Question 11.13: Water at 70^◦ F flows in a prototype 1700-ft-long concrete r...

(a) In order to determine the friction slope (and head loss) in flow of the water in the model, the major head loss equation, Equation 11.141 h_{f} = \frac{\tau _{w}L}{\gamma R_{h}} = S_{f}L = C_{D}\rho v^{2} \frac{L}{\gamma R_{h}} = \frac{v^{2}L}{C^{2}R_{h}} = f \frac{L}{(4R_{h})} \frac{v^{2}}{2g} = \left(\frac{vn}{R_{h}^{2/3}} \right)^{2}L , is applied as follows:

where the Manning’s roughness coefficient, n is used to model the flow resistance. Empirical calibration of the Manning’s roughness coefficient, n assumes turbulent flow, thus it is independent of R, and is only a function of ɛ/y. The absolute channel roughness, ɛ is indirectly modeled by the type of channel material, as illustrated in Table 8.6, which presents the Manning’s roughness coefficient, n for various channel materials. It important to note that although the derivation/formulation of the Manning’s equation assumes no specific units (SI or BG), the Manning’s roughness coefficient, n has dimensions of L^{-1/3}T . Furthermore, because the Manning’s roughness coefficient, n in Table 8.6 has been provided/calibrated in SI units m^{-1/3} sec, it must be adjusted when using BG units. In order to convert the Manning’s roughness coefficient, n with units of m^{-1/3} sec to units of ft^{-1/3} sec, note that 3.281 ft = 1m; thus:

Thus, for a concrete channel, assume n = 0.012 m^{-1/3} sec, which is converted to BG units as follows:

n_{m}: = 0.012 m^{\frac{-1}{3} } sec                           m: = 3.281 ft                          n_{m} = 8.076 \times 10^{-3} \frac{s}{ft^{0.333}}

Furthermore, in order to determine the length, depth of flow, and the absolute channel roughness of the model channel, the model scale, λ (inverse of the length ratio) is applied. The fluid properties for water are given in Table A.2 in Appendix A.

b_{p}: = 5 ft                           y_{p}: = 3 ft                           L_{p}: = 1700 ft                           \varepsilon _{p} : = 0.01 ft                           \lambda : = 0.25

Guess value:                           b_{m}: = 1 ft                           y_{m}: = 1 ft                           L_{m}: = 1 ft                           \varepsilon _{m}: = 0.01 ft

Given

\lambda = \frac{b_{m}}{b_{p}}                           \lambda = \frac{y_{m}}{y_{p}}                           \lambda = \frac{L_{m}}{L_{p}}                         \lambda = \frac{\varepsilon _{m}}{\varepsilon _{p}}
\left ( \begin{matrix} b_{m} \\ y_{m} \\ L_{m} \\ \varepsilon _{m} \end{matrix} \right ) : = Find (b_{m}, y_{m}, L_{m}, \varepsilon _{m}) = \left ( \begin{matrix} 1.25 \\ 0.75 \\ 425 \\ 2.5 \times 10^{-3} \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_{m}: = 60 \frac{ft}{sec}                            R_{m}: = \frac{\rho _{m} .V_{m} .y_{m}}{\mu _{m}} = 4.25 \times 10^{6}

A_{m}: = b_{m} . y_{m} = 0.938 ft^{2}                           P_{m}: = 2.y_{m} + b_{m} = 2.75 ft                           R_{hm}: = \frac{A_{m}}{P_{m} } = 0.341 ft

Guess value:                          h_{fm}: = 1 ft                           S_{fm}: = 0.01 \frac{ft}{ft}

Given

h_{fm} = \left(\frac{V_{m}n_{m}}{R_{hm}^{\frac{2}{3} }} \right)^{2} L_{m}                                                     S_{fm} = \frac{h_{fm}}{L_{m}}

h_{fm} = 418.998 ft                                                           S_{fm} = 0.986 \frac{ft}{ft}

It is important to note that for uniform flow, the channel bottom slope, S_{o} is equal to the friction slope, S_{f} . Furthermore, in order to satisfy the geometric similarity requirement when applying the model scale (and the dynamic similarity requirement), the prototype channel bottom slope, S_{op} must equal the model channel bottom slope, S_{om} ; thus, the prototype friction slope, S_{fp} must equal the model friction slope, S_{fm} (see (b), (c) below).
(b)–(c) To determine the velocity flow of the water in the prototype open channel flow in order to achieve dynamic similarity between the model and the prototype for turbulent open channel flow, and to determine the friction slope (and head loss) in the flow of the water in the prototype in order to achieve dynamic similarity between the model and the prototype, for turbulent open channel flow, the ɛ/y must remain a constant between the model and prototype as follows:

\left(\frac{\varepsilon}{y} \right)_{p} = \left(\frac{\varepsilon }{y} \right)_{m}
\frac{\varepsilon _{p}}{y_{p}} = 3.333 \times 10^{-3}                           \frac{\varepsilon _{m}}{y_{m}} = 3.333 \times 10^{-3}

However, because the Manning’s roughness coefficient, n is independent of R, R does not need to remain a constant between the model and the prototype.
(b)–(c) To determine the velocity flow of the water in the prototype open channel flow in order to achieve dynamic similarity between the model and the prototype for turbulent open channel flow, and to determine the friction slope (and head loss) in the flow of the water in the prototype, in order to achieve dynamic similarity between the model and the prototype for turbulent open channel flow, the Manning’s roughness coefficient, n must remain a constant between the model and the prototype (which is a direct result of maintaining a constant ɛ/y 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{\frac{h_{f}}{v^{2}L} }{R_{h}^{4/3}} \right]_{p} }_{n^{2}_{p}} = \underbrace{\left[\frac{\frac{h_{f}}{v^{2}L} }{R_{h}^{4/3}} \right]_{m} }_{n^{2}_{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_{p}: = b_{p} . y_{p} = 15 ft^{2}                           P_{p}: = 2.y_{p} + b_{p} = 11 ft                           R_{hp}: = \frac{A_{p}}{P_{p} } = 1.364 ft

Guess value:                           V_{p}: = 1 \frac{ft}{sec}                           h_{fp}: = 1 ft                           S_{fp}: = 0.01 \frac{ft}{ft}
n_{p}: = 0.01 ft^{\frac{-1}{3} }sec

Given

n_{p}^{2} = \frac{h_{fp}}{\left(\frac{V^{2}_{p}.L_{p}}{R_{hp}^{\frac{4}{3} }} \right) }                           \frac{V_{p}}{\sqrt{g.y_{p}} } = \frac{V_{m}}{\sqrt{g.y_{m}} }

n_{p} = n_{m}                           S_{fp} = \frac{h_{fp}}{L_{p}}                           S_{fp} = S_{fm}
\left ( \begin{matrix} V_{p} \\ h_{fp} \\ S_{fp} \\ n_{p} \end{matrix} \right ) : = Find (V_{p}, h_{fp}, S_{fp}, n_{p} )

V_{p} = 120 \frac{ft}{s}                           h_{fp} = 1.676 \times 10^{3} ft                           S_{fp} = 0.986 \frac{ft}{ft}
n_{p} = 8.077 \times 10^{-3}  \frac{s}{ft^{0.333}}

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

F_{m}: = \frac{V_{m}}{\sqrt{g.y_{m}} } = 12.214                           F_{p}: = \frac{V_{p}}{\sqrt{g.y_{p}}} = 12.214

Therefore, although the similarity requirements regarding the independent π term, \varepsilon /y ((\varepsilon /y)_{p} = (\varepsilon /y)_{m} = 3.333 \times 10^{-3}), the dependent π term, F (“gravity model”) ( F_{p} = F_{m} = 12.214 ), and the dependent π term, friction slope, S_{f} (S_{fp} = S_{fm} = 0.986 ft/ft ) are theoretically satisfied, the dependent π term (i.e., the friction factor, f ) will actually/practically remain a constant between the model and its prototype ( n_{p} = n_{m}= 8.076 \times 10^{-3} ft^{-1/3}sec ) only if it is practical to maintain/attain the model velocity, slope, fluid, scale, and cost. Furthermore, because the Manning’s roughness coefficient, n is independent of R, R does not need to remain a constant between the model and the prototype as follows:

R_{m} = 4.25 \times 10^{6}                           R_{p} : = \frac{\rho _{p} . V_{p}. Y_{p}}{\mu _{p}} = 3.4 \times 10^{7}