LTR: Drag in External Flows

In addition to the correlations for internal (pipe) flows discussed previously, there are a number of expressions and charts for drag on submersed objects (external flows) in the text.

One of the most famous drag formulas is that of Stoke's drag, used for a viscously dominated flow:

$F_d = F_P + F_\mu = 2\pi \mu R v_\infty + 4\pi \mu R v_\infty = 6\pi^2 \mu R v_\infty$


This expression is useful for viscous fluids, but also for small particles (recall the definition of the Re!). Can you show that the drag coefficient in Stoke's flow is:

$\displaystyle{C_D = \frac{24}{Re}}$

As the flow becomes "faster" the character of the wake behind the blunt body changes and thus the form drag increases On Page 141 a chart shows the behavior of $C_D$ as a function of Re. Why do you think it levels off?!


Use friction factors and/or drag coefficients to calculate drag on submersed objects (external flows) and for internal flows


Calculate the drag on a Nolan Ryan fastball.

As with just about all drag-related problem, we must first calculate the Reynolds number:

$\displaystyle{Re = \frac{VD}{\nu} = \frac{(100mph)(0.073m)(\frac{1609 m/mile}{3600s/hr})}{1.4x10^{-5}m^2/s} = 2.33x10^5}$

For flow past a sphere, we can get the drag coefficient from the chart on page 141: $C_D \approx 0.4$.

If we now want to calculate the drag force from this we use the definition of the drag coefficient:

$\displaystyle{C_D = \frac{F_d/A_p}{\frac12 \rho V^2}}$

Rearranging to get get the drag force, $F_d$:

$\displaystyle{F_d = \frac12 \rho V^2C_DA_p}$

which gives us:

$\displaystyle{F_d = (0.5)(1.2kg/m^3)(45m/s)^2(0.4)\frac{\pi(0.073m)^2}{4} = 2.4 N}$