Question 10.17: Figure 10.19 illustrates the variation of the drag coefficie...

(a)–(b) The frontal area is used to compute the drag force for the square cylinder, and the drag force is determined by applying Equation 10.12 F_{D} = C_{D} \frac{1}{2} \rho v^{2}A . The fluid properties for air at standard atmosphere at an altitude of 25,000ft above sea level is given in Table A.1 in Appendix A.
(a) The drag coefficient and the drag force for a velocity of flow over the cylinder of 400 ft/sec are determined as follows:

slug: = 1 lb \frac{sec^{2}}{ft}                               L: = 1500 ft                               D: = 6 ft                               A_{front}: = L.D = 9 \times 10^{3} ft^{2}

\rho : = 0.0010663 \frac{slug}{ft^{3}}                               c: = 1016.11 \frac{ft}{sec}                               E_{V}: = C^{2}. \rho = 1.101 \times 10^{3} \frac{lb}{ft^{2}}

V: = 400 \frac{ft}{sec}                                       M: = \frac{V}{\sqrt{\frac{E_{V}}{\rho } } } = 0.394                               C_{D}: = 2

Guess value:                              F_{D}: = 1 lb

Given

F_{D} = C_{D} \frac{1}{2} \rho.V^{2} . A_{front}
F_{D}: = Find (F_{D}) = 1.535 \times 10^{6} lb

(b) The drag coefficient and the drag force for a velocity of flow over the cylinder of 800 ft/sec are determined as follows:

V: = 800 \frac{ft}{sec}                               M: = \frac{V}{\sqrt{\frac{E_{V}}{\rho } } } = 0.787                               C_{D}: = 2.1

Guess value:                              F_{D}: = 1 lb

Given

F_{D} = C_{D} \frac{1}{2} \rho.V^{2} . A_{front}
F_{D}: = Find (F_{D}) = 6.449 \times 10^{6} lb

Therefore, although the drag coefficient, C_{D} increases (from 2 to 2.1) with an increase in the velocity, v (from 400 ft/sec to 800 ft/sec) and thus an increase in the Mach number, M (from 0.384 to 0.787), the drag force, F_{D} increases (from 1.535 \times 10^{10} lb to 6.449 \times 10^{11} lb) with an increase in velocity, as F_{D} is directly proportional to V^{2} .