(a) In order to determine the actual discharge in the model spillway, the actual discharge equation, Equation 11.158 Q_{a} = \underbrace{\left[\underbrace{\left(\frac{v_{a}}{\sqrt{2g\Delta y } } \right) }_{C_{v}} \underbrace{\left(\frac{A_{a}}{A_{i}} \right) }_{C_{c}} \right] }_{C_{d}} Q_{i} , is applied as follows:
Q_{a} = \underbrace{\left[\underbrace{\left(\frac{v_{a}}{\sqrt{2g\Delta y } } \right) }_{C_{v}} \underbrace{\left(\frac{A_{a}}{A_{i}} \right) }_{C_{c}} \right] }_{C_{d}} Q_{i}
where the C_{D} , or in the case of a spillway, the discharge coefficient, C_{d} , is used to model the flow resistance and is a function of the geometry of the flow-measuring device, L_{i}/L = H/P, as illustrated by the Rehbock formula (see Equation 9.303 C_{d} = 0.605 + \frac{1}{305H} + 0.08 \frac{H}{P} ) as follows:
c_{d} = 0.605 + \frac{1}{305 H} + 0.08 \frac{H}{P}
Furthermore, in order to determine the geometry H, and P of the model channel and spillway, the model scale, λ (inverse of the length ratio) is applied as follows:
B_{p}: = 10 ft H_{p}: = 2.95 ft P_{p}: = 3.9 ft \lambda : = 0.25
Guess value: B_{m}: = 1 ft H_{m}: = 1 ft P_{m}: = 1 ft
Given
\lambda = \frac{B_{m}}{B_{p}} \lambda = \frac{H_{m}}{H_{p}} \lambda = \frac{P_{m}}{P_{p}}
\left ( \begin{matrix} B_{m} \\ H_{m} \\ P_{m} \end{matrix} \right ) : = Find (B_{m}, H_{m}, P_{m}) = \left ( \begin{matrix} 2.5 \\ 0.738 \\ 0.975 \end{matrix} \right ) ft
slug: = 1 lb \frac{sec^{2}}{ft} \rho _{m} : = 1.936 \frac{slug}{ft^{3}} \mu _{m} : = 20.5 \times 10^{-6} lb \frac{sec}{ft^{2}}
Q_{im}: = 8.5 \frac{ft^{3}}{sec} A_{3im}: = B_{m} .H_{m} = 1.844 ft^{2} V_{3im}: = \frac{Q_{im}}{A_{3im}} = 4.61 \frac{ft}{s}
R_{m}: = \frac{\rho _{m} .V_{3im} .H_{m}}{\mu _{m}} = 3.211 \times 10^{5}
C_{dm}: = 0.605 + \frac{1}{305 H} + 0.08 \frac{H}{P} = 0.67 Q_{am}: =C_{dm}. Q_{im} = 5.695 \frac{ft^{3}}{sec}
(b)–(c) To determine the ideal discharge in the prototype spillway in order to achieve dynamic similarity between the model and the prototype, and to determine the actual discharge in the prototype spillway in order to achieve dynamic similarity between the model and the prototype, the geometry L_{i}/L must remain a constant between the model and prototype as follows:
\left(\frac{L_{i}}{L} \right)_{p} = \left(\frac{L_{i}}{L} \right)_{m}
where the geometry is modeled as follows:
\frac{H_{p}}{P_{p}} = 0.756 \frac{H_{m}}{P_{m}} = 0.756
However, because the discharge coefficient, C_{d} is independent of R for a spillway with a large head, R does not need to remain a constant between the model and the prototype.
(b)–(c) To determine the ideal discharge in the prototype spillway in order to achieve dynamic similarity between the model and the prototype, and to determine the actual discharge in the prototype spillway in order to achieve dynamic similarity between the model and the prototype, the discharge coefficient, C_{d} must remain a constant between the model and the prototype (which is a direct result of maintaining a constant L_{i}/L between the model and the prototype and applying the “gravity model” similitude scale ratio; specifically the velocity ratio, v_{r} given in Table 11.2) as follows:
\underbrace{\left[\frac{Q_{a}}{\sqrt{2g\Delta y} (A_{i}) } \right]_{p} }_{C_{D_{p}} = C_{d_{p}}} = \underbrace{\left[\frac{Q_{a}}{\sqrt{2g\Delta y} (A_{i}) } \right]_{m} }_{C_{D_{m}} = C_{d_{m}}}
v_{r} = \frac{v_{p}}{v_{m}} = \frac{\left(\sqrt{gL } \right)_{p} }{\left(\sqrt{ gL} \right)_{m} } = L_{r}^{\frac{1}{2} }
\rho _{p} : = 1.936 \frac{slug}{ft^{3}} \mu _{p} : = 20.5 \times 10^{-6} lb \frac{sec}{ft^{2}} g: = 32.174 \frac{ft}{sec^{2}}
A_{3ip}: = B_{p} .H_{p} = 29.5 ft^{2}
Guess value: V_{3ip}: = 1 \frac{ft}{sec} Q_{ap} : = 1 \frac{ft^{3}}{sec} Q_{ip} : = 1 \frac{ft^{3}}{sec} C_{dp} : = 0.1
Given
C_{dp} = \frac{Q_{ap}}{Q_{ip}} \frac{V_{3ip}}{\sqrt{g.H_{p}}} = \frac{V_{3im}}{\sqrt{g.H_{m}}}
C_{dp} = C_{dm} Q_{ip} = V_{3ip}. A_{3ip}
\left ( \begin{matrix} V_{3ip} \\ Q_{ap} \\ Q_{ip} \\ C_{dp} \end{matrix} \right ) : = Find (V_{3ip}, Q_{ap},Q_{ip} ,C_{dp})
V_{3ip} = 9.22 \frac{ft}{s} Q_{ap} = 182.229\frac{ft^{3}}{sec} Q_{ip} = 272 \frac{ft^{3}}{sec} C_{dp} = 0.67
Furthermore, the Froude number, F remains a constant between the model and the prototype as follows:
F_{m}: = \frac{V_{3im}}{\sqrt{g.H_{m}}} = 0.946 F_{p}: = \frac{V_{3ip}}{\sqrt{g.H_{p}}} = 0.946
Therefore, although the similarity requirements regarding the independent π term, L_{i}/L (H_{p}/P_{p}=H_{m}/P_{m} = 0.756) and the dependent π term, F (“gravity model”) ( F_{p} = F_{m} = 0.946 ) are theoretically satisfied, the dependent π term (i.e., the discharge coefficient, C_{d} ) will actually/practically remain a constant between the model and its prototype ( C_{dp} = C_{dm} = 0.67) only if it is practical to maintain/attain the model velocity, flow depth, fluid, scale, and cost. Furthermore, because the discharge coefficient, C_{d} is independent of R for a spillway with a large head, R does not need to remain a constant between the model and the prototype as follows:
R_{m} = 3.211 \times 10^{5} R_{p} : = \frac{\rho _{p} . V_{3ip}. H_{p}}{\mu _{p}} = 2.569 \times 10^{6}
Table 11.2 |
Similitude Scale Ratios for Physical Quantities for a Gravity Model |
Physical
Quantity |
FLT
System |
MLT
System |
Primary Scale Ratios |
Secondary/Similitude Scale Ratios for a Pressure Model |
|
|
|
F_{r} = \frac{F_{G_{p}}}{F_{G_{m}}} = \frac{F_{I_{p}}}{F_{I_{m}}} = constant |
\underbrace{\left[\left(\frac{ v}{\sqrt{gL} } \right)_{p} \right] }_{F_{p}} = \underbrace{\left[\left(\frac { v}{\sqrt{ gL} } \right)_{m} \right] }_{F_{m}} |
Geometrics
Length, L |
L |
L |
L_{r} = \frac{L_{p}}{L_{m}} |
L_{r} = \frac{L_{p}}{L_{m}} |
Area, A |
L^{2} |
L^{2} |
L_{r}^{2} = \frac{L_{p}^{2}}{L_{m}^{2}} |
L_{r}^{2} = \frac{L_{p}^{2}}{L_{m}^{2}} |
Volume, V |
L^{3} |
L^{3} |
L_{r}^{3} = \frac{L_{p}^{3}}{L_{m}^{3}} |
L_{r}^{3} = \frac{L_{p}^{3}}{L_{m}^{3}} |
Kinematics
Time, T |
T |
T |
T_{r} = \frac{L_{r}}{v_{r}} |
T_{r} = \frac{L_{r}}{v_{r}} = L_{r}^{1/2} |
Velocity, v |
LT^{-1} |
LT^{-1} |
v_{r} = \frac{v_{p}}{v_{m}} |
v_{r} = \frac{v_{p}}{v_{m}} = \frac{\left(\sqrt{gL}\right) _{p} }{\left(\sqrt{gL}\right) _{m}} = L_{r}^{1/2} |
Acceleration, a |
LT^{-2} |
LT^{-2} |
a_{r} = \frac{L_{r}}{T_{r}^{2}} = \frac{v_{r}^{2}}{L_{r}} |
a_{r} = \frac{v_{r}^{2}}{L_{r}} =1 |
Discharge, Q |
L^{3}T^{-1} |
L^{3}T^{-1} |
|
Q_{r} = v_{r}. L_{r}^{2} = L_{r}^{5/2} |
Dynamics
Mass, M |
FL^{-1}T^{2} |
M |
|
M_{r} = F_{r}a_{r}^{-1} = \rho _{r} L_{r}^{3} |
Force, F |
F |
MLT^{-2} |
F_{r} = \frac{F_{G_{p}}}{F_{G_{m}}} = \frac{F_{I_{p}}}{F_{I_{m}}} |
F_{r} = \rho_{r} L_{r}^{3} g_{r} = \rho _{r} v^{2}_{r} L_{r}^{2} |
Pressure, p |
FL^{-2} |
ML^{-1}T^{-2} |
|
p_{r} = F_{r} L_{r}^{-2} = \rho _{r} L_{r} |
Momentum, Mv
or Impulse, FT |
FT |
MLT^{-1} |
|
F_{r} T_{r} = \rho _{r} L_{r}^{7/2} |
Energy, E or
Work, W |
FL |
ML^{2}T^{-2} |
|
W_{r} = F_{r} L_{r}= \rho _{r} L_{r}^{4} |
Power, P |
FLT^{-1} |
ML^{2}T^{-3} |
|
p_{r} = W_{r} T_{r}^{-1} = \rho _{r} L^{7/2}_{r} |