, CHAPTER 1. Basic Principles
1.1. Classify each of the following flows as steady or unsteady from the viewpoint of the observer:
Flow Observer
(a) Flow of river around bridge piers (1) Standing on bridge
(2) In boat, drifting
(b) Movement of flood surge downstream (1) Standing on bank
(2) Moving with surge
Solution.
(a) Over relatively short time intervals, the observer standing on the bridge will see a steady flow
even though the flow may be unsteady over longer time intervals; however, the observer drifting
in the boat will see an unsteady flow as the boat passes under the bridge because the velocity
increases around the bridge piers even if the approach flow is steady.
(b) An observer standing on the river bank will see an unsteady flow as the surge passes but a
steady flow while riding on the surge if the flow is uniform in the direction of movement.
1.2. At the crest of an ogee spillway as shown in Figure 1.1c, would you expect the pressure on the
face of the spillway to be greater than, less than, or equal to the hydrostatic value? Explain your
answer.
Solution.
The convex curvature near the crest of the spillway results in a centripetal acceleration toward the
center of curvature and a corresponding pressure gradient with decreasing pressure toward the
center of curvature. The decreasing pressure reduces the equivalent hydrostatic pressure for a
parallel flow so that the pressure is less than hydrostatic on the face of the spillway. The decrease
in pressure can be so severe that vapor pressure is reached and cavitation occurs with pitting and
erosion of the concrete spillway surface (see Chapter 6).
1.3. The river flow at an upstream gauging station is measured to be 1500 m3/s, and at another
gauging station 3 km downstream, the discharge is measured to be 750 m3/s at the same instant of
time. If the river channel is uniform with a width of 300 m, estimate the rate of change in the
water surface elevation in meters per hour. Is it rising or falling?
Solution.
Use Equation 1.6, the continuity equation:
∂A ∂y ∂Q 750 − 1500
=B =− ≅− = 0.25 m 2 /s
∂t ∂t ∂x 3000
∂y 0.25 0.25
= = = 8.33 × 10 −4 m/s or 3.0 m/hr (rising)
∂t B 300
1
,Open Channel Hydraulics CHAPTER 1
1.4. A paved parking lot section has a uniform slope over a length of 100 m (in the flow direction)
from the point of a drainage area divide to the inlet grate, which extends across the lot width of 30
m. Rainfall is occurring at a uniform intensity of 10 cm/hr. If the detention storage on the paved
section is increasing at the rate of 60 m3/hr, what is the runoff rate into the inlet grate?
Solution.
Utilize the continuity equation for a finite control volume given by Equation 1.3 for an
incompressible fluid so that the fluid density ρ cancels on both sides of the equation. Then we
have
d∀
= − ∑ Qout + ∑ Qin
dt
10
60 = − Q runoff + × 100 × 30
100 cm/m
Q runoff = 300 − 60 = 240 m 3 /hr
1.5. A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The
channel undergoes a smooth, gradual contraction to a width of 4.5 m.
(a) Calculate the depth and velocity in the contracted section.
(b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.
In each case, identify any assumptions that you make.
Solution.
1
F 2
6m 4.5 m
(a) Apply the energy equation from the approach section 1 to the contracted section 2 with
negligible head losses and assuming a horizontal channel bottom:
V12 q22
y1 + = y2 +
2g 2 gy 22
where q 2 = V 2 y 2 = (6/4.5)q 1 = (6/4.5)(1.5)(3.0) = 6.0 m2/s. Substituting and solving, we have
2
, Open Channel Hydraulics CHAPTER 1
1.5 2 6.0 2
3.0 + = y2 +
19.62 19.62 × y 22
1.835
y2 + = 3.115
y 22
from which y 2 = 2.90 m by trial and error and V 2 = q 2 /y 2 = 6.0/2.90 = 2.07 m/s. Note that there
are two solutions, but this is the subcritical solution and the correct one as discussed in more
detail in Chapter 2.
(b) Apply the momentum equation in the flow direction in which F = the resultant force of the
walls and floor on the flow. Assume a hydrostatic pressure distribution at sections 1 and 2.
Because the transition is horizontal, there is no component of the gravity force in the flow
direction. The momentum equation becomes
y12 y2
− F + b1γ − b2γ 2 = ρQ (V2 − V1 )
2 2
3.0 2 2.90 2
− F + 6.0 × 9810 × − 4.5 × 9810 × = 1000 × (1.5 × 3.0 × 6.0) × (2.07 − 1.5)
2 2
from which F = 63.8 kN.
1.6. A bridge has cylindrical piers 1 m in diameter and spaced 15 m apart. Downstream of the bridge
where the flow disturbance from the piers is no longer present, the flow depth is 2.9 m and the
mean velocity is 2.5 m/s.
(a) Calculate the depth of flow upstream of the bridge assuming that the pier coefficient of
drag is 1.2.
(b) Determine the head loss caused by the piers.
Solution.
In part (a), apply the momentum equation with the control volume boundaries halfway between
the piers; then apply the energy equation in part (b).
D s = 15 m
Fp1 Fp2
1 2
3
1.1. Classify each of the following flows as steady or unsteady from the viewpoint of the observer:
Flow Observer
(a) Flow of river around bridge piers (1) Standing on bridge
(2) In boat, drifting
(b) Movement of flood surge downstream (1) Standing on bank
(2) Moving with surge
Solution.
(a) Over relatively short time intervals, the observer standing on the bridge will see a steady flow
even though the flow may be unsteady over longer time intervals; however, the observer drifting
in the boat will see an unsteady flow as the boat passes under the bridge because the velocity
increases around the bridge piers even if the approach flow is steady.
(b) An observer standing on the river bank will see an unsteady flow as the surge passes but a
steady flow while riding on the surge if the flow is uniform in the direction of movement.
1.2. At the crest of an ogee spillway as shown in Figure 1.1c, would you expect the pressure on the
face of the spillway to be greater than, less than, or equal to the hydrostatic value? Explain your
answer.
Solution.
The convex curvature near the crest of the spillway results in a centripetal acceleration toward the
center of curvature and a corresponding pressure gradient with decreasing pressure toward the
center of curvature. The decreasing pressure reduces the equivalent hydrostatic pressure for a
parallel flow so that the pressure is less than hydrostatic on the face of the spillway. The decrease
in pressure can be so severe that vapor pressure is reached and cavitation occurs with pitting and
erosion of the concrete spillway surface (see Chapter 6).
1.3. The river flow at an upstream gauging station is measured to be 1500 m3/s, and at another
gauging station 3 km downstream, the discharge is measured to be 750 m3/s at the same instant of
time. If the river channel is uniform with a width of 300 m, estimate the rate of change in the
water surface elevation in meters per hour. Is it rising or falling?
Solution.
Use Equation 1.6, the continuity equation:
∂A ∂y ∂Q 750 − 1500
=B =− ≅− = 0.25 m 2 /s
∂t ∂t ∂x 3000
∂y 0.25 0.25
= = = 8.33 × 10 −4 m/s or 3.0 m/hr (rising)
∂t B 300
1
,Open Channel Hydraulics CHAPTER 1
1.4. A paved parking lot section has a uniform slope over a length of 100 m (in the flow direction)
from the point of a drainage area divide to the inlet grate, which extends across the lot width of 30
m. Rainfall is occurring at a uniform intensity of 10 cm/hr. If the detention storage on the paved
section is increasing at the rate of 60 m3/hr, what is the runoff rate into the inlet grate?
Solution.
Utilize the continuity equation for a finite control volume given by Equation 1.3 for an
incompressible fluid so that the fluid density ρ cancels on both sides of the equation. Then we
have
d∀
= − ∑ Qout + ∑ Qin
dt
10
60 = − Q runoff + × 100 × 30
100 cm/m
Q runoff = 300 − 60 = 240 m 3 /hr
1.5. A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The
channel undergoes a smooth, gradual contraction to a width of 4.5 m.
(a) Calculate the depth and velocity in the contracted section.
(b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.
In each case, identify any assumptions that you make.
Solution.
1
F 2
6m 4.5 m
(a) Apply the energy equation from the approach section 1 to the contracted section 2 with
negligible head losses and assuming a horizontal channel bottom:
V12 q22
y1 + = y2 +
2g 2 gy 22
where q 2 = V 2 y 2 = (6/4.5)q 1 = (6/4.5)(1.5)(3.0) = 6.0 m2/s. Substituting and solving, we have
2
, Open Channel Hydraulics CHAPTER 1
1.5 2 6.0 2
3.0 + = y2 +
19.62 19.62 × y 22
1.835
y2 + = 3.115
y 22
from which y 2 = 2.90 m by trial and error and V 2 = q 2 /y 2 = 6.0/2.90 = 2.07 m/s. Note that there
are two solutions, but this is the subcritical solution and the correct one as discussed in more
detail in Chapter 2.
(b) Apply the momentum equation in the flow direction in which F = the resultant force of the
walls and floor on the flow. Assume a hydrostatic pressure distribution at sections 1 and 2.
Because the transition is horizontal, there is no component of the gravity force in the flow
direction. The momentum equation becomes
y12 y2
− F + b1γ − b2γ 2 = ρQ (V2 − V1 )
2 2
3.0 2 2.90 2
− F + 6.0 × 9810 × − 4.5 × 9810 × = 1000 × (1.5 × 3.0 × 6.0) × (2.07 − 1.5)
2 2
from which F = 63.8 kN.
1.6. A bridge has cylindrical piers 1 m in diameter and spaced 15 m apart. Downstream of the bridge
where the flow disturbance from the piers is no longer present, the flow depth is 2.9 m and the
mean velocity is 2.5 m/s.
(a) Calculate the depth of flow upstream of the bridge assuming that the pier coefficient of
drag is 1.2.
(b) Determine the head loss caused by the piers.
Solution.
In part (a), apply the momentum equation with the control volume boundaries halfway between
the piers; then apply the energy equation in part (b).
D s = 15 m
Fp1 Fp2
1 2
3
