Solutions Manual for Fundamentals of Open Channel
Flow, 2nd Edition by Glenn Moglen Latest
2025-2026 Update!!
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
x and the plane of the channel then,
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
: p v2 p v2
2 2
s 1
s
1s
z1 s s s s z h
2 L
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,
v2 2v 2
z h
s 1s
z1 s s
2 L
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-
, 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 9810N
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
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, Chapter 1: Introductory Material
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:
1.1
Fr
v 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.
-3-
Flow, 2nd Edition by Glenn Moglen Latest
2025-2026 Update!!
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
x and the plane of the channel then,
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
: p v2 p v2
2 2
s 1
s
1s
z1 s s s s z h
2 L
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,
v2 2v 2
z h
s 1s
z1 s s
2 L
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-
, 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 9810N
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
-2-
, Chapter 1: Introductory Material
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:
1.1
Fr
v 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.
-3-
