(a) In order to determine the drag force on the model circular cylinder, the drag force equation, Equation 11.91 F_{D} = C_{D} \frac{1}{2} \rho.v^{2} . A , is applied, where the frontal area of the circular cylinder is used to compute the drag force. Because the Reynolds number, R ≤ 200,000 (laminar flow) over a circular cylinder is assumed, the drag coefficient, C_{D} decreases with an increase in R, as illustrated in Figure 10.17 (Stokes law illustrated). Furthermore, in order to determine the geometry L and D of the model circular cylinder, the model scale, λ (inverse of the length ratio) is applied as follows:
D_{P}: = 6 ft L_{p}: = 2000 ft \lambda : = 0.25
Guess value: D_{m}: = 0.1 ft L_{m}: = 0.8 ft
Given
\lambda = \frac{D_{m}}{D_{p}} \lambda = \frac{L_{m}}{L_{p}}
\left(\frac{D_{m}}{L_{m}} \right) : = Find ( D_{m},L_{m}) = \left ( \begin{matrix} 1.5 \\ 500 \end{matrix} \right ) ft
slug: = 1 lb \frac{sec^{2}}{ft} s_{m}: = 0.86 \rho _{w}: = 1.94 \frac{slug}{ft^{3}}
\rho _{m}: =s_{m} . \rho _{w} = 1.668 \frac{slug}{ft^{3}} \mu _{m}: =150 \times 10^{-6} lb \frac{sec}{ft^{2}}
A_{m}: = L_{m} . D_{m} = 750 ft^{2} V_{m}: = 2 \frac{ft}{sec}
R_{m} : = \frac{\rho _{m} .V_{m} .D_{m}}{\mu _{m}} = 3.337 \times 10^{4} C_{Dm}: = 1.0
F_{Dm}: = C_{Dm} \frac{1}{2} \rho _{m} . v_{m}^{2}.A_{m} = 2.503 \times 10^{3} lb
(b) To determine the velocity of flow of the water over the prototype circular cylinder in order to achieve dynamic similarity between the model and the prototype for laminar flow over a circular cylinder, 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{L_{p}}{D_{p}} = 333.333 \frac{L_{m}}{D_{m}} = 333.333
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}}
\rho _{p}: = 1.936 \frac{slug}{ft^{3}} \mu _{p}: = 20.50 \times 10^{-6} lb \frac{sec}{ft^{2}}
Guess value V_{p}: = 10 \frac{ft}{sec} R_{p} : = 100
Given
R_{p} = \frac{\rho _{p} .V_{p} .D_{p}}{\mu _{p}}
R_{p} = R_{m}
\left ( \begin{matrix} V_{p} \\ R_{p} \end{matrix} \right ) : = Find ( V_{p}, R_{p})
V_{p} = 0.059 \frac{ft}{s} R_{p} = 3.337 \times 10^{4}
(c) To determine the drag force on the prototype circular cylinder in order to achieve dynamic similarity between the model and the prototype for laminar flow over a circular cylinder, the drag coefficient, C_{D} must remain a constant between the model and the prototype (which is a direct result of maintaining a constant R and a constant L_{i}/L between the model and the prototype) as follows:
\underbrace{\left[\frac{F_{D}}{\frac{1}{2} \rho v^{2}A} \right]_{p} }_{c_{D_{p}}} = \underbrace{\left[\frac{F_{D}}{\frac{1}{2} \rho v^{2}A} \right]_{m} }_{c_{D_{m}}}
Furthermore, the frontal area of the circular cylinder is used to compute the drag force as follows:
A_{p}: = L_{p} . D_{p} = 1.2 \times 10^{4} ft^{2}
Guess value: F_{Dp}: = 1 lb C_{Dp}: = 0.5
Given
C_{Dp} = \frac{ F_{Dp}}{\frac{1}{2} \rho _{p} . v_{p}^{2}.A_{p}}
C_{Dp} = C_{Dm}
\left ( \begin{matrix} F_{Dp} \\ C_{Dp} \end{matrix} \right ) : = Find (F_{Dp}, C_{Dp})
F_{Dp} = 40.282 lb C_{Dp} = 1
Therefore, although the similarity requirements regarding the independent π term, L_{i}/L and the independent π term, R (“viscosity model”) are theoretically satisfied (R_{p} = R_{m} = 3.337 \times 10^{4} ), the dependent π term (i.e., the drag coefficient, C_{D} ) will actually/practically remain a constant between the model and its prototype (C_{Dp} = C_{Dm} = 1.0) only if it is practical to maintain/attain the model velocity, drag force, fluid, scale, and cost.
