Flow past through submerged bodies

For a circular cylinder placed across a stream, the flow pattern and drag coefficient $C_D$ change a lot with the Reynolds number $Re = \rho V D/\mu$, because the nature of the wake and the position of separation change.

Sketch of $C_D$ against $Re$ (log scale)

 CD
 100|\
  10| \
   1|  \___          ____ plateau ~1.2
    |      \____/\__/        \
 0.3|                \        \ drag crisis
    |                 \___/ (CD ~0.3)
    +----+----+----+----+----+----+--> Re (log)
        1   10   10^2 10^3 10^5 3x10^5 10^6

Flow regimes

Explanation

• At low $Re$ drag is mostly viscous friction.
• At moderate $Re$ the drag is mainly pressure (form) drag caused by the wake.
• At the critical $Re$ the turbulent boundary layer carries more momentum, stays attached longer, the wake narrows, and pressure drag falls sharply.
• Vortex shedding frequency $f$ follows the Strouhal number $St = fD/V \approx 0.2$.

In level flight lift equals weight. The power used against drag gives the drag force ($P = F_D V$), and then both coefficients follow from $F = C\cdot\tfrac12\rho S V^2$.

Given

$W = 170$ kN, $S = 25\ \text{m}^2$, $V = 200$ km/h $= 55.56$ m/s, $\rho = 1.25\ \text{kg/m}^3$, engine power $= 580$ kW, of which 70% overcomes drag.

Drag force

$$\begin{aligned} P_D &= 0.7(580) = 406\ \text{kW} = 406\,000\ \text{W} \\ F_D &= \frac{P_D}{V} = \frac{406\,000}{55.56} = 7308\ \text{N} \end{aligned}$$

Dynamic pressure term

$$\tfrac12\rho S V^2 = 0.5(1.25)(25)(55.56)^2 = 48\,225\ \text{N}$$

Coefficients

$$\begin{aligned} C_L &= \frac{W}{\tfrac12\rho S V^2} = \frac{170\,000}{48\,225} = 3.525 \\[4pt] C_D &= \frac{F_D}{\tfrac12\rho S V^2} = \frac{7308}{48\,225} = 0.1515 \end{aligned}$$

Answer: $C_L \approx 3.53$ and $C_D \approx 0.152$. (The high $C_L$ is due to the low speed, so it implies flaps or high-lift devices.)

In level flight lift equals weight. The power used against drag gives the drag force ($P = F_D V$), and then both coefficients follow from $F = C\cdot\tfrac12\rho S V^2$.

Given

$W = 19\,920$ N, $S = 25\ \text{m}^2$, $V = 200$ km/h $= 55.56$ m/s, $\rho = 1.25\ \text{kg/m}^3$, engine power $= 588.5$ kW, of which 80% overcomes drag.

Drag force

$$\begin{aligned} P_D &= 0.8(588.5) = 470.8\ \text{kW} = 470\,800\ \text{W} \\ F_D &= \frac{P_D}{V} = \frac{470\,800}{55.56} = 8474\ \text{N} \end{aligned}$$

Dynamic pressure term

$$\tfrac12\rho S V^2 = 0.5(1.25)(25)(55.56)^2 = 48\,225\ \text{N}$$

Coefficients

$$\begin{aligned} C_L &= \frac{W}{\tfrac12\rho S V^2} = \frac{19\,920}{48\,225} = 0.413 \\[4pt] C_D &= \frac{F_D}{\tfrac12\rho S V^2} = \frac{8474}{48\,225} = 0.1757 \end{aligned}$$

Answer: $C_L \approx 0.413$ and $C_D \approx 0.176$.

At very low Reynolds number ($Re < 1$, creeping flow) inertia is negligible compared with viscous forces, and the drag on a sphere is given by Stokes' law:

$$F_D = 3\pi\mu V d$$

where $\mu$ is the dynamic viscosity, $V$ the velocity of the sphere relative to the fluid, and $d$ the diameter. This holds for $Re = \rho V d/\mu \lesssim 0.2$ to $1$.

Proof that $C_D = 24/Re$

By definition the drag coefficient is the drag divided by the dynamic pressure times the projected area:

$$C_D = \frac{F_D}{\tfrac12\rho V^2 A},\qquad A = \frac{\pi d^2}{4}$$

Substituting Stokes' drag:

$$\begin{aligned} C_D &= \frac{3\pi\mu V d}{\tfrac12\rho V^2\cdot\dfrac{\pi d^2}{4}} = \frac{3\pi\mu V d\times 8}{\rho V^2\pi d^2} \\ &= \frac{24\mu}{\rho V d} = \frac{24}{Re} \end{aligned}$$

Hence for $Re = 0.2$:

$$C_D = \frac{24}{0.2} = 120$$

Result: $F_D = 3\pi\mu V d$ and $C_D = 24/Re$ for the Stokes range.

For take-off, the lift must equal the weight at the take-off speed. This gives the required $C_L$, then the angle of attack from the given relation; the drag and power follow.

Given

$W = 13.5$ kN $= 13\,500$ N, $S = 30\ \text{m}^2$, $V = 30$ m/s, $\rho = 1.29\ \text{kg/m}^3$.

Required lift coefficient

$$L = W = C_L\cdot\tfrac12\rho S V^2 \;\Rightarrow\; C_L = \frac{2W}{\rho S V^2} = \frac{2(13\,500)}{1.29(30)(30)^2} = 0.7752$$

Angle of attack

$$0.35\,(1 + 0.2\alpha) = 0.7752 \;\Rightarrow\; 1 + 0.2\alpha = 2.215 \;\Rightarrow\; \alpha = 6.07^\circ$$

Drag and power

$$\begin{aligned} C_D &= 0.008(1 + 6.07) = 0.05659 \\ F_D &= C_D\cdot\tfrac12\rho S V^2 = 0.05659(0.5)(1.29)(30)(30)^2 = 985.6\ \text{N} \\ P &= F_D\,V = 985.6 \times 30 = 29\,567\ \text{W} \approx 29.6\ \text{kW} \end{aligned}$$

Answer: angle of attack $\alpha \approx 6.07^\circ$; power required for take-off $\approx 29.6$ kW.

The air flowing along both faces of the plate produces a friction (drag) force acting downward on the plate. The extra mass on the counterweight side must balance this force.

Assumptions

Air at 25 °C and 1 atm: $\rho = 1.184\ \text{kg/m}^3$, $\nu = 1.562\times10^{-5}\ \text{m}^2/\text{s}$. The plate is thin and parallel to the flow, so only skin friction acts.

        air 10 m/s
         |  |
     +---v--v---+
     |  plate   |      balance beam
   ==|==========|======o======[ counter weight 2 kg ]

Reynolds number

$$Re_L = \frac{VL}{\nu} = \frac{10(0.5)}{1.562\times10^{-5}} = 3.20\times10^5 < 5\times10^5$$

The boundary layer is laminar over the whole plate.

Drag coefficient (Blasius)

$$C_f = \frac{1.328}{\sqrt{Re_L}} = \frac{1.328}{\sqrt{3.201\times10^5}} = 0.002347$$

Drag force on both surfaces

$$\begin{aligned} F_D &= 2\,C_f\cdot\tfrac12\rho V^2 A,\quad A = 0.5\times0.5 = 0.25\ \text{m}^2 \\ &= 2(0.002347)(0.5)(1.184)(10)^2(0.25) = 0.0695\ \text{N} \end{aligned}$$

Mass to add

$$m = \frac{F_D}{g} = \frac{0.0695}{9.81} = 0.00708\ \text{kg} \approx 7.1\ \text{g}$$

Answer: about 7.1 g must be added to the 2 kg counterweight (total about 2.007 kg).

At terminal velocity, the weight minus buoyancy equals the drag. For a small sphere in a viscous liquid, assume Stokes' law, and check $Re$ afterwards.

Given

$d = 3$ mm $= 0.003$ m, $\gamma_s = 75\ \text{kN/m}^3$, $\gamma_f = 12.5\ \text{kN/m}^3$, $\mu = 0.893\ \text{N s/m}^2$.

Terminal velocity

Net downward force: $(\gamma_s - \gamma_f)\dfrac{\pi d^3}{6}$. Stokes' drag: $F_D = 3\pi\mu V d$.

$$\begin{aligned} 3\pi\mu V d &= (\gamma_s - \gamma_f)\frac{\pi d^3}{6} \\ V &= \frac{(\gamma_s - \gamma_f)\,d^2}{18\mu} = \frac{(75\,000 - 12\,500)(0.003)^2}{18(0.893)} = 0.0350\ \text{m/s} \end{aligned}$$

Check of Reynolds number

$\rho_f = 12\,500/9.81 = 1274\ \text{kg/m}^3$

$$Re = \frac{\rho_f V d}{\mu} = \frac{1274(0.0350)(0.003)}{0.893} = 0.15 < 1$$

So Stokes' law is valid.

Drag force

$$F_D = 3\pi\mu V d = 3\pi(0.893)(0.0350)(0.003) = 8.84\times10^{-4}\ \text{N}$$

Check: $(\gamma_s - \gamma_f)\pi d^3/6 = 62\,500 \times 1.414\times10^{-8} = 8.84\times10^{-4}$ N.

Answer: terminal velocity $\approx 0.035$ m/s (3.5 cm/s); drag force $\approx 0.88$ mN ($8.84\times10^{-4}$ N).

Total drag on a body in a fluid is the sum of pressure (form) drag and friction (skin) drag.

Pressure drag vs friction drag

Why an aerofoil is a streamlined body

 flow ->   bluff body            streamlined body
           ___                     ______
          |   |  wide wake       /      \____
  ---->   |   | ~~~~~~          (               >  narrow wake
          |___|                   \______/----

• A bluff body causes the boundary layer to separate early (the adverse pressure gradient on the rear is steep). This gives a wide, low-pressure wake and large pressure drag.
• An aerofoil has a rounded nose and a long, gradually tapering tail. The rear pressure rises slowly, so the adverse pressure gradient is mild.
• The boundary layer stays attached almost to the trailing edge, the wake is very narrow, and the pressure drag becomes very small.
• The total drag is then mostly friction drag, which is small, so the aerofoil gives high lift with low drag ($C_L/C_D$ is large).

At take-off, lift equals the total weight (aircraft plus cargo). The allowable cargo is the lift minus the empty weight. The power equals drag times speed.

Given

Empty weight $= 1000$ kN, $S = 220\ \text{m}^2$, $V = 300$ km/h $= 83.33$ m/s, $\rho = 1.2\ \text{kg/m}^3$, $C_L = 1.42$, $C_D = 0.17$ at $20^\circ$.

Dynamic pressure term

$$\tfrac12\rho S V^2 = 0.5(1.2)(220)(83.33)^2 = 916\,667\ \text{N}$$

Lift and cargo

$$\begin{aligned} L &= C_L\cdot\tfrac12\rho S V^2 = 1.42(916\,667) = 1\,301\,667\ \text{N} = 1301.7\ \text{kN} \\ W_{total} &= L = 1301.7\ \text{kN} \\ W_{cargo} &= 1301.7 - 1000 = 301.7\ \text{kN} \end{aligned}$$

Drag and power

$$\begin{aligned} F_D &= C_D\cdot\tfrac12\rho S V^2 = 0.17(916\,667) = 155\,833\ \text{N} \\ P &= F_D V = 155\,833(83.33) = 1.2986\times10^{7}\ \text{W} \approx 12.99\ \text{MW} \end{aligned}$$

Answer: allowable cargo weight $\approx 301.7$ kN (about 302 kN); engine power required $\approx 13.0$ MW.

An aerofoil (airfoil) is a streamlined body shaped to produce lift with low drag when placed in a flow, as in wings and blades.

              chord line  c
      leading  <------------------------->  trailing
       edge    ___________________          edge
         (    /       camber line  \___
          \  /_________ mean line ______>
           \/___________________________/
   angle of attack a: between chord line and flow
 flow ->

• Leading edge: the front, rounded edge of the aerofoil.
• Trailing edge: the rear, sharp edge where the flow leaves.
• Chord line: the straight line joining the leading and trailing edges.
• Chord length ($c$): the length of the chord line; it is the reference length for the Reynolds number.
• Camber line (mean line): the line midway between the upper and lower surfaces. Camber is its maximum distance from the chord line.
• Thickness: the maximum distance between upper and lower surfaces, usually given as a percentage of chord.
• Angle of attack ($\alpha$): the angle between the chord line and the direction of the undisturbed flow.
• Span ($b$): the length of the wing perpendicular to the flow.
• Aspect ratio: $AR = b^2/S$ (span squared over plan area).
• Stall angle: the angle of attack at which the flow separates and lift drops suddenly.