distribution of water
TRANSCRIPT
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