Reynolds Number Drag
Consider a road sign is placed in parallel with the direction of wind flowing 50 ft/s. If the sign’s Reynolds Number (Re) is 2 x 10^4 and its drag force is 0.1 lbf, what is the sign’s projected area in ft^2? Assume the density of air is 0.08 lb/ft^3.
Expand Hint
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D=\frac{1}{2}\rho U^{2}C_{D}A$$$
where
$$C_D$$
is the drag coefficient,
$$U$$
is the flowing fluid or moving object’s velocity,
$$\rho$$
is the fluid density, and
$$A$$
is the projected area of blunt objects with axes perpendicular to the flow.
Hint 2
For flat plates placed parallel with the flow and have
$$10^4 < Re<5\cdot10^5$$
:
$$$C_D=\frac{1.33}{\sqrt{Re}}$$$
where
$$Re$$
is the Reynolds Number.
First, let’s find the road sign’s coefficient of drag. For flat plates placed parallel with the flow and have
$$10^4 < Re<5\cdot10^5$$
:
$$$C_D=\frac{1.33}{\sqrt{Re}}$$$
where
$$Re$$
is the Reynolds Number.
$$$C_D=\frac{1.33}{\sqrt{2\cdot 10^4}}=\frac{1.33}{141.42}=0.0094$$$
Next, the drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D_f=\frac{1}{2}\rho U^{2}C_{D}A$$$
where
$$C_D$$
is the drag coefficient,
$$U$$
is the flowing fluid or moving object’s velocity,
$$\rho$$
is the fluid density, and
$$A$$
is the projected area of blunt objects with axes perpendicular to the flow.
$$$0.1lb_f=\frac{1}{2}(0.08\frac{lb}{ft^3})(50\frac{ft}{s})^{2}(0.0094)A$$$
Before solving for
$$A$$
, recall that
$$1\:lb_f=1\:lb \times gravity\approx 1\:lb \times 32.2 \frac{ft}{s^2}$$
. Thus,
$$$0.1lb \times 32.2\frac{ft}{s^2}=\frac{1}{2}(0.08\frac{lb}{ft^3})(2,500\frac{ft^2}{s^2})(0.0094)A$$$
$$$3.22ft=(\frac{0.94}{ft})A$$$
$$$A=\frac{3.22ft}{\frac{0.94}{ft}}=3.4\:ft^2$$$
$$$3.4\:ft^2$$$
Time Analysis
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