(See www.efm.leeds.ac.uk/CIVE/CIVE3400 for assistance)
Q1.
The flow in a rectangular channel of width 5m, slope 0.001 and a Chezy C of 25is uniform and has a depth of 1.5m. A gate at the downstream end increases the depth by 10%. How long will it take this change in depth to affect levels in the river 0.5 km upstream?
B = 5m i = 0.001
C = 25 d = 1.5m
Uniform flow v = C mi v = C
B
Bd
+ 2 d i v = 25
5
5 × 1 .
5
+ 2 × 1 .
5
0 .
001 = 0 .
765
Downstream end so have a forward characteristic v p
+ 2 c p
− v l
− 2 c l
= g
( i − j
) o as uniform flow i = j
10% increase in depth so c d p p
=
= c 2 p
9 .
81
/ g
×
=
(
1 .
5
1 .
65
× 1 .
1
)
Therefore v p
= v l
+ 2
( c l
− c p
)
=
= 4 .
023
0 .
391 m / s
To find the time for this depth change to move upstream we look at the equation of the characteristic travelling away from the boundary – i.e. a backward characteristic.
This has the equation: dx
= v − c dt dx = 500m, v = v p
, c = c p dt =
0 .
−
391
500
− 4 .
023
= 137 sec
1

Q2. At a certain instant in time the depth and velocities of flow at three points in a channel are known and give in the table below:
A 1.0 1.0 1.0
B 2.0 0.8 1.2
C 3.0 0.8 1.3
Using the method of characteristics calculate the flow conditions at point B at maximum stable time step forward in time. Channel Slope is 0.0005 and
Chezy C = 40.
NOTE: The channel slope has changed from the original handout.
P
∆ t t
W L O R E
∆ x
L
∆ x
R
∆ x ∆ x
Point A is equivalent to W, point B to O and point C to E
∆ X = 1000m
Calculate time step using dx
= v + c for point O. dt
Interpolate the values at L and R using the values at O to get ∆ X
L
and ∆ X
R
.
Approximate the energy terms for left and right characteristics at O
The solve the problem using: v
P
= v
L
+
2 v
R +
( c
L
− c
R
)
+ ∆ t c
P
= v
L
−
4 v
R + c
L
+
2 c
R
Following this procedure we get
∆ t = v o
∆ x
+ c o
= 236 .
35 s g
( i − j
)
O
2

∆ X
∆ X
R
L
=
=
(
( v v o o
+ c o
− c o
)
)
∆ t
∆ t
=
=
1000
− 621 .
8
Interpolating :
For the left points we do not need to interpolate as this is the same as the node W.
For the right v c
R
R
=
= v c o o
−
−
∆
∆ x
∆ x x
∆ x
R
R
(
( v c o o
−
− v c
E
E
)
)
=
=
0
3
.
.
8
518
The energy term at O is (assuming a wide channel so m = d)
E = g

 i −
C v o
2
2 d o


= 9 .
81


0 .
0005 −
0 .
8 2
40 2 1 .
2


= 0 .
00164
So v
P c
P d p
=
1 .
0
=
1 .
0
+ 0 .
8
2
− 0 .
8
4
= c 2 p
/ g
+
+
(
3
3 .
132
.
132
= 1 .
16 m
2
− 3 .
518
+ 3 .
518
)
+ 236 .
35 × 0 .
00164
= 3 .
372 m / s
= 0 .
9 m / s
3

Q3.
A long wide channel flows with uniform flow of 3.0 m 3 period the upstream depth increases by 25%. Calculate:
/s/m. Over a half hour i) the upstream flow at the time of this depth measurement and ii) estimate how long it will take for this depth change to reach 1km downstream.
NOTE: the original handout omitted the Chezy C = 40 and slope i = 0.001 q = 3.0 m 3 /s /m, i = 0.001
Chezy C = 40
Calculate normal depth
V
Q
= C
= AC mi mi
For wide channel m = d and Q given in flow per m width, q , so q = d n
C d n i
P q 2 t+t ∆ d n
=
3
C 2 i d n
= 1.7784 m d
R
= d n
= 1.7784 t
∆ t
O c
R
= gd = 4 .
1768 m / s ∆ x
R v
R
= 3.0/1.7784 = 1.687 m/s v
R
< c
R
so sub critical.
∆ x
Backward characteristic:
(v
P
– 2 c
P
) - (v
R
– 2 c
R
) = dt g (i-j)
Approximation the energy term at the know time level, i.e. at R then i = j
But as we have normal flow then the energy term is zero.
Therefore v p
= v
R
+ 2(c
P
– c
R
)
25% increase in depth so d p
= 1.25 d n
= 2.223 m c
P
= 4.67 m/s giving v
P
= 2.673 m/s which is still subcritical.
R E
4

q p
= d p
v p
= 5.94 m 3 /s /m
Time to reach 1km downstream:
∆ x
∆ t
= v + c
∆ t =
(
1000
2 .
673 + 4 .
67
)
= 136 .
18
136 seconds to travel downstream.
5