[1-2] hydraulic analysis - general background

1

Fall 2012 – 2013 [1] General Background Mohammad N. Almasri, PhD 1

[1-‐2] General Background

Hydraulic Analysis

Mohammad N. Almasri

http://sites.google.com/site/mohammadnablus/Home

Fall 2012 – 2013 [1] General Background Mohammad N. Almasri, PhD

What is the Analysis of WDNs?

§  Analysis of WDNs implies taking the following hydraulic parameters into considera9on:

o  Flow distribu9on and water veloci9es in pipes

o Water pressure at nodes

§  In order to be able to analyze a network, you must simulate the water behavior in the WDN

§  To do so, there are two main principles that are u9lized in the network analysis

2

2


Main Principles of WDN Analysis

Con9nuity: The algebraic sum of the flow rates in the pipes mee9ng at a node together with any external flows is zero

Q1

Q2

Q3

D (demand)

Q1 + Q2 = Q3 + D D = Q1 + Q2 -‐ Q3

A demand node The node is connected to three pipes


Main Principles of Network Analysis

Energy conserva9on: For all paths around closed loops, the accumulated energy loss including minor losses minus any energy gain or heads generated by pumps must be zero

hf1

+ hf3

hf2

hf4

A part of a looped network Closed loop Given total headloss for each link (pipe) as hf

Assume counterclockwise to be posiCve -‐hf1 – hf4 + hf3 + hf2 = 0

3


The Outcome of WDN Analysis

The idea from employing the two concepts of WDNs analysis is to determine both the flow in each pipe and the pressure head at each node by

simply developing and solving a set of equa9ons as a func9on of pipe flow

5


Branched WDNs

6

4


Branched WDNs with Two Supply Points

7


Looped Networks

8

5


Main Principles of Network Analysis

Pipe Head Loss (m) 1 6.753 2 13.238 3 4.504 4 18.728 5 3.737 6 4.998 7 9.994 8 9.992

What are the flow direc9ons in pipes 5, 6, and 8?


Example

Write the con9nuity and energy conserva9on equa9ons for the following network:

10

6


Example

§  First of all, keep in mind two issues: o Nodes provide water or receive water o  Links (pipes) convey water to nodes or from nodes

§  General data includes: o Ground surface eleva9on (z) o Nodal demand or supply (q) o  Pressure head at the source (P1/γ) o  Pipe diameter (D) o  Pipe length (L) o Hazen-‐Williams coefficients

11


Example

§  Label the nodal demands and pipe flows in the figure and assume flow direc9ons

§  This is necessary to be able to formulate the equa9ons needed to analyze the network

12

7


Example

13

Node ID Con4nuity 1 Q1 = q1 2 Q1 = Q2 + Q3 + q2 3 Q2 = q3 + Q5 4 Q3 = q4 + Q4 5 Q4 + Q5 = q5 + Q6 6 Q6 = q6

Loop Energy Conserva4on 2 – 5 – 4 – 3 hf2 + hf5 – hf4 – hf3 = 0

Solving these equa9ons provides the flow distribu9on in the network


Example

§  Let us take a numerical example that shows the outcome of the hydraulic analysis of this network (that of example [5])

§  The objec9ve is to inspect the solu9on, reflect on the results and make the necessary deduc9ons

§  The following tables summarize the related data

14

8


Example Input Data

15

Link ID Length (m) Diameter (in) CHW 1 [1→2] 1,000 10 120 2 [2→3] 1,000 6 120 3 [2→4] 1,000 6 120 4 [5→4] 1,000 6 120 5 [3→5] 1,000 6 120 6 [5→6] 1,000 4 120

Node ID ElevaCon (m) Demand (CMH) 1 100 (total head) -‐280 2 0 50 3 -‐5 75 4 3 25 5 8 85 6 12 45


Example Analysis

16

Node ID Pressure (m) Head (m) 1 -‐ 100 2 89.47 89.47 3 66.21 61.21 4 65.69 68.69 5 48.09 56.09 6 13.17 25.17

Link ID Flow (CMH) Velocity (m/s) Headloss (m/km) 1 280 1.53 10.53 2 124.51 1.9 28.26 3 105.49 1.61 20.79 4 -‐80.49 1.23 12.6 5 49.51 0.75 5.12 6 45 1.54 30.92

9


Hydraulics of Sewers Just a Simplified Overview

17


A Sewer

18

Sewers do not have a full flow and thus the hydraulics of the sewers compare well with the open flow channel

10


The Key Equa9on Manning Equa9on

19


Hydraulics of Sewers

§  Consider the circular-‐pipe cross sec9on in the figure, where h is the depth of flow and θ is the water-‐surface angle

§  The depth of flow, h, cross-‐sec9onal area, A, and weged perimeter, P, can be expressed in terms of θ by the following geometric rela9ons:

20

11


Example

Water flows at a rate of 4 m3/s in a circular concrete sewer of diameter 1,500 mm and a Manning n of 0.015. If the slope of the sewer is 1%, calculate the depth of flow and velocity in the sewer. θ = 3.5 radians

21


Calcula9on of Normal Depth Example

§  A 3-‐k-‐diameter circular culvert is laid on a slope of 0.005 k/k and is carrying a discharge of 25 cfs

§  If the roughness coefficient of the culvert is 0.012, what is the normal depth for this case?

22

12



23



24

13


Example

§  A sewer with a flow of 10 cfs enters manhole 6. The distance downstream to manhole 7 is 400 k

§  The finished street surface eleva9on at manhole 6 is 166.3 k and at manhole 7 it is 164.3 k

§  Note that the crown of the pipe must be at least 6 k below the street surface at each manhole

§  For a Manning coefficient of 0.013, find the required standard pipe size to carry the flow under (a) full flow and (b) one-‐half of full flow. The flow velocity must be at least 2 k/s and must not exceed 10 k/s

25


Example

§  The fall of the sewer in 400 k is 2 k if the street grade is used as the slope

§  The ini9al assump9on for S is 5 k/1000 k

§  If this slope is not sufficient to sustain a velocity of 2 fps, it will be increased later (to increase velocity)

§  In like manner, if the calculated velocity exceeds 10 fps, the slope will have to be reduced

26

14


Example

§  Using Manning’s equa9on for Q = 10 and S = 0.005, the closest diameter will be 21 inches

§  This pipe will support a discharge somewhat higher than the design value of 10 cfs, but it is the closest standard size

§  The velocity associated with a 21-‐in pipe flowing full at a slope of 0.005 is between 4 and 5 fps and thus meets the design criteria

27


Par9ally-‐Full Flow

28

This table helps in finding the design diameter of the pipe when the flow encountered is not full or when we desire that the flow is not to be full

dm d

15


Example

§  The pipe size for one-‐half flow is calculated from the equa9on:

§  From the table, we note that A×R2/3/(dm)8/3 = 0.1558 for d/dm = 0.5. Subs9tu9ng in Manning’s equa9on gives dm = 2.17 k and the closest standard pipe size is 27 inches

29

[1-2] hydraulic analysis - general background

Documents

general background

general background

full ow

main principles

branched wdns

normal depth

pressure head

full