The following points highlight the top two methods used for the analysis of flow in a pipe network. The methods are: 1. Hardy Cross Method 2. Equivalent Pipe Method.

Method # 1. Hardy-Cross Method:

In this method, the corrections are applied to the assumed flow in each successive trial. The head loss in each pipe is determined by pipe flow formula. The successive corrections are made in the flows in each pipe until the heads are balanced and the principle of continuity is satisfied at each junction.

Now if Qa be the assumed flow in a pipe and Q be the actual flow in that pipe, then correction will be given by the relation

If the head loss in the pipe under reference is Ht, it can be determined by the formula

In this equation the numerator is obtained by the algebraic sum of the head losses in the various pipes of the closed loop. As a common practice + ve sign is given to the head losses in clockwise direction and – ve signs to those in the anti-clockwise direction. The value of x is taken 1.85 (as per Hazen Williams formula) in this method known as Hardy Cross Method.

The minor losses are usually neglected, although they can be calculated by substituting an equivalent length of pipe. In case of network of pipes having many loops, the system must be divided into two or more loops, such that each pipe in the network is included in the circuit of one loop.

Following solved example will clearly illustrate the use of this method in the design of distribution system:

ADVERTISEMENTS:

Example 1:

Determine the sizes of the pipes in the network given in Fig. 20.4. The average water is to be supplied at 200 litres/day/capita. The maximum rate of supply is 2.7 times the average demand.

Solution:

The flow discharge with their direction and magnitude are now to be assumed in all the pipes, keeping in consideration the law of continuity (i.e., the input must be equal to the off take at each junction). The assumed flows have been shown in Fig. 20.5. Water demand at point D is satisfied by flowing water along line.4BD and ACD.

The sizes of the pipes are now to be chosen such that the velocity in each pipe remains between 0.9 to 1.15 m/sec. The total head loss in line ABD and ACD should remain approximately equal and less than the available head approx. 15 m.

The pipe diameters are now chosen by common sense as shown in Fig. 20.5 and table 20.3 (First Trial),

ADVERTISEMENTS:

Using Hardy Cross equation, and table 20.3,

The correction is small and hence the assumed diameters of the pipes are O.K.

Example 2:

Calculate the head losses and the corrected flows in the various pipes of a distribution network shown in fig. 20.6. The diameters and the lengths of the pipes used are given against each pipe. Make use of Hardy-cross method with Hazen-William’s formula. Compute the corrected flows after two corrections.

Solution:

First of all, the magnitudes as well as the directions of the possible flows in each pipe are assumed keeping in consideration the law of continuity at each junction (i.e., input equals the output at each junction). These assumed flows are shown in fig. 20.7. The two closed loops i.e., ABCDA and DCFED are then analysed by Hardy-cross method.

The analysis of the pipe loops will require computation of head loss (H2) in each pipe, which is to be computed by Hazen’s William’s formula as below:

For all the pipes of both the loops, we first compute values of k, as given in Table 20.4 below:

Computation of 'K' Values for Pipes of Network

Now, both the loops are analysed for Hardy Cross method procedure in Table 20.5.

Hardy Cross Procedure for First Correction

These discharges are now again used to re-analyse both the loops for the second correction in Table 20.6.

Handy Cross Procedure for Second Correction

Note:

The corrected flow in common pipe CD = (-) 8 + (-) 1.3 – (+) 0.5 = (-) 9.8 1/sec.

Hence, corrected flows after second correction, are:

These flows are marked on the given pipe network in fig. 20.18:

Method # 2. Equivalent Pipe Method:

This method is sometimes used as an aid in solving large networks of pipes, in which it becomes convenient to; first of all, replace the different small loops by single equivalent pipes having the same head loss.

In this method, pipe circuit can be reduced into a single equivalent pipe of using the following two principles of hydraulics:

(i) The loss of head caused by a given flow of water through the pipes connected in series, is additive.

(ii) The quantity of discharge flowing through the different pipes connected in parallel will be such as to cause equal head loss through each pipe.

The use of this method can be under stood with the help of example illustrated below:

Example 3:

Solve the following problem as illustrated in fig. 20.9 by equivalent pipe method using Hazen-William s formula.

Solution:

As Loop ABCD represents a pipe circuit with the pipe sizes and lengths, as indicated in fig. 20.9.

As a flow of 60 1/s be assumed to be flowing through the routes ABC and ADC. Using Hazen-William’s nomogram, the head losses into the two circuits will be,

Now, if the routes ABC and ADC are replaced by 300 mm. dia. pipes each, then the length of each pipe required can be obtained from the fact that a 300 mm dia. pipe offers a head loss of 4.5 m per 1000 m length.

... Length of 300 dia pipe required for replacing route ABC,

Similarly, length of 300 mm dia pipe required for replacing route ADC,

Moreover, the loss of head via both the routes should be the same, provided that the flows are properly divided. Hence, if a loss of head of 7.95 m is to occur via route ADC also (i.e., 300 mm dia pipe of 1865 m length)

The rate of loss of head per 1000 m will be,

In a 300 mm dia. pipe, such a loss (i.e. 4.25 m per 1000 m) will occur only if the discharge is 56 l/s (from Hazen-William’s monogram). Therefore, the circuits ABC and ADC can be replaced by an equivalent pipe carrying a total discharge of 60 + 56 = 116 l/s.

Now, for this equivalent pipe having a dia. of300 mm and a discharge of 116 l/s, the rate of loss of head per 1000 m is found from nomogram as 15 m. In order to ensure a head loss of 7.95 m in this equivalent pipe also, the length of this pipe should be,

Hence, the total loop ABCD can thus be replaced by a single equivalent pipe of300 mm dia and 530 m length, irrespective of the flow discharges.