College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
May 6 Open-Channel-Flow-Homework Solutions
10.2
The flowrate per unit width in a wide channel is q = 2.3 m2/s. Is the flow subcritical or
supercritical if the depth is (a) 0.2 m, (b) 0.8 m, or (c) 2.5 m?
Open-channel flows are subcritical, critical or supercritical if the Froude number is less than,
equal to, or greater than one. We can compute the Froude number from the data given using the
following relations, where b is the channel width and y is the channel depth
Fr 
V
Q 1
qb 1



gy A gy yb gy
q
gy 3
For the given flow per unit width, q = 2.3 m2/s, we find the following values for Froude number at
the different depths.
y  0.2 m
y  0.8 m
y  2.5 m
10.13
Fr 
Fr 
Fr 
q
gy 3
q
gy 3

2.3 m 2
s
 8.21
9.80655 m
3
0.15 m
s2
Flow is supercritical

2.3 m 2
s
 1.03
9.80655 m
3
0.8 m
s2
Flow is supercritical
q
gy 3

2.3 m 2
s
 0.186
9.80655 m
3
2.5 m
s2
Flow is subcritical
Water flows in a rectangular channel with a flowrate per unit width of q = 25 ft 2/s. Plot the
specific energy diagram for this flow. Determine the two possible depths of flow if E = 7 ft.
We can substitute the given data into the equation for the specific energy, E, to get the following
computational equation. Note that the units for the constant mean that the specific energy will be
in feet when y is in feet.
2
 25 ft 2 


s 
q
9.713 ft 3

E  y
 y
 y
32.174 ft 2
2 gy 2
y2
2
y
2
s
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
May 6 OCF homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
We can find the value of yc that makes E a minimum and the corresponding value of Emin as
follows.
13
 q2 
yc   
 g 
  25 ft 2  2 
 
 
 s  

32.174 ft 


s2


13
 2.688 ft
Emin 
3
3
yc  2.688 ft   4.03 ft
2
2
We can check this formula for the minimum energy by substituting yc into our computational
equation for the specific energy.
E  y
9.713 ft 3
y2

E min  yc 
Using the computational
equation for E as a function of y
we can determine the specific
energy for a range of y values to
obtain the plot at the right. The
plot shows the minimum energy
(and its location, yc) at the
values found above.
9.713 ft 3
7 ft  y 
y2
Noting that we have dimensional
consistency and the solution for
y will be in feet we can
rearrange this equation into a
cubic form as shown below.
Problem 10.13 Solution
14
12
Depth, y, ft
At the point where E = 7 ft, our
graph shows that the depth can
be approximately 1.5 ft or 6.5 ft.
For E = 7 ft, our computational
equation becomes
9.713 ft 3
9.713 ft 3

2
.
688
ft

 4.03 ft
yc2
2.688 ft 2
10
8
6
4
2
0
0
1
2
3
4
5
6
7
8
9 10 11 12
Specific Energy, E, ft
y 3  7 y 2  9.713  0
We can solve this equation numerically using the values found from the graph as initial guesses
to get the desired solution. The positive roots of this cubic equation are y = 1.306 ft for the
supercritical depth and y = 6.790 ft for the subcritical depth. The negative root, y = –1.095 ft is
ignored as being physically unrealistic.
May 6 OCF homework solutions
Page 3
Water flows in a rectangular channel with a
flowrate per unit width of q = 1.5 m2/s and a
depth of 0.5 m at section (1). The head loss
between sections (1) and (2) is 0.03 m. Plot
the specific energy diagram for this flow
and locate states (1) and (2) on this
diagram. Is it possible to have a head loss
of 0.06 m? Explain.
We can apply the energy equation to the
diagram shown above. Here we use the surface of the liquid as the points to apply the energy
equation. This means that p1 = p2 = 0. We assume that the slope is very small so that we can
use the fluid depth as the elevation. This gives z1 = y1 = 0.5 ft and z2 = y2. With these
simplifications our energy equation becomes.
z2 
p2


V22
p V2
 z1  1  1  hs  hL
2g
 2g

y2 
V22
V2
 y1  1  hL
2g
2g
We replace the velocity, V, by the flow per unit width: V = Q/A = (qb)/(yb) = q/y. This gives the
following result in our energy equation. Note that Q = qb is constant as the flow changes depth.
If the width, b, is constant, then q = Q/b, is also constant. Thus we do not have to subscript q in
the equation below.
q y2   E  y  q y1   h  E  h
V22
y2 
 y2 
2
1
L
1
L
2g
2g
2g
2
2
 E2  E1  hL
Although the specific energy changes in the flow as a result of head loss, the relationship
between specific energy and depth does not change. At any point in the flow we must have
2
E  y
q y 2
2g
 1.5 m 2 


s 
0.1147 m 3

 y
 y
9.80665 m 2
y2
2
y
s2
We can compute the values of E
for various values of y and plot
the results as shown in the
figure to the left. In particular,
we know the value of the depth,
y = 0.5 m at section (1) in the
flow. Thus, the specific energy
at this point can be found as
Problem 10.22 Solution
3
2.5
Depth, y, m
10.22
ME 390, L. S. Caretto, Spring 2008
2
0.1147 m 3

y12
0.1147 m 3
0.5 m 
 0.959 m
0.5 m2
E1  y1 
1.5
1
0.550
0.5
0.5
0
0.0
0.5
1.0
1.5
2.0
Specific Energy, E, m
2.5
3.0
This confirms the value that we
can read from the chart for y =
0.5 m. According to the energy
equation, the head loss means
that the energy at station 2 is
0.03 m less than the energy at
May 6 OCF homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
station (1). This gives E2 = 0.929 ft. This specific energy corresponds to the following a depth of
0.550 m. This is found by solving the final equation below, using a numerical root finder to find y.
Both points are shown on the energy curve with the value of their depts. indicated
E2  E1  0.03 m  0.929 m  y 2 
0.1147 m 3
y 22

y 23  0.929 y 22  0.1147  0
We see that the lower energy corresponds to a higher depth with the same flow.
If we had a head loss of 0.06 m from station 1 the energy at station (2) would be 0.959 m –
0.06 m = 0.859 m. This is only possible if the minimum energy is less than 0.859 m. To check
this we can compute the minimum energy by first finding the critical depth for this energy, yc. and
then applying the formula that Emin = 1.5 yc.
13
 q2 
yc   
 g 
  1.5 m 2  2 
 
 
 s  

9.80665 m 


s2


13
 0.6122 m
Emin 
3
3
yc  0.6122 m   0.9183 m
2
2
Since Emin = 0.9183 m we cannot have a head loss of 0.06 m because this would make the
specific energy less than its minimum value.
10.34
An old, rough-surfaced, 2-m-diameter concrete pipe with a Manning coefficient of 0.025
carries water at a rate of 5.0 m3/s when it is half full. It is to be replaced by a new pipe with
a Manning coefficient of 0.012 that is also to flow half full at the same flowrate. Determine
the diameter of the new pipe.
To solve this problem we apply the Manning equation to both the old and new pipes and set the
results equal to each other because that the two flowrates are the same: Q old = Qnew.
Qold
 AR 2 3 S 1 2 
 AR 2 3 S 1 2 
0
h

  Qnew   h 0 
n
n

 old

 new
We are given no data on the slope, but we can assume that the slope will be the same for the old
and the new pipes. In addition, we note that for a pipe half full, the cross sectional area for the
flow is A = D2/8 and the wetted perimeter is P = D/2, so the hydraulic radius Rh = A/P = D/4.
With these expressions for area and hydraulic radius our equation for the old and new flowrates
become (after canceling the slope that is the same for both flows)
 D 2  D  2 3 
 D 2  D  2 3 


  
  
 8  4   8  4 




n
n





 old 
 new
D 
  new 
 Dold 
83

nnew 0.012

 0.48
nold 0.025
Since we are dealing with ratios, the dimensions of the diameters and Manning coefficients will
cancel. Taking the 3/8th power of the second equation gives.
Dnew
 0.483 8  0.7594 
Dold
Dnew  0.7594 Dold  0.75942 m  = 1.52 m
May 6 OCF homework solutions
10.44
ME 390, L. S. Caretto, Spring 2008
Page 5
A trapezoidal channel with a
bottom width of 3 m and sides with
a horizontal:vertical slope of 2:1 is
lined with fine gravel (n = 0.020)
and is to carry 10 m3/s. Can this
channel be built with a slope of S0
= 0.00010 if it is necessary to keep
the velocity below 0.75 m/s to
prevent scouring of the bottom?
Explain.
To solve this problem we apply the Manning equation to determine the size, y. required with the
proposed slope and the given flow rate. Once we find the value of y we can solve for the velocity
to see if it is less than 0.75 m/s.
V
Rh2 3 S 01 2
n
Q
ARh2 3 S 01 2
n
The cross sectional area is the area of the rectangle, 3y, plus two times the areas of one
triangular side, (1/2)(2y)(y) = y2, so the total area is 3y + 2y2. The wetted perimeter is the 3 m
along the bottom, plus the two diagonal sides of length [(2y)2 + y2]1/2 = 51/2y. This gives a total
length of 3 + 2(51/2y). With this definition of the area and perimeter the flowrate is
3 y  2 y 
2
Q
3
10 m

s
ARh2 3 S 01 2
n

 3y  2 y2 


 3 2y 5 


0.020
23
0.000101 2
With the flow rate in m3/s and the default dimensions for the Manning coefficient, the length terms
are in meters, so the value of y will be in meters. Rearranging the equation to solve for y gives.


 3y  2 y2 

3y  2 y 
 3 2y 5 


10 
0.020
2
23
0.000101 2

100.020
0.000101 2
3 y  2 y 
 20 
2 53
3  2 y 5 2 3
Using a computer or calculator equation solver, the value of y is found to be 2.249 m. We can
then compute the area and the resulting velocity.
10 m 3
10 m 3
10 m 3
Q
0.593 m
s
s
s
V 



2
2
2
A 3 m  y  2 y
s
3 m2.249 m  22.249 m 16.86 m
So, this design will give a velocity less than the required maximum of 0.75 m/s.
May 6 OCF homework solutions
10.51
ME 390, L. S. Caretto, Spring 2008
Page 6
A 10-ft-wide rectangular channel is built to bypass a dam so that fish can swim upstream
during their migration. During normal conditions when the water depth is 4 ft, the water
velocity is 5 ft/s. Determine the velocity during a flood when the water depth is 8 ft.
In this problem, we can assume that the slope and the Manning coefficient will be the same for
both normal and flood conditions and we can write the equations for the velocity in these two
conditions as follows.
Vnormal
 R 2 3S1 2 
 h 0 
n

 normal
V flood
 R 2 3S1 2 
 h 0 
n

 flood
With the assumption that the slope and the Manning coefficient will be the same for both normal
and flood conditions, these equations give the following relation between the velocities in these
two conditions. In getting the final equation we use the definition of the hydraulic radius for a
rectangular channel as A/P = by/(2y + b) where y is the depth and b is the width. (Note that the
wetted perimeter does not include the open surface of the channel. Since the width, b = 10 ft for
both the normal and flood conditions, we do not use a subscript for b.
Vnormal
V flood

3
Rh2, normal
Rh2, 3flood
 by normal 


2 y normal  b 


V flood
 by flood 


 2 y flood  b 


23
23
 10 ft 4 ft  
 24 ft   10 ft 


 0.8050
23
 10 ft 8 ft  
 28 ft   10 ft 


5 ft
V
 normal  s  = 6.21 ft/s
0.8050 0.8050
23