Solutions Manual for Fundamentals of Open Channel Flow,
2e by Glenn Moglen (All Chapters)
Chapter 1: Introductory Material - Solutions
1.1. What slope would lead to a 1% difference between depth in the vertical plane rather than
depth measured perpendicular to the channel bottom? Compare this slope to the
observation that a channel slope of S0 = 0.01 m/m is generally considered quite steep for
open channel flow.
Solution:
If is the angle between the horizontal plane and the plane of the channel then,
x
= cos
1.01x
Thus,
= 8.1o
or, in terms of rise/run,
S = tan(8.1o) = 0.14 m/m
Comparing this number to a channel slope of S0=0.01 m/m we see that the slope
corresponding to a 1.0 percent difference between depths is more than an order of
magnitude larger.
1.2. Using Bernoulli’s equation, write the energy balance in general terms for flow in an open
channel from location 1 to 2 where hL is the head loss between these two locations.
Simplify the equation by taking the perspective of a point on the water surface at both
locations. Note: your solution should show that the pressure term from Bernoulli’s
equation is not relevant for open channel flow.
Solution:
p1v12 p 2 v22
+ + z1 = + + z 2 + hL
2g 2g
If we take a point on the water surface at both locations, the p1 equals p2 equals
atmospheric pressure, and thus these terms may be cancelled from both sides of the
equality,
v12 v2
+ z1 = 2 + z 2 + hL
2g 2g
The remaining equation if y is substituted for z and if hL is set to zero, forms the basis for
the specific energy equation which is the focus for Chapter 2.
-1-
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, Chapter 1: Introductory Material
1.3. Parts (a), (b), and (c) require simple multiplication/division and/or addition/subtraction to
solve. The reader is cautioned to pay special attention to significant digits when reporting
the final answer.
a. If the density of water is 1000 kg/m3 and gravitational acceleration is 9.81 m/s2, what
is the unit weight of water?
b. If the density of water is 1.0 103 kg/m3 and gravitational acceleration is 9.81 m/s2,
what is the unit weight of water?
c. The cross-sectional area of a channel is broken into three separate subareas with the
following sizes: 1.3 m2, 0.92 m2, and 15 m2. What is the total cross-sectional area of
the channel?
Solution:
a) The unit weight of water is the product of density and gravitational acceleration so,
= g = (1000) (9.81) = 9810 N
Since density is given with one significant figure. The answer has one significant figure
resulting in: 10,000 N.
b) The new statement gives density with two significant figures, so the answer becomes:
9800 N.
c) The calculator-based sum of the three provided numbers is 17.22. However, the number
“15” indicates uncertainty in the “ones” place of the number. This same uncertainty
needs to be conveyed in the answer, so the correct answer is 17 m2.
1.4. The mean or bulk velocity of flow in a stream is observed to be 1.1 m/s. A rock tossed
into this same flow sets up ripples that radiate outward in all directions. It is noted that
the ripples propagating directly upstream travel at a velocity of 0.67 m/s in the opposite
direction to the direction of the flowing stream.
a. What is the Froude number for this flow?
b. Estimate the depth of flow in this stream.
Solution:
a) The wave velocity (velocity of ripple propagation) provided is the net velocity, equal
to the velocity of wave propagation in a still pool of water minus the bulk velocity
downstream. The wave velocity is 1.1 + 0.67 = 1.8 m/s. Using the definition of the
Froude number:
v 1.1
Fr = = = 0.61
gy 1.8
b) The depth of flow in the stream can be estimated based on the wave velocity, vw = 1.8
m/s.
-2-
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, Chapter 1: Introductory Material
(v w )2 (1.8)
2
y= = = 0.33 m
g 9.81
1.5. In the final chapter of this book, we study sediment transport. In a particular stream, it is
found that the sediment transport rate can be approximated as
Qs = c( 0 − * ) p
where c, p, and * are positive constants.
a. Write an analytical expression for the sensitivity, dQs d 0 .
b. Let c = 1, p = 2.2, and * be 1.4.
i. Plot dQs d 0 for 0 1.4.
ii. Determine the value of the sensitivity at 0 = 1.5 and 0 = 1.7.
c. Briefly discuss how “Rule 3” as presented in Section 1.5 relates to your findings in
Part (b) of this problem.
Solution:
a) We take the derivative of the sediment transport rate with respect to 0:
dQs
=
d 0 d 0
d
( )
c 0 − * = p c 0 − *
p p −1
( )
b)
i. The figure below shows (from top to bottom) the sediment transport function
itself (not requested), the absolute sensitivity function (requested), and the relative
dQ 0
sensitivity function (not requested but defined as s ).
d 0 Qs
-3-
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, Chapter 1: Introductory Material
This figure shows that the absolute sensitivity increases gradually from zero once
0 exceeds *. In contrast, the relative sensitivity spikes just beyond * because,
the value of Qs has increased infinitely from zero to a number greater than zero.
Other changes at 0 increases further are not nearly so dramatic.
ii. At 0 =1.5 the absolute sensitivity is:
dQs
d 0
(
= p c 0 − *
p −1
)
= (2.2)(1.0)(1.5 − 1.4)1.2 = 0.139
While the relative sensitivity is:
dQs 0
= (0.139)
1.5
= 33
d 0 Qs 0.0063
In contrast, at 0=1.7, the absolute sensitivity is:
dQs
== (2.2)(1.0)(1.7 − 1.4)1.2 = 0.519
d 0
And the relative sensitivity is:
dQs 0
= (0.139)
1.7
= 13
d 0 Qs 0.071
The lesson here being that the absolute sensitivity (slope of the sediment transport
function) increases as 0 increases while the relative sensitivity is greatest just as
0 exceeds *.
c) Rule 3 is related to the findings above in that * is a threshold and we see how the
sensitivity shifts from zero to non-zero right at this threshold and how the relative
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2e by Glenn Moglen (All Chapters)
Chapter 1: Introductory Material - Solutions
1.1. What slope would lead to a 1% difference between depth in the vertical plane rather than
depth measured perpendicular to the channel bottom? Compare this slope to the
observation that a channel slope of S0 = 0.01 m/m is generally considered quite steep for
open channel flow.
Solution:
If is the angle between the horizontal plane and the plane of the channel then,
x
= cos
1.01x
Thus,
= 8.1o
or, in terms of rise/run,
S = tan(8.1o) = 0.14 m/m
Comparing this number to a channel slope of S0=0.01 m/m we see that the slope
corresponding to a 1.0 percent difference between depths is more than an order of
magnitude larger.
1.2. Using Bernoulli’s equation, write the energy balance in general terms for flow in an open
channel from location 1 to 2 where hL is the head loss between these two locations.
Simplify the equation by taking the perspective of a point on the water surface at both
locations. Note: your solution should show that the pressure term from Bernoulli’s
equation is not relevant for open channel flow.
Solution:
p1v12 p 2 v22
+ + z1 = + + z 2 + hL
2g 2g
If we take a point on the water surface at both locations, the p1 equals p2 equals
atmospheric pressure, and thus these terms may be cancelled from both sides of the
equality,
v12 v2
+ z1 = 2 + z 2 + hL
2g 2g
The remaining equation if y is substituted for z and if hL is set to zero, forms the basis for
the specific energy equation which is the focus for Chapter 2.
-1-
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, Chapter 1: Introductory Material
1.3. Parts (a), (b), and (c) require simple multiplication/division and/or addition/subtraction to
solve. The reader is cautioned to pay special attention to significant digits when reporting
the final answer.
a. If the density of water is 1000 kg/m3 and gravitational acceleration is 9.81 m/s2, what
is the unit weight of water?
b. If the density of water is 1.0 103 kg/m3 and gravitational acceleration is 9.81 m/s2,
what is the unit weight of water?
c. The cross-sectional area of a channel is broken into three separate subareas with the
following sizes: 1.3 m2, 0.92 m2, and 15 m2. What is the total cross-sectional area of
the channel?
Solution:
a) The unit weight of water is the product of density and gravitational acceleration so,
= g = (1000) (9.81) = 9810 N
Since density is given with one significant figure. The answer has one significant figure
resulting in: 10,000 N.
b) The new statement gives density with two significant figures, so the answer becomes:
9800 N.
c) The calculator-based sum of the three provided numbers is 17.22. However, the number
“15” indicates uncertainty in the “ones” place of the number. This same uncertainty
needs to be conveyed in the answer, so the correct answer is 17 m2.
1.4. The mean or bulk velocity of flow in a stream is observed to be 1.1 m/s. A rock tossed
into this same flow sets up ripples that radiate outward in all directions. It is noted that
the ripples propagating directly upstream travel at a velocity of 0.67 m/s in the opposite
direction to the direction of the flowing stream.
a. What is the Froude number for this flow?
b. Estimate the depth of flow in this stream.
Solution:
a) The wave velocity (velocity of ripple propagation) provided is the net velocity, equal
to the velocity of wave propagation in a still pool of water minus the bulk velocity
downstream. The wave velocity is 1.1 + 0.67 = 1.8 m/s. Using the definition of the
Froude number:
v 1.1
Fr = = = 0.61
gy 1.8
b) The depth of flow in the stream can be estimated based on the wave velocity, vw = 1.8
m/s.
-2-
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, Chapter 1: Introductory Material
(v w )2 (1.8)
2
y= = = 0.33 m
g 9.81
1.5. In the final chapter of this book, we study sediment transport. In a particular stream, it is
found that the sediment transport rate can be approximated as
Qs = c( 0 − * ) p
where c, p, and * are positive constants.
a. Write an analytical expression for the sensitivity, dQs d 0 .
b. Let c = 1, p = 2.2, and * be 1.4.
i. Plot dQs d 0 for 0 1.4.
ii. Determine the value of the sensitivity at 0 = 1.5 and 0 = 1.7.
c. Briefly discuss how “Rule 3” as presented in Section 1.5 relates to your findings in
Part (b) of this problem.
Solution:
a) We take the derivative of the sediment transport rate with respect to 0:
dQs
=
d 0 d 0
d
( )
c 0 − * = p c 0 − *
p p −1
( )
b)
i. The figure below shows (from top to bottom) the sediment transport function
itself (not requested), the absolute sensitivity function (requested), and the relative
dQ 0
sensitivity function (not requested but defined as s ).
d 0 Qs
-3-
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, Chapter 1: Introductory Material
This figure shows that the absolute sensitivity increases gradually from zero once
0 exceeds *. In contrast, the relative sensitivity spikes just beyond * because,
the value of Qs has increased infinitely from zero to a number greater than zero.
Other changes at 0 increases further are not nearly so dramatic.
ii. At 0 =1.5 the absolute sensitivity is:
dQs
d 0
(
= p c 0 − *
p −1
)
= (2.2)(1.0)(1.5 − 1.4)1.2 = 0.139
While the relative sensitivity is:
dQs 0
= (0.139)
1.5
= 33
d 0 Qs 0.0063
In contrast, at 0=1.7, the absolute sensitivity is:
dQs
== (2.2)(1.0)(1.7 − 1.4)1.2 = 0.519
d 0
And the relative sensitivity is:
dQs 0
= (0.139)
1.7
= 13
d 0 Qs 0.071
The lesson here being that the absolute sensitivity (slope of the sediment transport
function) increases as 0 increases while the relative sensitivity is greatest just as
0 exceeds *.
c) Rule 3 is related to the findings above in that * is a threshold and we see how the
sensitivity shifts from zero to non-zero right at this threshold and how the relative
-4-
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