B) Water flows at a depth of 1.70 m in a trapezoidal concrete-lined section (k,= 1.5 mm) with a bottom width of 3 m and side slopes of 2:1 (H: V). The slope of the channel is 0.0004, and the water temperature is 20 °C. Assuming uniform-flow conditions, estimate the average velocity and flow rate in the channel. y = 1.7m b=3m 0 = 50°

B) Water flows at a depth of 1.70 m in a trapezoidal concrete-lined section (k,= 1.5 mm) with a bottom width of 3 m and side slopes of 2:1 (H: V). The slope of the channel is 0.0004, and the water temperature is 20 °C. Assuming uniform-flow conditions, estimate the average velocity and flow rate in the channel. y = 1.7m b=3m 0 = 50°

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: Water is flowing in trapezoidal channel with 2V:3H at a discharge of Q=3.5m^3/s. If the channel top…

A: Solution:

Q: A right-angled triangular channel carries a flow of 16,000 l/s. The Manning's n = 0.020 and the…

Q: a rectangular concrete channel, the equilibrium depth is 3.2 m. The channel is 2 m wide and the bed…

A: Gievn data,y = Uniform equilibrium flow depth =3.2 mb = Channel width = 2 mS = Bed slope = 0.0006n =…

Q: Uniform flow (15.0 m³/s) occurs in a rectangular channel of width 3.50m at a normal depth of 3.80m.…

A: Given Width 3.5 mNormal depth before gate 3.8 mDepth at gate 0.5 mDischarge 15 m3/sTo find out Depth…

Q: This cross-section shows steady incompressible flow in an open channel. Copy the figure, then add…

A: To find: Draw the EGL and HGL.

Q: d )- A rectangular channel of 10 m width, 4 m depth and bed slope of 18 cm/km. The Q discharge is 18…

Q: Consider a concrete (n = 0.014) wide rectangular channel that discharges 2.7 m^3/s per unit width of…

A: Data given n = 0.014 Q = 2.7 m3/s / unit width S = 0.0002 Step rise, △z = 0.15 m Find: Flow depth…

Q: PROBLEM 3: Water flows at a steady rate in this vertical pipe. The pressure at A is 10 kPa, and at B…

Q: Complete the spreadsheat, by calculating the amount of cut, fill, and either excess or shortage of…

A: Spreadsheet is givennew elevation and existing elevation is givenFINDamount of cut, fill, and either…

Q: Find the maximum step height for a flow with a constant discharge, q = 12 m2 /s, and at 1 a water…

Concept explainers

Open Channel Flow

Open channel flow is a branch of fluid mechanics. It deals with the flow of fluid in an open conduit or free surface. In an open channel flow, gravity induces the flow of water whose surface is open to the air. This flow is free-falling rather than confined in rigid body conduits. The open channel flow depends upon two parameters that are the cross-section of the conduit and the velocity profile. Practical examples of open channel flow are streams, rivers, canals, and other drainage systems.In open channel flow, the cross-sectional area of at least one of the sides is exposed to air, and the surface of water faces atmospheric pressure. This flow is also known as free gravity flow. The cross-sectional area of the conduit can be either man-made or a natural channel. For instance, the flow in rivers and streams are natural conduits, and the flow in a canal and sanitary systems are man-made or engineered channels.

Question

B) Water flows at a depth of 1.70 m in a trapezoidal concrete-lined
section (ks = 1.5 mm) with a bottom width of 3 m and side slopes
of 2:1 (H: V). The slope of the channel is 0.0004, and the water
temperature is 20 °C. Assuming uniform-flow conditions,
estimate the average velocity and flow rate in the channel.
y = 1.7 m
b=3m
0 = 50°

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Step 5

Trending now

This is a popular solution!

Step by step

Solved in 5 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Fluid Mechanics 2 open channel flow
The base of the artificial channel has a bottom width of 3m and sides with a slope of 1 vertical to 2 horizontal. The depth of water being predicted/estimated is 1.2m when the discharge is 3.51m3/s. Using Manning's equation with n = 0.022 a. Calculate the slope of the channel bed.b. Calculate the discharge in m3/s when the depth of flow is 1.3m.
Hydraulic structure Two concrete pipes (n = 0.01) must carry the flow from an open channel half square section 2m wide and 1m deep (n = 0.01168) The slope of both structures is (0.0009). Determine the diameter of the pipes.
Water discharges in too wide rectangular channel slope =0.01 Just apstream the weir where the water depth is 2.5m the flow rate is 5m ^ 3/s As per meter width as shown below Assume n = 0.015 and neglected losses. 1. Find the normal depth, the critical depth and the critical slope? 2. Use the specific energy equation to find the weir height such that the flow over the wer is critical? 3. What water surface profile exists upstream the weir and locate the distance upstream the weit where the water depth equals the normal depth ?
4.7. Water flows at a depth of 2.20 m in a trapezoidal, concrete-lined section (ks of 3.6 m and side slopes of 2:1 (H:V). The longitudinal slope of the channel is 0.0006 and the water temperature is 20°C. Assuming uniform-flow conditions, estimate the average velocity and flow rate in the channel. Use : 2 mm) with a bottom width Manning equations
Water is flowing in a rectangular channel that has a width b-3 m. If the depth of flow y = 2 m and Froud Number NF33, then the flow rate (Q) in the channel is
Prob. # 5 (a): In a partially flowing channel having a isosceles-triangular cross section as shown below (Figure 5a), the rate of flow can be expressed as Q=KAR 2³, where k is a constant and A and R are respectively cross sectional area and the hydraulic radius. Determine the depth at which the discharge is maximum. For the channel, A= [B - (h/√3)]h, and P= B + (4h/3). 60 60° Figure section. 5a-Triangular channel cross
Water flows in a concrete rectangular channel (n = 0.013) that is 12.0 m wide and water depth of 2.5 m. The channel slope is 0.0028. Calculate the water velocity V and flow rate Q.
Calculate the depth d in which the discharge (Q = 1.5 m / sec) will flow through the channel with trapezoid - shaped cross section. Width of the channel bed is (b = 2 m), slope of bed (S = 0.05%), sides slope is (Z = 1.5). The channel is excavated in gravel sends in which the roughness coefficient is (n = 0.0247).
4. The water at the floodgate looks like in the picture with V1 = 0.2 m/s and y1 = 1m. Calculate: - depth at point 2 - flow velocity at point 2 - Froude Number value Gap = 20 cm (1) (2)