Piping system for residence

CVE 341 – Water Resources

Class No: 11-12

Announcements:

•PIPE NETWORK

Today’s subject:

Pipe Networks

The need to design the original network

to add additional nodes to an existing network

Two guiding principles (each loop):

1- Continuity must be maintained

2- Head loss btw 2 nodes must be

independent of the route.

0iQ

fCCfC hh

The problem is to determine

flow & pressure at each node

Pipe Networks

Consider 2-loop network: Procedure:

1- Taking ABCD loop first, Assume Q

in each line

2- Compute head losses in each pipe

& Express it in terms of Q

For the loop:

The difference is known:

Hardy Cross Method

3- If the first Q assumptions were incorrect, Compute a correction to

the assumed flows that will be added to one side of the loop & subtracted

from the other.

► Suppose we need to subtract a ∆Q from the clockwise side and add

it to the other side for balancing head losses. Then

► After applying Taylor`s series expansion & math manipulations:

4- After balancing of flows in the first loop, Move on to the next one

Hardy Cross Method --- Example 8.11

Given info:

Review (HARDY CROSS METHOD)

Express the energy loss in each pipe in the form h=rQ2

Number each of the various loops

Assume a flow direction (clockwise = positive ; counterclockwise =

negative)

and assume an initial flow through each pipe.

Calculate the head loss in each loop and h. Use the same sign

convention as above.

Calculate for each pipe and sum all values

Calculate the value of Q

For each pipe calculate a new estimate for Q from Q’=Q+ Q.

Repeat process until head losses converge to desired accuracy.

QrQh

Qr2h Qr2

Qr2

QrQ

rQ2

hQ

Iteration Loop Pipe r Q Q Corrected

Q QrQ Qr2

EXAMPLE (HARDY CROSS METHOD)

Compute the distribution of flows in the pipe network in

Figure below, where the head loss in each pipe is given

by QrQh

Ref: Water Resources Engineering by David A Chin

Solution

Assume an

initial flow

through each

pipe