Derive the relation for the discharge
per unit of width q under a sluice gate, as a function of upstream depth y_{1}
and downstream depth y_{3}.
Fig. 312 Discharge under a sluice gate.


The conservation of specific energy between the u/s flow and the d/s flow is:
y_{1} + v_{1}^{2}/(2g) = y_{3} + v_{3}^{2}/(2g)
y_{1} + q^{2}/(2gy_{1}^{2}) = y_{3} + q^{2}/(2gy_{3}^{2})
[q^{2}/(2g)] [ (1/y_{1}^{2})  (1/y_{3}^{2}) ] = y_{3}  y_{1}
q^{2}/(2g) = (y_{1}^{2} y_{3}^{2}) / (y_{1} + y_{3})
q = y_{1} y_{3} [ 2 g / (y_{1} + y_{3}) ] ^{1/2} ANSWER.
Prove that the equation derived in the previous problem (Problem 1) is mathematically equivalent
to the equation based on y_{1}
and y_{2} on used in ONLINE CHANNEL 13.
The two equations are mathematically the same.
Using the specific energy principle, derive the formula for the
dimensionless throat width of a channel constriction that forces critical flow through it [Henderson (1966), p. 267].
σ = b_{c} / b_{1} = (27)^{1/2} F_{1} / (2 + F_{1}^{2}) ^{3/2}
Fig. 313 Critical width constriction using specific energy.


The specific energy for the u/s flow is:
E_{1} = y_{1} + v_{1}^{2}/(2g) = y_{1} [ 1 + (F_{1}^{2}/2) ]
y_{c} = (2/3) E_{1} = (2/3) y_{1} [ 1 + (F_{1}^{2}/2) ]
y_{c} / y_{1} = (2/3) [ 1 + (F_{1}^{2}/2) ] = (1/3) (2 + F_{1}^{2})
From water continuity:
Q = v_{1} y_{1} b_{1} = v_{c} y_{c} b_{c}
σ = b_{c} / b_{1} = (v_{1} y_{1}) / (v_{c} y_{c})
σ = v_{1} / [ v_{c} (y_{c} / y_{1}) ]
σ = v_{1} / [ ( g y_{c} ) ^{1/2} (y_{c} / y_{1}) ]
σ = v_{1} / [ ( g y_{1} ) ^{1/2} (y_{c} / y_{1}) ^{3/2} ]
σ = F_{1} / (y_{c} / y_{1}) ^{3/2}
σ = F_{1} / [ (1/3) (2 + F_{1}^{2}) ] ^{3/2}
σ = (27)^{1/2} F_{1} / (2 + F_{1}^{2}) ^{3/2} ANSWER.
Use ONLINE CHANNEL 17 to calculate the required
throat width b_{c} for the following upstream conditions: y_{1} = 2.2 m, v_{1} = 1.2 m/s,
and b_{1} = 3.2 m. What would be the required throat width if the upstream channel width is b_{1} = 2.2 m?
Q = v_{1} y_{1} b_{1} = 1.2 × 2.2 × 3.2 = 8.448 m^{3}/s
Run ONLINE CHANNEL 17 to calculate b_{c} = 1.445 m. ANSWER.
Q = v_{1} y_{1} b_{1} = 1.2 × 2.2 × 2.2 = 5.808 m^{3}/s
Run ONLINE CHANNEL 17 to calculate b_{c} = 0.994 m. ANSWER.
Using the specific force principle, show that the force f_{o} (in kN/m) exerted by a blunt obstruction at the bottom of a wide rectangular channel is:
f_{o} =  γ y_{1}^{2} { [ (1  α^{ 2}) / 2 ] + [ 1  (1/α ) ]
F_{1}^{2} }
where γ = unit weight of water, F_{1} = upstream Froude number, y_{1} = upstream flow
depth, and α = y_{2}/y_{1}, where y_{2} = downstream flow depth (after the obstruction).
Given q = 1.5 m^{2}/s, v_{1} = 1.0 m/s, and α = 0.91, calculate the force f_{o}.
Assuming the flow is from left to right, in what direction is the force f_{o} acting?
The conservation of specific force leads to the force ON the obstruction:
f_{o}, ON = f_{1}  f_{2}
f_{o}, ON = γ [ q^{2}/(gy_{1}) + (y_{1}^{2}/2) ]  γ [ q^{2}/(gy_{2}) +
(y_{2}^{2}/2) ]
f_{o}, ON = γ [ q^{2}/(gy_{1}) + (y_{1}^{2}/2)  q^{2}/(gy_{2}) 
(y_{2}^{2}/2) ]
f_{o}, ON = γ y_{1}^{2} [ q^{2}/(gy_{1}^{3}) + 1/2
 q^{2}/(gy_{2}y_{1}^{2})  y_{2}^{2}/(2y_{1}^{2}) ]
f_{o}, ON = γ y_{1}^{2} [ q^{2}/(gy_{1}^{3}) + 1/2
 q^{2}/(g y_{1}^{3}α)  (α^{2}/2) ]
f_{o}, ON = γ y_{1}^{2} [ F_{1}^{2} + 1/2  (F_{1}^{2} / α)  (α^{2}/2) ]
f_{o}, ON = γ y_{1}^{2} { [ (1  α^{ 2}) / 2 ] + [ 1  (1/α ) ]
F_{1}^{2} }
The force exerted by the obstruction is: f_{o} =  f_{o}, ON
f_{o} =  γ y_{1}^{2} { [ (1  α^{ 2}) / 2 ] + [ 1  (1/α ) ]
F_{1}^{2} }
y_{1} = q/v_{1} = 1.5 / 1.0 = 1.5 m.
y_{1}^{2} = (1.5)^{2} = 2.25 m^{2}
y_{1}^{3} = (1.5)^{3} = 3.375 m^{3}
γ = 9.806 kN/m^{3}
α = 0.91
F_{1}^{2} = q^{2}/(g y_{1}^{3}) = (1.5)^{2} / (9.806 × 3.375 ) = 0.068
f_{o} =  9.806 (2.25) [ (1  0.8281)/2 + 0.068 (1  (1/0.91) ) ] =  1.748 kN/m ANSWER.
Because f_{o} is negative, it acts in a direction opposite to the flow, i.e., from right to left. ANSWER.
Using the specific force principle, derive the formula for the
dimensionless throat width of a channel constriction that forces critical flow through it [modified from Henderson (1966), p. 267].
σ = b_{c} / b_{3} = (3)^{3/4} F_{3} / (1 + 2 F_{3}^{2}) ^{3/4}
Fig. 314 Critical width constriction using specific force.


The momentumbased equation is derived by equating the specific force at critical flow at the
contraction (Section 2, or Section c) with the specific force at the downstream flow section (Section 3).
M_{3} = M_{2} = M_{c}
 (1) 
The specific force at the downstream flow section is:
q_{3}^{2} y_{3}^{2}
M_{3} = ^{_______} + ^{_______}
g y_{3} 2  (2) 
Replacing the discharge Q:
Q^{2} y_{3}^{2}
M_{3} = ^{___________} + ^{______}
g y_{3} b_{3}^{2} 2  (3) 
From continuity:
Q = v_{3} y_{3} b_{3} = v_{c} y_{c} b_{c}
 (4) 
Thus:
v_{3}^{2} y_{3}^{2} y_{3}^{2}
M_{3} = ^{__________} + ^{______}
g y_{3} 2  (5) 
The Froude number at the downstream section is defined as follows:
v_{3}^{2}
F_{3}^{2} = ^{_______}
g y_{3}
 (6) 
Thus, Eq. 5 reduces to:
1
M_{3} = y_{3}^{2} ( ^{____} + F_{3}^{2} )
2  (7) 
The specific force at the contraction is:
q_{c}^{2} y_{c}^{2}
M_{c} = ^{_______} + ^{_______}
g y_{c} 2 
(8) 
Replacing the discharge Q:
Q^{2} y_{c}^{2}
M_{c} = ^{___________} + ^{______}
g y_{c} b_{c}^{2} 2  (9) 
From continuity:
v_{3}^{2} y_{3}^{2} b_{3}^{2} y_{c}^{2}
M_{c} = ^{_______________} + ^{______}
g y_{c} b_{c}^{2} 2  (10) 
Or:
v_{3}^{2} y_{3}^{3} b_{3}^{2} y_{c}^{2}
M_{c} = ^{_______________} + ^{______}
g y_{c} b_{c}^{2} y_{3} 2  (11) 
Replacing Eq. 6 in Eq. 11:
F_{3}^{2} y_{3}^{3} y_{c}^{2}
M_{c} = ^{___________} + ^{______}
σ^{2} y_{c} 2  (12) 
By definition, the critical depth is:
q^{2}
y_{c} = ( ^{____} ) ^{1/3}
g  (13) 
Or:
Q^{2}
y_{c} = ( ^{________} ) ^{1/3}
b_{c}^{2}g  (14) 
From continuity:
v_{3}^{2} y_{3}^{2} b_{3}^{2}
y_{c} = ( ^{_____________} ) ^{1/3}
b_{c}^{2}g  (15) 
Or:
v_{3}^{2} y_{3}^{3} b_{3}^{2}
y_{c} = ( ^{_____________} ) ^{1/3}
b_{c}^{2}g y_{3}  (16) 
Replacing Eq. 6 in Eq. 16:
F_{3}^{2} y_{3}^{3}
y_{c} = ( ^{_________} ) ^{1/3}
σ^{2}  (17) 
Replacing Eq. 17 in Eq. 12:
F_{3}^{2} y_{3}^{2} σ^{2/3} F_{3}^{2} y_{3}^{3}
M_{c} = ^{______________} + (1/2) ( ^{_________} )^{2/3}
σ^{2} F_{3}^{2/3} σ^{2}  (18) 
Reducing terms:
F_{3}^{4/3}
M_{c} = (3/2) ( ^{________} ) y_{3}^{2}
σ^{4/3}  (19) 
Equating Eqs. 7 and 19:
1 F_{3}^{4/3}
( ^{ ___} + F_{3}^{2} ) = (3/2) ( ^{________} )
2 σ^{4/3}  (20) 
Reducing:
3^{3/4} F_{3}
σ = ^{_____________________}
( 1 + 2 F_{3}^{2} ) ^{3/4}  (21) 
ANSWER.
Using ONLINE LIMITING CONTRACTION,
calculate the limiting contraction ratios using both energy and momentum principles, for a Froude number F = 0.3.
Using the specific energy principle, for F = 0.3: σ = 0.516 ANSWER.
Using the specific force principle, for F = 0.3: σ = 0.604 ANSWER.
Using ONLINE LIMITING CONTRACTION SET,
calculate the limiting contraction ratios using both energy and momentum principles, for Froude numbers is the range 0.1 ≤ F ≤ 2.0, at intervals of 0.1.
Discuss the results.
Using the specific energy principle, σ varies between 0.182 and 1.0, for F = 0.1 to F = 1.0; then, it reduces to σ = 0.707 at F = 2. ANSWER.
Using the specific force principle, σ varies between 0.225 and 1.0, for F = 0.1 to F = 1.0; then, it reduces to σ = 0.877 at F = 2. ANSWER.
A submerged hydraulic jump
occurs immediately downstream of a sluice outlet in a rectangular channel.
Using the momentum principle, prove that the ratio of submerged depth y_{s}
to tailwater depth y_{2} is:
y_{s} y_{2}
^{ _____ } = [ 1 + 2 F_{2}^{2} ( 1  ^{ _____ } ) ]^{ 1/2}
y_{2} y_{1}  (22) 
Fig. 315 A submerged hydraulic jump at a sluice outlet.


The specific force principle applied between the usptream and downstream sections is:
V_{1}^{ 2}y_{1} y_{s}^{ 2}
V_{2}^{ 2}y_{2} y_{2}^{ 2}
^{ _________ } + ^{ ______ } = ^{ ________ } + ^{ ______ }
g 2 g 2
 (23) 
y_{s}^{ 2}
V_{2}^{ 2}y_{2} y_{2}^{ 2} V_{1}^{ 2}y_{1}
^{ ______ } = ^{ ________ } + ^{ ______ }  ^{ ________ }
2 g
2 g
 (24) 
y_{s}^{ 2}
2 V_{2}^{ 2} 2 2 V_{1}^{ 2}y_{1}
^{ ______ } = ^{ _______ } + ^{ _____ }  ^{ __________ }
y_{2}^{ 2} gy_{2}
2 g y_{2}^{ 2}
 (25) 
y_{s}^{ 2}
2 V_{2}^{ 2} 2
2 V_{1}^{ 2}y_{1} V_{2}^{ 2}
^{ ______ } = ^{ _______ } + ^{ _____ }  ^{ _______________ }
y_{2}^{ 2} gy_{2}
2 g y_{2}^{ 2} V_{2}^{ 2}
 (26) 
From continuity:
V_{1} y_{1} = V_{2} y_{2}
 (27) 
y_{s}^{ 2}
y_{2}
^{ ______ } = 2 F_{2}^{2} + 1  2 F_{2}^{2} ^{ _____ }
y_{2}^{ 2}
y_{1}
 (28) 
y_{s} y_{2}
^{ _____ } = [ 1 + 2 F_{2}^{2} ( 1  ^{ _____ } ) ]^{ 1/2}
y_{2} y_{1}  (29) 
ANSWER.