Learning Objectives:

• Define external flow and contrast it with internal flow.
• Explain the difference between developing flow and fully-developed flow.
• Solve simple engineering problems involving the effects of External (open) flows on bodies of various shapes.

• The stagnation point is the point where the free stream flow divides itself.
• The fluid at the body takes the velocity of the body (no-slip condition).
• A boundary layer is formed at the upper and lower surface of the airfoil.
• The flow in the boundary layer is initially laminar and the transition to turbulence takes place at downstream of the stagnation point, depending on the free stream conditions.
• The turbulent boundary layer grows more rapidly than the laminar layer, thus thickening the boundary layer on the body surface. So, the flow experiences a thicker body compared to original one.
• In the region of increasing pressure (adverse pressure gradient), the flow separation may occur. The fluid inside the boundary layer forms a viscous wake behind the separated points.

Orion 8, CC BY-SA 3.0, via Wikimedia Commons

Olivier Cleynen, CC BY 3.0, via Wikimedia Commons

Solved Example: 64-1-01

The displacement thickness is:

A. The distance measured perpendicular to the boundary by which the free stream is displaced because of formation of boundary layer.

B. The one which has minimum energy loss

C. The thickness at which the velocity is 99% of the free stream velocity.

D. The layer which represents reduction in momentum because of the boundary layer.

Correct Answer: A

Solved Example: 64-1-02

A streamlined body in which:

A. Flow separation is minimized

B. Flow is laminar

C. Drag is minimum

D. Flow is turbulent

Correct Answer: A

Solved Example: 64-1-03

A vehicle travels with a speed of 70 km/hr, with a drag coefficient of 0.24. If the cross-sectional area is 6 square meters, then find out the drag force. Density of air = 1.2 kg/m$^3$

A. 416.5 N

B. 444.6 N

C. 326.5 N

D. 706.2 N

70 kmph = 70 $\times$ $\dfrac{5}{18}$ = 19.44 m/s \begin{align*} F_D &= \dfrac{1}{2} C_D\rho v^2 A\\ &= \dfrac{1}{2} (0.24) \times 1.2 \times (19.44)^2 \times (6)\\ &= 326.5\ \mathrm{N} \end{align*}

Correct Answer: C

Solved Example: 64-1-04

The drag coefficient for flat plate placed parallel to the flow with Re = 200,000 is:

A. 0.000297

B. 0.00297

C. 0.0297

D. 0.297

Here the Re is between 10$^4$ and 5 $\times$ 10$^5$. \begin{align*} C_D &= \dfrac{1.33}{\mathrm{Re}^{0.5}}\\ &= \dfrac{1.33}{(200000)^{0.5}}\\ &= 0.00297 \end{align*}

Correct Answer: B

Solved Example: 64-1-05

A flat plate 1.5 m X 1.5 m moves at 45 km/hour in stationary air of specific weight 11.3 N/m$^3$. If the coefficient of lift is 0.75, then what is the lift force?

A. 120.4 N

B. 151.9 N

C. 180.4 N

D. 225.9 N

\begin{align*} F_L &= \dfrac{1}{2}C_L \rho AV^2\\ &= \dfrac{1}{2}(0.75) \left(\dfrac{11.3}{9.81}\right) (1.5 \times 1.5) (12.5)^2\\ &= 151.9\ \mathrm{N} \end{align*}

Correct Answer: B