Water at 70^◦ F flows over a prototype circular cylinder with a diameter of 6ft and a length of 2000 ft, as illustrated in Figure EP 11.6. A smaller model of the larger prototype is designed in order to study the flow characteristics of laminar flow over the circular cylinder. The model fluid is

Question

Water at [latex] 70^{ \circ} [/latex] flows over a prototype circular cylinder with a diameter of 6ft and a length of 2000 ft, as illustrated in Figure EP 11.6. A smaller model of the larger prototype is designed in order to study the flow characteristics of laminar flow over the circular cylinder. The model fluid is crude oil at [latex] 68^{ \circ} [/latex] (s= 0.86, [latex] \mu = 150 imes 10^{-6} lb - sec/ft^{2} [/latex]), the velocity of flow of the oil over the smaller model circular cylinder is 2 ft/sec, and the model scale, λ is 0.25. (a) Determine the drag force on the model circular cylinder.

(b) 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. (c) Determine the drag force on the prototype circular cylinder in order to achieve dynamic similarity between the model and the prototype.

Accepted Answer

(a) In order to determine the drag force on the model circular cylinder, the drag force equation, Equation 11.91 [latex] F_{D} = C_{D} \frac{1}{2} ho.v^{2} . A [/latex], 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, [latex]C_{D} [/latex] 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:

[latex] D_{P}: = 6 ft [/latex] [latex] L_{p}: = 2000 ft[/latex] [latex] \lambda : = 0.25 [/latex]

Guess value: [latex] D_{m}: = 0.1 ft [/latex] [latex] L_{m}: = 0.8 ft[/latex]

Given

[latex] \lambda = \frac{D_{m}}{D_{p}} [/latex] [latex] \lambda = \frac{L_{m}}{L_{p}} [/latex]
[latex] \left(\frac{D_{m}}{L_{m}} ight) : = Find ( D_{m},L_{m}) = \left ( \begin{matrix} 1.5 \ 500 \end{matrix} ight ) ft [/latex]

[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] s_{m}: = 0.86 [/latex] [latex] ho _{w}: = 1.94 \frac{slug}{ft^{3}} [/latex]

[latex] ho _{m}: =s_{m} . ho _{w} = 1.668 \frac{slug}{ft^{3}} [/latex] [latex] \mu _{m}: =150 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex]

[latex] A_{m}: = L_{m} . D_{m} = 750 ft^{2} [/latex] [latex] V_{m}: = 2 \frac{ft}{sec} [/latex]

[latex] R_{m} : = \frac{ ho _{m} .V_{m} .D_{m}}{\mu _{m}} = 3.337 imes 10^{4} [/latex] [latex] C_{Dm}: = 1.0 [/latex]

[latex] F_{Dm}: = C_{Dm} \frac{1}{2} ho _{m} . v_{m}^{2}.A_{m} = 2.503 imes 10^{3} lb [/latex]

(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 [latex] L_{i}/L [/latex] must remain a constant between the model and prototype as follows:

[latex] \left(\frac{L_{i}}{L} ight)_{p} = \left(\frac{L_{i}}{L} ight)_{m} [/latex]

where the geometry is modeled as follows:

[latex] \frac{L_{p}}{D_{p}} = 333.333 [/latex] [latex] \frac{L_{m}}{D_{m}} = 333.333 [/latex]

Furthermore, the R must remain a constant between the model and prototype as follows:

[latex]\underbrace{\left[\left(\frac{ ho vL}{\mu } ight)_{p} ight] }_{R_{p}} = \underbrace{\left[\left(\frac{ ho vL}{\mu } ight)_{m} ight] }_{R_{m}} [/latex]

[latex] ho _{p}: = 1.936 \frac{slug}{ft^{3}} [/latex] [latex] \mu _{p}: = 20.50 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex]

Guess value [latex] V_{p}: = 10 \frac{ft}{sec} [/latex] [latex] R_{p} : = 100 [/latex]

Given

[latex] R_{p} = \frac{ ho _{p} .V_{p} .D_{p}}{\mu _{p}} [/latex]
[latex] R_{p} = R_{m} [/latex]
[latex] \left ( \begin{matrix} V_{p} \ R_{p} \end{matrix} ight ) : = Find ( V_{p}, R_{p}) [/latex]

[latex] V_{p} = 0.059 \frac{ft}{s} [/latex] [latex] R_{p} = 3.337 imes 10^{4} [/latex]

(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, [latex]C_{D} [/latex] must remain a constant between the model and the prototype (which is a direct result of maintaining a constant R and a constant [latex] L_{i}/L [/latex] between the model and the prototype) as follows:

[latex] \underbrace{\left[\frac{F_{D}}{\frac{1}{2} ho v^{2}A} ight]_{p} }_{c_{D_{p}}} = \underbrace{\left[\frac{F_{D}}{\frac{1}{2} ho v^{2}A} ight]_{m} }_{c_{D_{m}}} [/latex]

Furthermore, the frontal area of the circular cylinder is used to compute the drag force as follows:

[latex] A_{p}: = L_{p} . D_{p} = 1.2 imes 10^{4} ft^{2} [/latex]

Guess value: [latex] F_{Dp}: = 1 lb [/latex] [latex] C_{Dp}: = 0.5 [/latex]

Given

[latex] C_{Dp} = \frac{ F_{Dp}}{\frac{1}{2} ho _{p} . v_{p}^{2}.A_{p}} [/latex]

[latex] C_{Dp} = C_{Dm} [/latex]
[latex] \left ( \begin{matrix} F_{Dp} \ C_{Dp} \end{matrix} ight ) : = Find (F_{Dp}, C_{Dp}) [/latex]

[latex] F_{Dp} = 40.282 lb [/latex] [latex] C_{Dp} = 1 [/latex]

Therefore, although the similarity requirements regarding the independent π term, [latex] L_{i}/L [/latex] and the independent π term, R (“viscosity model”) are theoretically satisfied ([latex]R_{p} = R_{m} = 3.337 imes 10^{4} [/latex]), the dependent π term (i.e., the drag coefficient, [latex]C_{D} [/latex]) will actually/practically remain a constant between the model and its prototype ([latex]C_{Dp} [/latex] = [latex]C_{Dm} [/latex] = 1.0) only if it is practical to maintain/attain the model velocity, drag force, fluid, scale, and cost.

Question 11.6: Water at 70^◦ F flows over a prototype circular cylinder wit...

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