ABE 527
Rabi H. Mohtar
Gradually Varied Flow (GVF)
Needed when normal depth cannot be maintained in a channel.
Assumption in GVF:
 Head loss within any given section calculation is the same as for uniform flow
 Velocity is the same across the entire cross-section
 Slope of conduit is <10%
 Roughness coefficient is constant in the section and is independent of the flow depth
(non-grassed waterway)
 Gradual changes in flow depth starting from controlling boundary condition. No
sudden changes.
 If section length is long, flow approaches normal depth.
Slope Classification:
Mild (M): yn  yc
Critical (C): yn  yc
Steep (S): yn  yc
Horizontal channel S=0
yn is undefined from Manning's Eq.
Adverse slope S=-ve 
Flow Zone Classification:
Zone 1:
flow depth y  yn , yc
yn , yc  y  yn , yc
Zone 2:
y  yn , yc
Zone 3:
Profile Classification:
1. Backwater analysis: y  yn M1 profile
2. Drawdown analysis: yn  y  yc between critical and normal depth. Profiles 1
and 2 had their hydraulic control at the downstream end (subcritical)
3. Front water analysis: y, yn  yc upstream control.
Energy Balance
v2
v2
Z1  1
 Z2  1
 H L 1,2 being two ends of the segment
2g
2g
Z1,2  hydraulic grade at upstream and downstream end
P1

drops, free surface water
V1 ,V2 = velocity m/sec
HL=head loss
g=m/s2 gravitational acceleration
S1  S2
x
2
S1, S2=friction slope at upstream and downstream end
Solution methods include: Standard step method or Direct step method
H L  Savg .x 
Mixed flow profiles:
Sealing conditions: such as part of the channel is flowing full (treated as pressure flow)
the other part is flowing partial and is treated as GVF
Rapidly varied flow
Abrupt changes in cross-section, slope, bends, bridge contractions, hydraulic jumps. They
are turbulent flows.
Hydraulic Jumb
S2
S1
yc
yn
Storm Sewer Applications
Sequence of storm sewer analysis:
1.
Hydrology: flow quantity
2.
Hydraulics: outlet condition (tailwater condition)
Compute hydraulic grades from outlets toward upstream inlets
Hydrology Example:
Tc=8 min
CA=1.23 acres
Inlet 1
Pipe
Tc=2 min
Inlet 2
What is the CA for the entire system?
What is D=storm duration?
Total CA=1.23 acres + 0.84 acres = 2.07 acres
Upstream time = 2 + 8 = 10 min
10> 9 min so 10 min will be used as the duration of the storm event
Hydraulic Model:
Tailwater conditions (Computations start at outlet)
Assumptions:
 Normal depth is at farthest downstream pipe (S2 profile)
 Flow depth at pipe outfall is critical as in subcritical flow to a free discharge
 The depth is set to crown top of the pipe for free outfall
 Fixed tailwater depth is specified pond/river
Tc=9 min
CA=0.84 acres
Non-Uniform Flow Computation
Equation of Motion
dE
 So  S f
dx
(1)
or
dy So  Sf

dx
1-Fr2
(1)
indifference form yields:

v2 
 y 

2g 
E
v2
 
 So  S f  So  2
X
X
C R
where
2
1.49  13
C 
 R
 n 
2
2
1
 1.49  13
C2  
 R  3530 R 3 , n=0.025
 0.025 
So = constant for uniform slope
C, R and V, and Sf are a function of y
Method of Solution (Step Method)
 Nominate a series of values of y between yc and yn, depending on the flow
classification
 Calculate E, V, C, R
 Calculate ΔX for intervals between successive values of y
Example
y
(ft)
A
(ft)
P
(ft)
R
(ft)
C2
V
2
(ft/sec ) (ft/sec)
v2
2 g (ft)
E
(ft)
v2
C2R
 v2 
 2 
 C R m
avg .1 2
Min (yc)
______
______
______
______
______
Max (yn)
.
 V2 
So   2 
 C R m
ΔE
(ft)
ΔX
(ft)
X=ΣΔΧ
(ft)
Example
Given:
Required:
Solution:
A rectangular channel that is 20 ft. wide carries a discharge of 800 ft3/s.
The channel slope is 0.003 and Manning’s n is 0.025. The channel outlets
to a free fall.
Calculate the water surface profile upstream of the outfall.
1.
We know that we have critical depth at the outfall and that for a
rectangular section:
1
 13
 Q2  3 
(800 ft )2
yc   2   
 3.68 ft
2
2 
b g 
 (20 ft) 32.2 ft / s 
2.
Now set up a tabular solution to Equation (21) starting with y=3.68
ft at the outfall.
NOTE: The calculation of ΔE is inherently imprecise since we are
calculating a small difference between two large numbers. For this
reason, the velocity head must be calculated to 3 decimal places.
ADDITIONAL EXERCISE
Calculate the backwater curve for the channel in the example above if a control structure
(dam) at the outlet maintains a flow depth of 7.5 ft for the 800 ft2/s discharge.