Question 11.16: Water at 70^◦ F flows in a prototype rectangular open channe...

(a) In order to determine the actual discharge in the model sluice gate, 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:

where the C_{D} , or in the case of a sluice gate, 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_{1}/L =y_{1}/a , as illustrated in Figure 9.37. Furthermore, in order to determine the geometry  y_{1} , y_{2} , and a of the model channel and sluice gate, the model scale, λ (inverse of the length ratio) is applied as follows:

b_{p}: = 10 ft                       y_{1p}: = 6.56 ft                       y_{2p}: = 3.11 ft                       a_{p}: = 3.28 ft                       \lambda : = 0.25

Guess value:                b_{m}: = 1 ft                       y_{1m}: = 1 ft                       y_{2m}: = 0.5 ft                       a_{m}: = 0.5 ft

Given

\lambda = \frac{b_{m}}{b_{p}}                       \lambda = \frac{y_{1m}}{y_{1p}}                       \lambda = \frac{y_{2m}}{y_{2p}}                       \lambda = \frac{a_{m}}{a_{p}}
\left ( \begin{matrix} b_{m} \\ y_{1m} \\ y_{2m} \\ a_{m} \end{matrix} \right ) : = Find (b_{m}, y_{1m},y_{2m} , a_{m}) = \left ( \begin{matrix} 2.5 \\ 1.64 \\ 0.778 \\ 0.82 \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}}

V_{2im}: = 8.46 \frac{ft}{sec}                       R_{m}: = \frac{\rho _{m} .V_{2im} .a_{m}}{\mu _{m}} = 6.551 \times 10^{5}

A_{2im}: = b_{m} .a_{m} = 2.05 ft^{2}                       Q_{im}: = V_{2im}. A_{2im} = 17.343 \frac{ft^{3}}{sec}

\frac{y_{1m}}{a_{m}} =2                       C_{dm}: = 0.51                        Q_{am}: =C_{dm}. Q_{im} = 8.845 \frac{ft^{3}}{sec}

(b)–(c) To determine the ideal downstream velocity of flow of the water in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype, and to determine the actual discharge in the prototype sluice gate 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:

where the geometry is modeled as follows:

\frac{y_{1p}}{a_{p}} = 2                       \frac{y_{1m}}{a_{m}} = 2                       \frac{y_{2p}}{a_{p}} = 0.948                       \frac{y_{2m}}{a_{m}} = 0.948

However, because the discharge coefficient, C_{d} is independent of R for a sluice gate, R does not need to remain a constant between the model and the prototype.
(b)–(c) To determine the ideal downstream velocity of flow of the water in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype, and to determine the actual discharge in the prototype sluice gate 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 proto type (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}}

Guess value:                       V_{2ip}: = 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_{2ip}}{\sqrt{g.a_{p}}} = \frac{V_{2im}}{\sqrt{g.a_{m}}}

C_{dp} = C_{dm}                       Q_{ip} = V_{2ip}. A_{2ip}
\left ( \begin{matrix} V_{2ip} \\ Q_{ap} \\ Q_{ip} \\ C_{dp} \end{matrix} \right ) : = Find (V_{2ip}, Q_{ap},Q_{ip} ,C_{dp})

V_{2ip} = 16.92 \frac{ft}{s}                       Q_{ap} = 283.038 \frac{ft^{3}}{sec}                       Q_{ip} = 554.976 \frac{ft^{3}}{sec}                       C_{dp} = 0.51   

Furthermore, the Froude number, F remains a constant between the model and the prototype as follows:

F_{m}: = \frac{V_{2im}}{\sqrt{g.a_{m}}} = 1.647                       F_{p}: = \frac{V_{2ip}}{\sqrt{g.a_{p}}} = 1.647

Therefore, although the similarity requirements regarding the independent π term, L_{i}/L (y_{1p}/a_{p}=y_{1m}/a_{m} = 2 and y_{2p}/a_{p}=y_{2m}/a_{m} = 0.948) and the dependent π term, F (“gravity model”) ( F_{p} = F_{m} = 1.647 ) 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} = 0.51) 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 sluice gate, R does not need to remain a constant between the model and the prototype as follows:

R_{m} = 6.551 \times 10^{5}                       R_{p} : = \frac{\rho _{p} . V_{2ip}. a_{p}}{\mu _{p}} =5.241 \times 10^{6}