Question 11.9: (Refer to Example Problem 11.8.) Water at 50^◦ F flows at a ...

(a) In order to determine the drag force on the prototype streamlined hull, the drag force equation, Equation 11.91 F_{D} = C_{D} \frac{1}{2} \rho.v^{2} . A , is applied. The drag coefficient, C_{D} is dependent on both R and F. Table 10.5 illustrates the variation of the drag coefficient, C_{D} with the R for two-dimensional axisymmetric bodies; note that a streamlined body is assumed to represent the streamlined hull, and the frontal area is assumed in the determination of the drag coefficient. Figure 10.21 illustrates the variation of the wave drag coefficient, C_{D} with the F for a streamlined hull; note that the planform area is assumed in the determination of the drag coefficient. The fluid properties for water are given in Table A.2 in Appendix A. Because in this example the model fluid is not specified, it is possible to simultaneously satisfy the dynamic requirements of R and F in the determination of the velocity ratio. However, typically it is difficult to find a practical fluid for the resulting model fluid property, as illustrated in (b) below.

slug: = 1 lb \frac{sec^{2}}{ft}                               \rho _{p} : = 1.94 \frac{slug}{ft^{3}}                               \mu _{p} : = 27.35 \times 10^{-6} lb \frac{sec}{ft^{2}}

g: = 32.174 \frac{ft}{sec^{2}}                               L_{p}: = 30 ft                               b_{p}: = 30 ft                               D_{p}: = 4 ft                               V_{p}: = 9 \frac{ft}{sec}

R_{p}: = \frac{\rho _{p} .V_{p} .D_{p}}{\mu _{p}} = 1.915 \times 10^{7}                               C_{DpR}: = 0.04                               A_{pfront}: = b_{p} . D_{p} = 120 ft^{2}

F_{p}: = \frac{V_{p}}{\sqrt{g.L_{p}} } = 0.29                               C_{DpF}: = 0.00125                               A_{pplan}: = b_{p} . L_{p} = 900 ft^{2}

F_{DpF}: = C_{DpF} .\frac{1}{2} .\rho _{p}.V^{2}_{p} . A_{pplan} = 88.391 lb
F_{Dp}: = F_{DpR} + F_{DpF} = 465.527 lb

(b) To determine the model fluid and the velocity of flow of the model fluid over the model streamlined hull in order to achieve dynamic similarity between the model and the prototype for the flow over a streamlined body, the geometry, L_{i}/L must remain a constant between the model and prototype as follows:

Furthermore, in order to determine the length, L; width, b; and depth, D of the model streamlined hull, the model scale, λ (inverse of the length ratio) is applied as follows:

Guess value:                               L_{m}: = 10 ft                               b_{m}: = 5 ft                               D_{m}: = 1 ft

Given
\lambda = \frac{L_{m}}{L_{p}}                               \lambda = \frac{b_{m}}{b_{p}}                               \lambda = \frac{D_{m}}{D_{p}}
\left ( \begin{matrix} L_{m} \\ b_{m} \\ D_{m} \end{matrix} \right ) : = Find (L_{m}, b_{m}, D_{m}) = \left ( \begin{matrix} 6 \\ 6 \\ 0.8 \end{matrix} \right ) ft

And, the geometry is modeled as follows:

\frac{b_{p}}{L_{p}} =1                               \frac{b_{m}}{L_{m}} =1                               \frac{D_{p}}{L_{p}} = 0.133                               \frac{D_{m}}{L_{m}} = 0.133

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

And, finally, the F must remain a constant between the model and prototype as follows:

Guess value:                               V_{m}: = 10 \frac{ft}{sec}                               V_{m}: = 1 \times 10^{-5} \frac{ft}{sec}                               F_{m}: = 0.1                               R_{m}: = 10,000

\mu _{m} : = 1 \times 10^{-5} lb \frac{sec}{ft^{2}}                               \rho _{m} : = 1 \frac{slug}{ft^{3}}

Given

F_{m} = \frac{V_{m}}{\sqrt{g.L_{m}} }                               R_{m} = \frac{ V_{m} .L_{m}}{V _{m}}                               V_{m} = \frac{\mu _{m}}{\rho _{m}}

F_{m} =F_{p}                               R_{m} =R_{p}
\left ( \begin{matrix} V_{m} \\ V_{m} \\ \mu _{m} \\ \rho _{m} \\ F _{m} \\ R_{m} \end{matrix} \right ) : = Find ( V_{m}, V _{m}, \mu _{m}, \rho _{m}, F _{m}, R_{m})

V_{m} = 4.025 \frac{ft}{s}                               V_{m} = 1.261 \times 10^{-6} \frac{ft^{2}}{sec}                               \mu _{m}= 1.77 \times 10^{-6} \frac{lb}{ft^{2}}

\rho _{m} = 1.403 \frac{slug}{ft^{3}}                               F_{m} =0.29                               R_{m} = 1.915 \times 10^{7}

However, it is difficult to find a practical/convenient model fluid with a kinematic viscosity, v = 1.261 \times 10^{-6} ft^{2}/sec .

(c) To determine the drag force on the model streamlined hull in order to achieve dynamic similarity between the model and the prototype for the flow over a streamlined body, although the dependence of the drag coefficient, C_{D} on R and F are simultaneously achieved through the determination of a common model velocity, v_{m} and a common model fluid, the drag coefficient, C_{DR} assumes the frontal area, while the drag coefficient, C_{DF} assumes the planform area. Therefore, the total drag is the sum of the two types of drag.

The drag coefficient, C_{DR} 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:

Guess  value:                               F_{DmR}: = 1 lb                               C_{DmR}: = 1

Given

C_{DmR} = C_{DpR}
\left ( \begin{matrix} F_{DmR} \\ C_{DmR} \end{matrix} \right ) : = Find (F_{DmR}, C_{DmR})

F_{DmR} = 0.089 lb                               C_{DmR} = 0.04

And, the drag coefficient, C_{DF} must remain a constant between the model and the prototype (which is a direct result of maintaining a constant F and a constant L_{i}/L between the model and the prototype) as follows:

Guess value:                               F_{DmF}: = 1 lb                               C_{DmF}: = 1

Given

C_{DmF} = C_{DpF}
\left ( \begin{matrix} F_{DmF} \\ C_{DmF} \end{matrix} \right ) : = Find (F_{DmF}, C_{DmF})

F_{DmF} = 2.79 \times 10^{-3} lb                               C_{DmF} = 1.25 \times 10^{-3}

with the total drag being the sum of the two types of drag as follows:

Therefore, although the similarity requirements regarding the independent π term, L_{i}/L ; the independent πterm, R(“viscosity model”); and the independent π term, F (“gravity model”) are theoretically satisfied (R_{p} = R_{m} = 1.915 \times 10^{6} , F_{p} = F_{m} = 0.29), the dependent π term (i.e., the drag coefficient, C_{D} ) will actually/practically remain a constant between the model and its prototype (C_{DpR} = C_{DmR} = 0.04, and C_{DpF} = C_{DmF} = 0.00125) only if it is practical to maintain/attain the model velocity, drag force, fluid, scale, and cost. However, because it is difficult to find a practical/convenient model fluid with a kinematic viscosity, v = 1.261 \times 10^{-6} ft^{2}/sec , the solution to this type of difficult situation is to realize that due to the wave action at the free surface, it becomes more important to satisfy the dynamic requirement for F. 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 = 1.915 \times 10^{6} , which is large; thus, not modeling the viscous effects may be insignificant. Thus, by not satisfying the dynamic requirement that the model fluid have a kinematic viscosity, v = 1.261 \times 10^{-6} ft^{2}/sec represents not satisfying the R dynamic requirement. Instead, typically, a practical/convenient model fluid such as water is used. It is interesting to compare this example to Example Problem 11.8 above. In Example Problem 11.8 above, the model fluid was assumed to be water, and the effects of R and F were modeled separately, which also resulted in a “distorted model.” However, there were two significantly different model velocities that separately satisfied R and F, and the resulting drag force on the model was significantly higher than in this example problem. Which one is a less “distorted model”?

Table 10.5
The Drag Coefficient, C_{D} for Three-Dimensional Axisymmetric Bodies at Given Orientation to the Direction of Flow, v for R > 10,000 Unless Otherwise Stated
	C_{D} = 1.05				θ	C_{D}
					10^{\circ }	0.30
					30^{\circ }	0.55
					60^{\circ }	0.80
					90^{\circ }	1.15
					R \leq 2 \times 10^{4}
Cube, A = D^{2}		C_{D} = 1.10 + 0.02(L/D + D/L) For 1/30 < (L/D) < 30 Rectangular plate, A = LD	Cone, A = \pi D^{2} /4
	Laminar R \leq  2 \times 10^{5} C_{D} = 0.5				C_{D}
	Turbulent: R \geq  2 \times 10^{6} C_{D} = 0.2			L/D	Laminar R \leq  2 \times 10^{5}	Turbulent R \geq  2 \times 10^{6}
				0.75	0.50	0.20
				1	0.47	0.20
				2	0.27	0.13
				4	0.25	0.10
				8	0.20	0.08
Sphere, A = \pi D^{2} /4 (See Fig. 10.18 for C_{D} vs. R for smooth and rough spheres)		Ellipsoid, A = \pi D^{2} /4
	C_{D} = 0.4					C_{D} = 1.1
	C_{D} = 1.2
Hemisphere, A =  \pi D^{2} /4		C_{D} = 0.04 Streamlined body, A = \pi D^{2} /4		Thin circular disk, A = \pi D^{2} /4
	L/D	C_{D}			L/D	C_{D}
	1	0.64			0.5	1.15
	2	0.68
	3	0.72			1	0.90
	5	0.74
	10	0.82			2	0.85
	20	0.91
	40	0.98			4	0.87
	∞	1.20
	Laminar flow (R ≤ 2 × 10^{5} )				8	0.99
Short cylinder, vertical, A = LD			Short cylinder, horizontal, A = \pi D^{2} /4