Fluid Mechanics · Chapter 11
Fluid Mechanics · Chapter 11

Flow over Bodies: Drag & Lift

When a body moves through a fluid — a car, a wing, a golf ball — the flow exerts forces along and across its path: drag and lift. This chapter separates friction drag from pressure drag, tabulates the coefficients engineers use, explains the boundary-layer behavior behind the drag crisis, and traces the origins of aerodynamic lift, stall, and induced drag.

Chapter Objectives

  • Resolve surface pressure and shear into drag and lift using a declared force convention.
  • Select reference area, properties, and coefficients consistent with geometry and Reynolds regime.
  • Compute bluff-body, flat-plate, and lifting-surface forces and required power.
  • Recognize separation, transition, drag crisis, stall, and induced-drag limitations.
  • Verify force estimates through units, momentum balance, regime checks, and uncertainty bounds.
Section 11–1

Introduction

Chapter 8 looked at flow inside pipes; here we look at flow over bodies immersed in a fluid — external flow — and the forces it produces. The boundary layer is free to grow, surrounded by an outer region of nearly inviscid flow.

Drag & Lift

The force component a flow exerts along the flow direction is drag; the component perpendicular to it is lift. Both arise from the combined action of pressure and wall shear stress integrated over the body surface.

Physical meaning

A body-centered system receives distributed pressure and shear from the fluid. An enclosing control volume sees the same net force through momentum-flux change and pressure on its control surface. Drag is defined positive opposite the relative free stream on the body; lift is positive in the chosen transverse direction.

Problem-solving strategy

  1. Sketch the body, relative free stream, force axes, wetted surfaces, and projected or planform reference area.
  2. State gauge or absolute pressure, property temperature, and characteristic length, then calculate \(\mathrm{Re}\) and \(\mathrm{Ma}\).
  3. Select a coefficient or correlation only when its shape, area definition, roughness, and regime match the givens.
  4. Compute forces from dynamic pressure and, when power is requested, project force onto body velocity.
  5. Check signs, dimensions, momentum conservation, sensitivity, and whether separation or transition invalidates the model.
Section 11–2

Drag & Lift

Dimensionless Coefficients

Drag and lift are reported as dimensionless coefficients, normalizing the force by the dynamic pressure and a reference area \(A\). This collapses data across size, speed, and fluid.

Drag & Lift Coefficients
\[ C_D = \frac{F_D}{\tfrac12\rho V^2 A}, \qquad C_L = \frac{F_L}{\tfrac12\rho V^2 A} \]

For drag, \(A\) is usually the frontal area (projected normal to flow); for lift on wings, it is the planform area. Always confirm which reference area a quoted coefficient uses.

Section 11–3

Friction & Pressure Drag

Two Contributions to Drag

Friction (skin) drag is the part due to wall shear stress — dominant for streamlined bodies aligned with the flow. Pressure (form) drag is the part due to the pressure difference between front and back, large when the flow separates and leaves a wide wake — dominant for bluff bodies.

Total Drag
\[ C_D = C_{D,\text{friction}} + C_{D,\text{pressure}} \]

Streamlining

Tapering the rear of a body delays separation and shrinks the wake, slashing pressure drag — though it adds surface area and thus friction drag. There is an optimum shape that minimizes the total, which is why airfoils and fish have their characteristic teardrop form.

Section 11–4

Drag Coefficients of Common Geometries

Tabulated Coefficients

For many shapes, drag coefficients are measured and tabulated as roughly constant over wide ranges of Reynolds number. A flat plate normal to flow has \(C_D\approx1.1\); a smooth sphere about 0.4 (subcritical); a streamlined body can be below 0.1.

Drag from a Tabulated Coefficient
\[ F_D = C_D\,\tfrac12\rho V^2 A \]
EXAMPLE 1Drag on a rooftop sensor enclosure
A service vehicle carries a compact rooftop sensor enclosure; estimate its aerodynamic drag and additional steady propulsion power at the specified road speed.

Recognition strategy

A supplied enclosure coefficient and frontal area identify a direct external-flow coefficient problem; required power follows from force times relative speed.

Given information

  • Supplied wind-tunnel data: \(C_D=0.72\), based on frontal area \(A=0.420\ \text{m}^2\), valid at the operating Reynolds number.
  • Supplied air density at \(30^\circ\text{C}\) and local absolute pressure: \(\rho=1.16\ \text{kg/m}^3\); \(V=14.0\ \text{m/s}\), negligible crosswind.

Model and interpretation

The enclosure is the system; \(+x\) is vehicle motion. Ambient pressure is the zero-gauge reference and cancels over a closed surface. The steady incompressible coefficient model is approximate.

Step-by-step solution
1. Form dynamic pressure. \(q=\rho V^2/2=(1/2)(1.16\ \text{kg/m}^3)(14.0\ \text{m/s})^2=113.68\ \text{Pa}\).
2. Apply the matched coefficient. \(F_D=C_DqA=(0.72)(113.68\ \text{Pa})(0.420\ \text{m}^2)=34.38\ \text{N}\).
3. Assign direction. Fluid drag on the enclosure is \(-34.38\ \text{N}\) in the selected \(+x\) vehicle direction.
4. Determine required power. Steady propulsion must supply \(\dot W=F_DV=(34.38\ \text{N})(14.0\ \text{m/s})=481\ \text{W}\).

Verification

Units: Dynamic pressure is in pascals, pressure times area gives newtons, and force times metres per second gives watts.

Signs and direction: Aerodynamic force opposes \(+x\); positive propulsion power is required to maintain speed, consistent with energy transfer to the wake.

Plausibility: The force is below \(qA=47.7\ \text{N}\) because \(C_D<1\), and the predicted power varies as \(V^3\) while the coefficient remains valid.

Final result: Drag is \(34.4\ \text{N}\) opposite travel, requiring approximately \(0.481\ \text{kW}\) of added propulsion power.

Engineering interpretation

This isolates enclosure drag; vehicle interaction, yaw, mounting interference, density variation, and drivetrain efficiency require separate data before estimating fuel or battery impact.

Section 11–5

Parallel Flow over Flat Plates

Flat-Plate Friction Drag

A plate aligned with the flow experiences only friction drag. The average friction coefficient depends on whether the boundary layer is laminar, turbulent, or a combination, set by the plate-length Reynolds number \(\mathrm{Re}_L\).

Average Friction Coefficient
\[ C_f = \frac{1.33}{\sqrt{\mathrm{Re}_L}}\ \text{(laminar)}, \qquad C_f = \frac{0.074}{\mathrm{Re}_L^{1/5}}\ \text{(turbulent)} \]

The transition from laminar to turbulent on a flat plate typically occurs near \(\mathrm{Re}_x \approx 5\times10^5\). A turbulent boundary layer produces more friction drag but, on a bluff body, resists separation better — a key trade-off revisited below.

The quoted laminar relation assumes steady incompressible flow, negligible pressure gradient, a smooth plate, and a sharp leading edge. Transition is disturbance-sensitive, so a nominal Reynolds threshold is not a universal standard. The turbulent correlation shown is restricted to its stated smooth-plate Reynolds range and should not be extrapolated.

EXAMPLE 2Shear load on an immersed optical window
Water flows along one face of a flush rectangular optical window; estimate average skin-friction coefficient and total tangential force under a forced-laminar test condition.

Recognition strategy

The plate is parallel to a uniform stream, so pressure drag is absent in the ideal geometry and a flat-plate boundary-layer correlation supplies skin friction.

Given information

  • Supplied water properties at \(20^\circ\text{C}\): \(\rho=998\ \text{kg/m}^3\), \(\mu=1.002\times10^{-3}\ \text{Pa}\!\cdot\!\text{s}\).
  • \(U=1.50\ \text{m/s}\), streamwise \(L=0.400\ \text{m}\), width \(b=0.600\ \text{m}\), one wetted face; pressure is \(0\ \text{Pa}\) gauge in the outer stream.

Model and interpretation

Use \(+x\) downstream. The supplied test condition maintains a laminar, zero-pressure-gradient layer despite sensitivity near transition; the Blasius average correlation is approximate and incompressible.

Step-by-step solution
1. Evaluate Reynolds number. \(\mathrm{Re}_L=\rho UL/\mu=5.98\times10^5\), which flags that natural transition uncertainty must be reported.
2. Use the stated laminar model. \(\bar C_f=1.328/\sqrt{\mathrm{Re}_L}=0.001718\), based on one face and length \(L\).
3. Establish reference load. \(q=\rho U^2/2=1122.75\ \text{Pa}\) and wetted reference area is \(A=bL=0.240\ \text{m}^2\).
4. Resolve tangential force. \(F_f=\bar C_fqA=(0.001718)(1122.75)(0.240)=0.463\ \text{N}\), acting downstream on the window.

Verification

Units: Reynolds number and coefficient are dimensionless; pascals times square metres reduce exactly to newtons.

Signs and direction: The fluid pulls the window in \(+x\), while the equal-and-opposite wall shear removes downstream momentum from the fluid.

Plausibility: The shear is a small fraction of \(qA=269\ \text{N}\), but the Reynolds number near transition makes the laminar prediction a lower-confidence estimate.

Final result: \(\bar C_f\approx0.00172\) and the approximate one-sided friction force is \(F_f\approx0.463\ \text{N}\).

Engineering interpretation

Surface contamination or inlet turbulence could trigger transition and increase drag. A professional test should measure transition location and repeat the estimate with a mixed-layer model.

Section 11–6

Flow over Cylinders & Spheres

The Drag Crisis

For a cylinder or sphere, the drag coefficient falls sharply at a critical Reynolds number (~\(2\times10^5\) for a sphere). A turbulent boundary layer clings to the surface longer than a laminar one, moving separation rearward, narrowing the wake, and cutting pressure drag — the famous "drag crisis."

Why golf balls have dimples: the dimples trip the boundary layer to turbulent at a lower Reynolds number, triggering the drag drop early. The ball experiences less pressure drag and flies markedly farther than a smooth ball would.
V → separation point low-pressure wake pressure drag comes from the wake
Figure 11–1. Flow separates from a bluff body and forms a low-pressure wake; the front–back pressure difference is the pressure drag. A turbulent boundary layer delays separation and shrinks the wake. (Original schematic.)
Section 11–7

Lift

How Lift Is Generated

An airfoil at a small angle of attack turns the oncoming flow downward; by Newton’s third law the flow pushes the wing up. Equivalently, the airfoil shape makes the flow faster (lower pressure) over the top than the bottom, and the net pressure difference is the lift.

Lift Force
\[ F_L = C_L\,\tfrac12\rho V^2 A_{\text{planform}} \]

Stall

The lift coefficient rises with angle of attack — until the boundary layer separates from the upper surface. Past this critical angle the wing stalls: lift drops abruptly and drag climbs. Avoiding stall sets the low-speed handling limits of every aircraft.

Induced Drag

On a finite wing, high-pressure air below spills around the tips into the low-pressure region above, forming trailing vortices. The energy shed into these vortices appears as induced drag, the unavoidable cost of producing lift — reduced by high-aspect-ratio wings and winglets.

EXAMPLE 3Preliminary wing operating point
Determine the minimum steady level-flight speed and aerodynamic power at a supplied lift-coefficient limit for an unmanned survey aircraft configuration.

Recognition strategy

Steady level flight requires lift equal to weight; the same dynamic pressure then combines with the supplied total drag coefficient to estimate power.

Given information

  • Supplied at \(25^\circ\text{C}\) and operating altitude: \(\rho=1.18\ \text{kg/m}^3\), weight \(W=18.4\ \text{kN}\), planform \(S=16.0\ \text{m}^2\).
  • Supplier test values at this configuration and Reynolds regime: \(C_{L,\max}=1.45\), total \(C_D=0.092\); ambient pressure is the zero-gauge reference.

Model and interpretation

Take \(+x\) forward and \(+z\) upward. The aircraft is the system, steady and level, with incompressible approximate coefficients; thrust balances drag and lift balances weight.

Step-by-step solution
1. Impose vertical momentum balance. With no vertical acceleration, \(W=C_{L,\max}\rho V_{\min}^2S/2\).
2. Solve the limiting speed. \(V_{\min}=\sqrt{2W/(\rho C_{L,\max}S)}=36.66\ \text{m/s}\).
3. Evaluate drag at that state. \(D=C_D\rho V_{\min}^2S/2=1.167\times10^3\ \text{N}\).
4. Determine ideal propulsive power. \(P_D=DV_{\min}=42.8\ \text{kW}\), excluding propeller and drivetrain inefficiencies.

Verification

Units: The speed expression reduces to metres per second, coefficient force gives newtons, and drag times speed gives watts.

Signs and direction: Lift is positive \(+z\), weight is negative \(z\), drag is negative \(x\), and required thrust and shaft output are positive.

Plausibility: \(D/W=C_D/C_L=0.0634\), matching \(1.167/18.4\); this independent ratio checks both force calculations.

Final result: \(V_{\min}\approx36.7\ \text{m/s}\), \(D\approx1.17\ \text{kN}\), and ideal aerodynamic power is \(42.8\ \text{kW}\).

Engineering interpretation

This is not an approved operating limit: gust margin, stall variability, Reynolds and Mach effects, control authority, propulsion efficiency, mass uncertainty, regulations, and flight-test evidence remain necessary.

Instructor Note

Keep the two reference areas straight: drag uses frontal area, wing lift uses planform area. And resist the popular "equal transit time" myth for lift — the honest explanations are flow turning (Newton) and the pressure field (Bernoulli applied to the faster upper flow). Both are correct; they are two views of the same physics.

Chapter recap

  • Pressure and shear integrate to one resultant that may be resolved into drag and lift.
  • Every coefficient is inseparable from its area definition, geometry, roughness, Reynolds number, and Mach regime.
  • Transition may raise skin friction yet delay separation and reduce bluff-body pressure drag.
  • Before moving on, be able to choose axes and data, calculate force and power, and verify regime, momentum, signs, and uncertainty.