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

Parachute ek bluff body hai: flow uski cupped shape follow nahi kar sakta, woh turant separate ho jaata hai aur ek enormous wake chhod jaata hai. Front (high pressure) aur back (low pressure) cancel nahi hote → almost 100% form (pressure) drag. Edge-on rakhi plate almost koi frontal area present nahi karti; flow uske faces ke saath slide karta hai, toh almost saari resistance viscous rub hai → almost 100% skin friction drag. Rule of thumb: bada wake ⇒ form drag; bada wetted area with attached flow ⇒ skin friction.

${\tau }_w=\mu \frac{du}{dy}{|}_{y=0}$ use karo, phir $D_{\text{friction}}={\tau }_{w}A$. Kyun: ek aligned flat plate essentially pure skin friction hai, aur tumhe wall par velocity gradient diya gaya tha — yeh exactly aur sirf wahi hai jo shear stress ko chahiye. $\frac{1}{2}\rho v^{2}A{C}_D$ form woh lumped, experiment-fitted package hai jab tum microscopic profile nahi jaante.

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

$u$ ka slope constant hai: $\frac{du}{dy}=500{\text{s}}^{-1}$, toh wall par ${\tau }_w=\mu \frac{du}{dy}=(2.0\times 10^{-5})(500)=1.0\times 10^{-2}\text{Pa}.$ Phir $D_{\text{friction}}={\tau }_{w}A=(1.0\times 10^{-2})(1.5)=1.5\times 10^{-2}\text{N}.$ Sirf slope kyun? Shear stress sirf dekh sakti hai ki wall par velocity kitni steeply change hoti hai — no-slip condition $u=0$ wahan pin karti hai, aur $\mu$ steepness ko force per area mein convert karta hai.

Kyunki $D=\frac{1}{2}\rho v^{2}A{C}_D$ aur $v$ ke alawa sab fixed hai, $D\propto v^2$. Toh $\frac{D_2}{D_1}={\left(\frac{v_2}{v_1}\right)}^2={\left(\frac{40}{25}\right)}^2=(1.6{)}^2=\mathrm{2.56.}$ $D_2=300\times 2.56=\boxed{768\text{N}}.$ 60% speed increase se 156% drag increase hota hai — square bahut kaatda hai. (Valid hai kyunki car high $Re$ par hoti hai, jahan $C_D$ roughly constant hota hai.)

Maano skin friction $=f$. Toh form $=4f$, aur $f+4f=50\Rightarrow 5f=50\Rightarrow f=10$. $D_{\text{friction}}=10\text{N}(20\%),D_{\text{form}}=40\text{N}(80\%).$ Yeh body bluff-dominated hai (zyaadatar drag wake se hai), cars aur blunt shapes ke liye typical.

Terminal velocity par, drag weight ko balance karta hai: $D=W$, toh $\frac{1}{2}\rho v_t^{2}A{C}_D=W\Rightarrow v_t=\sqrt{\frac{2W}{\rho A{C}_D}}.$ Square root kyun? Drag $v^2$ ki tarah badhta hai; balancing $v$ ke liye undo karne ke liye hum root lete hain — isliye tool  ​ hai na ki division. $v_t=\sqrt{\frac{2(6)}{(1.2)(0.02)(0.5)}}=\sqrt{\frac{12}{0.012}}=\sqrt{1000}\approx 31.6\text{m/s}.$ Regime check karo: is speed aur size par, $Re=\rho vL/\mu$ comfortably thousands mein hai, toh constant $C_D$ use karna (aur thus $D\propto v^2$) justified hai.

$\frac{1}{2}\rho v^{2}A{C}_D$ pehle hi form + friction ko $C_D$ mein lump kar deta hai, aur $A$ shared frontal reference hai, toh hum sirf $C_D$ compare karte hain. 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: $\frac{24-2.88}{24}=0.88=\boxed{88\%\text{ less drag}}.$ Haan, bahut zyada worth it hai. Extra wetted area thodi skin friction add karta hai, lekin wake ko khatam karne se dominant form-drag term slash ho jaata hai. Tum sum optimize karte ho, aur yahan sum plummet karta hai. Neeche ki figure ise visible banati hai.

Master formula invert karo: $C_D=\frac{2D}{\rho v^{2}A}=\frac{2(5.4)}{(1.2)(15^2)(0.08)}=\frac{10.8}{(1.2)(225)(0.08)}=\frac{10.8}{21.6}=\mathrm{0.5.}$ $v=30=2\times 15$ par, drag $v^2$ ki tarah scale karta hai: $D_{30}=5.4\times 2^2=\boxed{21.6\text{N}}$. Yahi to drag coefficient ka kaam hai: ek baar measure karo, kisi bhi speed par predict karo (jab tak $Re$ itna high rahe ki $C_D$ steady rahe).

(a) Skin friction. Ek face par wall shear: ${\tau }_w=\mu \frac{du}{dy}=(1.0\times 10^{-3})(8000)=8.0\text{Pa}.$ Ek face: $8.0\times 0.6=4.8\text{N}$. Do 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 karo: $\frac{1}{2}(1000)(4)=2000$; $2000\times 0.004=8$; $8\times 0.02=0.16\text{N}$. (c) Total. $D=9.6+0.16=9.76\text{N}$. Friction fraction $=9.6/9.76=0.9836\approx \boxed{98.4\%}$. Ek aligned plate overwhelmingly skin friction hai — exactly parent note ka claim, ab raw numbers se derive kiya gaya.

(a) Master formula ko $C_D$ ke liye solve karo: $C_D^{\max }=\frac{2D}{\rho v^{2}A}=\frac{2(400)}{(1.2)(35^2)(2.0)}=\frac{800}{(1.2)(1225)(2.0)}=\frac{800}{2940}\approx \mathrm{0.272.}$ (b) Fixed $A$ par, drag area $C_D$ ke saath scale karta hai. Needed reduction: $\frac{0.45-0.272}{0.45}=\frac{0.178}{0.45}\approx 0.395=\boxed{\approx 39.5\%}.$ Ek stiff lekin achievable streamlining target.

Inviscid fluid ka matlab hai $\mu =0$, jiska matlab yeh bhi hai $Re=\rho vL/\mu \to \infty$ — koi stickiness nahi ka idealised limit. $\mu =0$ ke saath hume milta hai ${\tau }_w=0$: skin friction term directly vanish ho jaata hai. Lekin deeper point form term hai. Viscosity ke bina koi boundary layer nahi hoti jo separate ho, toh flow perfectly attached rehti hai aur sphere ke around symmetrically wrap karti hai. Front par pressure back par pressure ke barabar hota hai, aur $\oint p(\hat{n}\cdot \hat{x})dA=0$: form drag term bhi vanish ho jaata hai. Total drag $=0$ — yahi d'Alembert's paradox hai. Lesson: viscosity dono drags ka hidden source hai. Yeh directly skin friction banati hai, aur indirectly form drag banati hai separation cause karke. Viscosity khatam karo toh wake bhi khatam ho jaata hai.

Har problem ya toh ${\tau }_w=\mu du/dy$ hai (skin friction, wall slope chahiye) ya $D=\frac{1}{2}\rho v^{2}A{C}_D$ (lumped total), jisme $D\propto v^2$ high $Re$ par, aur dono drags ultimately viscosity se paida hote hain.