Question 11.17: Water at 70^◦ F flows in a prototype rectangular open channe...

Water at 70^{\circ } F flows in a prototype rectangular open channel with a width of 10 ft, and a spillway with a large head is inserted in the channel, as illustrated in Figure EP 11.17.The head on the spillway is 2.95 ft, the height of the spillway is 3.9 ft, and the width of the crest of the spillway is 10ft. A smaller model of the larger prototype is designed in order to study the flow characteristics of open channel flow in the flow-measuring device. The model fluid is also water at 70^{\circ } F , the ideal discharge (see Equation 9.297 Q_{i} = \int{dQ_{i}} = \int_{0}^{H}{B\sqrt{2g}y^{1/2} dy } = b \sqrt{2g} \left[\frac{y^{3/2}}{\frac{3}{2} } \right]^{H}_{0} = \frac{2}{2} \sqrt{2g} BH^{3/2} ) in the smaller model channel fitted with a spillway is 8.5 ft^{3}/sec , and the model scale, λ is 0.25. (a) Determine the actual discharge in the model spillway. (b) Determine the ideal discharge in the prototype spillway in order to achieve dynamic similarity between the model and the prototype. (c) Determine the actual discharge in the prototype spillway in order to achieve dynamic similarity between the model and the prototype.

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(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}

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