piping system for residence
DESCRIPTION
Piping for residential areaTRANSCRIPT
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