JHGFD
SOLUTION MANUAL
Fundamentals of Open Channel Flow
by Glenn E. Moglen
N
2nd Edition
U
R
SE
D
O
C
S
KJHGFDS
, JHGFD
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 andx the plane of the channel then,
= cos
1.01x
N
Thus,
U
= 8.1o
or, in terms of rise/run,
S = tan(8.1o) = 0.14 m/m
R
Comparing this number to a channel slope of S0=0.01 m/m we see that the slope
SE
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
D
locations. Note: your solution should show that the pressure term from Bernoulli’s
equation is not relevant for open channel flow.
O
Solution:
p v2 p2 v22
+ z1 = + +z +h
C
1
+ 1
2g 2g 2 L
If we take a point on the water surface at both locations, the p1 equals p2 equals
S
atmospheric pressure, and thus these terms may be cancelled from both sides of the
equality,
v12 v22 + z + h
+ z1 = 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.
KJHGFDS
, JHGFD
-1-
N
U
R
SE
D
O
C
S
KJHGFDS
, 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?
N
Solution:
a) The unit weight of water is the product of density and gravitational acceleration so,
U
= g = (1000) (9.81) = 9810N
Since density is given with one significant figure. The answer has one significant figure
R
resulting in: 10,000 N.
b) The new statement gives density with two significant figures, so the answer becomes:
9800 N.
SE
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.
D
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.
O
a. What is the Froude number for this flow?
b. Estimate the depth of flow in this stream.
C
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
S
downstream. The wave velocity is 1.1 + 0.67 = 1.8 m/s. Using the definition of the
Froude number:
v = 1.1 = 0.61
Fr =
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-
SOLUTION MANUAL
Fundamentals of Open Channel Flow
by Glenn E. Moglen
N
2nd Edition
U
R
SE
D
O
C
S
KJHGFDS
, JHGFD
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 andx the plane of the channel then,
= cos
1.01x
N
Thus,
U
= 8.1o
or, in terms of rise/run,
S = tan(8.1o) = 0.14 m/m
R
Comparing this number to a channel slope of S0=0.01 m/m we see that the slope
SE
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
D
locations. Note: your solution should show that the pressure term from Bernoulli’s
equation is not relevant for open channel flow.
O
Solution:
p v2 p2 v22
+ z1 = + +z +h
C
1
+ 1
2g 2g 2 L
If we take a point on the water surface at both locations, the p1 equals p2 equals
S
atmospheric pressure, and thus these terms may be cancelled from both sides of the
equality,
v12 v22 + z + h
+ z1 = 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.
KJHGFDS
, JHGFD
-1-
N
U
R
SE
D
O
C
S
KJHGFDS
, 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?
N
Solution:
a) The unit weight of water is the product of density and gravitational acceleration so,
U
= g = (1000) (9.81) = 9810N
Since density is given with one significant figure. The answer has one significant figure
R
resulting in: 10,000 N.
b) The new statement gives density with two significant figures, so the answer becomes:
9800 N.
SE
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.
D
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.
O
a. What is the Froude number for this flow?
b. Estimate the depth of flow in this stream.
C
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
S
downstream. The wave velocity is 1.1 + 0.67 = 1.8 m/s. Using the definition of the
Froude number:
v = 1.1 = 0.61
Fr =
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-
