Water at 70^{\circ } F flows in a prototype rectangular open channel with a width of 10 ft, and a spillway with a small head is inserted in the channel, as illustrated in Figure EP 11.18. The head on the spillway is 0.95 ft, the height of the spillway is 3.9 ft, and the width of the crest of the spillway is 10 ft. 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 1.5 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.
Question 11.18: Water at 70^◦ F flows in a prototype rectangular open channe...
(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 (and R and W for a small head, H) 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}where the head of the weir/spillway, H term is a direct measure of the Reynolds number, R and the Weber number, W, as the effects of viscosity and surface tension are significant only when H is small, as assumed in this example problem. 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}: = 0.95 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 \\ 0.237 \\ 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}}
\sigma _{m} : = 0.00498 \frac{lb}{ft} Q_{im}: = 1.5 \frac{ft^{3}}{sec} A_{3im}: = B_{m} .H_{m} = 0.594 ft^{2}
V_{3im}: = \frac{Q_{im}}{A_{3im}} = 2.526 \frac{ft}{s}R_{m}: = \frac{\rho _{m} .V_{3im} .H_{m}}{\mu _{m}} = 5.666 \times 10^{4} W_{m}: = \frac{\rho _{m} .V_{3im}^{2} .H_{m}}{\sigma _{m}} = 589.271
C_{dm}: = 0.605 + \frac{1 ft}{305. H_{m}} + 0.08 \frac{H_{m}}{P_{m}} = 0.638 Q_{am}: = C_{dm}. Q_{im} = 0.957 \frac{ft^{3}}{sec}
(b) To determine the ideal 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}} = 0.244 \frac{H_{m}}{P_{m}} = 0.244
Furthermore, the R must remain a constant between the model and prototype as follows:
\underbrace{\left[\left(\frac{\rho vL}{\mu } \right)_{p} \right] }_{R_{p}} = \underbrace{\left[\left(\frac{\rho vL}{\mu } \right)_{m} \right] }_{R_{m}}And, finally, the W must remain a constant between the model and prototype as follows:
\underbrace{\left[\left(\frac{\rho v^{2}L}{\sigma } \right)_{p} \right] }_{W_{p}} = \underbrace{\left[\left(\frac{\rho v^{2}L}{\sigma } \right)_{m} \right] }_{W_{m}}Because in this example the model fluid is specified to be water, it will become impossible to simultaneously satisfy the dynamic requirements of R and W in the determination of the velocity ratio. Specifically, the “viscosity model” requires that the velocity ratio be defined as follows:
v_{r} = \frac{v_{p}}{v_{m}} = \frac{\left(\frac{\mu }{\rho L} \right)_{p} }{\left(\frac{\mu }{\rho L} \right)_{m} } = \mu _{r} \rho _{r}^{-1} L_{r}^{-1 } = v_{r} L_{r}^{-1}However, the “surface tension model” requires that the velocity ratio be defined as follows:
v_{r} = \frac{v_{p}}{v_{m}} = \frac{\left(\sqrt{\frac{\sigma }{\rho L}} \right)_{p} }{\left(\sqrt{\frac{\sigma }{\rho L} } \right)_{m} } = \sigma _{r}^{\frac{1}{2} } \rho _{r}^{\frac{-1}{2} } L_{r}^{\frac{-1}{2}}Thus, equating the two velocity ratios yields the following dynamic requirement:
v_{r} = \frac{\sigma _{r}^{1/2}}{\rho _{r}^{1/2}} L _{r}^{1/2}However, in fact v_{r} \neq (\sigma _{r}^{1/2}/\rho _{r}^{1/2} ) L _{r}^{1/2}as follows:
\rho _{p} : = 1.936 \frac{slug}{ft^{3}} \mu _{p} : = 20.5 \times 10^{-6} lb \frac{sec}{ft^{2}} \sigma _{p} : = 0.00498 \frac{lb}{ft}
V_{m}: = \frac{\mu _{m}}{\rho _{m}} = 1.059 \times 10^{-5} \frac{ft^{2}}{sec} V_{p}: = \frac{\mu _{p}}{\rho _{p}} = 1.059 \times 10^{-5} \frac{ft^{2}}{sec}
L_{r}: = \frac{1}{\lambda } = 4 \sigma _{r}: = \frac{\sigma _{p}}{\sigma _{m}} = 1 \rho _{r}: = \frac{\rho _{p}}{\rho _{m}} = 1
V_{r}: = \frac{V_{p}}{V_{m}} = 1 \frac{\sigma _{r}^{\frac{1}{2} }}{\rho _{r}^{\frac{1}{2} }} . L _{r}^{\frac{1}{2} } = 2
which does not satisfy the dynamic requirement. The solution to this type of difficult situation is to note that because the flow over a weir/spillway typically involves water, the Reynolds number, R is usually high where the viscous effects are small. Thus, it becomes more important to satisfy the dynamic requirement for W. Although ignoring the dynamic requirement for R will result in a “distorted model” (which can be accommodated for by proper interpretation of the model results), the model will be less distorted and more of a “true model” when the R is large, which reduces the dependence of the drag coefficient, C_{d} on R. In this example R_{m} = 5.666 \times 10^{4} , which is large; thus, not modeling the viscous effects may be insignificant.
A_{3ip}: = B_{p} .H_{p} = 9.5 ft^{2}Guess value: V_{3ip}: = 1 \frac{ft}{sec} Q_{ip} : = 1 \frac{ft^{3}}{sec} W_{p} : = 500
Given
W_{p}= \frac{\rho _{p} .V_{3ip}^{2} .H_{p}}{\sigma _{p}}
W_{p} = W_{m} Q_{ip} = V_{3ip}. A_{3ip}
\left ( \begin{matrix} V_{3ip} \\ Q_{ip} \\ W_{p} \end{matrix} \right ) : = Find (V_{3ip}, Q_{ip} ,W_{p})
V_{3ip} = 1.263 \frac{ft}{s} Q_{ip} = 12 \frac{ft^{3}}{sec} W_{p} = 589.271
(c) 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 W, and a constant L_{i}/L between the model and the prototype) 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}}}Guess value: Q_{ap} : = 1 \frac{ft^{3}}{sec} C_{dp} : = 0.1
Given
C_{dp} = \frac{Q_{ap}}{Q_{ip}}
C_{dp} = C_{dm}
\left ( \begin{matrix} Q_{ap} \\ C_{dp} \end{matrix} \right ) : = Find (Q_{ap},C_{dp})
Q_{ap} = 7.66 \frac{ft^{3}}{sec} C_{dp} = 0.638
Therefore, although the similarity requirements regarding the independent π term, L_{i}/L (H_{p}/P_{p}=H_{m}/P_{m} = 0.244 ), and the independent π term, W (“surface tension model”) ( W_{p} = W_{m} = 589.271 ) 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} = 0.638) only if it is practical to maintain/attain the model velocity, flow depth, fluid, scale, and cost. And, thus, application of the actual discharge equation theoretically yields a “true model” only with respect to W. Furthermore, the similarity requirements regarding the independent π term, R (“viscosity model”) ( R_{p} = 1.133 \times 10^{5} \neq R_{m} = 5.666 \times 10^{4} ) are not satisfied and thus application of the actual discharge equation yields a “distorted model” with respect to R as illustrated by this example.
R_{m} = 5.666 \times 10^{4} R_{p} : = \frac{\rho _{p} . V_{3ip}. H_{p}}{\mu _{p}} = 1.133 \times 10^{5}