(Refer to Example Problem 11.8.) Water at 50^◦ F flows at a velocity of 9 ft/sec over aprototype 30-ft-long streamlined hull with a width of 30 ft and a depth of 4 ft, as illustrated in Figure EP 11.9. A smaller model of the larger prototype is designed in order to study the flow characteristics of

Question

(Refer to Example Problem 11.8.) Water at [latex] 50^{ \circ} F [/latex] flows at a velocity of 9 ft/sec over aprototype 30-ft-long streamlined hull with a width of 30 ft and a depth of 4 ft, as illustrated in Figure EP 11.9. A smaller model of the larger prototype is designed in order to study the flow characteristics of the flow over the streamlined body with wave action at the free surface. The model scale, λ is 0.20. (a) Determine the drag force on the prototype streamlined hull. (b) 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. (c) Determine the drag force on the model streamlined hull in order to achieve dynamic similarity between the model and the prototype.

Accepted Answer

(a) In order to determine the drag force on the prototype streamlined hull, the drag force equation, Equation 11.91 [latex] F_{D} = C_{D} \frac{1}{2} ho.v^{2} . A [/latex], is applied. The drag coefficient, [latex]C_{D} [/latex] is dependent on both R and F. Table 10.5 illustrates the variation of the drag coefficient, [latex]C_{D} [/latex] 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, [latex]C_{D} [/latex] 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.

[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] ho _{p} : = 1.94 \frac{slug}{ft^{3}} [/latex] [latex] \mu _{p} : = 27.35 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex]

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

[latex] R_{p}: = \frac{ ho _{p} .V_{p} .D_{p}}{\mu _{p}} = 1.915 imes 10^{7} [/latex] [latex] C_{DpR}: = 0.04 [/latex] [latex] A_{pfront}: = b_{p} . D_{p} = 120 ft^{2} [/latex]

[latex] F_{DpR}: = C_{DpR} .\frac{1}{2} . ho _{p}.V^{2}_{p} . A_{pfront} = 377.136 lb [/latex]

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

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

(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, [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]

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:

[latex] \lambda : = 0.2 [/latex]

Guess value: [latex] L_{m}: = 10 ft [/latex] [latex] b_{m}: = 5 ft [/latex] [latex] D_{m}: = 1 ft [/latex]

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

And, the geometry is modeled as follows:

[latex] \frac{b_{p}}{L_{p}} =1 [/latex] [latex] \frac{b_{m}}{L_{m}} =1 [/latex] [latex] \frac{D_{p}}{L_{p}} = 0.133 [/latex] [latex] \frac{D_{m}}{L_{m}} = 0.133 [/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]

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

[latex] \underbrace{\left[\left(\frac{v}{\sqrt{gL} } ight)_{p} ight] }_{F_{p}} = \underbrace{\left[\left(\frac{v}{\sqrt{gL} } ight)_{m} ight] }_{F_{m}} [/latex]

Guess value: [latex] V_{m}: = 10 \frac{ft}{sec} [/latex] [latex] V_{m}: = 1 imes 10^{-5} \frac{ft}{sec} [/latex] [latex] F_{m}: = 0.1 [/latex] [latex] R_{m}: = 10,000 [/latex]

[latex] \mu _{m} : = 1 imes 10^{-5} lb \frac{sec}{ft^{2}} [/latex] [latex] ho _{m} : = 1 \frac{slug}{ft^{3}} [/latex]

Given

[latex] F_{m} = \frac{V_{m}}{\sqrt{g.L_{m}} } [/latex] [latex] R_{m} = \frac{ V_{m} .L_{m}}{V _{m}} [/latex] [latex] V_{m} = \frac{\mu _{m}}{ ho _{m}} [/latex]

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

[latex] V_{m} = 4.025 \frac{ft}{s} [/latex] [latex] V_{m} = 1.261 imes 10^{-6} \frac{ft^{2}}{sec} [/latex] [latex] \mu _{m}= 1.77 imes 10^{-6} \frac{lb}{ft^{2}} [/latex]

[latex] ho _{m} = 1.403 \frac{slug}{ft^{3}} [/latex] [latex] F_{m} =0.29 [/latex] [latex] R_{m} = 1.915 imes 10^{7} [/latex]

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

(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, [latex]C_{D} [/latex] on R and F are simultaneously achieved through the determination of a common model velocity, [latex]v_{m} [/latex] and a common model fluid, the drag coefficient, [latex]C_{DR} [/latex] assumes the frontal area, while the drag coefficient, [latex]C_{DF} [/latex] assumes the planform area. Therefore, the total drag is the sum of the two types of drag.

The drag coefficient, [latex]C_{DR} [/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_{DR}}{\frac{1}{2} ho v^{2} A_{front} } ight]_{p} }_{C_{D_{pR}}} = \underbrace{\left[\frac{F_{DR}}{\frac{1}{2} ho v^{2} A_{front} } ight]_{m} }_{C_{D_{mR}}} [/latex]

[latex] A_{mfront}: = b_{m}. D_{m} = 4.8 ft^{2} [/latex]

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

Given

[latex] C_{DmR} = \frac{ F_{DmR}}{\frac{1}{2} . ho _{m} . V_{m}^{2}.A_{m}} [/latex]

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

[latex] F_{DmR} = 0.089 lb [/latex] [latex] C_{DmR} = 0.04 [/latex]

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

[latex] \underbrace{\left[\frac{F_{DF}}{\frac{1}{2} ho v^{2} A_{plan} } ight]_{p} }_{C_{D_{pF}}} = \underbrace{\left[\frac{F_{DF}}{\frac{1}{2} ho v^{2} A_{plan} } ight]_{m} }_{C_{D_{mF}}} [/latex]

[latex] A_{mplan}: = b_{m}. L_{m} = 36 ft^{2} [/latex]

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

Given

[latex] C_{DmF} = \frac{ F_{DmF}}{\frac{1}{2} . ho _{m} . V_{m}^{2}.A_{m}} [/latex]

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

[latex] F_{DmF} = 2.79 imes 10^{-3} lb [/latex] [latex] C_{DmF} = 1.25 imes 10^{-3} [/latex]

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

[latex] F_{Dm}: = F_{DmR} + F_{DmF} = 0.092 lb [/latex]

Therefore, although the similarity requirements regarding the independent π term, [latex] L_{i}/L [/latex]; the independent πterm, R(“viscosity model”); and the independent π term, F (“gravity model”) are theoretically satisfied ([latex]R_{p} = R_{m} = 1.915 imes 10^{6} [/latex], [latex]F_{p} [/latex] = [latex]F_{m} [/latex] = 0.29), 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_{DpR} [/latex] = [latex]C_{DmR} [/latex] = 0.04, and [latex]C_{DpF} [/latex] = [latex]C_{DmF} [/latex] = 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, [latex]v = 1.261 imes 10^{-6} ft^{2}/sec [/latex], 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, [latex]C_{D} [/latex] on R. In this example, R = 1.915 [latex] imes 10^{6} [/latex], 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, [latex]v = 1.261 imes 10^{-6} ft^{2}/sec [/latex] 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, [latex] C_{D} [/latex] for Three-Dimensional Axisymmetric Bodies at Given Orientation to the Direction of Flow, v for R > 10,000 Unless Otherwise Stated
	[latex] C_{D} = 1.05 [/latex]				θ	[latex] C_{D} [/latex]
					[latex] 10^{\circ } [/latex]	0.30
					[latex] 30^{\circ } [/latex]	0.55
					[latex] 60^{\circ } [/latex]	0.80
					[latex] 90^{\circ } [/latex]	1.15
					[latex] R \leq 2 imes 10^{4} [/latex]
Cube, A = [latex] D^{2} [/latex]		[latex] C_{D} [/latex] = 1.10 + 0.02(L/D + D/L) For 1/30 < (L/D) < 30 Rectangular plate, A = LD	Cone, A = [latex] \pi D^{2} [/latex] /4
	Laminar [latex] R \leq 2 imes 10^{5} [/latex] [latex] C_{D} [/latex] = 0.5				[latex] C_{D} [/latex]
	Turbulent: [latex] R \geq 2 imes 10^{6} [/latex] [latex] C_{D} [/latex] = 0.2			L/D	Laminar [latex] R \leq 2 imes 10^{5} [/latex]	Turbulent [latex] R \geq 2 imes 10^{6} [/latex]
				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 = [latex] \pi D^{2} [/latex]/4 (See Fig. 10.18 for [latex] C_{D} [/latex] vs. R for smooth and rough spheres)		Ellipsoid, A = [latex] \pi D^{2} [/latex]/4
	[latex] C_{D} [/latex] = 0.4					[latex] C_{D} [/latex] = 1.1
	[latex] C_{D} [/latex] = 1.2
Hemisphere, A = [latex] \pi D^{2} [/latex]/4		[latex] C_{D} [/latex] = 0.04 Streamlined body, A = [latex] \pi D^{2} [/latex]/4		Thin circular disk, A = [latex] \pi D^{2} [/latex]/4
	L/D	[latex] C_{D} [/latex]			L/D	[latex] C_{D} [/latex]
	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 × [latex] 10^{5} [/latex] )				8	0.99
Short cylinder, vertical, A = LD			Short cylinder, horizontal, A = [latex] \pi D^{2} [/latex]/4

Table A.2
Physical Properties for Water at Standard Sea-Level Atmospheric Pressure as a Function of Temperature
Temperature (θ) [latex]^{\circ } F[/latex]	Density (ρ) [latex]slug/ft^{3} [/latex]	Specific Weight (γ) [latex]Ib/ft^{3}[/latex]	Absolute (Dynamic) Viscosity (μ) [latex]10^{-6} Ib-sec/ft^{3}[/latex]	Kinematic Viscosity (ν) [latex]10^{-6} ft^{2}/sec[/latex]	Surface Tension (σ) lb=ft	Vapor Pressure [latex]( ho _{ u } )[/latex] psia	Bulk Modulus of Elasticity [latex](E_{\upsilon } )[/latex] psi
32	1.940	62.42	37.46	19.31	0.00518	0.0885	293,000
40	1.940	62.43	32.29	16.64	0.00514	0.1220	294,000
50	1.940	62.41	27.35	14.10	0.00509	0.1780	305,000
60	1.938	62.37	23.59	12.17	0.00504	0.2560	311,000
70	1.936	62.30	20.50	10.59	0.00498	0.3630	320,000
80	1.934	62.22	17.99	9.30	0.00492	0.5070	322,000
90	1.931	62.11	15.95	8.26	0.00486	0.6980	323,000
100	1.927	62.00	14.24	7.39	0.00480	0.9490	327,000
110	1.923	61.86	12.84	6.67	0.00473	1.2750	331,000
120	1.918	61.71	11.68	6.09	0.00467	1.6920	333,000
130	1.913	61.55	10.69	5.58	0.00460	2.2200	334,000
140	1.908	61.38	9.81	5.14	0.00454	2.8900	330,000
150	1.902	61.20	9.05	4.76	0.00447	3.7200	328,000
160	1.896	61.00	8.38	4.42	0.00441	4.7400	326,000
170	1.890	60.80	7.80	4.13	0.00434	5.9900	322,000
180	1.883	60.58	7.26	3.85	0.00427	7.5100	318,000
190	1.876	60.36	6.78	3.62	0.00420	9.3400	313,000
200	1.868	60.12	6.37	3.41	0.00413	11.5200	308,000
212	1.860	59.83	5.93	3.19	0.00404	14.6900	300,000
[latex]^{\circ } C[/latex]	[latex]kg/m^{3} [/latex]	[latex]KN/m^{3} [/latex]	[latex]N-sec/m^{2} [/latex]	[latex]10^{-6} m^{2} /sec[/latex]	[latex]N/m[/latex]	[latex]KN/m^{2} abs[/latex]	[latex]10^{6} KN/m^{2} [/latex]
0	999.8	9.805	0.001781	1.785	0.0756	0.611	2.02
5	1000.0	9.807	0.001518	1.519	0.0749	0.872	2.06
10	999.7	9.804	0.001307	1.306	0.0742	1.230	2.10
15	999.1	9.798	0.001139	1.139	0.0735	1.710	2.14
20	998.2	9.789	0.001002	1.003	0.0728	2.340	2.18
25	997.0	9.777	0.000890	0.893	0.0720	3.170	2.22
30	995.7	9.765	0.000798	0.800	0.0712	4.240	2.25
40	992.2	9.731	0.000653	0.658	0.0696	7.380	2.28
50	988.0	9.690	0.000547	0.553	0.0679	12.330	2.29
60	983.2	9.642	0.000466	0.474	0.0662	19.920	2.28
70	977.8	9.589	0.000404	0.413	0.0644	31.160	2.25
80	971.8	9.530	0.000354	0.364	0.0626	47.340	2.20
90	965.3	9.467	0.000315	0.326	0.0608	70.100	2.14
100	958.4	9.399	0.000282	0.294	0.0589	101.330	2.07

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

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