Exercises — Drag — pressure (form) drag, skin friction drag

A parachute is a bluff body: the flow cannot follow its cupped shape, it separates immediately and leaves an enormous wake. The front (high pressure) and back (low pressure) do not cancel → almost 100% form (pressure) drag. A plate held edge-on presents almost no frontal area; the flow slides along its faces, so nearly all the resistance is the viscous rub → almost 100% skin friction drag. Rule of thumb: big wake ⇒ form drag; big wetted area with attached flow ⇒ skin friction.

Reach for ${\tau }_w=\mu \frac{du}{dy}{|}_{y=0}$, then $D_{\text{friction}}={\tau }_{w}A$. Why: an aligned flat plate is essentially pure skin friction, and you were handed the velocity gradient at the wall — that is exactly and only what shear stress needs. The $\frac{1}{2}\rho v^{2}A{C}_D$ form is the lumped, experiment-fitted package for when you don't know the microscopic profile.

$D=\frac{1}{2}\rho v^{2}A{C}_D=\frac{1}{2}(1.2)(12^2)(0.4)(0.9)$ Step by step: $12^2=144$; $\frac{1}{2}(1.2)(144)=86.4$ (this is the dynamic pressure, $\frac{1}{2}\rho v^2=86.4\text{Pa}$); then $86.4\times 0.4\times 0.9=31.104$. $\boxed{D\approx 31.1\text{N}}$

The slope of $u$ is constant: $\frac{du}{dy}=500{\text{s}}^{-1}$, so at the wall ${\tau }_w=\mu \frac{du}{dy}=(2.0\times 10^{-5})(500)=1.0\times 10^{-2}\text{Pa}.$ Then $D_{\text{friction}}={\tau }_{w}A=(1.0\times 10^{-2})(1.5)=1.5\times 10^{-2}\text{N}.$ Why only the slope? Shear stress "sees" nothing but how steeply velocity changes at the wall — the no-slip condition pins $u=0$ there, and $\mu$ converts steepness into force per area.

Because $D=\frac{1}{2}\rho v^{2}A{C}_D$ and everything except $v$ is fixed, $D\propto v^2$. So $\frac{D_2}{D_1}={\left(\frac{v_2}{v_1}\right)}^2={\left(\frac{40}{25}\right)}^2=(1.6{)}^2=\mathrm{2.56.}$ $D_2=300\times 2.56=\boxed{768\text{N}}.$ A 60% speed increase gives a 156% drag increase — the square bites hard. (Valid because a car sits at high $Re$, where $C_D$ is roughly constant.)

Let skin friction $=f$. Then form $=4f$, and $f+4f=50\Rightarrow 5f=50\Rightarrow f=10$. $D_{\text{friction}}=10\text{N}(20\%),D_{\text{form}}=40\text{N}(80\%).$ This body is bluff-dominated (most drag is the wake), typical of cars and blunt shapes.

At terminal velocity, drag balances weight: $D=W$, so $\frac{1}{2}\rho v_t^{2}A{C}_D=W\Rightarrow v_t=\sqrt{\frac{2W}{\rho A{C}_D}}.$ Why the square root? Drag grows as $v^2$; to undo that and solve for the balancing $v$ we take the root — that's why the tool is  ​ and not division. $v_t=\sqrt{\frac{2(6)}{(1.2)(0.02)(0.5)}}=\sqrt{\frac{12}{0.012}}=\sqrt{1000}\approx 31.6\text{m/s}.$ Check the regime: at this speed and size, $Re=\rho vL/\mu$ is comfortably in the thousands, so using a constant $C_D$ (and thus $D\propto v^2$) is justified.

The $\frac{1}{2}\rho v^{2}A{C}_D$ already lumps form + friction into $C_D$, and $A$ is the shared frontal reference, so we just compare $C_D$. Dynamic-pressure block: $\frac{1}{2}\rho v^{2}A=\frac{1}{2}(1.2)(20^2)(0.10)=\frac{1}{2}(1.2)(400)(0.10)=24$. $D_{\text{blunt}}=24\times 1.0=24\text{N},D_{\text{fair}}=24\times 0.12=2.88\text{N}.$ Reduction: $\frac{24-2.88}{24}=0.88=\boxed{88\%\text{ less drag}}.$ Yes, hugely worth it. The extra wetted area adds a little skin friction, but killing the wake slashes the dominant form-drag term. You optimise the sum, and here the sum plummets. The figure below makes this visible.

Invert the master formula: $C_D=\frac{2D}{\rho v^{2}A}=\frac{2(5.4)}{(1.2)(15^2)(0.08)}=\frac{10.8}{(1.2)(225)(0.08)}=\frac{10.8}{21.6}=\mathrm{0.5.}$ At $v=30=2\times 15$, drag scales as $v^2$: $D_{30}=5.4\times 2^2=\boxed{21.6\text{N}}$. This is exactly what the drag coefficient is for: measure once, predict at any speed (as long as $Re$ stays high enough that $C_D$ holds steady).

(a) Skin friction. Wall shear on one face: ${\tau }_w=\mu \frac{du}{dy}=(1.0\times 10^{-3})(8000)=8.0\text{Pa}.$ One face: $8.0\times 0.6=4.8\text{N}$. Two faces: $D_{\text{friction}}=2\times 4.8=9.6\text{N}$. (b) Form drag. $D_{\text{form}}=\frac{1}{2}\rho v^2{A}_f{C}_{D,\text{form}}=\frac{1}{2}(1000)(2^2)(0.004)(0.02).$ Compute: $\frac{1}{2}(1000)(4)=2000$; $2000\times 0.004=8$; $8\times 0.02=0.16\text{N}$. (c) Total. $D=9.6+0.16=9.76\text{N}$. Friction fraction $=9.6/9.76=0.9836\approx \boxed{98.4\%}$. An aligned plate is overwhelmingly skin friction — exactly the parent note's claim, now derived from raw numbers.

(a) Solve the master formula for $C_D$: $C_D^{\max }=\frac{2D}{\rho v^{2}A}=\frac{2(400)}{(1.2)(35^2)(2.0)}=\frac{800}{(1.2)(1225)(2.0)}=\frac{800}{2940}\approx \mathrm{0.272.}$ (b) At fixed $A$, drag area scales with $C_D$. Reduction needed: $\frac{0.45-0.272}{0.45}=\frac{0.178}{0.45}\approx 0.395=\boxed{\approx 39.5\%}.$ A stiff but achievable streamlining target.

An inviscid fluid means $\mu =0$, which also means $Re=\rho vL/\mu \to \infty$ — the idealised limit of no stickiness. With $\mu =0$ we get ${\tau }_w=0$: the skin friction term vanishes outright. But the deeper point is the form term. With no viscosity there is no boundary layer to separate, so the flow stays perfectly attached and wraps symmetrically around the sphere. Pressure on the front equals pressure on the back, and $\oint p(\hat{n}\cdot \hat{x})dA=0$: the form drag term also vanishes. Total drag $=0$ — that is d'Alembert's paradox. Lesson: viscosity is the hidden source of both drags. It directly makes skin friction, and it indirectly makes form drag by causing separation. Kill viscosity and you kill the wake too.

Every problem is either ${\tau }_w=\mu du/dy$ (skin friction, needs the wall slope) or $D=\frac{1}{2}\rho v^{2}A{C}_D$ (the lumped total), with $D\propto v^2$ at high $Re$, and both drags ultimately born from viscosity.