WATER RESOURCES ENGINEERING
3rd Edition
ISBN: 9781119490579
Author: Mays
Publisher: WILEY
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Chapter 5, Problem 5.3.4P
To determine
The flow depth downstream of the step assuming no transition losses.
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A rectangular channel 1.25m wide and 30m long with a flat bottom carries a flow of 0.5m³/s. The channel terminates in a free discharge. Plot a sketch of the water profile in the channel. Take n = 0.011.
Water flows at Q = 10 m³/s in a rectangular channel with b =4 m, n = 0.015, and So = 0.0009. If the flow depth at a gaging station is measured as 1 m, classify the gradually varied water surface profile, and use the differential equation of gradually varied flow to state whether the depth increases or decreases in the downstream direction. Assume the energy slope S/= 0.0025.
1. Consider a rectangular channel section having a length of 500 feet. The width of the channel is constant at
b = 25 feet. The slope of the channel bottom is S, = 0.003. The flow in the channel is Q = 2,000 cubic feet
per second (cfs) and the velocity at section 1 is v; = 10 feet per second (fps). Neglecting energy losses,
apply the energy eauation between sections 1 and 2 to determine:
a) the depths of flow yi and y2 in feet at sections 1 and 2, respectively.
b) the cross-sectional areas Az and Az of the flow at sections 1 and 2, respectively.
c) the volumetric flowrate Qz at section 2 in cubic feet per second.
d) the velocity v2 at section 2 in feet per second.
Chapter 5 Solutions
WATER RESOURCES ENGINEERING
Ch. 5 - Prob. 5.1.1PCh. 5 - Prob. 5.1.2PCh. 5 - Prob. 5.1.3PCh. 5 - Prob. 5.1.4PCh. 5 - Prob. 5.1.5PCh. 5 - Prob. 5.1.6PCh. 5 - Prob. 5.1.7PCh. 5 - Prob. 5.1.8PCh. 5 - Prob. 5.1.9PCh. 5 - Prob. 5.2.1P
Ch. 5 - Prob. 5.2.2PCh. 5 - Prob. 5.3.1PCh. 5 - Prob. 5.3.2PCh. 5 - Prob. 5.3.3PCh. 5 - Prob. 5.3.4PCh. 5 - Prob. 5.3.5PCh. 5 - Prob. 5.3.6PCh. 5 - Prob. 5.4.1PCh. 5 - Prob. 5.4.2PCh. 5 - Prob. 5.5.1PCh. 5 - Prob. 5.5.2PCh. 5 - Prob. 5.5.3PCh. 5 - Prob. 5.5.4PCh. 5 - Prob. 5.5.5PCh. 5 - Prob. 5.5.6P
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- Determine the Froude number for the flow in a trapezoidal channel with the depth of water as 2 m and flow rate as (15+x) m³/s. The width of the bed is 5 m and side slope is 1:1 as shown in Figure 3. Classify the type of flow. x=8 2 m 5 marrow_forwardConsider a concrete (n = 0.014) wide rectangular channel that discharges 2.7 m^3/s per unit width of flow. The channel bottom slope is 0.0002. There is a step rise of 0.15 m. Determine the flow depth downstream of the step assuming no transition losses. Does the water rise or fall at the step?arrow_forwardIn a circular channel 3.0 meters in diameter, flowing over half full, the water surface width is 1.268 meter in length. a. Compute the hydraulic radius. b. The channel is laid on a grade of 16.10 meters per mile; compute its discharge using Manning’s formula with n = 0.030. c. For flow as in (b), compute the C value in the Chezy formula. d. Using C in (c), and for flow as in (b), compute the value of Kutter’s n in the Kutter’s C formula. Simplify your solution, no solution by calculator or shift solve. e. Compute the maximum velocity that may occur for the given circular channel using n as in (d) and Manning’s formulaarrow_forward
- (7) (8) Water flows in a rectangular channel with base width B=5 m and depth D=1.5 m. It then flows under a sluice gate that has a width equal to that of the channel base and is raised so that the depth of flow under the gate is d-0.5 m. Calculate the flow rate in the channel and the force on the sluice gate Downstream of the sluice gate in question (7) there is a hydraulic jump. Calculate the downstream depth of flow and the head loss in the jump.arrow_forward13. Compute the critical depth for a rectangular canal having a critical velocity of 4 m/s.arrow_forward542..A hydraulic jump was established in the river. The depth before the jump measured 3.38 cm and the depth after the jump was 11.46 cm. The width of the river is 31.2 cm. Write the type of the jump when the discharge is 0.0162 m3/s.arrow_forward
- The bottom width and bed slope of a long rectangular channel are 4.0 m and 0 0008, respectively. The volumetric steady-state discharge is 1.50 m/s. The flow depth at a specific location is 0 30 m. Ifn=0.016, determine: (a) The type of water surface profile expected Show all work and state your reasoning in support of your work Pleasell!! (b) Sketch the water surface profile with all necessary information and labels shownarrow_forwardExample. The discharge in a channel with bottom width 3 m is 12 m s. If Manning's n is 0.013 and the channel slope is 1 in 200, find the normal depth if: the channel has vertical sides (i.e. rectangular channel): (a) (b) the channel is trapezoidal with side slopes 2H:1V.arrow_forwardo T ☺ ť Water is flowing in a compound open channel as shown in Figure. Given the channel bed slope as 0.0016 and Manning's n = 0.03, nz = 0.025 and ns = 0.03. calculate the flow rate using equivalent Manning's n method. Determine the equivalent n: i. ii. Determine the flow rate. 0.8 m 1.8 m 1.4 m 1.2 0.6 m مرّ للأعلى لاستعراض الفلاتر إضافة شرح. .. mum <arrow_forward
- In a circular channel 2.5meters in diameter, flowing below half full, the water surface width is 2.415meter in length. a.Compute the hydraulic radius. b.The channel is laid on a grade of 24.146meters per mile; compute its discharge using Manning’s formula withn= 0.025. c.For flow as in (b), compute the C value in the Chezy formula. d.Using C in (c), and for flow as in (b), compute the value of Kutter’s nin the Kutter’s C formula. Simplify your solution, no solution by calculator or shift solve. e.Compute the maximum velocity that may occur for the given circular channel using as in (d) and Manning’s formula.arrow_forwardDischarge in a concrete channel with bottom width of 3m is 10 cumecs. Bed slope of a given channel is equal to 1 in 664 . Find the normal, critical and hydraulic depths of flow if: 1. the channel has vertical sides 2. the channel is trapezoidal with 3H:2V 3. the channel is triangular with apex angle equal to 1000arrow_forwardExample 5: Water flows with a velocity of 3m/s, and a depth of 3 m in a rectangular channel of 2 m. Then there is an upward step of 30 cm as shown below. Compute the depth of flow over the step. Vj=3 m/s y2=? I _Az=0.30 m Yı=3 m Datum (1) (2) 83arrow_forward
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