Analysis of Water Distribution Systems

ANALYSIS OF DISTRIBUTION SYSTEMS

Conditions to be satisfied in pipe networks:

• The Algebraic sum of pressure drops around a closed loop must be zero. (i.e. no discountinuity in pressure)

• The flow entering a junction must be equal to the flow leaving the same junction. i.e. law of continuity must be satisfied.

Pipe networks are solved by the following methods:

(1) Hardy - Cross Method

(2) Equivalent Pipe Method

 

1. HARDY-CROSS METHOD

• This method is based on the principle of "Law of Continuity". i.e. at any junction, Inflow = Outflow

• The flow in each pipe is assumed.

• A correction to the assumed flow is computed successively for each pipe loop in the network, until the correcton is reduced to an acceptable magnitude.

If Q = assumed flow, Q = actual flow, then correction A is given by sub divendr

∆ = Q – Q

• Algebraic sum of head losses in various pipes in closed loop is computed with assumed flow.

• Correction A is found for each loop

• Assumed flows in each pipe are corrected.

• Pipes common to two loops will receive both corrections with due attention to sign.

 

The Procedure adopted in Hardy Cross method is as follows:

• The Distribution system is divided into 2 or more loops, such that each pipe in network is included in atleast one loop.

• Assume any internally consistent distribution of flow. The sum of flows entering 02 2 bor any junction must be equal to the sum of flows leaving that junction. no (INFLOW OUTFLOW)

• Compute head loss in each pipe by equation or diagram. Conventionally, clockwise flows are positive and vice versa.

• With due attention to sign, compute total head loss around the circuit. EKQ2

• Compute without regard to sign, for the circuit, the sum of 

• Apply corrections obtained from Equation 1 and 2 to the flow in each pipe. Pipes common to two loops will receive both corrections with due regard to sign.

 

Problem 3.3:

Determine the distribution of flow in the pipe network. The Head loss H1 may be assumed as KQ". The flow is turbulent and pipes are rough. The Value of K for each pipe is shown in figure. Use Hardy Cross method.

Solution:

 

Problem 3.4:

Calculate the head losses and the corrected flows in the various pipes of a distribution network shown in figure. The diameters and the lengths of the pipes used are given against each pipe. Make use of Hardy-Cross method with William Hazen's formula. Compute the corrected flows after two corrections.

Solution:

 

EQUIVALENT PIPE METHOD:

• In this method, a complex network of pipes is replaced by a single hydraulically equivalent pipe.

• The equivalent pipe is one which will replace a system of pipes with equal head loss for a given flow.

The two hydraulic principles used in this method are :

(i) The Head loss through the pipes connected in series (AB-BD) is additive.

(ii) The flows (discharge) through the pipes connected in parallel (ABD-ACD) is distributed such that the head losses are identical.

For solving the pipe network, any head loss formula can be used.

Let D_E - Diameter of equivalent pipe

L_E=Length of equivalent pipe

L, D = represent length and diameter of pipes in the network

Note: For Hazen williams formula x = 4.87, a = 2.63, b = 0.54

For the flow network shown in figure.

Pipes in Series

Length of equivalent pipe for pipe AB and BD in series is given by:

Total head loss = summation of individual head loss.

Pipes in parallel (pipes ABD and ACD)

 

Problem 3.5

Find the equivalent length of 30 cm dia pipe for the network shown in figure using (a) Darcy's formula, (b) Hazen Williams formula.

Solution:

(a) Darcy's formula

(b) Hazen Williams formula

Problem 3.6:

Find the equivalent length of 30 cm dia pipe for the network shown in Figure. Use Hazen Williams formula.

Solution:

Using Hazen Williams formula [x = 4.87, n=1.85]

Let the discharge in loop BCD be Q1 and BED be _Q2

Q = Q₁ + Q₂

(a) For the equivalent pipe BD,

(b) Equivalent length for AB,

(c) Equivalent length for DF,