# Hydraulics Part: Open Channel Flow

Size: px
Start display at page:

Transcription

1 Hydraulics Part: Open Channel Flow Tutorial solutions -by Dr. K.N. Dulal Uniform flow 1. Show that discharge through a channel with steady flow is given by where A 1 and A 2 are the sectional areas of flow at the section 1 and 2 and h is the drop in water surface level between the sections and h f is the loss of energy head between two sections. V 1 2 /2g HGL EGL h f V 2 2 /2g Y 1 1 Z 1 Channel Datum Y 2 2 Z 2 Applying Bernoulli's equation between points 1 and 2 [ ] (a) From continuity equation

2 Substituting V 2 in a 2. The velocity distribution in a wide river of 3m deep (y 0 =3m) is given by Find α and β. [ ] [ ] [ ] [ ] [ ] = [ ] [ ] [ ] = 1.04

3 3. For the velocity distribution given in the figure below, find energy and momentum correction factors. 0.4 m/s 0.4 m 2.4 m 1.4 m/s y 0 = 2.4m Mean velocity = 0.55 m/s = 6.84 = The velocity profile in a circular pipe is given by, where v = velocity at any radius r, v max = maximum velocity and R = radius of pipe. Find average velocity, α and β. Mean velocity = (Considering an elementary strip of thickness dr at a radial distance r from pipe center.)

4 [ ] = 2 [ ] = Water flows in a rectangular channel that is 10m wide at a depth of 1.5m. The channel slope is Taking Chezy s C = 120, compute water velocity, flow rate and bed shear stress. Width (b) =10m Flow depth (y) =1.5m Channel slope (S) = C = 120 Velocity (V) =? Flow rate (Q) =? Bed shear stress (τ 0 ) =? A = by = 10x1.5 = 15 m 2 P = b + 2y = 10+2x1.5 = 13 m R =A/P = 15/13 = 1.15m = 6.43 m/s Q = A V = 15x6.43 = m 3 /s = 9810x1.15x = 28.2 N/m 2

5 6. A trapezoidal channel with bottom width 2m and side slope 2:1, flows with a depth of 0.7m in a slope of If n = 0.015, compute bed shear stress, shear velocity, Chezy s C and friction factor f. Bottom width (b) = 2m Z:1 =2:1 Flow depth (y) = 0.7m Slope (S) = Manning s n = Bed shear stress (τ 0 ) =? Shear velocity (V * ) =? Chezy s C =? Friction factor (f) =? = 2.38m 2 = 5.13m R =A/P = 2.38/5.13 = 0.464m = 9810x0.464x = 1.82 N/m 2 = m/s = 59 ( = A trapezoidal channel has a bottom width of 2.5m and depth of flow of 0.8m. The side slopes are 1.5:1. The channel is lined with bricks (K = 3mm). If the longitudinal slope of the channel is , estimate (a) the average shear stress, (b) the hydrodynamic nature of the surface, (c) Chezy C by using f, (d) Manning s n, (e) the uniform flow discharge for cases c and d. Take = 10-6 m 2 /s. Bottom width (b) = 2.5m Z:1 =1.5:1 Flow depth (y) = 0.8m Slope (S) = k = 3mm = 3x10-3 m Bed shear stress (τ 0 ) =? Chezy s C =? Manning s n =? Discharge (Q) =?

6 = 2.96m 2 = 5.384m R =A/P = 2.96/5.384 = m a) = 9810x0.5947x = 1.75 N/m 2 b) = m/s = 125 As R n >60, the boundary is hydrodynamically rough. C) For rough boundary f = 0.02 = 63 d) = (e) Using Chezy = m 3 /s Using Manning, = m 3 /s 8. The cross-section of a stream could be approximated to a rectangular section of 6m bottom width. The stream is in a mountainous region and is formed by cobbles (d 90 = 300mm). Estimate the discharge if the depth of flow is 1.5m and the bed slope is Width (b) =6m Flow depth (y) =1.5m Channel slope (S) = d 90 = 300mm = 0.3m Flow rate (Q) =? Manning s n = = A = by = 6x1.5 = 9 m 2 P = b + 2y = 6+2x1.5 = 9 m R =A/P = 9/9 = 1m Using Manning s equation

7 = 9.03 m 3 /s 9. A triangular channel of side slope 3:1 carries water at a flow rate of 10m 3 /s. If the depth of the flow is 1.32m and the channel is laid on slope of , find Manning s n. Z:1 = 3:1 Water depth (y) =1.32m Flow rate (Q) = 10m 3 /s Channel slope (S) = Manning s n =? = 5.227m 2 = 8.348m R = A/P = 5.227/8.348 = 0.626m n = A circular pipe, for which n =0.014, is laid on a slope of and is to carry 2.56 m 3 /s when the pipe flows at 80% of full depth. Determine the required diameter of pipe. Discharge (Q) = 2.56 m 3 /s Manning s n = Channel slope (S) = Flow depth (y) = 0.8D Where D = Dia. of pipe A O C α B y OC =0.8D-0.5D = 0.3D <AOC = <BOC = arccos(0.3d/0.5d) = α = 360-2x53.1 = = 0.673D 2

8 = 2.214D R = A/P = 0.673D 2 /2.214D = 0.304D D = 2.16m 11. A trapezoidal channel having bottom width = 6.5m and side slope 1.5:1, carries a discharge of 10.4 m 3 /s. If the depth of flow is 1.15m and n = 0.013, compute the slope of the channel. Bottom width (b) = 6.5m Z:1 =1.5:1 Discharge (Q) = 10.4 m 3 /s Flow depth (y) = 1.15m Manning s n = Slope of channel (S) =? = 9.46m 2 = 10.65m R =A/P = 9.46/10.65 = 0.888m S = A rectangular channel carries water at a rate of 7.3 m 3 /s, which is laid on a slope of If the depth of flow is 1.9m, compute the width of the channel. Take n = Discharge (Q) = 7.3 m 3 /s Flow depth (y) =1.9m Channel slope (S) = n = Width (b) =? A = by = 1.9b m 2 P = b + 2y = b+2x1.9 = b+3.8 m

9 R =A/P = 1.9b/b+3.8 Solving for b b = 4.23m 13. A sewer pipe is proposed to be laid on a slope of 1 in 2500 to carry a discharge of 1.5 m 3 /s. What size of circular channel is used if the pipe runs half full. Take n = Discharge (Q) =1.5 m 3 /s Slope (S) = 1/2500 n = Diameter (D) =? D/2 When the pipe is half full Flow depth (y) = D/2 C/S area (A) = Wetted perimeter (P) = Hydraulic radius (R) = D = 2.1m

10 14. A circular drainage pipe 0.8m in diameter conveys a discharge at a depth of 0.25m. If the pipe is laid on a slope of 1 in 900, calculate discharge. Take n = Diameter (D) = 0.8m Flow depth (y) = 0.25m Slope (S) = 1/900 n = Discharge (Q) =? O θ OA=OB=OC = 0.4m OD = OC-CD = = 0.15m θ = arccos(0.15/0.4) = rad = 68 0 A D C B y Flow area (A) = Area of sector OACB- Area of triangle AOB = 0.134m 2 Wetted perimeter (P) = =1.186x0.8 = 0.949m Hydraulic radius (R) = A/P = 0.134/0.949 = 0.141m = 0.08 m 3 /s 15. Compute the normal depth in a triangular channel with bottom angle 60 0 when it carries a discharge of 3m 3 /s. Take n = and S = Bottom angle (<AOB) = 60 0 Discharge (Q) = 3m 3 /s n = S = Normal depth (y n ) =? A C O B y n AC =BC =y n tan30 =0.577 y n Area (A) = P =2y n /cos30 = 2.31y n R = A/P = y n

11 From Manning s equation y n = 2.63m 16. A rectangular channel of 6.1m wide is laid on a slope of The channel carries discharge of 6.8m 3 /s. Compute the normal depth of the channel and maximum shear stress on the bed taking n = Width (b) = 6.1m Discharge (Q) = 6.8 m 3 /s Channel slope (S) = n = Normal depth (y n ) =? Maximum bed shear stress (τ 0 ) =? A = b y n = = 6.1 y n m 2 P = b + 2 y n = y n m From Manning s equation Solving for y n y n = 1.465m = = 9810x0.989x = 1.26 N/m 2

12 17. A channel with a trapezoidal cross-section is to carry 25m 3 /s. If S = , n = 0.015, b =6m and Z:1 =1.5:1, determine the normal depth of flow. Bottom width (b) = 6m Z:1 =1.5:1 Discharge (Q) = 25 m 3 /s S = Manning s n = Normal depth (y n ) =? From Manning s equation Solving for y n y n =2.42m 18. A circular channel, 2.5m in diameter, is made of concrete (n = 0.014) and is laid on a slope of 1 in 200. Calculate the depth of uniform flow for a discharge of 15 m 3 /s. Discharge (Q) = 15 m 3 /s Manning s n = Channel slope (S) = 1/200 Dia. of channel (D) = 2.5m Depth of flow (y) =? A O C α B y

13 ( ) Solving rad <AOC = ( )/2 =67 0 OC=1.25cos67 = 1.22m y = = 2.47m 19. A rectangular channel (n = 0.02) carries flow of 25 m 3 /s with the depth of flow equal to the width of channel. If bed slope = , find the Froude number of the flow. Discharge (Q) = 25 m 3 /s n = 0.02 Depth of flow = y Width = b b = y Bed slope (S) = Froude number (Fr) =? A = by =yxy = y 2 P = b+2y =y+2y = 3y R= A/P= y/3 Using Manning s equation y = 4.4m b = 4.4m A = 4.4x4.4 = m 2 V = Q/A = 25/19.36 = 1.3m/s = A triangular channel of apex angle 90 0 and a rectangular channel of the same material have the same bed slope. If the rectangular channel has the depth of flow equal to the width and the flow areas in both channels are the same, find the ratio of discharges in the rectangular and triangular channels respectively.

14 Bed slope = S and Manning n are same for both channel Triangular channel Bottom angle (<AOB) = 90 0 A C B AC =BC =y t tan45 =y t Area (A t ) = y t P t =2y t /cos45 = 2.828y t R t = A t /P t = y t O Rectangular channel Depth of flow = y Width = b b = y A = by =yxy = y 2 P = b+2y =y+2y = 3y R= A/P= y/3 A t = A y t = y = An open channel is to be designed to carry 1 m 3 /s at a slope of Find the most economical cross-section for (a) a rectangular section, (b) a trapezoidal section, (c) a triangular section and (d) a semicircular section. Take n = 0.01.

15 Discharge (Q) = 1 m 3 /s S = n = 0.01 a. For most economical Rectangular section, b = 2y A = by =2y 2 P = b+2y = 4y R= A/P =0.5y y = 0.42m b = 2x0.42 = 0.84m b. For most economical Triangular section, Z:1 = 1:1 A= Zy 2 = y 2 R= A/P = = 0.353y y = 0.59m c. For most economical Trapezoidal section Z:1 =,, P = 3b A = (b+zy)y = R = A/P = 0.5y y = 0.44m =0.51m d. For most economical semi-circular section, R = A/P = 0.25D, D = 0.92m

16 y = D/2 = 0.46m 22. Find the slope at which a circular sewer of 1.5m diameter should be laid to provide the maximum velocity at a discharge of 0.75 m 3 /s. Take n = Radius (R) = 0.75m Discharge (Q) = 0.75 m 3 /s n = Slope (S) =? For maximum velocity = = = =2.247 rad r 2θ = m 2 = 3.37m Hydraulic radius (R) = A/P= 0.45m S = The wetted perimeter of a most efficient rectangular channel is 3.8m. If the channel has a value of n = 0.012, and slope of 1 in 2000, find the discharge and the state of flow (Subcritical/supercritical) Wetted perimeter of most efficient rectangular channel (P) = 3.8m n = Bed slope (S) = 1/2000 Discharge (Q) =? State of flow =? For most efficient rectangular channel, b = 2y where b = width, y = depth of flow P = b+2y = 2y+2y = 4y y = P/4 = 3.8/4 = 0.95m A = by = 2y 2 = 2x = 1.805m 2 R = y/2 = 0.95/2 = 0.475m Using Manning s equation = 2.05 m 3 /s V = Q/A = 2.05/1.805 = m/s

17 Froude no. As Fr<1, the flow is subcritical. = The area of cross-section of flow in a channel is 6m 2. Calculate the dimensions of the most efficient section if the channel is (a) triangular, (b) rectangular and (c) trapezoidal (2:1). Which has the least perimeter? Flow area (A) = 6 m 2 a. For most efficient Triangular section, Z:1 = 1:1 A= Zy 2 = y 2 6 = y 2 y = 2.45m b. For most economical Rectangular section, b = 2y A = by =2y 2 6 =2y 2 y = 1.732m b = 2x1.732 = 3.464m c. For most economical Trapezoidal section For most economical trapezoidal channel, Given Z:1 = b = 0.47y A = (b+zy)y = (0.47y+2y)y = 2.47 y 2 6 = 2.47 y 2 y = 1.558m b= 0.47x1.558=0.732m Computing wetted perimeter Triangular = 6.929m Rectangular P = b+2y = 4y = 4x1.732 = 6.928m Trapezoidal = 7.699m The rectangular channel has the least wetted perimeter.

18 25. Determine the most economical section of a trapezoidal channel with side slope of 2:1, carrying a discharge of 11.5m 3 /s with a velocity of 0.75m/s. What should be the bed slope of the channel? Take n = Z:1 = 2:1 n= Discharge (Q) = 11.5m 3 /s Velocity (V) = 0.75m/s For most economical trapezoidal channel, b = 0.472y A = (b+zy)y = (0.472y+2y)y = 2.472y 2 = 4.944y R =A/P = 0.5y Q = AV 11.5 = 2.472y 2 x0.75 y = 2.49m b = 0.472x2.49 = 1.17m A = 2.472x = m 2 R = 0.5x2.49 = 1.245m S = In a wide rectangular channel, if the normal depth is increased by 25%, find the increase in discharge. For wide rectangular channel, R = y n For 25% increase in normal depth Taking the ratio

19 Q 2 = 1.45Q 1 Hence discharge increases by 45%. 27. Two concrete pipes (Chezy s C = 100) carries flow from an open channel of 1.8m wide and 0.9m deep (Chezy s C = 120). The slope of both structures is (a) Determine the diameter of pipe. (b) Find the depth of water in the rectangular channel after it has become stabilized, if the slope is changed to (Use C = 120). C/s Area of channel (A 1 ) = 1.8x0.9 = 1.62m 2 Wetted perimeter of channel (P 1 ) = 1.8+2*0.9 = 3.6m Hydraulic radius of channel (R 1 ) = 1.62/3.6 = 0.45m Slope for both (S 1 = S 2 ) = C/S of pipe (A 2 ) = Wetted perimeter of pipe (P 2 ) = Hydraulic radius of pipe (R 2 ) = 0.25D a)chezy s C of pipe (C 2 ) = 100 Chezy s C of channel (C 1 ) = 120 Diameter of pipes (D) =? Qchannel = = 3.91m 3 /s Qchannel = Q pipe1+qpipe2 D =1.22m b) Q = 3.91m 3 /s Slope of channel (S) = C = 120 Depth (y) =? A = by = 1.8y P = b+2y = 1.8+2y Solving for y y = 0.716m

20 28. For the compound section shown in the fig., find discharge when n = 0.015, S = for all parts of section. 1.5: m 1.5:1 1:1 1:1 3m 10m 5m 10m C/s area1 (A 1 ) [ ] = m 2 C/S area 2 (A 2 ) [ ] = 35 m 2 C/S area 3 (A 3 ) [ ] = m 2 Wetted perimeter for sub-area1 (P 1 ) = ( ) = 11.8m Wetted perimeter for sub-area2 (P 2 ) = ( ) = 13.48m Wetted perimeter for sub-area1 (P 3 ) = ( ) = 11.8m Hydraulic radius for sub-area1 (R 1 ) =10.75/11.8 = 0.911m Hydraulic radius for sub-area1 (R 1 ) =35/13.48 = 2.6m Hydraulic radius for sub-area1 (R 3 ) =10.75/11.8 = 0.911m n = n 1 = n 2 = n 3 = S = S 1 = S 2 = S 3 = Discharge (Q) = Q 1 + Q 2 + Q 3 = = m 3 /s

21 29. Show that the discharge formula for a trapezoidal channel having n = carrying maximum discharge is For most economical trapezoidal channel, A = (b+zy)y R = 0.5y From a, From a and c (a) (b) (c) 30. A lined rectangular channel with n = is 4.5m wide and has a flow depth of 2m with bed slope of 1 in Retaining the rectangular shape of the channel section and the same total area of lining, to what maximum extent can discharge be increased without changing the slope. Width (b) = 4.5m Flow depth (y) = 2m n = Bed slope (S) = 1/1600 A 1 = by = 4.5x2 = 9 m 2 P 1 = b + 2y = 4.5+2x2 = 8.5 m R 1 =A/P = 9/8.5 = 1.058m Discharge through the channel = m 3 /s Second case is the case of the economical channel. In this case b = 2y Area of lining per m length is same in both cases. i.e. wetted perimeter = constant So, P 2 = 8.5m P 2 = b 2 +2y 2

22 8.5 =2y 2 + 2y 2 y 2 = 2.125m b 2 = 2y 2 =4.25 A 2 = b 2 y 2 = 4.25x2.125 = 9.03 m 2 P 2 = b 2 + 2y 2 = x2.125 = 8.5 m R 2 =A 2 /P 2 = 9.03/8.5 = 1.062m Discharge through the channel Increase in discharge = = 0.09 m 3 /s = m 3 /s 31. A channel has a vertical wall 1.4m apart and semi-circular of radius 0.7m invert. If the depth of flow is 1m, bed slope is 1/2500, find the discharge using Chezy s formula. Take C = m 0.7m Radius (r) = 0.7m Bed slope (S) = 1/2500 C = 50 Discharge (Q) =? 1.4m C/S area (A) = = m 2 Wetted perimeter (P) = = 2.799m Hydraulic radius (R) = A/P = /2.799 = 0.425m = m 3 /s

23 32. Find the hydraulic exponent for trapezoidal channel. [ ] For trapezoidal channel, Substituting the values in the equation of N [ ] ( ) 33. Obtain an expression for the normal depth of the most efficient (a) rectangular channel and (b) trapezoidal channel. a) For the most efficient rectangular channel b = 2y n A = by = 2 y n 2 P = b+2y = 4 y n R = A/P = y n /2 Using Manning s equation

24 ( ) b) For the most efficient trapezoidal channel (a) From a and b, (b) (( ) ) Wetted perimeter (P) = (( ) ) R = A/P = = ( ) Using Manning s equation ( ) 34. A trapezoidal channel having side slope of 1:1 has to carry a flow of 15m 3 /s. The bed slope is 1 in Chezy s C is 45 if the channel is unlined and 70 if the channel is lined with concrete. The cost per m 3 of excavation is 3 times cost per m 2 of lining. Find which arrangement is economical. Discharge (Q) = 15m 3 /s Z:1 = 1:1 Bed slope (S) = 1/1000 For channel of best section, b = y A = (b+zy)y = (0.8284y+y)y = y 2

25 cost per m 2 of lining = p, cost per m 3 of excavation = 3p case a: Unlined channel C = 70 y = 2.31m A = x = 9.756m 2 Cost of excavation for 1 m length = 9.756x1x3p = p case b: Lined canal C = 70 y = 1.94m A = x = 6.881m 2 = 7.094m For 1m length cost of excavation = 6.881x1x3p = p cost of lining = 7.094x1xp = 7.094p Total cost = p (<29.268p) The lined channel is more economical.

26 Specific energy 1. A rectangular channel 3.5m wide carries a discharge of 11.4 m 3 /s and has its specific energy of 2m of water. Calculate alternate depths and corresponding Froude numbers. Width (b) = 3.5m Discharge (Q) = 11.4 m 3 /s Specific energy (E) = 2m Alternate depths (y 1, y 2 ) =? Froude no. (Fr 1, Fr 2 ) =? Solving for y y 1 = 0.62m, y 2 = 1.84m = 5.25m/s = 1.77m/s = 2 = A rectangular channel 9m wide carries 5.73 m 3 /s when flowing 1.41m deep. Compute specific energy. Is the flow subcritical or supercritical? Width (b) = 9m Discharge (Q) = 5.73 m 3 /s Flow depth (y) = 1.41m Specific energy (E) =? = 1.42m

27 q = Q/b = 5.73/9 = m 3 /s/m Critical depth for rectangular channel is = = 0.345m As y>y c, the flow is subcritical. 3. Water flows at 3.5 m/s in a rectangular channel at a depth of 0.6m. Find the critical depth for (a) this specific energy, (b) this rate of discharge, (c) type of flow and alternate depth for this discharge. Velocity (V) = 3.5m/s Flow depth (y) = 0.6m Critical depth (y c ) =? Alternate depth (y 1 ) =? a)specific energy is = 1.224m = 0.82m b) Discharge per unit width (q) = Vy = 3.5x0.6 = 2.1 m 3 /s/m = = 0.766m c) As y<y c, the flow is supercritical. For q = 2.1 m 3 /s/m = 1.224m Solving for by y 1 Alternate depth (y 1 ) = 0.999m 4. For a constant specific energy of 1.98m, what maximum flow can occur in a rectangular channel 3m wide? Specific energy (E) = 1.98m Width (b) = 3m

28 Maximum flow (Q max ) =? Critical depth = 1.32m Critical velocity (V c ) = = 3.6m/s Q max = A c V C = (3x1.32)x3.6 = m 3 /s 5. Calculate the critical depth (y c ) and corresponding specific energy (E C ) for the following different shapes of channel when Q = 7.5 m 3 /s. (a) Rectangular channel with b = 4m (b) Triangular channel with Z:1 = 0.5:1 (c) Trapezoidal channel with b = 3.6m, Z:1 =2:1 Discharge (Q) = 7.5 m 3 /s a) Rectangular channel b= 4m q = Q/b =7.5/4 = m 3 /s/m = = 0.71m = 1.5x0.71 = 1.065m b) Triangular channel Z:1 = 0.5:1 At critical depth, = 2.15m = 1.25x2.15 = 2.68m c) Trapezoidal channel b = 3.6m, Z:1 =2:1 At critical depth, [ ] [ ]

29 [ ] Solving for y c by trial y c = 0.669m = (3.6+2x0.669)x0.669 = 3.3 m 2 V c = Q/A c = 7.5/3.3 = 2.27m/s = 0.931m 6. A rectangular channel 7.5m wide carries 12m 3 /s with a velocity 1.5 m/s. Compute specific energy. Also find depth of flow when specific energy is minimum. What will be the value of critical velocity as well as minimum specific energy? Channel width (b) = 7.5m Discharge (Q) = 12m 3 /s Velocity (V) = 1.5m/s Specific energy (E) =? Critical depth (y c ) =? Critical velocity (V c ) =? Minimum specific energy (E C ) =? A = Q/V = 12/1.5 = 8 m 2 Depth of flow (y) = A/b = 8/7.5 = 1.067m = 1.18m q= Q/b = 12/7.5 = 1.6 m 3 /s/m = = m = 2.503m/s E c = 1.5 y c = 1.5x0.639 = m 7. Water flows down a wide rectangular channel of concrete (n = 0.014) laid on a slope of 1 in 500. Find the depth and flow rate for critical condition. Slope (S) = 1/500 n = 0.014

30 Critical depth (y c ) =? Discharge (Q) =? For wide rectangular channel, R = y Also, Equating y c = 0.886m q = 2.6 m 3 /s/m 8. Calculate the critical depth of the parabolic channel as shown in the fig., if Q = 4 m 3 /s. y y =2x 2 dy x da = xdy y = 2x 2 T = 2x = At critical depth, y c = 1.33m

31 9. A trapezoidal channel has a bottom width of 6m, side slopes 1:1, and flows at a depth of 1.8m. For n = and a discharge of 10.1 m 3 /s, calculate (a) the normal slope, (b) the critical depth and critical slope for 10.1 m 3 /s, and (c) the critical slope at normal depth of 1.8m. Bottom width (b) = 6m Z:1 = 1:1 Flow depth (y) = 1.8m n= Discharge (Q) = 10.1 m 3 /s a) Normal slope (S) =? = 14.04m 2 = 11.09m R =A/P = 14.04/11.09 = 1.266m S = b)critical slope (S c ) =? Critical depth (y c ) =? At critical depth, [ ] [ ] [ ] Solving for y c y c = = (6+1x0.637)x0.637 = m 2 = 7.8m R C = 4.227/7.8 = 0.542m

32 S c = c) Normal depth (y n ) = 1.8m From a A =14.04 m 2, R =1.266m T = b+2zy c = 6+2x1x1.8 =9.6m = 3.78 m/s S c = A rectangular channel (n = 0.012) is laid on a slope of and carries m 3 /s. For critical flow conditions, what width is required? Slope (S) = Discharge (Q) = m 3 /s n = Width (b) =? Solving for b b = 0.568m 11. Water flows through a rectangular channel, 3m wide, with a velocity of 1.25m/s. The depth of flow is 0.5m. Compute (a) specific energy, (b) specific force, (c) alternate depth, (d) conjugate depth, (e) critical depth for the discharge and (f) type of flow.

33 Width of channel (b) = 3m Velocity (V) = 1.25m/s Depth (y) = 0.5m (a) Specific energy = 0.579m (b) Q = AV = (3x0.5)X1.25 = 1.875m 3 /s Specific force (c) Alternate depth (y 1 ) =? For y = 0.5m, Froude number = 0.614m = The flow is subcritical. The alternate depth should be in supercritical flow, i.e. it is less than 0.5m. Solving for y 1 y 1 = m (d) Conjugate depth (y 2 ) =? Solving for y 2 y 2 = m (e) Critical depth = m (f) As y>y c, the flow is subcritical. 12. At what depths may 1m 3 /s flow in a trapezoidal channel 2m wide and having side slopes 1:1 if the specific energy is 0.5m? What would be the corresponding slopes required to sustain uniform flow if n = 0.015? Also find the minimum specific energy required to carry this discharge. Discharge (Q) = 1m 3 /s Width (b) = 2m Z:1 = 1:1, n = Specific energy (E) = 0.5m Depths =? Slope (S) =?

34 Minimum specific energy (E c ) =? A = (b+zy)y = (2+y)y = y [ ] Solving for y for alternate depths y = m, 0.46m For y = m A = (2+y)y = ( ) = m 2 P = y = x = R = A/P = / = S = For y = 0.46m A = (2+y)y = (2+0.46)0.46 = m 2 P = y = x0.46 = 3.3 R = A/P = /3.3 = S = Finding critical depth At critical depth, [ ] [ ] [ ] Solving y c = m =(2+1x0.2803) = m 2

35 = 0.405m 13. Find at what bed slope a 4m wide rectangular channel be led so that the flow is critical at a normal depth of 1.25m, with Manning s coefficient (n) = Channel width (b) = 4m Normal depth (y) =1.25m n = Channel slope (S) =? A = by = 4x1.25=5 m 2 P = b + 2y = 4+2x1.25 = 6.5 m R =A/P = The flow is critical at depth y = 1.25m. For rectangular channel, the critical depth is given by Q =17.5 m 3 /s From Manning s equation S = Show that alternate depths and critical depth in a rectangular channel is related by the following equation. and (a) (b) Similarly, From a, b and c (c)

36 [ ] [ ] Also, (d) (e) From d and e 15. If F n is the Froude no. for normal depth y n of a rectangular channel, show that Dividing by y n ( ) 16. In a flow through rectangular channel, show that the Froude numbers at alternate depths is related by As, above expression reduces to

37 (a) Also, and Taking ratio (b) From a and b 17. Determine the d/s depth in a horizontal rectangular channel in which the bottom rises 0.15m, if discharge is 8.4 m 3 /s, the width is 3.6m and the u/s depth is 1.2m. What would be the d/s depth if the rise is 0.3m. Discharge (Q) = 8.4m 3 /s Width (b) = 3.6m Rise of bottom = 0.15m u/s depth (y 1 ) = 1.2m d/s depth (y 2 ) =? u/s velocity (V 1 ) = Q/A = 8.4/(3.6x1.2) = 1.94m/s Froude no. at 1 = 0.56 As Fr 1 <1, the u/s flow is subcritical and the transition will cause a drop in water level. q=q/b = 8.4/3.6 = 2.33 m 3 /s/m = 0.82m a) E min = 1.5y c =1.5x0.82 = 1.23m Specific energy at u/s (E 1 ) = = 1.39m D/s specific energy = E 2 = E 2 =1.24>E min. The u/s depth will remain unchanged.

38 Solving for y 2 (y 2 >y c and y 2 <y 1 - ) y 2 = 0.89m D/s depth = 0.89m b) Rise of bottom = 0.3m As computed in a, E 1 = 1.39m, E min = 1.23m = E 2 =1.09<E min, which is not possible. So E 2 = E min and y 2 = y c = 0.82m 18. A rectangular channel contracts from a width of 3m to 2.5m in a short transition section. If the discharge is 6.5 m 3 /s, and the u/s depth is 0.4m, determine the d/s depth. Width at 1 (b 1 ) = 3m Width at 2 (b 2 ) =2.5m Discharge (Q) = 6.5 m 3 /s u/s depth (y 1 ) = 0.4m u/s velocity (V 1 ) = Q/A 1 = 6.5/(3X0.4) = 1.587m/s Froude no. at 1 = 5.41 As Fr 1 >1, the u/s flow is supercritical and the transition will cause a rise in water level. a) d/s depth (y 2 ) =? q 1 = Q/b 1 = 6.5/3 = m 3 /s/m q 2 = Q/b 2 = 6.5/2.5 = 2.6 m 3 /s/m = 0.883m E c2 = 1.5 y c2 = 1.5x0.883 = 1.324m. Specific energy at u/s (E 1 ) = = 1.896m D/s specific energy = E 2 = E 1 = 1.896m As E 2 >E c2, the u/s depth remains unchanged.

39 E 2 = Solving for y 2 y 2 = 0.496m D/s depth = 0.496m 19. A wide rectangular channel carries a flow of 2.7m 3 /s/m, the depth of flow being 1.5m. (a) Calculate the minimum rise in the floor at a section required to produce critical flow condition. (b) What is the corresponding fall in water level? If the loss of energy over the rise is 10% of upstream velocity head, what will be the minimum rise in the floor to produce critical flow? Flow (q) = 2.7m 3 /s/m Depth of flow (y 1 ) = 1.5m Velocity (V 1 ) =q/y 1 = 2.7/1.5 = 1.8m/s a) Rise in floor level =? Critical depth (y c ) = = 0.906m Specific energy at u/s (E 1 ) = = 1.665m Critical depth occurs at section 2. At section 2, specific energy (E 2 ) = E min = 1.5y c = 1.5x0.906 = 1.359m = = 0.306m b) Fall in water level =? = 0.288m Head loss over hump (h L ) = = m Minimum height of hump = = m

40 20. In a 3.4m wide rectangular channel laid at a slope of , uniform flow occurs at a depth of 2m. Find how high a hump can be raised on the channel bed without causing a change in the u/s depth. If the u/s depth is to be raised to 2.35m, what should be the height of hump? Take n = u/s depth (y 1 ) = 2m Channel width (b) = 3.4m Bed slope (S) = n = C/s area (A) = by 1 = 3.4x2 = 6.8 m 2 Wetted perimeter (P) = b+2y 1 = 3.4+2x2 = 7.4m R = A/P = 6.8/7.4 = 0.92m Discharge (Q) = = 7.43 m 3 /s q = Q/b = 7.43/3.4 = m3/s/m a) Height of hump =? Critical depth (y c ) = = 0.786m Specific energy at u/s (E 1 ) = = 2.06m Critical depth occurs at section 2. At section 2, specific energy (E 2 ) = E min = 1.5y c = 1.5x0.786 = 1.18m = = 0.88m b)u/s depth (y 1a ) is increased to 2.35m keeping Q same. Height of hump =? Specific energy at u/s (E 1 ) = = 2.394m From above, y c = 0.786m Critical depth occurs at section 2. At section 2, specific energy (E 2 ) = E min = 1.5y c = 1.5x0.786 = 1.18m = = 1.214m

41 21. A rectangular channel is 2.1m wide and carries a flow of 3m 3 /s at a depth of 0.9m. A contraction of the channel is proposed at a certain section. Compute the depth at contraction and the smallest allowable contracted width that will not affect the u/s flow condition. Channel width (b) = 2.1m Flow rate (Q) = 3m 3 /s u/s flow depth (y 1 ) = 0.9m Depth at contraction (y c ) =? Minimum width at contraction (B m ) =? C/s area (A) = by 1 = 2.1x0.9 = 1.89m 2 u/s velocity (V 1 ) = Q/A = 3/1.89 = 1.587m/s Froude no. at 1 = 0.53 As Fr 1 <1, the u/s flow is subcritical and the transition will cause a drop in water level. Specific energy at u/s (E 1 ) = = 1.028m At the contacted section, specific energy (E 2 ) = E 1 = 1.028m E c = E 2 Critical depth at contracted section (y c )= = 0.685m B m = 1.69m 22. Water flows in a 4m wide rectangular channel at a depth of 2.1m and velocity of 2 m/s. The channel is contracted to 1.9m and bed raised by 0.55m in a given reach. What will be the depth in the u/s reach? Width of channel (b) = 4m Depth of flow (y 1 ) = 2.1m Velocity (V 1 ) = 2m/s Width of contraction (B c ) = 1.9m Rise =0.55m Upstream depth (y 1a ) =? Discharge (Q) = 4x2.1x2 = 16.8 m 3 /s q 1 = 16.8/4 = 4.2 m 3 /s/m

42 Froude number = 0.44 As the flow is subcritical, the depth in transition will decrease. Specific energy (E 1 ) = = 2.304m Discharge/width at contraction (qa) = 16.8/1.9 = m 3 /s/m If the flow in contraction is critical, then critical depth at contraction (y c ) is = 1.997m E min = 1.5 y c = 1.5x1.997 = 2.995m = 1.754m Since E 2 <E min, the flow is not possible. So, the flow at 2 is critical and the u/s depth will rise. Solving for y 1a y 1a = 3.47m 23. A 6m wide rectangular channel carries a discharge of 30m 3 /s at a depth of 2.7m. (a) Calculate the width at which the channel should be contracted so that the depth in the contracted section is critical. (b) What will be the depth at the contracted section if width there is 5m? (c) What will be the depth of flow in the u/s and in the contracted section if the width of channel is reduced to 3.1m? Channel width (b) = 6m Flow rate (Q) = 30m 3 /s Flow depth (y 1 ) = 2.7m Velocity at 1 = 1.85m/s Froude no. at 1 = 0.36 As Fr 1 <1, the u/s flow is subcritical and the transition will cause a drop in water level. a) Minimum width at contraction (b m ) =? Discharge per unit width at 1 (q 1 ) = 30/6 = 5m 3 /s/m Specific energy at u/s (E 1 ) = = 2.875m At the contacted section, specific energy (E 2 ) = E 1 = 2.875m E c = E 2 Critical depth at contracted section (y c )= = 1.916m

43 B m = 3.61m b) Width (b 2 ) = 5m Depth at contracted section (y 2 ) =? As b 2 >b m (critical width), the flow in the contracted section will not be critical. At u/s, = 1.365m Discharge per unit width at 2 (q 2 ) = Q/b 2 = 30/5 = 6m 3 /s/m E 1 =E 2 Solving for y 2 (y 2 >y c, y 2 <y 1 ) y 2 = 2.605m (>y c ) c) Width (b 2 ) = 3.1m Depth at contracted section (y 2 ) =? Depth at u/s section (y 1a ) =? Discharge per unit width (q 2 ) = Q/b 2 = 30/3.1 = 9.677m 3 /s/m As b 2 <b m, the flow cannot take place with the given specific energy (E 2 <E min ). So the flow in the contracted section should be critical. = 2.121m E min = E 2 = 1.5y c = 1.5x2.121 = 3.181m However, since E 1 = 2.875m, the flow cannot occur unless E 1a =E min = 3.181m Solving for y 1a (y 1a >y 1 ) U/s depth = 3.043m Depth at contracted section = critical depth = 2.121m

44 24. A 4.9m wide rectangular channel carries a discharge of 20m 3 /s at a depth of 2m. (a) Calculate the depth of flow over a hump of 0.3m on the bed. (b) What will be the minimum rise in the bed level required to obtain critical depth over the rise? (c) What will be the water depths u/s and over the hump if the bed level is raised by 0.7m? Channel width (b) = 4.9m Discharge (Q) = 20m 3 /s U/s depth (y 1 ) = 2m Discharge per unit width (q) = 20/4.9 = 4.08m 3 /s/m Velocity at 1 Froude no. at 1 = 2.04 m/s = 0.46 As Fr 1 <1, the u/s flow is subcritical and the transition will cause a drop in water level. Specific energy at u/s (E 1 ) = = 2.212m = 1.193m Critical depth occurs at section 2. At section 2, specific energy (E 2 ) = E min = 1.5y c = 1.5x1.193 = m a) Rise =0.3m = = 1.912m E 2 >E min, the flow is sub-critical. Depth of flow over hump (y 2 ) =? Solving for y 2 (y 2 >y c, y 2 <y 1 ) y 2 =1.566m b) Minimum rise in bed level =? Critical depth occurs at section 2. As calculated above, Specific energy at u/s (E 1 ) = 2.212m and minimum specific energy at section 2 (E 2 ) = E min = m = = m c) Rise =0.7m Water depth over hump (y 2 ) =?

45 U/s water depth (y 1a ) =? = = 1.512m As E 2 <E min, specific energy at 1 is inadequate for water to flow at critical depth at section 2. Hence the depth at section 1 increases to y 1a and depth at section 2 is critical. y 2 = y c = 1.193m and E 2 = E min Solving y 1a by trial and error (y 1a >y 1 ) Solving for y 1a y 1a = 2.333m U/s depth = 2.333m 25. The width of a rectangular channel is reduced gradually from 3m to 2m and the floor is raised by 0.3m at a given section. When the approaching depth of flow is 2.05m, what rate of flow will be indicated by a drop of 0.2m in the water surface elevation at the contracted section? U/s depth (y 1 ) = 2.05m = 0.3m Depth at contracted section (y 2 ) = = 1.55m Width at u/s (b 1 ) = 3m Width at contracted section (b 2 ) = 2m Discharge (Q) =? Q = 7.11 m 3 /s 26. A flow of 30m 3 /s is carried in a 5m wide rectangular channel at a depth of 1.0m. Find the slope necessary to sustain uniform flow at this depth if n = What change in roughness would produce uniform critical flow at this discharge on this given slope?

46 Discharge (Q) = 30m 3 /s Width (b) =5m Flow depth (y) =1.0m n= Channel slope (S) =? Change in roughness for uniform critical condition =? A = by = 5x1.0 = 5 m 2 P = b + 2y = 5+2x1.0 = 7 m R =A/P = 5/7 = 0.714m S = For critical flow Critical depth (y c ) = = 1.54m A c = by c = 5x1.54 = 7.7 m 2 P c = b + 2yc = 5+2x1.54 = 8.08 m R c =A/P = 7.7/8.08 = 0.953m Finding Manning s n with S = Change in roughness = 83% increase n = A 3m wide rectangular channel carries 3 m 3 /s of water at a depth of 1m. If the width is to be reduced to 2m and bed raised by 10cm, what would be the depth of flow in the contracted section? What maximum rise in the bed level of the contracted section is possible without affecting the depth of flow upstream of transition? Neglect loss of energy in transition. What would be the change in water surface elevations if the rise in bed is 30cm? Width of channel at 1 (b 1 ) = 3m Width of channel at 1 (b 2 ) = 2m

47 Discharge (Q) = 3 m 3 /s Depth of channel at 1 (y 1 ) = 1m Rise in bed level at 2 =0.1m Velocity at 1 Froude no. at 1 = 1m/s = 0.32 As Fr 1 <1, the u/s flow is subcritical and the transition will cause a drop in water level. Critical depth at 2 (y c2 ) = = m Minimum specific energy (E c2 ) = 1.5 y c2 = 1.5x0.6121= m (a) Depth of flow in the contracted section (y 2 ) =? Specific energy at u/s (E 1 ) = ( = 1.051m ) = = As E 2 >E c2, the u/s depth will remain unchanged Solving for y 2 (y 2 <y 1, y 2 >y c ) y 2 = m b) Maximum rise in bed level at 2 =? At critical state, the rise in bed level will be maximum, without affecting the u/s depth. E 2 = E C2 = = 0.133m c) Rise in bed level at 2 =0.3m = = As E 2 <E c2, the flow is not possible. So, the d/s depth should be critical and the u/s depth will increase to y 1a. y 2 = y c2 =0.6121m, E 2 = E c2 = m

48 Solving for y 1a (y 1a >y 1 ) y 1a = 1.174m 28. A rectangular channel 2m wide has a flow of 2.4 m 3 /s at a depth of 1.0m. Determine if critical depth occurs (a) a section where a hump of = 20cm high is installed across the bed, (b) a side wall constriction (no hump) reducing the channel width to 1.7m, and (c) both the hump and side wall constriction combined. Will the upstream depth be affected for case (c)? If so, to what extent? Neglect head losses of the hump and constriction cause by friction, expansion and contraction. Width of channel (b) = 2m Discharge (Q) = 2.4 m 3 /s Flow depth (y 1 ) = 1.0m Specific energy at section 1 (E 1 ) = = m (a) = 20cm = 0.2m Critical depth is given by = 0.527m E c =1.5 y c =1.5x0.527 = m At critical condition, E 2 = E C The height of hump for critical condition As (b) b 2 = 1.7m Critical depth is given by, the critical depth does not occur. = 0.588m E c2 =1.5 y c =1.5x0.588 = 0.882m For constricted flow E 1 = E 2 E 2 = m As E 2 >E c2, the critical depth does not occur. = = m (c) = 20cm = 0.2m and b 2 = 1.7m As computed in (b), E c2 = 0.882m At critical condition, E 2 = E C2 The height of hump for critical condition = = m As, the flow is not possible and the depth at 2 should be critical.

49 In case (c), = = m As E 2 <E c2, specific energy at 1 is inadequate for water to flow. Hence the u/s will be affected. The u/s depth should rise to y 1a and depth at section 2 should be critical. y 2 = y c2 = 0.588m and E 2 = E c2 = 0.882m. E1a = u/s specific energy Solving for y 1a y 1a = 1.01m 29. A discharge of 16 m 3 /s flows with a depth of 2m in a 4m wide rectangular channel. At a downstream section, the width is reduced to 3.5m and the channel bed is raised by 0.35m. To what extent will the surface elevation be affected by these changes? Width of channel at 1 (b 1 ) = 4m Width of channel at 1 (b 2 ) = 3.5m Discharge (Q) = 16 m 3 /s Depth of channel at 1 (y 1 ) = 2m Rise in bed level at 2 =0.35m Velocity at 1 Froude no. at 1 = 2m/s = 0.45 As Fr 1 <1, the u/s flow is subcritical and the transition will cause a drop in water level. Critical depth at 2 (y c2 ) = = 1.287m Minimum specific energy (E c2 ) = 1.5 y c2 = 1.5x1.287= m Specific energy at u/s (E 1 ) = ( = m ) = = m As E 2 <E c2, the flow is not possible. So, the d/s depth should be critical and the u/s depth will increase to y 1a. y 2 = y c2 =1.287m, E 2 = E c2 = m

50 Solving for y 1a (y 1a >y 1 ) y 1a = m Upstream elevation will rise by m. 30. A venturiflume in a rectangular channel of width B has the throat width of b. The depth of liquid at the entry is H and at the throat is h. Prove that the following relation exists for the discharge and width ratio: 1 2 B b H h At the contraction, the flow is critical. So, Discharge (Q) = Rearranging (a) Expression for b/b E 1 = E 2 E 2 = E C = 1.5h So, E 1 = 1.5h

51 h (I) From continuity equation A 1 V 1 = A 2 V 2 BH V 1 = bh V 2 Substituting the value of V 1 in I [ ] (Expanding by neglecting higher order terms) (b)

52 Gradually varied flow 1. A rectangular channel 7.5m wide has a uniform depth of flow of 2m and has a bed slope of 1 in If due to weir constructed at the downstream end of the channel, water surface at a section is raised by 0.75m, determine the slope of water surface (a) with respect to channel bottom and (b) with respect to horizontal at this section. Take n = 0.02 and use dynamic equation of GVF. Width (b) = 7.5m Uniform flow depth (y) = 2m Bed slope (S 0 ) = 1/3000 n = 0.02 A = by = 7.5x2= 15 m 2 P = b + 2y = 7.5+2x2 = 11.5 m R =A/P = 15/11.5 = 1.304m T = b = 7.5m Discharge through the channel = m 3 /s Computing energy slope (S f ) at the weir y1 = =2.75m A1 = by1 = 7.5x2.75= m 2 V= Q/A1 = 16.34/ = 0.8 m/s P1 = b + 2y 1= 7.5+2x2.75 = 13 m R1 =A1/P1 = /13 = = Slope of water surface with respect to channel bottom = For rising water surface profile, Slope of water surface with respect to horizontal = (1/3000) =

53 2. In a 1.8m wide rectangular channel (bottom slope = 0.002, n = 0.014), water flows at 6.72 m 3 /s. A temporary dam increases the depth to 2.73m. What type of flow profile is created behind the dam? Width of channel (b) = 1.8m Bottom slope (S) = n = Discharge (Q) = 6.72 m 3 /s Depth at dam (y) = 2.73m q= Q/b = 6.72/1.8 = 3.73 m 3 /s/m Finding critical depth (y c ) and normal depth (y n ) = 1.123m A = by n = 1.8y n P = b+2y n = 1.8+2y n For normal depth Solving for y n y n = 1.67m y = 2.73m, y c = 1.123m, y n = 1.67m As y n >y c, the slope is mild. For y> y n >y c, the profile is M Water flows at a rate of 7.5m 3 /s in a 2m wide rectangular channel having bottom slope = If a dam at d/s increases the depth to 4m, examine the water surface profile u/s from the dam. Take n = Width of channel (b) = 2m Bottom slope (S) = n = Discharge (Q) = 7.5 m 3 /s Depth at dam (y) = 4m q= Q/b = 7.5/2 = 3.5 m 3 /s/m

54 Finding critical depth (y c ) and normal depth (y n ) = 1.076m A = by n = 2y n P = b+2y n = 2+2y n For normal depth Solving for y n y n = 0.955m y = 4m, y c = 1.076m, y n = 0.955m As y n <y c, the slope is steep. For y> y c >y n, the profile is S A 3m wide rectangular channel (n = ) carries water at a rate of 9.53 m 3 /s. There is an abrupt change of slope from (mild) to (steep). Flow depth is 1.3m at u/s and 0.7m at d/s. Identify the type of water surface profile u/s and d/s. Also sketch the profile. Width of channel (b) = 3m Bottom slope u/s (S1) = Bottom slope d/s (S2) = n = Discharge (Q) = 9.53 m 3 /s q= Q/b = 9.53/3 = m 3 /s/m Finding critical depth (y c ) and normal depth (y n ) = 1.009m A = by n = 3y n P = b+2y n = 3+2y n

55 Normal depth at u/s Solving for y n1 y n1 = 1.38m Normal depth at d/s Solving for y n2 y n2 = 0.62m y n1 At u/s, y = 1.3m, y n1 =1.38m y c = 1.009m At d/s, y = 0.7m, y n2 =0.62m y c = 1.009m y c At u/s y n1 >y>y c, the profile is M 2. At d/s y c >y>y n2, the profile is S 2. y c y n2 The flow is subcritical u/s and supercritical d/s. At the break, critical flow occurs. 5. A 1.8m rectangular channel carries a discharge of 1.8m 3 /s at a bed slope of At a certain section, the depth of flow is 1m. Predict the type of profile. Take n = Channel width (b) = 1.8m Discharge (Q) = 1.8m 3 /s Bed slope (S 0 ) = n= Flow depth (y) = 1m Type of profile =? Length of the profile =?

56 q= Q/b = 1.8/1.8 = 1 m 3 /s/m Finding critical depth (y c ) and normal depth (y n ) = 0.467m A = by n = 1.8y n P = b+2y n = 1.8+2y n For normal depth Solving for y n y n = 0.255m Here y>y c >y n, the profile is S Sketch the flow profile having sluice gate as shown in the figures. M1 Mild slope Jump M1 or M2 S1 Jump M3 Jump H3 S3 Steep slope Horizontal (I) (II) (III) H2

57 7. Sketch the flow profile having series of channels as shown in the figures. I. CDL NDL M1 M3 J M2 Mild slope S2 Steep NDL II. CDL NDL S2 J Reservoir H3 Steep Horizontal III. NDL CDL Mild slope M2 S2 J S1 J NDL M2 Steep S3 M3 Mild

58 IV. CDL NDL NDL S2 J M2 Steep slope M3 S2 V. NDL Mild Steep NDL CDL M2 NDL S2 Mild Steeper S3 NDL Steep 8. A rectangular channel is 4.5m wide and is to carry 14m 3 /s of water. The bottom slope of the channel is It discharges into a stream at which the depth of flow is 3m. Calculate the distance from the channel outlet to the point where normal depth would occur under this condition. Take n = Use direct step method and solve in one step. Channel width (b) = 4.5m Discharge (Q) = 14m 3 /s Bed slope (S 0 ) = n= Flow depth at outlet (y) = 3m Distance from outlet to point at which depth = normal depth (y n ) =? Finding normal depth A = by n = 4.5y n

59 P = b+2y n = 4.5+2y n For normal depth Solving for y n y n = 1.22m Here y n = y 1 =1.22m and y 2 = y = 3m Direct step method in tabular form y A P R V E Sf x Formulae used in the table A= by, P = b+2y, R=A/P V= Q/A, E = y +V 2 /2g,, [ ], x = cumulative sum of The distance from outlet to point at which depth is equal to normal depth = 2198m 9. A rectangular channel of 2m wide carries a discharge of 2.35 m 3 /s. The depth of flow in the channel varies from 1.2m at one section to 0.9m at a section 200m d/s. Find the bed slope of the channel. n = Use direct step method. Channel width (b) = 2m Discharge (Q) =2.35 m 3 /s y 1 =1.2m y 2 = 0.9m = 200m n = Bed slope (S 0 ) =? y A P R V E S f Formulae used in the table A= by, P = b+2y, R=A/P

60 V= Q/A, E = y +V 2 /2g,, [ ] The variables A to are computed in the table above S 0 = A rectangular concrete channel 4m wide has a slope of 9x10-4. It carries a flow of 18 m 3 /s and has a depth of 2.3m at one section. By using direct step method and taking one step, compute the depth 300m downstream. Take n = What is the type of surface curve obtained? Channel width (b) = 4m Discharge (Q) =18m 3 /s y 1 =2.3m = 300m n = Bed slope (S 0 ) = 9x10-4 y 2 =? Finding critical depth (y c ) = 1.27m Finding normal depth Solving for y n y n = 1.85m y n >y c : Mild slope Section 1 A 1 = by 1 = 4x2.3 = 9.2 m 2 P 1 = b+2y 1 = 4+2x2.3 = 8.6m R 1 =A 1 /P 1 = 9.2/8.6 = V 1 = Q/A 1 = 18/9.2 = m/s = 2.495

61 = Section2 A 2 = by 2 = 4y 2 P 2 = b+2y 2 = 4+2y 2 R 2 =A 2 /P 2 =4y 2 /4+2y 2 V 2 = Q/A 2 = 18/4y 2 [ ] [ ] [ ] [ ] Solving for y 2 (y 2 >y 1 for backwater) y 2 = 2.46m As y> y n >y c : the profile is M A trapezoidal channel has a constant bed slope of , b=3m and Z:1 = 1:1. A control gate increases the depth immediately u/s to 4m when discharge is 19m 3 /s. Compute GVF profile from the gate to the point at which the depth = 3m. Use direct step method and perform computation for water depth at 0.1m interval. Take n = Bottom width (b) = 3m Z:1 = 1:1 Bed slope (S 0 ) = Discharge (Q) = 19m 3 /s n =0.017

62 GVF Computation using direct step method y A P R V E Sf Sfb dx x Formulae used in the table,, R =A/P V= Q/A, E = y +V 2 /2g,, [ ], x = cumulative sum of 12. A long rectangular channel 8m wide, bed slope 1:5000 and n = conveys discharge of 40m 3 /s. A sluice gate raises depth immediately u/s to 5.0m. Compute GVF profile from the gate to a distance of 1000m at 500m interval using standard step method. Take bed elevation at sluice gate = 100m, next station = 100.5m and last station = 100.7m Channel width (b) = 8m Bed slope (S 0 ) = 1/5000 = n = Discharge (Q) = 40m 3 /s Formulae to be used A= by, P = b+2y, R=A/P H= Z+ V 2 /2g (where Z = water surface elevation), [ ] H(2) = (H(2)) i-1 + h f +h e

63 Computation of GVF by standard step method st Trial Z y A P R V H(1) S f h f h e H(2) A B C Final values Station Water surface elev. A 105 B C

64 13. A rectangular channel (n = 0.011) is 1.5m wide and carries 1.7 m 3 /s of water. The bed slope is , and at section 1 the depth is 0.9m. Find the distance to section 2, where the depth is 0.75m. Use direct integration method (Bresse s method). Channel width (b) = 1.5m Discharge (Q) = 1.7 m 3 /s Bed slope (S 0 ) = n = Flow depth at section 1 (y1) = 0.9m Flow depth at section 2 (y2) = 0.75m Distance to section 2 (L) =? q = Q/b = 1.7/1.5 = m 3 /s/m Finding critical depth (y c ) and normal depth (y n ) = 0.508m A = by n = 1.5y n P = b+2y n = 1.5+2y n For normal depth Solving for y n y n = 0.985m u 1 = y 1 /y n = 0.9/0.985 = u 2 = y 2 /y n = 0.75/0.985 = = =

65 [ ( ) ( )] = 16.4m [ ( ) ] 14. A very wide rectangular channel with horizontal bed terminates in a vertical drop. If it discharges 3.5 m 3 /s/m of water, estimate depth at the drop. How long on upstream depth will be 2.5m? Take n = Use direct step method and solve in 3 steps. Discharge per unit width (q) = 3.5 m 3 /s/m Depth at drop =? Finding critical depth (y c ) = m Depth at drop is assumed to be equal to critical depth, although it is a brink depth slightly less than critical depth. Hence, depth at drop = m Computing length by direct step method (S 0 = 0) y R V E S f x Distance at which water depth becomes 2.5 m is m Formulae used in the table R=y (For wide rectangular channel) V= q/y, E = y +V 2 /2g,, [ ], 15. A wide rectangular channel having a bottom slope of 5x10-3 has a constant value of Chezy s coefficient equal to 76. If discharge per unit width is 4.65 m 3 /s and if the channel joins a reservoir in which water depth is 3m above channel bed, find the length of surface curve. Use direct step method taking maximum of 5 steps.

66 Discharge per unit width (q) = 4.65 m 3 /s/m Bottom slope (S 0 ) = 5x10-3 C = 76 R=y (For wide rectangular channel) Finding critical depth (y c ) = 1.3m Finding normal depth (y n ) ( ) ( ) = 0.9m y c >y n : Steep slope As the channel joins a reservoir, the profile is backwater curve (S 1 ). So the depth decreases from d/s to u/s. Computation is done from d/s to the point at which critical depth occurs. Computation by Direct step method taking 5 steps y R V E S f x Length of surface curve = m Formulae used in the table V= q/y, E = y +V 2 /2g,, [ ], 16. A rectangular channel conveying a discharge of 30 m 3 /s is 12m wide with a bed slope 1 in 6000 and having Manning s n = The depth of flow at a section is 1.5m. Find how far upstream or downstream of this section the depth of flow will be 2m. Find also the type of profile. Use direct step method for calculation and take only two steps for calculation. Discharge (Q) = 30 m 3 /s Channel width (b) = 12m Bed slope (S 0 ) = 1/6000 n = 0.025

### CE 6403 APPLIED HYDRAULIC ENGINEERING UNIT - II GRADUALLY VARIED FLOW

CE 6403 APPLIED HYDRAULIC ENGINEERING UNIT - II GRADUALLY VARIED FLOW Dynamic equations of gradually varied and spatially varied flows - Water surface flow profile classifications: Hydraulic Slope, Hydraulic

### Presented by: Civil Engineering Academy

Presented by: Civil Engineering Academy Open-Channel Flow Uniform Flow (See CERM Ch. 19) Characterized by constant depth volume, and cross section. It can be steady or unsteady Non-uniform Flow *Not on

### DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING Urban Drainage: Hydraulics. Solutions to problem sheet 2: Flows in open channels

DEPRTMENT OF CIVIL ND ENVIRONMENTL ENGINEERING Urban Drainage: Hydraulics Solutions to problem sheet 2: Flows in open channels 1. rectangular channel of 1 m width carries water at a rate 0.1 m 3 /s. Plot

### Closed duct flows are full of fluid, have no free surface within, and are driven by a pressure gradient along the duct axis.

OPEN CHANNEL FLOW Open channel flow is a flow of liquid, basically water in a conduit with a free surface. The open channel flows are driven by gravity alone, and the pressure gradient at the atmospheric

### Water Flow in Open Channels

The Islamic Universit of Gaza Facult of Engineering Civil Engineering Department Hdraulics - ECIV 33 Chapter 6 Water Flow in Open Channels Introduction An open channel is a duct in which the liquid flows

### OPEN CHANNEL FLOW. One-dimensional - neglect vertical and lateral variations in velocity. In other words, Q v = (1) A. Figure 1. One-dimensional Flow

OPEN CHANNEL FLOW Page 1 OPEN CHANNEL FLOW Open Channel Flow (OCF) is flow with one boundary exposed to atmospheric pressure. The flow is not pressurized and occurs because of gravity. Flow Classification

### NPTEL Quiz Hydraulics

Introduction NPTEL Quiz Hydraulics 1. An ideal fluid is a. One which obeys Newton s law of viscosity b. Frictionless and incompressible c. Very viscous d. Frictionless and compressible 2. The unit of kinematic

### Uniform Flow in Open Channels

1 UNIT 2 Uniform Flow in Open Channels Lecture-01 Introduction & Definition Open-channel flow, a branch of hydraulics, is a type of liquid flow within a conduit with a free surface, known as a channel.

### Open Channel Flow Part 2. Ch 10 Young, notes, handouts

Open Channel Flow Part 2 Ch 10 Young, notes, handouts Uniform Channel Flow Many situations have a good approximation d(v,y,q)/dx=0 Uniform flow Look at extended Bernoulli equation Friction slope exactly

### Closed duct flows are full of fluid, have no free surface within, and are driven by a pressure gradient along the duct axis.

OPEN CHANNEL FLOW Open channel flow is a flow of liquid, basically water in a conduit with a free surface. The open channel flows are driven by gravity alone, and the pressure gradient at the atmospheric

### Hydromechanics: Course Summary

Hydromechanics: Course Summary Hydromechanics VVR090 Material Included; French: Chapters to 9 and 4 + Sample problems Vennard & Street: Chapters 8 + 3, and (part of it) Roberson & Crowe: Chapter Collection

### Gradually Varied Flow I+II. Hydromechanics VVR090

Gradually Varied Flow I+II Hydromechanics VVR090 Gradually Varied Flow Depth of flow varies with longitudinal distance. Occurs upstream and downstream control sections. Governing equation: dy dx So Sf

### Chapter 4: Non uniform flow in open channels

Chapter 4: Non uniform flow in open channels Learning outcomes By the end of this lesson, students should be able to: Relate the concept of specific energy and momentum equations in the effect of change

### Dr. Muhammad Ali Shamim ; Internal 652

Dr. Muhammad Ali Shamim ali.shamim@uettaxila.edu.pk 051-904765; Internal 65 Channel Tranistions A channel transition is defined as change in channel cross section e.g. change in channel width and/or channel

### UNIFORM FLOW CRITICAL FLOW GRADUALLY VARIED FLOW

UNIFORM FLOW CRITICAL FLOW GRADUALLY VARIED FLOW Derivation of uniform flow equation Dimensional analysis Computation of normal depth UNIFORM FLOW 1. Uniform flow is the flow condition obtained from a

### CEE 3310 Open Channel Flow, Nov. 26,

CEE 3310 Open Channel Flow, Nov. 6, 018 175 8.10 Review Open Channel Flow Gravity friction balance. y Uniform Flow x = 0 z = S 0L = h f y Rapidly Varied Flow x 1 y Gradually Varied Flow x 1 In general

### LECTURE 9: Open channel flow: Uniform flow, best hydraulic sections, energy principles, Froude number

LECTURE 9: Open channel flow: Uniform flow, best hydraulic sections, energy principles, Froude number Assist. Prof. Neslihan SEMERCİ Marmara University Department of Environmental Engineering Open channel

### 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

### We will assume straight channels with simple geometries (prismatic channels) and steady state flow (in time).

56 Review Drag & Lift Laminar vs Turbulent Boundary Layer Turbulent boundary layers stay attached to bodies longer Narrower wake! Lower pressure drag! 8. Open-Channel Flow Pipe/duct flow closed, full,

### P10.5 Water flows down a rectangular channel that is 4 ft wide and 3 ft deep. The flow rate is 15,000 gal/min. Estimate the Froude number of the flow.

P10.5 Water flows down a rectangular channel that is 4 ft wide and ft deep. The flow rate is 15,000 gal/min. Estimate the Froude number of the flow. Solution: Convert the flow rate from 15,000 gal/min

### P = 2Rθ. The previous Manning formulas are used to predict V o and Q for uniform flow when the above expressions are substituted for A, P, and R h.

Uniform Flow in a Partly Full, Circular Pipe Fig. 10.6 shows a partly full, circular pipe with uniform flow. Since frictional resistance increases with wetted perimeter, but volume flow rate increases

### 1.060 Engineering Mechanics II Spring Problem Set 8

1.060 Engineering Mechanics II Spring 2006 Due on Monday, May 1st Problem Set 8 Important note: Please start a new sheet of paper for each problem in the problem set. Write the names of the group members

### 3.2 CRITICAL DEPTH IN NONRECTANGULAR CHANNELS AND OCCUR- RENCE OF CRITICAL DEPTH

3.2 CRITICAL DEPTH IN NONRECTANGULAR CHANNELS AND OCCUR- RENCE OF CRITICAL DEPTH Critical Depth in Non-Rectangular Channels Consider an irregular channel: da w dd dd d Specific energy is defined as: E

### 1. Open Channel Hydraulics

Open Channel Flow. Open Channel Hydraulics.... Definition and differences between pipe flow and open channel flow.... Types of flow.... Properties of open channels...4.4 Fundamental equations... 5.4. The

### Pressure Head: Pressure head is the height of a column of water that would exert a unit pressure equal to the pressure of the water.

Design Manual Chapter - Stormwater D - Storm Sewer Design D- Storm Sewer Sizing A. Introduction The purpose of this section is to outline the basic hydraulic principles in order to determine the storm

### Lecture Note for Open Channel Hydraulics

Chapter -one Introduction to Open Channel Hydraulics 1.1 Definitions Simply stated, Open channel flow is a flow of liquid in a conduit with free space. Open channel flow is particularly applied to understand

### 28.2 Classification of Jumps

28.2 Classification of Jumps As mentioned earlier, the supercritical flow Froude number influences the characteristics of the hydraulic jump. Bradley and Peterka, after extensive experimental investigations,

### Hydraulics for Urban Storm Drainage

Urban Hydraulics Hydraulics for Urban Storm Drainage Learning objectives: understanding of basic concepts of fluid flow and how to analyze conduit flows, free surface flows. to analyze, hydrostatic pressure

### y 2 = 1 + y 1 This is known as the broad-crested weir which is characterized by:

CEE 10 Open Channel Flow, Dec. 1, 010 18 8.16 Review Flow through a contraction Critical and choked flows The hydraulic jump conservation of linear momentum y = 1 + y 1 1 + 8Fr 1 8.17 Rapidly Varied Flows

### R09. d water surface. Prove that the depth of pressure is equal to p +.

Code No:A109210105 R09 SET-1 B.Tech II Year - I Semester Examinations, December 2011 FLUID MECHANICS (CIVIL ENGINEERING) Time: 3 hours Max. Marks: 75 Answer any five questions All questions carry equal

### STEADY UNIFORM FLOW IN OPEN CHANNEL

11/4/018 School of Environmental Engineering STEY UNIFORM FLOW IN OEN CHNNEL ZULKRNIN BIN HSSN COURSE OUTCOMES CO1: ble to analyze and design the steady flow in pipeline (O1) CO: ble to analyze and design

### Created by Simpo PDF Creator Pro (unregistered version)

2 nd Semester TRANSITIONS STRUCTURES 1 of 9 A transition is a local change in cross-section which produces a variation of flow from one uniform state to another due to the change in cross sections of channels.

### VARIED FLOW IN OPEN CHANNELS

Chapter 15 Open Channels vs. Closed Conduits VARIED FLOW IN OPEN CHANNELS Fluid Mechanics, Spring Term 2011 In a closed conduit there can be a pressure gradient that drives the flow. An open channel has

### New Website: M P E il Add. Mr. Peterson s Address:

Brad Peterson, P.E. New Website: http://njut009fall.weebly.com M P E il Add Mr. Peterson s Email Address: bradpeterson@engineer.com If 6 m 3 of oil weighs 47 kn calculate its If 6 m 3 of oil weighs 47

### Fluid Mechanics Prof. S.K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Fluid Mechanics Prof. S.K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 42 Flows with a Free Surface Part II Good morning. I welcome you to this session

### Open Channel Hydraulics

30 Open Channel Hydraulics Aldo Giorgini (deceased) Donald D. Gray West Virginia University 30. Definitions and Principles Classification of Flows Flow Regimes 30. Balance and Conservation Principles Conservation

### CIE4491 Lecture. Hydraulic design

CIE4491 Lecture. Hydraulic design Marie-claire ten Veldhuis 19-9-013 Delft University of Technology Challenge the future Hydraulic design of urban stormwater systems Focus on sewer pipes Pressurized and

### EXPERIMENTAL STUDY OF BACKWATER RISE DUE TO BRIDGE PIERS AS FLOW OBSTRUCTIONS

Tenth International Water Technology Conference, IWTC1 6, Alexandria, Egypt 19 EXPERIMENTAL STUDY OF BACKWATER RISE DUE TO BRIDGE PIERS AS FLOW OBSTRUCTIONS Kassem Salah El-Alfy Associate Prof., Irrigation

### Lecture 10: River Channels

GEOG415 Lecture 10: River Channels 10-1 Importance of channel characteristics Prediction of flow was the sole purpose of hydrology, and still is a very important aspect of hydrology. - Water balance gives

### 5. ROADSIDE AND MEDIAN CHANNELS

5. ROADSIDE AND MEDIAN CHANNELS Roadside and median channels are open-channel systems which collect and convey stormwater from the pavement surface, roadside, and median areas. These channels may outlet

### conservation of linear momentum 1+8Fr = 1+ Sufficiently short that energy loss due to channel friction is negligible h L = 0 Bernoulli s equation.

174 Review Flow through a contraction Critical and choked flows The hydraulic jump conservation of linear momentum y y 1 = 1+ 1+8Fr 1 8.1 Rapidly Varied Flows Weirs 8.1.1 Broad-Crested Weir Consider the

### PROPERTIES OF FLUIDS

Unit - I Chapter - PROPERTIES OF FLUIDS Solutions of Examples for Practice Example.9 : Given data : u = y y, = 8 Poise = 0.8 Pa-s To find : Shear stress. Step - : Calculate the shear stress at various

### NON UNIFORM FLOW IN OPEN CHANNEL

For updated version, please click on http://ocw.ump.edu.my HYDRAULICS NON - UNIFORM FLOW IN OPEN CHANNEL TOPIC 3.3 by Nadiatul Adilah Ahmad Abdul Ghani Faculty of Civil Engineering and Earth Resources

### Open Channel Flow I - The Manning Equation and Uniform Flow COURSE CONTENT

Open Channel Flow I - The Manning Equation and Uniform Flow Harlan H. Bengtson, PhD, P.E. COURSE CONTENT 1. Introduction Flow of a liquid may take place either as open channel flow or pressure flow. Pressure

### Open-channel hydraulics

Open-channel hydraulics STEADY FLOW IN OPEN CHANNELS constant discharge, other geometric and flow characteristics depended only on position Uniform flow Non-uniform flow S; y; v const. i i 0 i E y 1 y

### EFFECT OF BAFFLE BLOCKS ON THE PERFORMANCE OF RADIAL HYDRAULIC JUMP

Fourth International Water Technology Conference IWTC 99, Alexandria, Egypt 255 EFFECT OF BAFFLE BLOCKS ON THE PERFORMANCE OF RADIAL HYDRAULIC JUMP O. S. Rageh Irrigation & Hydraulics Dept., Faculty of

### presented by Umut Türker Open Channel Flow

presented by Umut Türker Open Channel Flow What is open channel flow? Open channel flow is a flow which has a free surface and flows due to the gravitational effect What is open channel flow? Open channel

### OPEN CHANNEL FLOW. Computer Applications. Numerical Methods and. Roland Jeppson. CRC Press UNIVERSITATSB'BUOTHEK TECHNISCHE. INFORMATlONSBiBUOTHEK

OPEN CHANNEL FLOW Numerical Methods and Computer Applications Roland Jeppson TECHNISCHE INFORMATlONSBiBUOTHEK UNIVERSITATSB'BUOTHEK HANNOVER Si. i. CRC Press Taylor &.Francis Group Boca Raton London New

### SOE2156: Fluids Lecture 7

Weirs and SOE2156: Fluids Lecture 7 Vee Vee Last lecture { assumed the channel was uniform (constant depth, shape, slope etc.) { steady uniform Found that : location of free surface to be determined 2

### Open Channel Hydraulics I - Uniform Flow

PDHonline Course H138 (2 PDH) Open Channel Hydraulics I - Uniform Flow Instructor: Harlan H. Bengtson, Ph.D., PE 2012 PDH Online PDH Center 5272 Meadow Estates Drive Fairfax, VA 22030-6658 Phone & Fax:

### UNIT I FLUID PROPERTIES AND STATICS

SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code : Fluid Mechanics (16CE106) Year & Sem: II-B.Tech & I-Sem Course & Branch:

### Chapter (3) Water Flow in Pipes

Chapter (3) Water Flow in Pipes Water Flow in Pipes Bernoulli Equation Recall fluid mechanics course, the Bernoulli equation is: P 1 ρg + v 1 g + z 1 = P ρg + v g + z h P + h T + h L Here, we want to study

### FE Fluids Review March 23, 2012 Steve Burian (Civil & Environmental Engineering)

Topic: Fluid Properties 1. If 6 m 3 of oil weighs 47 kn, calculate its specific weight, density, and specific gravity. 2. 10.0 L of an incompressible liquid exert a force of 20 N at the earth s surface.

### Lab 7: Nonuniform Flow and Open Channel Transitions

CE 3620: Water Resources Engineering Spring 2015 Lab 7: Nonuniform Flow and Open Channel Transitions BACKGROUND An open channel transition may be defined as a change either in the direction, slope, or

### Q = α n AR2/3 h S1/2 0. bd 2d + b d if b d. 0 = ft1/3 /s

CEE 330 Open Channel Flow, Nov., 00 7 8.8 Review Open Channel Flow Gravity friction balance. In general we take an energy equation approach. y Uniform Flow x = 0 z = S 0L = h f where we could find h f

### Guo, James C.Y. (1999). "Critical Flow Section in a Collector Channel," ASCE J. of Hydraulic Engineering, Vol 125, No. 4, April.

Guo, James C.Y. (1999). "Critical Flow Section in a Collector Channel," ASCE J. of Hydraulic Engineering, Vol 15, No. 4, April. CRITICAL FLOW SECTION IN A COLLECTOR CHANNEL By James C.Y. Guo, PhD, P.E.

### If a stream of uniform velocity flows into a blunt body, the stream lines take a pattern similar to this: Streamlines around a blunt body

Venturimeter & Orificemeter ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 5 Applications of the Bernoulli Equation The Bernoulli equation can be applied to a great

### 1.060 Engineering Mechanics II Spring Problem Set 4

1.060 Engineering Mechanics II Spring 2006 Due on Monday, March 20th Problem Set 4 Important note: Please start a new sheet of paper for each problem in the problem set. Write the names of the group members

### Uniform Channel Flow Basic Concepts Hydromechanics VVR090

Uniform Channel Flow Basic Concepts Hydromechanics VVR090 ppt by Magnus Larson; revised by Rolf L Feb 2014 SYNOPSIS 1. Definition of Uniform Flow 2. Momentum Equation for Uniform Flow 3. Resistance equations

### Computation of gradually varied flow in compound open channel networks

Sādhanā Vol. 39, Part 6, December 014, pp. 153 1545. c Indian Academy of Sciences Computation of gradually varied flow in compound open channel networks 1. Introduction H PRASHANTH REDDY 1,, M HANIF CHAUDHRY

### Chapter 3.8: Energy Dissipators. By Dr. Nuray Denli Tokyay

Chapter 3.8: Energy Dissipators By Dr. Nuray Denli Tokyay 3.1 Introduction A stilling basin is a short length of paved channel placed at the foot of a spillway or any other source of supercritical flow

### Hydraulics Prof. Dr. Arup Kumar Sarma Department of Civil Engineering Indian Institute of Technology, Guwahati

Hydraulics Prof. Dr. Arup Kumar Sarma Department of Civil Engineering Indian Institute of Technology, Guwahati Module No. # 04 Gradually Varied Flow Lecture No. # 07 Rapidly Varied Flow: Hydraulic Jump

### The University cannot take responsibility for any misprints or errors in the presented formulas. Please use them carefully and wisely.

Aide Mémoire Suject: Useful formulas for flow in rivers and channels The University cannot take responsiility for any misprints or errors in the presented formulas. Please use them carefully and wisely.

### Beaver Creek Corridor Design and Analysis. By: Alex Previte

Beaver Creek Corridor Design and Analysis By: Alex Previte Overview Introduction Key concepts Model Development Design Accuracy Conclusion Refresh v = Beaver Creek Site = Wittenberg Introduction Low head

### Block 3 Open channel flow

Numerical Hydraulics Block 3 Open channel flow Markus Holzner Contents of the course Block 1 The equations Block Computation of pressure surges Block 3 Open channel flow (flow in rivers) Block 4 Numerical

### 9. Flood Routing. chapter Two

9. Flood Routing Flow routing is a mathematical procedure for predicting the changing magnitude, speed, and shape of a flood wave as a function of time at one or more points along a watercourse (waterway

### Basic Hydraulics. Rabi H. Mohtar ABE 325

Basic Hydraulics Rabi H. Mohtar ABE 35 The river continues on its way to the sea, broken the wheel of the mill or not. Khalil Gibran The forces on moving body of fluid mass are:. Inertial due to mass (ρ

### Experiment 7 Energy Loss in a Hydraulic Jump

Experiment 7 Energ Loss in a Hdraulic Jump n Purpose: The purpose of this experiment is to examine the transition from supercritical (rapid) flow to subcritical (slow) flow in an open channel and to analze

### Open Channel Flow - General. Hydromechanics VVR090

Open Channel Flow - General Hydromechanics VVR090 ppt by Magnus Larson; revised by Rolf L Jan 2014, Feb 2015 SYNOPSIS 1. Introduction and Applications 2. The History of Open Channel Flow 3. Flow Classification

### Fluvial Dynamics. M. I. Bursik ublearns.buffalo.edu October 26, Home Page. Title Page. Contents. Page 1 of 18. Go Back. Full Screen. Close.

Page 1 of 18 Fluvial Dynamics M. I. Bursik ublearns.buffalo.edu October 26, 2008 1. Fluvial Dynamics We want to understand a little of the basic physics of water flow and particle transport, as so much

### Department of Hydro Sciences, Institute for Urban Water Management. Urban Water

Department of Hydro Sciences, Institute for Urban Water Management Urban Water 1 Global water aspects Introduction to urban water management 3 Basics for systems description 4 Water transport 5 Matter

### CIVL4120/7020 Advanced open channel hydraulics and design - Tutorial (1) Unsteady open channel flows

School of Civil Engineering at the University of Queensland CIVL4120/7020 Advanced open channel hydraulics and design - Tutorial (1) Unsteady open channel flows Attendance to tutorials is very strongly

### INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad AERONAUTICAL ENGINEERING QUESTION BANK : AERONAUTICAL ENGINEERING.

Course Name Course Code Class Branch INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad - 00 0 AERONAUTICAL ENGINEERING : Mechanics of Fluids : A00 : II-I- B. Tech Year : 0 0 Course Coordinator

### 5 ENERGY EQUATION OF FLUID MOTION

5 ENERGY EQUATION OF FLUID MOTION 5.1 Introduction In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics. The pertinent laws

### A note on critical flow section in collector channels

Sādhan ā, Vol. 26, Part 5, October 2001, pp. 439 445. Printed in India A note on critical flow section in collector channels 1. Introduction SUBHASISH DEY Department of Civil Engineering, Indian Institute

### CHAPTER 07 CANAL DESIGN

CHAPTER 07 CANAL DESIGN Dr. M. R. Kabir Professor and Head, Department of Civil Engineering University of Asia Pacific (UAP), Dhaka LECTURE 17 Canal Design Types Canal Design Drainage Channel Design Irrigation

### Mass of fluid leaving per unit time

5 ENERGY EQUATION OF FLUID MOTION 5.1 Eulerian Approach & Control Volume In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics.

### BACKWATERRISE DUE TO FLOW CONSTRICTION BY BRIDGE PIERS

Thirteenth International Water Technology Conference, IWTC 1 009, Hurghada, Egypt BACKWATERRISE DUE TO FLOW CONSTRICTION BY BRIDGE PIERS Kassem Salah El-Alfy Prof. Dr., Irrigation &Hydraulics Dept., Faculty

### Engineering Hydrology (ECIV 4323) CHAPTER FOUR. Stream flow measurement. Instructors: Dr. Yunes Mogheir Dr. Ramadan Al Khatib

Engineering Hydrology (ECIV 4323) CHAPTER FOUR Stream flow measurement Instructors: Dr. Yunes Mogheir Dr. Ramadan Al Khatib -١ 4.1 Introduction - Surface water hydrology deals with the movement of water

### Prof. B.S. Thandaveswara. Superelevation is defined as the difference in elevation of water surface between inside (1)

36.4 Superelevation Superelevation is defined as the difference in elevation of water surface between inside and outside wall of the bend at the same section. y=y y (1) 1 This is similar to the road banking

### Formation Of Hydraulic Jumps On Corrugated Beds

International Journal of Civil & Environmental Engineering IJCEE-IJENS Vol:10 No:01 37 Formation Of Hydraulic Jumps On Corrugated Beds Ibrahim H. Elsebaie 1 and Shazy Shabayek Abstract A study of the effect

### Cavitation occurs whenever the pressure in the flow of water drops to the value of the pressure of the saturated water vapour, pv (at the prevailing

Cavitation occurs whenever the pressure in the flow of water drops to the value of the pressure of the saturated water vapour, pv (at the prevailing temperature); cavities filled by vapour, and partly

### FORMATION OF HYDRAULIC JUMPS ON CORRUGATED BEDS

International Journal of Civil & Environmental Engineering IJCEE-IJENS Vol: 10 No: 01 40 FORMATION OF HYDRAULIC JUMPS ON CORRUGATED BEDS Ibrahim H. Elsebaie 1 and Shazy Shabayek Abstract A study of the

### STEADY FLOW THROUGH PIPES DARCY WEISBACH EQUATION FOR FLOW IN PIPES. HAZEN WILLIAM S FORMULA, LOSSES IN PIPELINES, HYDRAULIC GRADE LINES AND ENERGY

STEADY FLOW THROUGH PIPES DARCY WEISBACH EQUATION FOR FLOW IN PIPES. HAZEN WILLIAM S FORMULA, LOSSES IN PIPELINES, HYDRAULIC GRADE LINES AND ENERGY LINES 1 SIGNIFICANCE OF CONDUITS In considering the convenience

### THE EFFECTS OF OBSTACLES ON SURFACE LEVELS AND BOUNDARY RESISTANCE IN OPEN CHANNELS

Manuscript submitted to 0th IAHR Congress, Thessaloniki, 4-9 August 00 THE EFFECTS OF OBSTACLES ON SURFACE LEVELS AND BOUNDARY RESISTANCE IN OPEN CHANNELS J. D. FENTON Department of Civil and Environmental

### Advanced Hydraulics Dr. Suresh A. Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati

Advanced Hydraulics Dr. Suresh A. Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati Module - 3 Varied Flows Lecture - 7 Gradually Varied Flow Computations Part 1 (Refer Slide

5/6/18 3. Gradually-aried Flow Normal Flow vs Gradually-aried Flow Normal Flow /g EGL (energy grade line) iction slope Geometric slope S Normal flow: Downslope component of weigt balances bed friction

### Chapter 6 The Impulse-Momentum Principle

Chapter 6 The Impulse-Momentum Principle 6. The Linear Impulse-Momentum Equation 6. Pipe Flow Applications 6.3 Open Channel Flow Applications 6.4 The Angular Impulse-Momentum Principle Objectives: - Develop

### The Impulse-Momentum Principle

Chapter 6 /60 The Impulse-Momentum Principle F F Chapter 6 The Impulse-Momentum Principle /60 Contents 6.0 Introduction 6. The Linear Impulse-Momentum Equation 6. Pipe Flow Applications 6.3 Open Channel

### Advanced Hydraulics Prof. Dr. Suresh A. Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati

Advanced Hydraulics Prof. Dr. Suresh A. Kartha Department of Civil Engineering Indian Institute of Technology, Guwahati Module - 5 Channel Transitions Lecture - 1 Channel Transitions Part 1 Welcome back

### NOTES ON OPEN CHANNEL FLOW

NOTES ON OPEN CANNEL FLOW Prof. Marco Pilotti Facoltà di Ingegneria, Università degli Studi di Brescia Profili di moto permanente in un canale e in una serie di due canali - Boudine, 86 OPEN CANNEL FLOW:

### Part 2: Introduction to Open-Channel Flow SPRING 2005

Part : Introduction to Open-Cannel Flow SPRING 005. Te Froude number. Total ead and specific energy 3. Hydraulic jump. Te Froude Number Te main caracteristics of flows in open cannels are tat: tere is

### Uniform Channel Flow Basic Concepts. Definition of Uniform Flow

Uniform Channel Flow Basic Concepts Hydromechanics VVR090 Uniform occurs when: Definition of Uniform Flow 1. The depth, flow area, and velocity at every cross section is constant 2. The energy grade line,

### Open Channel Flow - General. Open Channel Flow

Open Channel Flow - General Hydromechanics VVR090 Open Channel Flow Open channel: a conduit for flow which has a free surface Free surface: interface between two fluids of different density Characteristics

### 1 Branch: CIVIL ENGINEERING Sub: HYDRAULICS AND HYDRAULIC MACHINERY Date: 11/03/2017

G.PULLAIAH COLLEGE OF ENGINEERING & TECHNOLOGY (AT) II B.Tech Objective Paper I MID EXAM 1 Branch: CIVIL ENGINEERING Sub: HYDRAULICS AND HYDRAULIC MACHINERY Date: 11/03/2017 Time: 20 min Max.Marks:10 Roll

### vector H. If O is the point about which moments are desired, the angular moment about O is given:

The angular momentum A control volume analysis can be applied to the angular momentum, by letting B equal to angularmomentum vector H. If O is the point about which moments are desired, the angular moment

### Comparison of Average Energy Slope Estimation Formulas for One-dimensional Steady Gradually Varied Flow

Archives of Hydro-Engineering and Environmental Mechanics Vol. 61 (2014), No. 3 4, pp. 89 109 DOI: 10.1515/heem-2015-0006 IBW PAN, ISSN 1231 3726 Comparison of Average Energy Slope Estimation Formulas