Question 11.4: (Refer to Example Problem 10.17.) Air at standard atmosphere...

(a) In order to determine the drag force on the prototype square 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 square cylinder is used to compute the drag force. Because the Reynolds number, R > 10,000 (turbulent flow) over a blunt body is assumed, the drag coefficient, C_{D} becomes independent of R, as illustrated in Figure 10.12 for two-dimensional bodies. However, Table 10.4 for two-dimensional bodies presents an easier-to-read magnitude for the drag coefficient, C_{D} for the square cylinder, assuming turbulent flow. Furthermore, Figure 10.19 illustrates that for blunt (square-shaped) two-dimensional bodies, the drag coefficient, C_{D} is independent of M for M ≤ 0.5, while the drag coefficient, C_{D} is dependent on M for M>0.5. The fluid properties for air are given in Table A.1 in Appendix A. The R and M = C^{0.5} are computed as follows:

slug: = 1 lb \frac{sec^{2}}{ft}                                     \rho _{p} : =0.0010663 \frac{slug}{ft^{3}}                                   \mu _{p} : = 0.32166 \times 10^{-6} lb \frac{sec}{ft^{2}}

C_{p}: = 1016.11 \frac{ft}{sec}                                   E_{vp}: = c_{P}^{2} \rho _{P} = 1.101 \times 10^{3} \frac{lb}{ft^{2}}

D_{P}: = 6 ft                                               L_{p}: = 1500 ft                                              V_{P}: = 800 \frac{ft}{sec}

R_{p}: = \frac{\rho _{p} .V_{p} .D_{p}}{\mu _{p}} = 1.591 \times 10^{7}                                   C_{p}: = \frac{\rho _{p} .V_{p}^{2} }{E_{vp}} = 0.62                                   M_{P}: = \sqrt{C_{p}} = 0.787

Thus, since R >10,000, flow is turbulent and thus the drag coefficient, C_{D} is independent of R, and since M> 0.5, the drag coefficient, C_{D} is dependent on M; thus, Figure 10.19 is used to determine the drag coefficient, C_{D} as follows:

C_{Dp}: = 2.1                                   A_{p}: = L_{p} . D_{p} = 9 \times 10^{3} ft^{2}
F_{Dp}: = C_{Dp} \frac{1}{2} \rho _{p}.V^{2}_{p} . A_{p} = 6.449 \times 10^{6} lb

(b) To determine the velocity of flow of the carbon dioxide over the model square cylinder and the required pressure of the carbon dioxide in order to achieve dynamic similarity between the model and the prototype, for turbulent flow over a blunt body, the geometry, L_{i}/L must remain a constant between the model and prototype as follows:

Furthermore, in order to determine the geometry L and D of the model square cylinder, the model scale, λ (inverse of the length ratio) is applied as follows:

Guess value:                                   D_{m}: = 1 ft                                   L_{m}: = 1 ft

Given

\lambda = \frac{D_{m}}{D_{p}}                                   \lambda = \frac{L_{m}}{L_{p}}
\left ( \begin{matrix} D_{m} \\ L_{m} \end{matrix} \right ) : = Find ( D_{m},L_{m}) = \left ( \begin{matrix} 1.2 \\ 300 \end{matrix} \right ) ft

And, the geometry is modeled as follows:

\frac{L_{p}}{D_{p}} = 250                                   \frac{L_{m}}{D_{m}} = 250

And, although the relative roughness, ɛ/L should remain a constant between the model and prototype as follows:

Figure 10.18is for a sphere of diameter, D and not for a square shaped cylinder. As such, the dependence of the drag coefficient, C_{D} on the relative roughness, ɛ/L will not be modeled in the example problem herein. Furthermore, the C (or M) must remain a constant between the model and prototype as follows:

The fluid properties (gas constant and specific heat ratio, which are in dependent of the gas pressure and are only a function of the temperature of the gas) for carbon dioxide are given in TableA.5 in Appendix A. Furthermore, E_{v} for the car bon dioxide is determined by assuming isentropic conditions for the compression and expansion of the gas; thus, E_{v}   = kp, where the pressure is determined by applying the ideal gas law, p = ρRT (see Chapter 1). The sonic velocity for the carbon dioxide, c = \sqrt{E_{v}/\rho } = \sqrt{kRT} , which is independent of the gas pressure and is only a function of the temperature of the gas.

T_{m}: = 68^{\circ } F = 527 \times 67^{\circ } R                                   Rgasc_{m}: = 1123 \frac{ft^{2}}{sec^{2} {^{\circ}}R}                                   k_{m}: = 1.28

Guess value:                                  V_{m}: = 0.1 \frac{ft}{sec}                                   P_{m}: = 1 \frac{lb}{ft^{2}}                                   E_{vm}: = 1 \frac{lb}{ft^{2}}

\rho _{m} : =0.00354 \frac{slug}{ft^{3}}                                   c_{m}: = 1000 \frac{ft}{sec}                                   c_{m}: = 0.5

Given

C_{m} = \frac{\rho _{m} .V_{m}^{2} }{E_{vm}}                                   P_{m} = \rho _{m} . Rgasc_{m} . T_{m}                                   E_{vm}= k_{m}.P_{m}

C_{m} = C_{p}                                   c_{m} = \sqrt{\frac{E_{vm}}{ \rho _{m}} }                                   c_{m} = \sqrt{k_{m}.Rgasc_{m}. T_{m}} 
\left ( \begin{matrix} V_{m} \\ P_{m} \\ E_{Vm} \\ \rho _{m} \\ c_{m} \\ C_{m} \end{matrix} \right ) : = Find (V_{m}, P_{m}, E_{Vm}, \rho _{m}, c_{m}, C_{m})

V_{m} = 685.686 \frac{ft}{s}                                   P_{m} = 79.432 \frac{lb}{ft^{2}}                                   E_{vm} = 101.673 \frac{lb}{ft^{2}}

\rho _{m} = 1.34 \times 10^{-4} \frac{slug}{ft^{3}}                                   c_{m} = 870.916 \frac{ft}{s}                                   C_{m} = 0.62

(c) To determine the drag force on the model square cylinder in order to achieve dynamic similarity between the model and the prototype for turbulent flow over a blunt body, the drag coefficient, C_{D} must remain a constant between the model and the prototype (which is a direct result of maintaining a constant C, and a constant L_{i}/L between the model and the prototype) as follows:

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

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

Given

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

F_{Dm} = 2.382 \times 10^{4} lb                                   C_{Dm} = 2.1

Therefore, although the similarity requirements regarding the independent π term, L_{i}/L and the independent π term, C (“elastic model”) are theoretically satisfied ( C_{p} = C_{m} = 0.62), 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} = 2.1) only if it is practical to maintain/attain the model velocity, drag force, fluid, scale, and cost. Furthermore, because the drag coefficient, C_{D} is independent of R for blunt bodies, R does not need to remain a constant between the model and the prototype as follows:

R_{m}: = \frac{\rho _{m} .V_{m} .D_{m}}{\mu _{m}} = 3.558 \times 10^{5}                                   R_{p}= 1.591 \times 10^{7}

One may ask: Is the model speed too high to be able to maintain for the model? If it is too high, then this is a “distorted model” that needs to be adjusted in order to achieve a “true model” or as close to it as possible. The answer is no, the speed of carbon dioxide under pressure is not too high for the model.

Table 10.4
The Drag Coefficient, C_{D} for Two-Dimensional Bodies at Various Orientations to the Direction of Flow, v for R > 10,000
	Sharp corners:				C_{D}
	L/D	C_{D}		L/D	Laminar	Turbulent
	0.0 *	1.9		1	1.2	0.30
	0.1	1.9		2	0.60	0.20
	0.4	2.3		4	0.35	0.15
	0.5	2.5		8	0.25	0.10
	0.7	2.7
	1.0	2.2
	1.2	2.1
	2.0	1.7
	2.5	1.4
	3.0	1.3
	*Corresponds to thin plate
Rectangular rod			Elliptical rod
	Round front edge:			Sharp corners: C_{D} = 2.2
	L/D	C_{D}
	0.5	1.16
	1.0	0.90		Round corners (r/D = 0.2): C_{D} = 1.2
	2.0	0.70
	4.0	0.68
	6.0	0.64
Rectangular rod			Square rod
Semicircular rod	Semicircular shell		Equilateral triangular rod		Hexagonal rod