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
10.60
Two canals join to form a lager canal as
shown in Figure P10.60 copied at the right.
Each of three rectangular canals is lined wit
the same material and has the same bottom
slope. The water depth in each is to be 2 m.
Determine the width of the merged canal, b3.
Explain physically (i. e., without using any
equations) why it is expected that the width
of the merged canal is less than the
combined widths of the two original canals
(i.e., b3 < 4 m + 8 m = 12 m).
The merged flow is the sum of the two original flows: Q 3 = Q1 + Q2. Each of these flows is given
by the Manning equation so that we can write.
 AR 2 3 S 1 2 
 AR 2 3 S 1 2   AR 2 3 S 1 2 
0
h
Q3  
   h 0    h 0   Q1  Q2
n
n
n

 3 
1 
 2
Since each canal has the same bottom slope and is lined with the same material the values of S0
and n are the same in all three canals. Thus we can divide the equation above by the common
values of S01/2/n and substitute A/P for Rh to obtain the following equation.
13
A3 Rh2,33

A1Rh2,13

A2 Rh2,23
 A5 
  32 
P 
 3 
13
 A5 
  12 
P 
 1 
13
 A5 
  22 
P 
 2 
If y is the depth and b is the width then the area is yb and the wetted perimeter is 2y + b. Making
these substitutions in the last equation gives.
13
 A35 


 P2 
 3 
  y3b3 5 

2
 b3  2 y3  
13
13
  y1b1 5 

2
 b1  2 y1  
13
  y2b2 5 

2
 b2  2 y2  
13
 A5 
  12 
P 
 1 
13
 A5 
  22 
P 
 2 
Since y1 = y2 = y3 = 2 m we can cancel the common y values from the numerator of the central
equation above and insert the values of b1 = 4 m and b2 = 8 m obtain the following equation.


b35


2
 b3  22 m 
13

4 m5 

2
 4 m  22 m 
13

8 m5 

2
 8 m  22 m 
13
 8.625 m
We see that the equations will be dimensionally consistent if b3 is in meters. Omitting the
dimensions, we can obtain the following computational equation.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
May 6 OCF homework solutions
 b35 


2
 b3  4 
ME 390, L. S. Caretto, Spring 2008
13
b35
 8.625 
Page 2
 8.6253  641.6
b3  4
b35  641.6b3  4 2  0
2
Using a computer root finder to solve this equation gives the new width, b3 = 10.66 m .
The open channel flow has an equilibrium between the gravity driving force and the resistance of
the viscous forces. The single channel has less wall space to generate viscous forces, so the
viscous forces are less, but the gravity force is the same, because the depth is the same. To
maintain the same depth, with the same slope, we have to have a final width that is less than the
original width to maintain a balance between the gravity driving force of depth and the viscous
forces retarding the flow.
10.63
The cross section of a long tunnel carrying water through a
mountain is shown in Figure P10.63, copied at the right. Plot a
graph of flowrate as a function of water depth, y, for 0  y 18
ft. The slope is 2 ft/mi and the surface of the tunnel is rough
rock (equivalent to rubble masonry). At what depth is the
flowrate maximum? Explain.
We can use the Manning equation to find the flow rate.
Q
ARh2 3 S 01 2
n
For BG units k = 1.49. For rubble masonry, n = 0.025 from Table
10.9. The slope of 2 ft/mi is equivalent to (2 ft) / (5280 ft) = 0.000379 = S0. For depths less than
12 ft, the area is the depth, y, times the width, 12 ft and the wetted perimeter is the 12-ft width
plus twice the depth. This gives the hydraulic radius and resulting flow rate as follows.
12 ft y
A
Rh  
P 12 ft   2 y
Q
1.49
12 y  6 y 
Q
0.025
6 y
23
ARh2 3 S 01 2
n
 12 y 


 12 y 
n
 12  2 y 
0.000379 
12
23
 y 

 45.835 y
6 y
S 01 2
23
The flow is in ft3/s when the depth, y, is in feet.
Above a depth of 12 ft, the circular top gives a different area and
wetted perimeter. The filled area below 12 ft is a square whose
area is (12 ft)2 = 144 ft2. The total area for a depth above this 12ft height is the area of a semicircle whose radius is 6 ft; this area
is (6 ft)2/2. The flow area is the area of this semicircle minus the
empty area above the depth, y. The calculation of this unfilled
area was done in Example 10.5 and the relation between this
unfilled area, A(1) and the depth, y, is linked through the angle .
A(1) 
D2
  sin  where   2 cos 1 y  12 ft
8
6 ft
Thus, the flow area is given by the following equation.
May 6 OCF homework solutions
A  144 ft 2 
ME 390, L. S. Caretto, Spring 2008
Page 3
2

6 ft 2  A(1)  200.5 ft 2  12 ft    sin  where   2 cos 1 y  12 ft
2
8
6 ft
The wetted perimeter when y > 12 ft is the wetted perimeter of the filled lower area which is 3(12
ft) = 36 ft plus the wetted part of the perimeter of the semicircular area. The entire perimeter of
the semicircle is 2R/2 = (6 ft). The perimeter of the semicircle that is not wetted is R = (6 ft).
So, the total wetted perimeter when the depth, y, is greater than 12 ft is 36 ft + (6 ft)( – ) =
54.85 – 6. We can now construct the following set of equations to compute the flow for depths,
y, above 12 ft.
Q
ARh2 3 S 01 2
n
  A
 A 
n P
23
S 01 2
1.49  A 

A 
0.025  P 
23
0.000379 
12
A  200.5  18  sin  P  54.85  6   2 cos 1
Qmax = 581 ft3/s .
For depths greater
than 17.1 ft, the wall
perimeter retarding the
flow increases much
more rapidly than the
flow area and the flow
rate decreases.
23
Solution to Problem 10.63
600
.
500
FLow, Q, (ft3/s)
The flow rates at
various depths were
computed on a
spreadsheet using the
equations derived
above. The
appropriate equations
were used for depths
below 12 ft and depths
above 12 ft. The
maximum flow rate
was determined using
the Excel Solver and
found to occur at a
depth of 17.1 m. This
maximum flow rate is
 A
 1.1568 A 
P
y  12
6
400
300
200
100
0
0
2
4
6
8
10
Depth (ft)
12
14
16
18
May 6 OCF homework solutions
10.72
ME 390, L. S. Caretto, Spring 2008
Page 4
Water flows in a rectangular channel with a bottom slope of 4.2 ft/mi and a head loss of 2.3
ft/mi. At a section where the depth is 5.8 ft and the average velocity is 5.9 ft/s, does the
flow depth increase of decrease in the direction of flow? Explain.
The change in depth with distance, dy/dx, is given in terms of the friction slope for head loss, S f,
the bottom slope, S0 and the Froude number, Fr, by the following equation.
dy S f  S 0

dx 1  Fr 2
Fr 2 
V2
gy
For the data of this problem S0 = 4.2 ft/mi, Sf = 2.3 ft/mi, y = 5.8 ft, and V = 5.9 ft/s. We can then
compute dy/dx as follows.
2
 5.9 ft 


2
V
s 

2
Fr 

 0.1865
gy 32.174 ft
5.8 ft 
s2
S f  S0
dy

dx 1  Fr 2
2.3 ft 4.2 ft

mi
mi  –2.34 ft/mi

1  0.1865
The head loss retarding the flow is less than the slope giving the gravity force. This force
imbalance accelerates the fluid so the level becomes shallower to maintain the same flow rate.
10.76
A 2.0-ft standing wave is produced at the bottom of the rectangular channel in an
amusement park ride. If the water depth upstream of the wave is estimated to be 1.5 ft,
determine how fast the boat is traveling when it passes through this standing wave
(hydraulic jump) for its final “splash.”
We can use solve the equation for the depth ratio (y2 /y1) across a hydraulic jump in terms of the
upstream Froude number, Fr1 = V1/(gy1)1/2, which can be used to find the upstream velocity, V1. The boat
will be moving at this water velocity when it passes through this standing wave.
2


y2 1 
V1
1  y 2
2

 2


  1  8Fr1  1  Fr1 


1

1


y1 2 
8  y1

gy1



 V1 
2


gy1  y 2
 2



1

1

8  y1




From the data given for this problem y1 = 1.5 ft and y2 = y1 + 2 ft so y2/y1 = (3.5 ft) /(1.5 ft) = 7/3.
We can use the equation above to find V1.
V1 
2


gy1  y 2
 2



1

1

8  y1




32.174 ft
s2
8
1.5 ft  
2

 7 
2

1

  1  13.7 ft/s
 3 

May 6 OCF homework solutions
10.44
ME 390, L. S. Caretto, Spring 2008
Page 5
Under appropriate conditions, water
flowing from a faucet, onto a flat place,
and over the edge of the place can
produce a circular hydraulic jump as
shown in Figure P10.78 copied at the
right. Consider a situation where a
jump forms 3.0 in from the center of the
plate with depths upstream and
downstream of the jump of 0.05 in and
0.20 in, respectively. Determine the
flowrate from the faucet.
We can use solve the equation for the depth ratio (y2 /y1) across a hydraulic jump in terms of the Froude
number for the upstream Froude number, Fr1 = V1/(gy1)1/2, which can be used to find the upstream
velocity. This water velocity can then be used to find the flow rate.
2


y2 1 
V1
1  y 2
2

 2


  1  8Fr1  1  Fr1 


1

1


y1 2 
8  y1

gy1



 V1 
2


gy1  y 2
 2



1

1

8  y1




From the data given for this problem y1 = 0.05 in = 0.004167 ft and y2 = 0.20.
V1 
2


gy1  y 2
 2



1

1

8  y1




32.174 ft
s2
0.04167 ft  
8
2

 0.20 in

 2
 1  1  1.158 ft/s
 0.05 in




The area of the flow before the jump will be Dy1 = (6 in)(0.05 in)(1 ft2/144 in2) = 0.006545 ft2.
The flow rate, Q = V1A1.
Q  V1 A1 
10.90


1.158 ft
0.006545 ft 2  0.00758 ft3/s
s
Water flows from a storage tank, over two
triangular weirs, and into two irrigation
channels as shown in Figure P10.90, copied
at the right. The head for each weir is 0.4 ft
and the flowrate in the channel fed by the 90degree V-notch weir is to be twice the
flowrate in the other channel. Determine the
angle  for the second weir.
The flow rate over a triangular weir is given by
equation 10.32.
Q  C wt
8

tan   2 g H 5 2
15  2 
In this problem we have two channels flows such that Q2 = 0.5Q1, where H2 = H1 = 0.4 ft and 1 =
90o. Using the weir equation for both flows and setting Q 2 = 0.5Q1 gives
 90 o
Q1 1
8
8
 2 
52
Q2  C wt ,2 tan   2 g H

 C wt ,1 tan 
15  2 
2 2
15  2

C
 2 g H 5 2  wt ,1 8 2 g H 5 2

2 15

May 6 OCF homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 6
In the last step above we used the result that tan(45o) = 1.
The weir coefficients depend on both H and . For the 90o
weir with 1 = 90o and H = 0.4 ft we see that Cwt,1` = 0.59
from Figure 10.25 on page 599, copied at the left. Using
this value in the equation above and canceling common
factors in the two streams gives the following result.
   C wt ,1 0.59
C wt , 2 tan  2  

2
2
 2 
  2  2 tan 1
0.295
C wt , 2
We cannot solve this equation explicitly since Cwt,2 depends
on 2. To solve this problem with the sparse data relating 
and Cwt,2, we can find Cwr,2 for the four values of  shown on
the figure and determine the relationship between the value
of  on the figure versus the computed value of  = 2tan-1(0.295/Cwt,2). The results are shown in
the table at the right. Here the first column represents the
angle for which there is a curve on Figure 10.25 and the
Cwt,2
Input 2
Computed 2
value of Cwt is read from that figure. Because there is such
20
0.626
50.47
a small range of Cwt values the computed values of 2 (from
45
0.607
51.84
the tan-1 equation above) are in a narrow range from 50.47o
60
0.597
52.59
to 53.13o. The plot of the computed values versus the input
90
0.590
53.13
values shown below uses the narrow range of the computed
angle results. The intersection of the input versus computed q line with a 1:1 line marks the
correct answer:  = 52.3o .
Problem 10.90 Solution
90
Input angle (degrees)
80
70
60
50
40
Calculations
1:1 line
30
20
50
51
52
53
Computed angle (degrees)
54
55