Transcript Design of a Pipe

Design of a Pipe
7
101
1
Runoff 102
8
2
3
4
102
9
10
103
11
5
6
In a tree network, each node can have only one outflow
link. Therefore we use the convention that link
numbers are the same as the upstream node number.
Get the Maximum Inflow
If no inflow hydrograph exists the user
can specify a peak flow for the design
Use Hydrograph|Add Runoff to update
Inflow hydrograph
Uniform Flow in Pipes
2
1
M
3
Q
AR S 0 2
n
D2
  sin  
A
8
PD

2
D

y  1  cos 
2
2
Solve for y0 using
3
 2 Q  5
 
f      sin   
Q 
 full 
2
5
0
Critical Depth in Pipes
Solution for Ycr is based on the
minimum energy criterion
A3 Q 2
f y 

0
T
g
T  2 Dy  y 2 

 T 

 2 tan 
 D  2y 
1
Q 2T
1
3
gA
A Trial Pipe Design
Table of feasible
designs for given Q
and ‘n’
Double click on a row
to test trial design
Click [Design] to
get results of partfull flow analysis
Surcharged Pipes
Due to closed top boundary resistance increases
as depth y approaches diameter D.
At y = 0.81963 D
Q = Qfull
1.2
Q/Qfull
1
0.8
When y = 0.93815 D
Q = 1.07571 Qfull.
0.6
0.4
0.2
0
0
0.5
y/D
1
Surcharged Pipes
Energy line
Q > Qfull
Water surface
Q = Qfull
Q < Qfull
MIDUSS 98 assumes uniform flow for
part-full pipes
Exercise 4
Design a pipe to carry 2 c.m/s when running
75% full with a gradient of 0.4% and n =
0.013
Check for surcharged hydraulic grade line if
discharge increases to 3 c.m/s