answers to questions in the notes autumn...
TRANSCRIPT
Hydraulics 3 Answers - 1 Dr David Apsley
ANSWERS TO QUESTIONS IN THE NOTES AUTUMN 2017
Section 1.2
Example.
The discharge in a channel with bottom width 3 m is 12 m3 s
–1. If Manning’s n is
0.013 m-1/3
s and the streamwise slope is 1 in 200, find the normal depth if:
(a) the channel has vertical sides (i.e. rectangular channel);
(b) the channel is trapezoidal with side slopes 2H:1V.
b = 3 m (base width)
Q = 12 m3 s
–1
n = 0.013 m–1/3
s
S = 0.005
(a)
Discharge:
VAQ
where, in normal flow,
2/13/21SR
nV h , bhA ,
bh
h
hb
bhRh
/212
Hence,
2/1
3/2
3/5
)/21(
1S
bh
bh
nQ
Rearranging as an iterative formula for h:
5/2
5/3
)/21( bhSb
nQh
Here, with lengths in metres,
5/2)3/21(8316.0 hh
Iteration (from, e.g., h = 0.8316) gives
hn = 1.024 m
Answer: normal depth = 1.02 m.
(b) Geometry: trapezoidal cross-section with base width b, surface width )2(2 hb and
two sloping side lengths 5)2( 22 hhh .
Area and wetted perimeter:
)/21()2()4(21 bhhbhbhhhbbA
52hbP
Hydraulic radius:
bh
bhh
hb
hbh
P
ARh
/521
/21
52
)2(
Hydraulics 3 Answers - 2 Dr David Apsley
Discharge:
ASRn
VAQ h
2/13/21
Hence,
)/21(/521
/211 2/1
3/2
3/2 bhhbSbh
bhh
nQ
3/2
3/53/5
)/521(
)/21(
bh
bhh
Sb
nQ
bh
bh
Sb
nQh
/21
)/521( 5/25/3
Here, with lengths in metres,
3/21
)491.11(8316.0
5/2
h
hh
Iteration (from, e.g., h = 0.8316) gives
hn = 0.7487 m
Answer: normal depth = 0.749 m.
Hydraulics 3 Answers - 3 Dr David Apsley
Section 1.4
Example.
The discharge in a rectangular channel of width 6 m with Manning’s n = 0.012 m–1/3
s is
24 m3 s
–1. If the streamwise slope is 1 in 200 find:
(a) the normal depth;
(b) the Froude number at the normal depth;
(c) the critical depth.
State whether the normal flow is subcritical or supercritical.
b = 6 m
n = 0.012 m–1/3
Q = 24 m3 s
–1
S = 0.005
(a)
Discharge:
VAQ
where, in normal flow,
2/13/21SR
nV h , bhA ,
bh
h
hb
bhRh
/212
Hence,
2/1
3/2
3/5
)/21(
1S
bh
bh
nQ
or, rearranging as an iterative formula for h:
5/2
5/3
)/21( bhSb
nQh
Here, with lengths in metres,
5/2)3/1(7926.0 hh
Iteration (from, e.g., h = 0.7926) gives
hn = 0.8783 m
Answer: normal depth = 0.878 m.
(b) At the normal depth, h = 0.8783 m:
1sm554.48783.06
24
A
QV
551.18783.081.9
554.4Fr
gh
V
Answer: Froude number = 1.55.
Hydraulics 3 Answers - 4 Dr David Apsley
(c) The critical depth is that depth (at the given flow rate) for which Fr = 1. It is not normal
flow, and does not depend on the slope S or the roughness n.
gh
VFr
where
A
QV (in general)
or
h
q
bh
QV (for a rectangular channel; q is the flow per unit width)
Hence, for a rectangular channel,
3
222 )/(
Frgh
q
gh
hq
For critical flow, Fr = 1 and so
3/12
g
qhc
Here, the flow per unit width is 12 sm46/24 q , so that
m177.181.9
43/1
2
ch
Answer: critical depth = 1.18 m.
The normal depth is supercritical because, when h = hn, then Fr > 1 (part (b)).
Alternatively (and often more conveniently), the normal depth is supercritical because
hn < hc; then speed V is larger, and depth h is smaller in normal flow than critical flow, so that
ghV /Fr must be greater than 1.
Hydraulics 3 Answers - 5 Dr David Apsley
Section 2.2
Example.
A 3-m wide channel carries a total discharge of 12 m3 s
–1. Calculate:
(a) the critical depth;
(b) the minimum specific energy;
(c) the alternate depths when E = 4 m.
b = 3 m
Q = 12 m3 s
–1
(a)
Discharge per unit width:
12 sm43
12 b
Then, for a rectangular channel:
m177.181.9
43/1
23/1
2
g
qhc
Answer: critical depth = 1.18 m.
(b) For a rectangular channel,
m766.1177.12
3
2
3 cc hE
Answer: minimum specific energy = 1.77 m.
(c) As E > Ec, there are two possible depths for a given specific energy.
g
VhE
2
2
, where h
q
A
QV (for a rectangular channel)
2
2
2gh
qhE
Substituting values in metre-second units:
2
8155.04
hh
For the subcritical (slow, deep) solution, the first term, associated with potential energy,
dominates, so rearrange as:
2
8155.04
hh
Iteration (from, e.g., h = 4) gives h = 3.948 m.
For the supercritical (fast, shallow) solution, the second term, associated with kinetic energy,
dominates, so rearrange as:
Hydraulics 3 Answers - 6 Dr David Apsley
h
h
4
8155.0
Iteration (from, e.g., h = 0) gives h = 0.4814 m.
Answer: alternate depths are 3.95 m and 0.481 m.
Hydraulics 3 Answers - 7 Dr David Apsley
Section 2.3.1
Example. (Exam 2009 – reworded and extended)
(a) Define specific energy and explain its relevance in determining critical conditions in a
channel flow.
A long, wide channel has a slope of 1:1000, a Manning’s n of 0.015 m–1/3
s and a discharge of
5 m3 s
–1 per metre width.
(b) Calculate the normal depth.
(c) Calculate the critical depth.
(d) In a region of the channel the bed is raised by a height of 0.5 m over a length
sufficient for the flow to be parallel to the bed over this length. Determine the depths
upstream, downstream and over the raised bed, ignoring any friction losses. Sketch
the flow, including gradually-varied flow upstream and downstream.
(e) In the same channel, the bed is lowered by 0.5 m from its original level. Again,
determine the depths upstream, downstream and over the lowered bed, ignoring any
friction losses. Sketch the flow.
(a) Specific energy is the head relative to the local bed of the channel and is given by
g
VhE
2
2
(Compare the total head, where the potential-energy part is the level of the free surface
relative to an absolute datum: g
VzH s
2
2
.)
Critical conditions (where Fr = 1) correspond to a minimum of specific energy at the given
discharge.
(b)
S = 0.001
n = 0.015 m–1/3
s
q = 5 m2 s
–1
Discharge per unit width:
Vhq where 2/13/21SR
nV h (Manning), hRh (“wide” channel)
hShn
q 2/13/21
n
Shq
3/5
m679.1001.0
5015.05/35/3
S
nqhn
Answer: normal depth = 1.68 m.
Hydraulics 3 Answers - 8 Dr David Apsley
(c) Critical depth:
m366.181.9
53/1
23/1
2
g
qhc
Answer: critical depth = 1.37 m.
(d) To determine the type of behaviour over the raised bed, compare the total head under
critical conditions (the minimum energy necessary to get over the weir at this flow rate) with
that available in the approach flow.
Critical
m366.1ch
m049.223 cc hE
m5.0weirz
m549.2 cweirc EzH
Approach Flow
Because the channel is described as “long” it will have sufficient fetch to develop normal
flow; hence the approach-flow head is that for the normal depth (h = 1.679 m):
m131.2
679.181.92
5679.1
2
2
2
2
2
2
2
n
n
n
naa
gh
qh
g
VhEH
At the normal depth the available head (Ha) is less than the minimum required to get over the
weir (Hc). Hence the water depth must increase upstream (“backing up”), to raise the head
immediately upstream of the weir. Thus:
critical conditions do occur;
the total head through the device is the critical head (H = Hc = 2.549 m).
Over the weir there is a critical flow transition, so the depth here is critical: h = hc = 1.366 m.
Just up- or downstream,
g
VhEH
2
2
where h
qV
2
2
2gh
qhH
2
274.1549.2
hh
Hydraulics 3 Answers - 9 Dr David Apsley
Upstream, rearrange for the deep, subcritical, solution:
2
274.1549.2
hh
Iteration (from, e.g., h = 2.549) gives h = 2.310 m.
Downstream, rearrange for the shallow, supercritical solution:
h
h
549.2
274.1
Iteration (from, e.g., h = 0) gives h = 0.8715 m.
normal
normal
hydraulicjump
hnch
1h
2h hn
GVF
GVF
RVF
Answer: depths upstream, over, downstream of the weir: 2.31 m, 1.37 m, 0.871 m.
(e) The flow does not require additional energy to pass a depressed section; hence, the total
head throughout is that supplied by the approach flow (H = Ha = 2.131 m) and the flow
remains subcritical. The depths just upstream and downstream of the lowered section are
those in the approach flow; i.e. normal depth.
As bed height zb decreases, specific energy E must increase to maintain the same total head.
In the lowered section:
EzH b
E 5.0131.2
m631.2E
Then
g
VhE
2
2
where h
qV
2
2
2gh
qhE
2
274.1631.2
hh
As we require the subcritical solution, rearrange as
2
274.1631.2
hh
Iteration (from, e.g., h = 2.631) gives h = 2.412 m.
(Note that this is the depth of the water column. The actual surface level here is
Hydraulics 3 Answers - 10 Dr David Apsley
m912.15.0 hzs
so the overall water level also rises in this section.)
normal
normal
hn
hn
RVF
Answer: depths upstream, within, downstream of the lowered section: 1.68 m, 2.41 m,
1.68 m.
Hydraulics 3 Answers - 11 Dr David Apsley
Section 2.3.1
Example.
A long channel of rectangular cross-section with width 3.5 m and streamwise slope 1 in 800
carries a discharge of 15 m3 s
–1. Manning’s n may be taken as 0.016 m
–1/3 s. A broad-crested
weir of height 0.7 m is constructed at the centre of the channel. Determine:
(a) the depth far upstream of the weir;
(b) the depth just upstream of the weir;
(c) whether or not a region of supercritical gradually-varied flow exists downstream of
the weir.
b = 3.5 m
S = 0.00125
Q = 15 m3 s
–1
n = 0.016 m–1/3
s
zweir = 0.7 m
(a) The depth far upstream is normal since the channel is described as “long”. For normal
flow in a rectangular channel:
VAQ
where:
2/13/21SR
nV h , bhA ,
bh
h
hb
bhRh
/212
Hence,
2/1
3/2
3/5
)/21(
1S
bh
bh
nQ
or, rearranging as an iterative formula for h:
5/2
5/3
)/21( bhSb
nQh
Here, with lengths in metres,
5/2)5714.01(488.1 hh
Iteration (from, e.g., h = 1.488) gives
h = 2.023 m
Answer: depth far upstream = 2.02 m.
(b) To establish depths near the weir we need to know the flow behaviour at the weir.
Compare the energy in the approach flow with that under critical conditions.
Approach flow
h = 2.023 m (from part (a))
1sm118.2023.25.3
15
bh
Q
A
QV
Specific energy in the approach flow:
Hydraulics 3 Answers - 12 Dr David Apsley
m252.281.92
118.2023.2
2
22
g
VhEa
Referring heads to the undisturbed bed near the weir:
m252.2 aa EH
Critical conditions
3/12
g
qhc where 12 sm286.4
5.3
15 b
m233.181.9
286.43/1
2
ch
m850.1233.12
3
2
3 cc hE
m550.2850.17.0 cweirc EzH
Since the head required to flow over the weir (Hc = 2.550 m) exceeds that in the approach
flow (Ha = 2.252 m), the depth just upstream of the weir must increase and the flow back up.
The total head at any position in the vicinity of the weir is H = Hc = 2.550 m.
Just upstream and downstream of the weir (i.e. at undisturbed bed level):
g
VhEH
2
2
, with bh
QV
22
2
2 hgb
QhH
2
9362.0550.2
hh (*)
The depth just upstream is the deep, subcritical solution. Hence, rearrange as
2
9362.0550.2
hh
Iteration (from, e.g., h = 2.550) gives
h = 2.385 m
Answer: depth just upstream of the weir = 2.39 m.
(c) Since the normal flow is subcritical, the flow must return to it via a hydraulic jump on the
downstream side of the weir.
If the flow in the vicinity of the weir is unaffected by the hydraulic jump the flow goes
smoothly supercritical on the downstream side, with total head H = 2.550 m (equation (*)).
Rearranging to get an iterative formula for the supercritical solution:
h
h
550.2
9362.0
Iteration (from, e.g., h = 0) gives
Hydraulics 3 Answers - 13 Dr David Apsley
m7141.0h
Denote by subscripts A and B respectively the conditions upstream and downstream of the
hydraulic jump. On the downstream side conditions may be assumed normal, since the
channel is “long” and hence there is sufficient fetch to develop the preferred depth:
m023.2Bh
1sm118.2 BV (from part (b))
4754.0023.281.9
118.2Fr
B
BB
gh
V
Hence, from the hydraulic-jump relation for the sequent depths:
m6835.0)4754.0811(2
023.2)Fr811(
2
22 BB
A
hh
Any gradually-varied supercritical flow downstream of the weir would increase in depth until
a hydraulic jump occurred (see the lectures on GVF). Since the depth downstream of the weir
is already greater than any sequent depth upstream of the hydraulic jump, no such increasing-
depth GVF is possible and the hydraulic jump must actually occur at (or just before) the
downstream end of the weir.
Hydraulics 3 Answers - 14 Dr David Apsley
Section 2.3.1
Example.
A reservoir has a plan area of 50 000 m2. The outflow passes over a broad-crested weir of
width 8 m and discharge coefficient 0.9. Calculate:
(a) the discharge when the level in the reservoir is 0.6 m above the top of the weir;
(b) the time taken for the level of water in the reservoir to fall by 0.3 m.
(a)
Total head:
Ezg
VzH bs
2
2
where levels z can be measured relative to any convenient datum. Relative to the top of the
weir, assuming constant head, still water in the reservoir and critical conditions over the weir:
head upstream = head over weir
3/12
02
3
g
qh
Hence, in ideal flow,
2/3
0
2/32/1 )3/2( hgq
2/3
0
2/3
0
2/32/1 705.1)3/2( bhbhgqbQideal (in metre-second units)
Representing non-ideal behaviour via a discharge coefficient cd, and taking b = 8 m,
2/3
08705.19.0 hQcQ ideald
2/3
028.12 hQ
When h0 = 0.6 m,
132/3 sm707.56.028.12 Q
Answer: initial discharge = 5.71 m3 s
–1.
(b) Drop the subscript 0 and write the freeboard as h.
Consider the change in volume of a tank, water surface area Aws. When the water level
changes by dh the change in volume is
hAvolume wsd)(d
Hence, by continuity,
outin QQvolumet
)(d
d
2/328.120d
dh
t
hAws
2/328.12d
d50000 h
t
h
Separating variables,
Hydraulics 3 Answers - 15 Dr David Apsley
th
hd28.12
d50000
2/3
Apply boundary conditions h = 0.6 when t = 0 and h = 0.3 when t = T, and integrate:
T
thh0
3.0
6.0
2/3 d28.12d50000
Th
28.12)2/1(
50000
3.0
6.0
2/1
T
6.0
1
3.0
1
28.12
250000
T = 4355 s
Answer: time = 4360 s (about 73 min).
Hydraulics 3 Answers - 16 Dr David Apsley
Section 2.3.2
Example (Exam 2008 – modified, including a change in the value of n)
A venturi flume is placed near the middle of a long rectangular channel with Manning’s
n = 0.012 m–1/3
s. The channel has a width of 5 m, a discharge of 12.5 m3 s
–1 and a slope of
1:2500.
(a) Determine the critical depth and the normal depth in the main channel.
(b) Determine the venturi flume width which will just make the flow critical at the
contraction.
(c) If the contraction width is 2 m find the depths just upstream, downstream and at the
throat of the venturi flume (neglecting friction in this short section).
(d) Sketch the surface profile.
n = 0.012 m
–1/3 s
b = 5 m (main channel)
Q = 12.5 m3 s
–1
S = 410–4
(a)
In the main channel,
12 sm5.25
5.12 b
Critical Depth
m8605.081.9
5.23/1
23/1
2
g
qhc
Normal Depth
VAQ
where:
2/13/21SR
nV h , bhA ,
bh
h
hb
bhRh
/212
Hence,
2/1
3/2
3/5
)/21(
1S
bh
bh
nQ
or, rearranging as an iterative formula for h:
5/2
5/3
)/21( bhSb
nQh
Here, with lengths in metres,
5/2)4.01(275.1 hh
Iteration (from, e.g., h = 1.275) gives
hn = 1.546 m
Answer: critical depth = 0.860 m; normal depth = 1.55 m.
Hydraulics 3 Answers - 17 Dr David Apsley
(b) The flow will just go critical if the head in the throat (Hc) is exactly equal to that in the
approach flow (Ha). Measure heads relative to the bed of the channel in the vicinity of the
venturi.
Critical Head
3/12
2
3
2
3
g
qhEH m
ccc where m
mb
(Note that the critical depth is different at the throat to that in the main channel, due to the
narrower width.)
3/1
2
2
2
3
m
cgb
QH
Approach Flow
The approach flow is normal, since the channel is “long”. Hence,
m546.1ah
1sm617.1546.15
5.12
a
abh
QV
m679.181.92
617.1546.1
2
22
g
VhEH a
aaa
For the flow just to go critical at the throat,
ac HH
679.12
33/1
2
2
mgb
Q
m370.3
81.9679.13
2
5.122/3
mb
Answer: throat width = 3.37 m.
(c) If the throat width is reduced further, then the flow will back up and undergo a critical
transition at the throat.
At the throat,
12 sm25.62
5.12 m
mb
m585.181.9
25.6
2
33/1
23/1
2
g
qh m
c
m378.2585.123
23 cc hE
m378.2378.20 cbc EzH
Hydraulics 3 Answers - 18 Dr David Apsley
The head throughout the venturi will be the critical head (H = Hc = 2.378 m).
Anywhere in the flume,
g
VzH s
2
2
where hzs , bh
QV
22
2
2 hgb
QhH
At the throat the depth will be the critical depth there; i.e. h = hc = 1.585 m.
Just upstream and downstream, b = 5 m; hence,
2
3186.0378.2
hh
Upstream
Rearrange for the deep, subcritical solution:
2
3186.0378.2
hh
Iteration (from, e.g., h = 2.378) gives h = 2.319 m.
Downstream
Rearrange for the shallow, supercritical solution:
h
h
378.2
3186.0
Iteration (from, e.g., h = 0) gives h = 0.4015 m.
Answer: depths upstream, in the throat, downstream = 2.32 m, 1.59 m, 0.401 m.
(d)
bmin
critical
PLAN VIEW
WATER PROFILE
Hydraulics 3 Answers - 19 Dr David Apsley
Section 2.3.3
Example.
The water depth upstream of a sluice gate is 0.8 m and the depth just downstream (at the vena
contracta) is 0.2 m. Calculate:
(a) the discharge per unit width;
(b) the Froude numbers upstream and downstream.
h1 = 0.8 m
h2 = 0.2 m
(a) Assuming total head the same on either side of the gate:
g
Vz
g
Vz ss
22
2
22
2
11
Substituting zs = h and h
qV :
2
2
2
22
1
2
122 gh
qh
gh
qh
From the given data, in metre-second units:
22 2742.12.00796.08.0 qq
21946.16.0 q
12 sm7087.0 q
Answer: discharge per unit width = 0.709 m2 s
–1.
(b) Use, on each side of the gate,
h
qV
gh
VFr
to get
1
1 sm8859.0 V
1
2 sm544.3 V
and then
3162.0Fr1
530.2Fr2
Answer: Froude numbers upstream, downstream = 0.316, 2.53.
Hydraulics 3 Answers - 20 Dr David Apsley
Section 2.3.3
Example.
A sluice gate controls the flow in a channel of width 2 m. If the discharge is 0.5 m3 s
–1 and
the upstream water depth is 1.5 m, calculate the downstream depth and velocity.
b = 2 m
Q = 0.5 m3 s
–1
h1 = 1.5 m
Use upstream conditions to get total head. Then, assuming no losses, find the supercritical
flow with the same head.
Total head (either side):
g
VzH s
2
2
where hzs and bh
QV
2
3
22
2 10186.3
2 hh
hgb
QhH
The upstream depth h1 = 1.5 m gives
m501.15.1
10186.35.1
2
3
H (dominated by h1)
Hence,
2
310186.3501.1
hh
Rearrange for the shallow, supercritical solution:
h
h
501.1
10186.3 3
Iteration (from, e.g., h = 0) gives
m04681.02 h
1
2
2 sm341.504681.02
5.0
bh
QV
Answer: downstream depth = 0.0468 m; velocity = 5.34 m s–1
.
Hydraulics 3 Answers - 21 Dr David Apsley
Section 2.4
Example. (Exam 2007)
Water flows down a wide channel with a steep slope of 1:50 on to a mild slope of 1:1000.
The flow rate is 0.645 m3 s
–1 per metre width and the Manning’s n on both slopes is
0.015 m-1/3
s. Blocks are placed just downstream of the change in slope to induce a hydraulic
jump with a sequent depth equal to the normal depth of the smaller slope.
(a) Making the wide channel assumption, determine the force per metre width which the
blocks must impart.
(b) If the blocks are not present, explain how the hydraulic jump will change position.
Sketch the surface profile (but do not calculate the position of the jump).
S1 = 0.02; S2 = 0.001
q = 0.645 m2 s
–1
n = 0.015 m–1/3
s
(a) Find depths (assumed normal) and velocities; then use the momentum principle to deduce
the force on the blocks.
Normal flow:
Vhq where 2/13/21SR
nV h (Manning), hRh (“wide” channel)
hShn
q 2/13/21
n
Shq
3/5
5/3
S
nqh
For the two slopes this gives
h1 = 0.2000 m
h2 = 0.4913 m
(Note that the critical depth is
m3487.081.9
645.03/1
23/1
2
g
qhc
This confirms that the normal flows are supercritical on the first slope and subcritical on the
second; i.e. the first slope is mild and the second one steep.)
The corresponding velocities are deduced from h
qV , whence:
V1 = 3.225 m s–1
V2 = 1.313 m s–1
Hydraulics 3 Answers - 22 Dr David Apsley
Let f be the magnitude of the force per unit width exerted by the fluid on the blocks and, by
reaction, the blocks on the fluid. The force on the fluid clearly acts in the upstream direction.
On each side of the blocks the hydrostatic pressure force is given by areapressureaverage
or
hghρ21 (per unit width)
Hence, from the steady-state momentum principle:
force = rate of change of momentum (mass flux change in velocity)
)(ρρρ 12
2
2212
121 VVqghghf
2.12337.987 f
f = 245.5 N
Answer: force (per metre width) = 246 N.
(b) If blocks are not present a hydraulic jump will still occur (as the flow must get from
supercritical to subcritical); however, it will move further downstream. To see this, observe
the following.
The depth at the change of slope is supercritical. On a mild slope, any supercritical flow must
increase in depth; (you will prove this in the lectures on GVF). To show that the hydraulic
jump takes place downstream (as opposed to immediately) we must therefore show that the
sequent depth just upstream of the hydraulic jump (call it hA) exceeds the depth at the change
of slope (0.2000 m), so that there is room for GVF of increasing depth.
change of slope
hydraulic jump
GVF
Downstream of the jump (normal flow for the milder slope):
hB = 0.4913 m
1sm313.14913.0
645.0 B
Bh
qV
5981.04913.081.9
313.1Fr
B
BB
gh
V
The sequent depth upstream of the jump is then, by formula,
m2371.0)5981.0811(2
4913.0)Fr811(
2
22 BB
A
hh
This exceeds the depth at the change of slope (0.2000 m), so the hydraulic jump takes place
further downstream.
Hydraulics 3 Answers - 23 Dr David Apsley
Section 2.4
Example.
A downward step of height 0.5 m causes a hydraulic
jump in a wide channel when the depth and velocity
of the flow upstream are 0.5 m and 10 m s–1
,
respectively.
(a) Find the downstream depth.
(b) Find the head lost in the jump.
(a) The downstream depth can be deduced from the momentum principle if the reaction force
from the step is known. The approximation is that this is the same as would occur if it were in
equilibrium with a hydrostatic pressure distribution here.
Flow rate per unit width:
12
11 sm55.010 hVq
Steady-state momentum principle
force = rate of change of momentum (mass flux change in velocity)
Per unit width:
)(ρρ)Δ(ρ 12
2
2212
121 VVqghhg
Since 22
2
5
hh
qV this gives, in metre-second units,
)105
(5000)1(49052
2
2 h
h
)105
(019.112
2
2 h
h
2
2
2
095.519.11
hh
Rearrange for the deep, subcritical solution:
2
2
095.519.11
hh
Iterating (from, e.g., 19.112 h ) gives h2 = 3.089 m.
Answer: downstream depth = 3.09 m.
(b) Head either side is given by:
g
VzH s
2
2
The datum is not important as it is only the difference in head that is required. For
convenience, measure z relative to the bed of the expanded part. Then,
h12h
Hydraulics 3 Answers - 24 Dr David Apsley
zs1 = 1 m (note: water surface level, not depth), V1 = 10 m s–1
H1 = 6.097 m
zs2 = 3.089 m, 1
2
2 sm619.1 h
qV H2 = 3.223 m
Hence,
m874.2223.3097.6losthead
Answer: head lost = 2.87 m.
Hydraulics 3 Answers - 25 Dr David Apsley
Section 3.6.2
Example. (Exam 2008 – modified)
A long, wide channel has a slope of 1:2747 with a Manning’s n of 0.015 m–1/3
s. It carries a
discharge of 2.5 m3 s
–1 per metre width, and there is a free overfall at the downstream end.
An undershot sluice is placed a certain distance upstream of the free overfall which
determines the nature of the flow between sluice and overfall. The depth just downstream of
the sluice is 0.5 m.
(a) Determine the critical depth and normal depth.
(b) Sketch, with explanation, the two possible gradually-varied flows between sluice and
overfall.
(c) Calculate the particular distance between sluice and overfall which determines the
boundary between these two flows. Use one step in the gradually-varied-flow
equation.
S0 = 1/2747 = 3.64010–4
n = 0.015 m–1/3
s
q = 2.5 m2 s
–1
(a)
Critical Depth
m8605.081.9
5.23/1
23/1
2
g
qhc
Normal Depth
Vhq where 2/13/21SR
nV h (Manning), hRh (“wide” channel)
hShn
q 2/13/21
n
Shq
3/5
m500.12747/1
5.2015.05/35/3
S
nqh
Answer: critical depth = 0.860 m; normal depth = 1.50 m.
(b) The depth just downstream of the sluice is supercritical (0.5 m < hc). However, the
preferred depth is subcritical (hn > hc). Hence, if the channel is long enough then there will be
a downstream hydraulic jump, with the flow depth then decreasing to pass through critical
again near the overfall.
Hydraulics 3 Answers - 26 Dr David Apsley
If the channel is too short, however, the region of supercritical flow from the sluice will
extend to the overfall.
normal
critical
hydraulic jump
normal
critical
supercritical
supercritical
(c) As the channel shortens, the depth change across the hydraulic jump diminishes. The
boundary between the two possible flow behaviours occurs when the supercritical GVF just
reaches critical depth at the overfall (i.e. the limiting depth change across the hydraulic jump
is zero).
As the flow is supercritical, integrate the GVF equation forward from the downstream side of
the sluice gate (where h = 0.5 m) to the overfall (where h = hc = 0.8605 m). Use 1 step.
GVF equation:
2
0
Fr1d
d
fSS
x
h
For the direct-step method invert the GVF equation:
fSSh
x
0
2Fr1
d
d and h
h
xx Δ
d
dΔ
For the working, write the derivative as a function of h; (all lengths in metres).
3
Frgh
q
gh
V
33
22 6371.0
Frhgh
q
3/10
32
3/5
10406.1
hh
nqS f
3605.05.08605.0Δ h
Hydraulics 3 Answers - 27 Dr David Apsley
Working formulae:
hh
xx
mid
Δd
dΔ
where
3/10
34
3
10406.110640.3
6371.01
d
d
h
h
h
x
, Δh = 0.3605
i hi xi hmid
midh
x
d
d
Δx
0 0.5 0
0.6803 217.1 78.26
1 0.8605 78.26
Answer: length to overfall = 78 m.
Hydraulics 3 Answers - 28 Dr David Apsley
Example.
A sluice gate discharges water at 9 m3 s
–1 into a 6 m wide rectangular channel laid on a slope
of 0.0004 with n = 0.015 m–1/3
s. The depth at the vena contracta is 0.15 m.
(a) Find the normal and critical depths.
(b) Compute the position of the hydraulic jump, assuming normal depth downstream. Use
one step in the GVF equation.
Q = 9 m3 s
–1
b = 6 m
S0 = 4×10–4
n = 0.015 m–1/3
s
(a)
Normal Depth
VAQ where 2/13/21SR
nV h
and
bhA
bh
h
hb
bhRh
/212
(rectangular channel)
Hence,
bhSbh
h
nQ 2/1
3/2
/21
1
3/2
3/5
)/21( bh
h
Sb
nQ
(*)
For the given slope, rearrange as an iterative formula for h:
5/2
5/3
0
)/21( bhSb
nQh
or, with numerical values:
5/2)3/1(073.1 hh
Iteration (from, e.g., h = 1.073) gives
m231.1nh
Critical Depth
3/12
g
qhc
Here,
12 sm5.16
9 b
Hydraulics 3 Answers - 29 Dr David Apsley
Hence,
m6121.081.9
5.13/1
23/1
2
g
qhc
Answer: normal depth = 1.23 m; critical depth = 0.612 m.
(b) At the hydraulic jump the downstream depth is the normal depth; i.e.
m231.12 h
1
2
2 sm219.1231.16
9
bh
QV
3508.0231.181.9
219.1Fr
2
2
gh
V
Just upstream of the hydraulic jump then
m2516.0)3508.0811(2
231.1)Fr811(
2
22
22
1 h
h
We must therefore do a GVF calculation from just downstream of the sluice gate (where
h = 0.15 m) to just upstream of the hydraulic jump (where h = 0.2516 m).
GVF equation:
2
0
Fr1d
d
fSS
x
h
For the direct-step method invert the GVF equation:
fSSh
x
0
2Fr1
d
d and h
h
xx Δ
d
dΔ
For the working, write the derivative as a function of h; (all lengths in metres).
3
Frgh
q
gh
V
33
22 2294.0
Frhgh
q
3/10
3/443/4
2
3/5
)3/1(100625.5)/21(
h
hbh
bh
nQS f
1016.015.02516.0Δ h
Hydraulics 3 Answers - 30 Dr David Apsley
Working formulae:
hh
xx
mid
Δd
dΔ
where
4
3/10
3/4
3
10])3/1(
0625.54[
2294.01
d
d
h
h
h
h
x, Δh = 0.1016
i hi xi hmid midhx )d/d( Δx
0 0.15 0
0.2008 235.6 23.9
1 0.2516 23.9
Answer: distance from sluice gate to hydraulic jump = 24 m.
Using the specific energy as dependent variable and avfSS
Ex
)(
ΔΔ
0 leads to a distance of
17.8 m. An accurate answer using a very small step size and many steps is 23.7 m.
Hydraulics 3 Answers - 31 Dr David Apsley
Example. (Exam 2014)
An undershot sluice is used to control the flow of water in a long wide channel of slope 0.003
and Manning’s roughness coefficient 0.012 m–1/3
s. The flow rate in the channel is 2 m3 s
–1
per metre width.
(a) Calculate the normal depth and critical depth in the channel and show that the channel
is hydrodynamically “steep” at this flow rate.
(b) The depth of flow just downstream of the sluice is 0.4 m. Assuming no head losses at
the sluice calculate the depth just upstream of the sluice.
(c) Sketch the depth profile along the channel, indicating clearly any flow transitions
brought about by the sluice and indicating where water depth is increasing or
decreasing.
(d) Use 2 steps in the gradually-varied flow equation to determine how far upstream of
the sluice a hydraulic jump will occur.
S0 = 0.003
n = 0.012 m–1/3
s
q = 2 m2 s
–1
(a)
Normal depth
hSRn
Vhq h
2/13/21 where Rh = h (wide channel)
3/5hS
nq (*)
m6095.0003.0
2012.05/35/3
0
S
nqhn
Critical Depth
m7415.081.9
23/1
23/1
2
g
qhc
The normal depth is smaller than the critical depth. Hence the normal flow is supercritical;
i.e. the channel is steep at this discharge.
Answer: hn = 0.610 m; hc = 0.742 m.
(b) Assuming the same total head on either side of the gate:
g
Vz
g
Vz ss
22
2
22
2
11
Hydraulics 3 Answers - 32 Dr David Apsley
2
2
2
22
1
2
122 gh
qh
gh
qh
2
2
2
1
2
14.02
24.0
2
2
gghh
674.12039.0
2
1
1 h
h
We require the subcritical solution, so rearrange for iteration as
2
1
1
2039.0674.1
hh
Iteration (e.g. from 1.674) gives h1 = 1.594 m
Answer: h1 = 1.59 m.
(c) The depth increases in each of the GVF regions (S1 and S3) shown below.
normalS1
2h S3
CP
nh
normal
nh
h1
(d) Upstream of the sluice there is GVF between the hydraulic jump and the sluice. This
starts at the subcritical sequent depth of the hydraulic jump (to be found below) and ends at
the depth h1 found in part (b).
For the hydraulic jump, on the upstream side (normal flow):
hA = 0.6095 m
1sm281.36095.0
2 A
Ah
qV
342.16095.081.9
281.3Fr
A
Agh
V
On the downstream side of the hydraulic jump:
m8915.0)342.1811(2
6095.0)Fr811(
2
22 AA
B
hh
Do a GVF calculation (subcritical, so physically
it should start at the fixed downstream control
and work upstream, although mathematically it
can be done the other way) from just upstream of
the sluice gate (where h = 1.594 m) to just
downstream of the hydraulic jump (where
h = 0.892 m). Using two steps the depth
Sluice
hydraulic jump
x0x1x2
Step 1Step 2
h = 1.5940
h = 1.2431
h = 0.8922
Hydraulics 3 Answers - 33 Dr David Apsley
increment per step is
m351.02
594.1892.0Δ
h
GVF equation:
2
0
Fr1d
d
fSS
x
h
For the direct-step method rewrite the GVF equation “the other way up”:
fSSh
x
0
2Fr1
d
d and h
h
xx Δ
d
dΔ
For the working, write the derivative as a function of h; (all lengths in metres).
3
Frgh
q
gh
V
33
22 4077.0
Frhgh
q
3/10
42
3/5
1076.5
hh
nqS f
Working formulae:
hh
xx
mid
Δd
dΔ
where
4
3/10
3
10]76.5
30[
4077.01
d
d
h
h
h
x, Δh = –0.351
i hi xi hmid midhx )d/d( Δx
0 1.594 0
1.419 303.9 –106.7
1 1.243 –106.7
1.068 262.2 –92.0
2 0.892 –198.7
Answer: upstream distance from sluice to hydraulic jump = 199 m.
Hydraulics 3 Answers - 34 Dr David Apsley
Section 4.2
Example. (From White, 2006)
A pencil point piercing the surface of a wide rectangular channel flow creates a wedgelike
25° half-angle wave, as in the figure right. If the channel has a Manning’s n of 0.014 m–1/3
s
and the depth is 350 mm, determine: (a) the Froude number; (b) the critical depth; and (c) the
critical slope.
25o
α = 25º
n = 0.014 m–1/3
s
h = 0.35 m
(a)
Fr
1αsin
366.225sin
1
αsin
1Fr
Answer: Froude number = 2.37.
(b)
gh
VFr
1sm384.435.081.9366.2Fr ghV
12 sm534.135.0384.4 Vhq
m6213.0
3/12
g
qhc
Answer: critical depth = 0.621 m.
(c) The critical slope is that at which the normal depth is equal to the critical depth.
Normal flow:
Vhq where 2/13/21SR
nV h (Manning), hRh (“wide” channel)
Hydraulics 3 Answers - 35 Dr David Apsley
hShn
q 2/13/21
Rearranging, and setting a depth h equal to the critical depth (hc = 0.6213 m),
3
3/10
2
3/10
2
10254.26213.0
)534.1014.0()(
h
nqS
Answer: critical slope = 2.2510–3
.
Hydraulics 3 Answers - 36 Dr David Apsley
Sediment Transport; Section 2.1 Note: removed from the present curriculum because of the reduction in teaching hours.
Example. (Exam 2007 – part)
An undershot sluice is placed in a channel with a horizontal bed covered by gravel with a
median diameter of 5 cm and density 2650 kg m–3
. The flow rate is 4 m3 s
–1 per metre width
and initially the depth below the sluice is 0.5 m. Assuming a critical Shields parameter τ*crit
of 0.06 and friction coefficient cf of 0.01:
(a) find the depth just upstream of the sluice and show that the bed there is stationary;
(b) show that the bed below the sluice will erode and determine the depth of scour.
d = 0.05 m
ρs = 2650 kg m–3
(s = 2.65)
q = 4 m2 s
–1
06.0τ* crit
cf = 0.01
(a) The bed is mobile if and only if the bed shear stress exceeds the critical shear stress for
incipient motion. (Alternatively, one could find the velocity at which incipent motion occurs.)
First find the critical shear stress for incipient motion. The local shear stress can be found by
finding velocities using open-channel flow theory and using the friction coefficient.
The critical shear stress for incipient motion is
2mN56.4805.081.9)10002650(06.0)ρρ(ττ * gdscritcrit
In the accelerated flow just below the sluice gate,
h = 0.5 m
1sm85.0
4 h
qV
The total head relative to the undisturbed bed is thus (since zs = h initially):
m762.381.92
85.0
2
22
g
VzH s
Upstream of the sluice one must have the same total head:
2
22
22 gh
qh
g
VzH s
but now we look for the subcritical (deep) solution:
2
2
2gh
qHh
Substituting numerical values, in metre-second units:
2
8155.0762.3
hh
Hydraulics 3 Answers - 37 Dr David Apsley
Iterating (from, e.g., h = 3.762) gives
m703.3h
The velocity upstream of the sluice is
1sm080.1703.3
4 h
qV
and the bed shear stress upstream of the sluice is
22
212
21 m N832.5080.1100001.0)ρ(τ Vc fb
Since critb ττ the bed here is stationary.
Answer: depth upstream of sluice = 3.70 m; as demonstrated, the bed here is stationary.
(b) Initially, beneath the sluice,
1sm8 V
22
212
21 m N3208100001.0)ρ(τ Vc fb
Since critb ττ the bed here is initially mobile.
The bed will continue to erode, the flow depth h increasing and velocity V decreasing until
critb ττ
At this point,
critf Vc τ)ρ( 2
21
whence
1sm116.31000
56.48
01.0
2
ρ
τ2 crit
fcV
m284.1116.3
4
V
qh
This is the depth of flow. Since the downstream water level is set (at 0.5 m above the
undisturbed bed) by the gate the depth of the scour hole is
m784.05.0284.1
Answer: depth of scour hole = 0.784 m.