1. PHYS207 Assignment 5 Open Channel Hydraulics
Q1. An old railway line near Nyngan is drained by a weed-overgrown winding earth channel roughly
0.8 m wide and with bank slopes of about 1:16. The channel is long and shallow, with a bedslope of
about 0.1 mm/m and only 400 mm deep at the lowest point. Engineers estimate the channel will need
to convey a flow of about 0.23 m3
/s in a five year storm, to avoid creating a drainage problem when
the line reopens next year. Does the channel have sufficient capacity or will it need to be cleared and
re-lined?
With no further information on channel length and downstream constraints we have to assume steady
uniform flow, and go to the Manning equation.:
n
S
P
A
n
S
ARVAQ o
/
/
o/
32
35
32
Define the terms:
A = flow area = Wy + ky2
where W is the base width (0.8 m) and k the side slope (1:16 = 1:k ; k = 16)
P = wetted perimeter = W + 2y√(1+k2
)
So = bed slope = 0.0001 m/m
n = channel roughness. Looking up the roughness for a weedy winding channel (study guide) we have
a range of 0.075 to 0.15 to choose from. Assuming the channel is at least well maintained we can use
the lower end of that range: say n = 0.075
A and P depend on the flow depth, y. We can use the maximum depth of the channel as a flow depth,
thus y = 0.4 m
A = Wy + ky2
= 0.8(0.4) + 16(0.42
) = 2.88 m2
P = W + 2y√(1+k2
) = 0.8 + 2(0.4)(1 + 162
)0.5
= 13.62 m
So at a flow depth of 400 mm the channel flowrate would be:
And Q = 2.885/3
0.00010.5
/(13.622/3
0.075) = 0.136 m3
/s
This is too small. Adjusting the roughness coefficient, we find the flowrate reaches 0.23 m3
/s only
when the roughness is at most 0.045, equivalent to a winding natural channel or a grassy formed
channel. The engineer should recommend that the channel be cleared of weeds and maintained with a
hardy native grass bed.
Q2. A concrete channel constructed from inverted box culverts is to be 250 m long and 2.7 m wide. A
maximum flowrate of 22.8 m3
/s is expected. The channel slope is 2 mm/m. Assuming steady uniform
flow, how deep does the channel need to be constructed if a freeboard of 300 mm is required?

2. Assuming steady uniform flow means “use Manning’s equation”:
n
S
P
A
n
S
ARVAQ o
/
/
o/
32
35
32
However in this case we don’t know the flow depth in advance, so we can’t calculate the area or
wetted perimeter.
Let’s use the approximate depth for wide square channels as a first guess, and see how close it is:
The roughness of manufactured concrete pipes and box sections is tabulated as 0.012 (p.161):
53/
oSW
n.Q
y = (22.8(0.012)/(2.7(0.002)0.5
))3/5
= 1.634 m
Let’s just check the flowrate that this depth gives:
A = Wy = 2.7(1.634) = 4.411 m2
P = W + 2y = 2.7 + 2(1.634) = 5.968 m
And Q = 4.4115/3
0.0020.5
/(5.9682/3
0.012) = 13.44 m3
/s
This is not even nearly right, because the channel cannot be called wide or shallow. It will do as a first
estimate though. To increase the flowrate we need to increase the depth: Say y = 2.5 m
A = Wy = 2.7(2.5) = 6.75 m2
P = W + 2y = 2.7 + 2(2.5) = 7.7 m
Q = 6.755/3
0.0020.5
/(7.72/3
0.012) = 23 m3
/s
That’s a pretty good second guess! Another iteration gives y = 2.49 m. With a freeboard of 300 mm
the channel needs to by 2.49 + 0.3 = 2.79 m deep. Pretty close to square.
Q3. At what critical depth will a 7 m wide spillway with vertical sides flow when the flowrate is 40.1
m3
/s? What depth above the spillway is the still water behind the weir?
Spillways and many other types of depth controls flow at the critical depth for a given discharge:
B
gA
Qc
3
Where A is the flow area and B is the surface flow width. For channels with vertical sides, this
reduces to
32
3
c
c
c ygB
B
Byg
Q

3. The critical discharge equals the flowrate when the flow depth y = yc:
3
2
2
gB
Q
yc = (40.12
/(9.8(72
))1/3
= 1.496 m
The still water behind the spillway has the same total energy as the spillway itself, but it is all
expressed as potential energy. The kinetic energy of the spillway is expressed as a velocity head:
H = yc + V2
/2g
Where V = Q/A = Q/(Byc) = 40.1/(7(1.496)) = 3.829 m/s
Thus H = yc + V2
/2g = 1.496 + 3.8292
/19.62 = 2.243 m
As a check, try the broad-crested weir equation:
23
LH67.1Q = 1.67(7)2.2431.5
= 39.27 m3
/s OK
Q4. A council garden is to be drained by a grassy v-shaped channel with side slopes of 1:4 and 0.7 m
deep. Assuming a maximum Froude number of 0.8, what is the channel’s maximum bedslope for
subcritical flow? What is the maximum flow velocity within the channel? Is this safe for children?
The Froude number governs the transition from subcritical to supercritical flow. Grassy channels must
be designed for subcritical flow or they will fail due to erosion. A Froude number of 0.8 is a safe
maximum:
gy
V
c
V
Fr
Where V is the normal flow velocity:
n
S
P
A
V o
/ 32
So 80
32
.
gyn
S
P
A
gy
V o
/
For a 1:4 v-shaped drain 0.7 m deep,
A = ky2
= 4(0.72
) = 1.96 m2
P = 2y√(1+k2
) = 2(0.7) √(1+42
) = 5.772 m
The roughness of a short grassy swale is about 0.03:
So
232
80
/
o
A
P
gyn.S = (0.8(0.03)(9.8(0.7))0.5
(5.772/1.96)2/3
)2
= 0.0167 m/m

4. The velocity of the channel is then
n
S
P
A
V o
/ 32
= (1.96/5.772)2/3
0.01670.5
/0.03 = 2.097 m/s
This fairly fast, and the channel is rather deep. To decide if it is unsafe for children, use the common
formula V.d < 0.4 (p.183 study guide)
V.d = 2.097(0.7) = 1.468
This is too fast and deep for a child to fall into. Try reducing the depth and putting additional flows
underground.
Q5. A rip-rap lined channel 1.4 m wide and with 1:4 bank slopes has two reaches: a 50 m long reach
at bedslope of 0.0017 m/m upstream of a 90 m long reach of bedslope 0.0003 m/m. The channel
outfalls over a stone apron of the same channel shape into a town pond which after ten years of
drought is very low. If the channel carries a flow of 1.45 m3
/s, calculate the flow depth at the apron
and at the upstream point 140 m from the weir.
The standard step method allows us to calculate flow depths in non-uniform flow situations such as
this: beginning with a critical depth calculation at the apron outfall we proceed upstream to determine
the depths at 90 m and 140 m upstream:
Let’s collate some data and formulae:
The roughness of a rip-rap channel is about 0.025
The area, wetted perimeter and surface width of a flow y are given by:
A = Wy + ky2
P = W + 2y√(1+k2
)
B = W + 2ky
At the outfall, critical depth conditions prevail:
B
gA
Qc
3
= Q
We can trial values of y that produce this critical flowrate:
Try y = 0.5:
A = 1.4(0.5) + 4(0.52
) = 1.7 m2
B = 1.4 + 2(4)0.5 = 5.4 m
Qc = (9.8(1.73
)/5.4)0.5
= 2.99 m3
/s
Try y = 0.2:

5. A = 1.4(0.2) + 4(0.22
) = 0.44 m2
B = 1.4 + 2(4)0.2 = 3.0 m
Qc = (9.8(0.443
)/3.0)0.5
= 0.527 m3
/s
Try y = 0.3:
A = 1.4(0.3) + 4(0.32
) = 0.78 m2
B = 1.4 + 2(4)0.3 = 3.8 m
Qc = (9.8(0.843
)/4.0)0.5
= 1.1 m3
/s
Try y = 0.35:
A = 1.4(0.35) + 4(0.352
) = 0.98 m2
B = 1.4 + 2(4)0.31 = 4.2 m
Qc = (9.8(0.983
)/4.2)0.5
= 1.48 m3
/s
This is close enough given the other uncertainties in the calculation.
So we have a starting depth of 0.35 m. The flow velocity at the weir is
V = Q/A = 1.45/0.98 = 1.48 m/s
The wetted perimeter is then P = W + 2y√(1+k2
) = 1.4 + 2(0.35)(1+42
)0.5
= 4.29 m
And the friction slope at the outfall is
2
32 /f
AR
nQ
S = (0.025(1.45)/(0.98(0.98/4.29)2/3
)2
= 0.0098 m/m
We can now proceed upstream to the next step. The flow will be somewhat deeper than 0.35, tending
towards the normal depth:
By trial and error, normal depth for 0.0003 m/m bedslope occurs at y = 0.78 m (solve Manning’s
equation with this value to confirm)
We can locate where y2 = 0.78 by the step method:
A = 1.4(0.78) + 4(0.782
) = 3.553 m2
P = 1.4 + 2(0.78)(1+42
)0.5
= 7.861 m
V = Q/A = 1.45/3.553 = 0.408 m/s
Sf = (0.025(1.45)/(3.553(3.553/7.861)2/3
)2
= 0.0003 m/m (= bedslope for normal depth)
The average friction slope over the section is
Sf = (0.0098 + 0.0003)/2 = 0.005 m/m

6. And the Step length is
of SS
y
g
V
y
g
V
L
1
2
1
2
2
2
22
= ((0.4082
/19.62 + 0.78) – (1.482
/19.62 + 0.35))/(0.005 – 0.0003) = 69.6 m
This is less than the reach length of 90 m, so we find the channel has established its normal depth at
69.6 m and will have this same depth at the channel section change at 90 m.
We can perform a similar calculation for the second reach:
By trial and error, normal depth for 0.0017 m/m bedslope occurs at y = 0.530 m (solve Manning’s
equation with this value to confirm)
We can locate where y2 = yo = 0.530 by the step method:
A = 1.4(0.53) + 4(0.532
) = 1.866 m2
P = 1.4 + 2(0.53)(1+42
)0.5
= 5.771 m
V = Q/A = 1.45/1.866 = 0.777 m/s
Sf = (0.025(1.45)/(1.866(1.866/5.771)2/3
)2
= 0.0017 m/m (= bedslope for normal depth)
The average friction slope over the section is
Sf = (0.0017 + 0.0003)/2 = 0.001 m/m
And the Step length is
L = ((0.7772
/19.62 + 0.53) – (0.4082
/19.62 + 0.78))/(0.001 – 0.0017) = 325 m
That is it will take 325 m for the flow to establish at normal depth. We only have 50 m, so we should
find the actual depth somewhere between 0.53 and 0.78 m depth.
Try y2 = 0.7:
A = 1.4(0.7) + 4(0.72
) = 2.94 m2
P = 1.4 + 2(0.7)(1+42
)0.5
= 7.17 m
V = Q/A = 1.45/2.94 = 0.493 m/s
Sf = (0.025(1.45)/(2.94(2.94/7.17)2/3
)2
= 0.0005 m/m
The average friction slope over the section is
Sf = (0.0005 + 0.0003)/2 = 0.0004 m/m
And the Step length is
L = ((0.4932
/19.62 + 0.7) – (0.4082
/19.62 + 0.78))/(0.0004 – 0.0017) = 61 m

7. Try y2 = 0.71:
A = 1.4(0.71) + 4(0.712
) = 3.01 m2
P = 1.4 + 2(0.71)(1+42
)0.5
= 7.255 m
V = Q/A = 1.45/3.3 = 0.482 m/s
Sf = (0.025(1.45)/(3.01(3.01/7.255)2/3
)2
= 0.00047 m/m
The average friction slope over the section is
Sf = (0.00047 + 0.0003)/2 = 0.00038 m/m
And the Step length is
L = ((0.4822
/19.62 + 0.71) – (0.4082
/19.62 + 0.78))/(0.00038 – 0.0017) = 53 m
This is near enough. The flow depth at the top of the 50 m reach is thus 0.71 m.
This laborious calculation can easily be transmitted to a spreadsheet for much easier trial and error.
Q6. A large dam has a spillway leading into a smooth concrete lined rectangular channel 70 m wide
with shallow bedslope. The spillway releases 170 m3
/s into the channel at a flow depth of 0.30 m. The
normal flow depth in the channel is 1.7 m. Locate the hydraulic jump that must occur in the channel,
the depth before the jump and the power dissipated by the channel and jump.
In this case we need to perform a supercritical standard step method and calculate the momentum
function of the flow until it equals the momentum of the downstream normal flow depth, at which
point a hydraulic jump occurs. This is a complex procedure that will benefit from application of a
spreadsheet:
Create a table of the provided data and calculate the momentum function of the normal depth:
Q 170
W 70
y1 0.3
yo 1.7
Ao 119 =W*yo
Vo 1.429 =Q/Ao
Mo 125.9 =Vo^2*Ao/9.8+yo/2*Ao
Note that the momentum function uses the centroid depth, which is half the total depth.
To start the standard step method, calculate the area, wetted perimeter, velocity and friction slope
associated with y1 (the initial channel depth):
A1 21 =W*y1
P1 70.6 =W+2*y1
V1 8.1 =Q/A1

8. Sf1 0.056 =(0.013*Q/(A1*(A1/P1)^(2/3)))^2
Now add a guessed y2 (the depth of flow after frictional losses down the channel), and calculate A, P,
V, Sf and the average Sf:
y2 0.5
A2 35 =W*y2
P2 71 =W+2*y2
V2 4.86 =Q/A2
Sf2 0.01 =(0.013*Q/(A2*(A2/P2)^(2/3)))^2
Sf 0.033 =(Sf1+Sf2)/2
Lastly include the step length calculation and the momentum function of the guessed y2:
L 58.7 =-((V2^2/19.62+y2)-(V1^2/19.62+y1))/(Sf)
M1 93 =V2^2*A2/9.8+y2/2*A2
Note that the step length function has a minus sign because in a supercritical step we move from
upstream to downstream, the reverse of the direction (downstream to upstream) by which the formula
was derived. Also note that there is no bedslope So in the denominator, as this is specified only as
“shallow” and can be treated as zero.
Now we can gradually adjust the value of y2 until the momentum function approaches that of the
normal depth yo:
y2 0.35
A2 24.5 =W*Y2
P2 70.7 =W+2*Y2
V2 6.939 =Q/A2
Sf2 0.033 =(0.013*Q/(A2*(A2/P2)^(2/3)))^2
Sf 0.045 =(SF1+SF2)/2
L 18.74 =-((V2^2/19.62+Y2)-(V1^2/19.62+Y1))/(Sf)
M1 124.7 =V2^2*A2/9.8+Y2/2*A2
Here M1 = Mo at a step length L of 18.7 m, where the flow depth is 0.35 m
The power dissipated by the apron and jump is a function of the difference in total head between the
initial depth and the normal depth:
Power = gQ H
H1 3.64 =y1+V1^2/19.62
Ho 1.80 =yo+Vo^2/19.62
Power 3058924 =9800*Q*(H1-Ho)
The total power dissipated is 3.06 MW