PIPE NETWORK

PIPE NETWORK

3 CONDITIONS NEED TO STATISFY:

1. At any junction, the flow into a junction equals the flow out of the junction.

2. Between any two junctions, the head loss is independent of the path taken

3. Each flow must satisfy the friction law in pipe

         

-First, we need to GUESS the values for the flow in network. For example, if Q7 enters a junction, and Q6 and Q4 leave the same junction. The initial guess Q7=Q6+Q4. 

First condition

A loop is considered. Next, we need to evaluate our second condition.

Given a starting node, we work our way around the loop in a clockwise fashion, as illustrated by Loop 1.

We add up the head losses according to the Darcy–Weisbach equation for each pipe if Q is in the same direction as our loop like Q1, and subtract the head loss if the flow is in the reverse direction, like Q4.

In order to satisfy the second condition, we should end up with 0 about the loop if the network is completely solved. If the actual sum of our head loss is not equal to 0, then we will adjust all the flows in the loop by an amount given by the following formula, where a positive adjustment is in the clockwise direction.

Where

where

• n is 1.85 for Hazen-Williams and
• n is 2 for Darcy–Weisbach.

The clockwise specifier (c) means only the flows that are moving clockwise in our loop, while the counter-clockwise specifier (cc) is only the flows that are moving counter-clockwise.

EXAMPLE 1

For this example, the in and out flows will be 10 liters per second. We will consider n to be 2, and the head loss per unit flow r, and initial flow guess for each pipe as follows:

Pipe	Q12	Q13	Q23	Q24	Q34
r	1	5	1	5	1
Q guess (L/s)	5	5	0	5	5

1. The initial guesses are set up so that continuity of flow is maintained at each junction in the network.

2. The loops of the system are identified as loop 1-2-3 and loop 2-3-4.

3. The head losses in each pipe are determined.

Loop 1-2-3	Q12	Q13	Q23
Head loss =	25	125	0
Direction	Clockwise	Counter-clockwise	Clockwise

For loop 1-2-3, the sum of the clockwise head losses is 25 and the sum of the counter-clockwise head losses is 125.

Loop 2-3-4	Q23	Q24	Q34
Head loss =	0	125	25
Direction	Counter-clockwise	Clockwise	Counter-clockwise

Pipe	Q12	Q13	Q23	Q24	Q34
Original Q	5	5	0	5	5
Change in flow	1.66	-1.66	Total of change in flow :	-1.66	+1.66
Q (L/s)	6.66	3.33	3.33	3.33	6.66

The process then repeats from step 3 until the change in flow becomes sufficiently small or goes to zero.

4.     The total lead loss in Loop 1-2-3 is

Loop 1-2-3	Q12	Q13	Q23
Head loss =	44.4	55.5	11.1
Direction	Clockwise	Counter-clockwise	Clockwise

Total head loss in clockwise direction = 44.1+11.1 =55.5

Total head loss in counter-clockwise direction=55.5

Clockwise head loss= Counter clockwise head loss

This means that the flow in this loop is balanced and the flow rates are correct. The total head loss in loop 2-3-4 will also be balanced (again due to symmetry).

Loop 2-3-4	Q23	Q24	Q34
Head loss =	11.1	55.5	44.4
Direction	Counter-clockwise	Clockwise	Counter-clockwise

Total head loss in clockwise direction= 55.5

Total head loss in counter-clockwise direction= 11.1 + 44.4 =55.5

In this case, the method found the correct solution in one iteration. For other networks, it may take multiple iterations until the flows in the pipes are correct or approximately correct.

EXAMPLE 2

Find discharge Q in every water pipes for the water supply system by using Hardy-Cross method and Darcy Weisbach equation. Neglect minor losses in pipes:

Pipe	AB	BC	CA
Length (mm)	600	600	200
Diameter (mm)	250	150	100
Friction factor, f	0.0039	0.0044	0.0051

Since dQ₂ =  0, the Q₂ is correct. Pipe AB has 8.4566 m3/s flow. Pipe AC has 1.5466 m3/s flow (counter-clockwise direction) and pipe CB has 1.4566 m3/s flow.