 m/s (a) = 0.00689 m 3 /s Considering loss due to valve Total loss = entry loss + friction loss in pipe 1+ loss due to sudden expansion +loss due to valve+ friction loss in pipe 2+exit loss

V2 = 1.86 m/s = 0.00526 m 3 /s Percentage reduction inflow = = 23.6% 13. For a head loss of 10m between A and B, determine the flow rate of water in each of the pipes lines, made of Cast Iron. Take kinematic viscosity of water = 1x10-6 m 2 /s. (Use Moody chart to determine f). A L 1 = 300m, D 1 = 30cm L 2 = 250m, D 2 = 20cm 1 2 B D 1 = 0.3m, L 1 = 300m D 2 = 0.2m, L 2 = 250m h f = 10m Q 1 =?, Q 2 =? For parallel pipes, h f1 = h f2 = h f (a) (b) For C.I. pipe roughness height (k) = 0.25mm k/d 1 = 0.0008, k/d 2 = 0.00125 Trial 1 Assume f 1 = 0.025 and f 2 = 0.02

Q 1 = 0.198 m 3 /s and Q 2 = 0.088 m 3 /s = 8.4x10 5 = 5.6x10 5 For Re1 =8.4x10 5, k/d 1 = 0.0008, f 1 = 0.022 (from Moody Chart) For Re2 =5.6x10 5, k/d 1 = 0.00125, f 2 = 0.0255 (from Moody Chart) Trial 2 With f 1 = 0.022 and f 2 = 0.0255 Q 1 = 0.211 m 3 /s and Q 2 = 0.078 m 3 /s = 9x10 5 = 5x10 5 For Re1 =9x10 5, k/d 1 = 0.0008, f 1 = 0.0221 (from Moody Chart) For Re2 =5x10 5, k/d 1 = 0.00125, f 2 = 0.0258 (from Moody Chart) As the difference in value of f is negligible, the trial is stopped here. With f 1 = 0.0221 and f 2 = 0.02585 Q 1 = 0.211 m 3 /s and Q 2 = 0.078 m 3 /s Simple pipe network 14. A pipe network is shown in figure in which Q and h f represents the discharge and head loss respectively. Determine head losses and discharges indicated by question mark, for this pipe network. Q A = 20 A Q 1 = 30, h f1 = 60 B Q B =? Q 4 =?, h f4 =? Q 5 =?, h f5 =? Q 2 =?, h f2 = 40 C D Q 3 = 40, h f3 = 120 Q D = 100 Q C = 30