Distribution of water

BY WAQAS AHMAD

Distribution of water

Methods of distribution system:

water is distributed to the consumers by1. Gravity Distribution2. Pumping with storage3. Direct pumping (without storage)

1) Gravity distribution:• Used when source of water supply is situated

at some elevation a2) Pumping with storage:

• used in plain areas & designed on the basis of pressure available at the end

• Excess of water is pumped during period of low consumption is stored in on elevated or overhead reservoir & consumed again during high consumption

3) Direct Pumping without storage:• Water is directly pumped to the distribution

system without any storage it is least desirable system as:i. Power failure would cause complete

interruption in water supplyii. as consumption varies, pressure is likely to

fluctuate, hence several pumps are needediii. High rate of pumping cause high power

consumption & high cost

Layout of distribution systemThere are 4 methods:

1. Dead end system or Tree system2. Grid iron system3. Circle or belt system4. Radial system

1) Dead end system:• Consist of one main taking water from the plant

to town from which submains are taken out• The service connections are given to houses from

submain• Mains & submains are laid parallel along the

road, layout is in the form of a tree

2) Grid iron system:• Also known as re-circulation or interlaced

system• In this mains, submains & branches are

interconnected with each other which prevents stagnation of water

• This system is designed by HARDY CROSS method which is based on the fact that some of head loses of a pipe forming a loop is zero

• Used in new cities having well planned roads

3) Circle or belt system: In this method • A loop/ring of main is formed • Distribution area is divided into a no of rectangular

or circular blocks• Submains are laid out on the periphery of these

blocks • Suitable for well planned cities

4) Radial system:• Water is taken from main or well & pumped into

distribution reservoirs located at centre of various blocks of the town, the water is then supplied through radially laid pipe

Design of distribution systems:

It means to find the diameters of various pipes to carry certain discharge under necessary pressure

various methods of analysis & design of distribution systems are:

1. Equivalent pipe method2. The cut circle or cut contour method3. Freemann graphical method4. The Cobb 3 dimensional method5. Hardy Cross method of analysis

Design Formula:

Hazen William Formula:

Where v= velocity of water R= hydraulic radiusS= slope of hydraulic gradient line (S=H/L)C= Hazen William roughness coeff= 100 for cast iron pipeK= unit correction constt = 0.849 for MKS system= 1.382 for KPS system

Modified formula for circular pipe:

where

In english system of units:

Sometimes nomograms are used a nomogram is a graph relating Q, hl/L, d and v

if 2 parameters are known, the other 2 can be taken from the graph

Hardy Cross method:Used on the fact that the sum of head

losses for closed loop or network is = 0Used for grid iron system (network of

distribution pipes)o Assumptions:

1) Sum of inflows at a node/point is equal to outflow ∑inflow = ∑outflow or ∑total= 0

2) Algebric sum of head loses in close loop = 03) Clockwise flows are positive4) Counter clockwise flows are negative

Derivation:Any formula of pipe (manning, chazy,

Hazen) can be expressed generally as:• ………………………(i)

Where H= head loss, Q=discharge through pipeK= constant depending on the dia, length of pipex= exponent whose value is generally 1.85

From Hazen William formula:

=>

For any pipe in a close loop ……………………………(ii)

Where Q= Actual flow=assumed flow𝜟= required flow correctionFrom (i) & (ii)=> =As 𝜟 is very small as compared to Q we can neglect 𝜟=-𝛴H/x.(𝛴H/Q))………………………(iii)eq (iii) is used in Hardy Cross method

Procedure for Hardy Cross method:Assume dia of each pipe in the loopAssume the flow in the pipe such that

sum of inflows= sum of outflowsAt any junction/node (V= V1 + V2) or (Q= Q1 + Q2)Compute head loses in each pipe by Hazen William eq

Conventionally clockwise flows are +ive & hence head loses are also positive & vice versa

With attention to sign compute the total head loses around each loop

i.e. ∑H= ∑KQ^x Compute without regard to the sign for the

same loop ∑H/Q1 or ∑Kx Q^x-1 Apply the correction obtained from the

equation

Compound pipe system

Compound pipe

Series comp pipeQ1=Q2=Q3

H=H1+H2+H3

Parallel comp pipeQ=Q1+Q2+Q3

H1=H2=H3

Thank you