4. Given a portion of a water supply pipe network system below. Determine the actualflow in each pipe and the respective head losses. Assume flow in pipes AB and BC inloop Il and I respectively equal to 0.142 m3/s clockwise and 0.038 m3/s clockwise forthe first trial. Use Manning's Equation with n0.011.%3DPipesАВLength (m)700Diameter (cm)30ВС80015CD110025DE650353001500EA10BD10Q, = 0.15 m/s%3DQA = 0.10 m/sQc = 0.10 m/sВLoop ILoop IIEAQ = 0.15 m/s%3D

Solution Given Data Flow in pipe AB 0.142 m3/s Flow in pipe BC is 0.038 m3/s For Mannings equation…

4. Given a portion of a water supply pipe network system below. Determine the actual flow in each pipe and the respective head losses. Assume flow in pipes AB and BC in loop IlI and I respectively equal to 0.142 m3/s clockwise and 0.038 m/s clockwise for the first trial. Use Manning's Equation with n = 0.011. Length (m) 700 Diameter (cm) Pipes АВ 30 BC CD 800 1100 650 15 25 DE 35 EA 300 10 BD 1500 10

4. Given a portion of a water supply pipe network system below. Determine the actual flow in each pipe and the respective head losses. Assume flow in pipes AB and BC in loop IlI and I respectively equal to 0.142 m3/s clockwise and 0.038 m/s clockwise for the first trial. Use Manning's Equation with n = 0.011. Length (m) 700 Diameter (cm) Pipes АВ 30 BC CD 800 1100 650 15 25 DE 35 EA 300 10 BD 1500 10

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

The three-cistern system in the figure contains water. If L1= 2000 m, f 1= 0.013, D1 = 1m; = 2500 m, f = 0.023, D3 = 1.20m. L2 E 2300 m, f2 = 0.02, D2 = 0.60m; and L3 100 m Datum Compute 02 in m³/s. Express your answer in 3 decimal places. 20 m 120 m K
4. What is the ground water travel time from well MW1 to well MW2 (days) according to the following figure's condition? (Hint: Solve by Darcy's Law) - TOC stands for Top of Casing, and is an elevation with respect to a common datum DTW stands for Depth to water, which is the distance from the TOC to the water table MW1 MWI MW2 L = K= 900 ft TOC (ft) 18 ft/day 0.24 1052.6 1045.3 DTW (ft) 12.5 14.2 MW2
Answer asap
Given below are pipe lengths, sizes, and connections. All pipes have a friction factor of 0.020. If the flow from A to B is 8 cfs, Determine the head loss in feet in PIPE 5. *23.8062*9.0254*1.0666*2.6570 Determine the flow in the upper branch pipe Q2 in cubic feet per second. *2.6667*2.3809*4.8840*12.4747 Determine the head loss in feet in PIPE 2. *3.4373*8.0165*9.2849*10.4720 Determine the total head loss in feet between the inlet A and the outlet D. *9.2849*13.1625*17.3474*35.9163
do not round off everything plsss
Explain and give complete solution and clear
For the water distribution system given below, find the discharge at each pipe. The roughness for all pipes is 0.02. Pipe Diameter Length El. = 126 m (cm) (m) AJ 15 350 El. = 109 m BJ 10 200 CJ 10 250 A El. = 100 m
In the figure shown pipes are in parallel connection with properties given. If the flow into the system is 850lps with total head loss of 10 J/N. For the value of f=0.020 for all pipes, find the diameter of pipe 4. Pipe No. Length - m Diameter - em 1 1000 90 500 50 3 600 40 4 800 5 700 75
Question 4 Pipe network is given bellow (Figure 2). All pipes are 1 km long, 500 mm in diameter and Hazen-Williams Coefficient of 100.
Question 1: The PVC pipeline (ε = 0.0015) depicted in figure below is 150 m. Determine the diameter of the pipe that can provide a flow rate is 0.8 m³/s if the height h is 12 m. Ignore minor losses. water temperature is 4°C. U= h 497 x 10-6 (T + 42.5)1.5 2 m
The total discharge (flowrate) from A to B in the figure below is 300 liters/s. Pipe lennthe are: L1= 150 m, and L2 = 450 m. Diameters, are: D1 = 500 mm, and D2 = 250 mm. L1:D1 Q1 Q Q2 L2; D2 Determine [Using Darcy's equation] the headloss [m] and the flowrates (/s] across A-B in pipes 1 and 2 if riyated steel iron pipes are used in the system and flow is FULLY TURBULENT. Flowrate pipe #1 [I/s] =- Flowrate pipe #2 [l/s] =- Headloss pipe #1 [m] =. %3D Headloss pipe #2 [m]
Problem 4: Approximating Discharge from Velocity Consider steady flow in a stream d = 0.50 m deep and w = 3.0 m wide ("into the page"): Part A Part B h X X The velocity distribution is u = Umax(1- h²/d²), where Umax = 0.20 m/s. A few sample values are shown in the following table: 0.25 h[m] 0 u [m/s] 0.20 0.15 ▼ Part A On the figure, sketch the velocity profile using the equation. No need to calculate points. Part B On the figure, sketch an approximate velocity profile, with an upper area and a lower area. Plot the average velocity for the upper area and the lower area (hint, recall the wedding cake). Part C Using your approximation, calculate the discharge Q [m³/s]. Part D Using your approximation, calculate the average velocity V[m³/s].