Drag — pressure (form) drag, skin friction drag: Worked examples

2.2.24 · D3 · Physics › Fluid Mechanics › Drag — pressure (form) drag, skin friction drag

Yeh page ek drill sheet hai. Parent note ne do drag sources build kiye the; yahan hum har tarah ke number cover karte hain jo yeh topic throw kar sakta hai — chhoti speeds, badi speeds, zero viscosity, flat plates, spheres, terminal fall, creeping crawls, aur ek nasty exam twist.

Pehle, chalte hain list karte hain ki "har tarah" ka matlab kya hai.

The scenario matrix

Drag ek master formula pe rehta hai (parent se):

$D = \frac{1}{2} ρ v^{2} A C_{D}$

Symbols ko ek shopping list ki tarah padho:

$ρ$ (Greek letter "rho") = fluid ki density, har cubic metre mein kitna mass hai. Units $kg/m^{3}$ .
$v$ = body aur fluid ke beech ki relative speed. Units $m/s$ .
$A$ = ek reference area (usually frontal area — woh shadow jo body flow mein daalta hai). Units $m^{2}$ .
$C_{D}$ = drag coefficient, ek pure number (koi units nahi) jo shape ki complexity ko pack karta hai.

Drag ke baare mein har question ek knob change karta hai aur poochta hai kya hota hai. Yeh hai knobs aur edge-cases ka full table jo hume cover karne hain:

Cell	Kya vary karta hai / degenerate case	Kaunsa drag dominate karta hai	Example
A	Speed doubled ( $v \to 2 v$ )	either	Ex 1
B	Flat plate, do orientations (form vs friction split)	dono, compared	Ex 2
C	Wall shear from a velocity slope ( $τ_{w} = μ d u / d y$ )	friction only	Ex 3
D	Degenerate: $v = 0$ (body at rest)	neither → $D = 0$	Ex 4
E	Limiting: inviscid fluid ( $μ \to 0$ )	d'Alembert, $D_{form} \to 0$	Ex 4
F	Real word problem: skydiver at terminal velocity	form (parachute)	Ex 5
G	Density change (sea level vs altitude)	either	Ex 6
H	Non-linear velocity profile (curved $u (y)$ )	friction only	Ex 7
I	Exam twist: tumhe drag diya, $C_{D}$ maango (formula invert karo)	either	Ex 8
J	Limiting: creeping (Stokes) flow, tiny slow body, $D \propto μv R$	friction-like, linear in $v$	Ex 9

Ab hum har cell walk karte hain. Prerequisites jo tum khule rakhna chahoge: Viscosity & Newton's Law of Viscosity, Boundary Layer & No-Slip Condition, Terminal Velocity, Dimensional Analysis & Drag Coefficient.

Intuition "High Reynolds number" ka matlab, ek baar, poori page ke liye

Reynolds number $R e$ ek pure number hai jo compare karta hai ki fluid ki inertia (uski tendency to keep barrelling forward, $ρ v^{2}$ se scale hoti hai) uski viscosity ( $L$ size ke body ke liye $μv / L$ se scale hoti hai) ko kitna beat karti hai. Dekho Reynolds Number. Jab $R e$ bada ho (fast flow, bada body, patla fluid) toh inertia jeet jaati hai: flow wake mein separate ho jaata hai aur drag coefficient $C_{D}$ roughly constant value pe settle ho jaata hai jo speed ke saath ab nahi badalti. Yahi exactly woh regime hai jahan humara master formula $D = \frac{1}{2} ρ v^{2} A C_{D}$ " $D \propto v^{2}$ with fixed $C_{D}$ " ki tarah behave karta hai — woh assumption jo Examples 1, 2, 5, 6, 8 quietly lete hain. Jab $R e$ tiny ho (slow, small, gooey — Example 9) toh viscosity jeet jaati hai, aur poori picture badal ke $D \propto v$ ho jaati hai. Yeh split apne dimag mein rakho; yeh decide karta hai ki kaunsa formula legal bhi hai.

Example 1 — Cell A: speed double karna

Step 1 — Sirf woh cheez isolate karo jo change hui. Kyunki $ρ, A, C_{D}$ constant hain, poora formula collapse hota hai $D \propto v^{2}$ mein ( $\propto$ ka matlab hai "proportional to" — saath-saath badhta hai). Yeh step kyun? Hum high Reynolds number par hain (cyclist in air fast aur large hai — upar callout dekho), toh $C_{D}$ essentially constant hai aur speed change hone par nahi badalti. Yahi exactly woh cheez hai jo humein $\frac{1}{2} ρ A C_{D}$ ko ek frozen constant treat karne deti hai aur drags ko sirf speeds se compare karne deti hai.

Step 2 — Ratio lo. $\frac{D _{2}}{D _{1}} = \frac{v _{2}^{2}}{v _{1}^{2}} = (\frac{16}{8})^{2} = 2^{2} = 4$ Yeh step kyun? Unknown constants $\frac{1}{2} ρ A C_{D}$ top-and-bottom cancel ho jaate hain, toh humein unki zaroorat hi nahi thi.

Step 3 — Scale up karo. $D_{2} = 4 \times 30 = 120 N$ Yeh step kyun? Ratio ne bataya ki naya drag purane ka $4$ guna hai, toh hum simply known old drag ( $30$ N) ko us factor se multiply karte hain actual number par land karne ke liye.

Example 2 — Cell B: same plate, do orientations

Neeche ki figure dono cases side by side draw karti hai — numbers se pehle isko study karo:

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

Step 1 — Dynamic-pressure block ek baar compute karo. $\frac{1}{2} ρ v^{2} A = \frac{1}{2} (1.2) (1 0^{2}) (0.5) = 30 (yeh ek force hai, N mein, per unit C_{D})$ Yeh step kyun? $C_{D}$ ke alawa sab kuch dono cases mein shared hai, toh ek baar compute karo.

Step 2 — Har $C_{D}$ se multiply karo. $D_{a} = 30 \times 0.005 = 0.15 N D_{b} = 30 \times 1.2 = 36 N$ Yeh step kyun? $C_{D}$ hi ek matra knob hai jo orientation information carry karta hai — yeh silently encode karta hai ki "wake kitna bada hai" jo figure ke right mein drawn hai.

Step 3 — Ratio. $\frac{D _{b}}{D _{a}} = \frac{1.2}{0.005} = 240$

Example 3 — Cell C: linear profile se wall shear

Figure yeh profile plot karti hai aur ek slope mark karti hai jo matter karta hai:

Step 1 — Newton's law of viscosity padho. $τ_{w} = μ \frac{d u}{d y}_{y = 0}$ Yahan $τ_{w}$ wall shear stress hai (force per area jo fluid surface ke saath rub karta hai, Pa mein). Dekho Viscosity & Newton's Law of Viscosity. Yeh step kyun? Shear stress defined hai viscosity times rate of shear. Koi slope nahi, koi rub nahi.

Step 2 — Profile differentiate karo. $u = 200 y$ ke liye, slope constant hai: $\frac{d u}{d y} = 200 s^{- 1}$ . Figure mein isliye lavender line straight hai — har jagah ek hi slope. Yeh step kyun? Straight line ka har jagah ek slope hota hai, toh $y = 0$ par value sirf $200$ hai (mint dot figure mein no-slip point mark karta hai jahan $u (0) = 0$ , lekin drag slope use karta hai, value nahi).

Step 3 — Stress assemble karo, phir force. $τ_{w} = (1.8 \times 1 0^{- 5}) (200) = 3.6 \times 1 0^{- 3} Pa$ $D_{friction} = τ_{w} A = (3.6 \times 1 0^{- 3}) (2) = 7.2 \times 1 0^{- 3} N$ Yeh step kyun? Newton's law ek stress deta hai (force per area), lekin question ek force maangta hai; stress ko wetted area $A$ se multiply karna us rub ko poori surface par sum karta hai — exactly $\oint τ_{w} d A$ for uniform stress.

Example 4 — Cells D & E: do "kuch nahi hota" cases

Definition Pressure

$p$ , aur form-drag piece $D_{form}$ Pressure $p$ woh normal (seedha-andar) push hai jo fluid surface par per unit area exert karta hai, pascals (Pa) mein. Jab fluid ek body ke around flow karta hai toh uska pressure uniform nahi hota: front stagnation point par flow rest mein aa jaata hai aur pressure high hota hai ( $p_{front}$ ); sides aur rear ke around yeh drop kar sakta hai ( $p_{back}$ ). Pressure (form) drag component $D_{form}$ sum karta hai ki front kitna zyada push karta hai back se: $D_{form} = \oint (p_{front} - p_{back}) d A_{⊥}$ jahan $d A_{⊥}$ body ki frontal (cross-section) area ka tiny patch hai — woh area jo flow mein face-on dikhta hai, $v$ ke perpendicular. Hum $d A_{⊥}$ use karte hain (full surface area nahi) kyunki sirf woh part survive karta hai drag ke roop mein jo flow ke along point karta hai; frontal plane par project karna woh bookkeeping automatically karta hai.

Step 1 — Case D, $v = 0$ plug karo. $D = \frac{1}{2} ρ (0)^{2} A C_{D} = 0 N$ Yeh step kyun? Koi relative motion nahi → fluid mein koi momentum change nahi → kuch push back karne ke liye nahi. $v^{2}$ drag ko quadratically vanish karta hai jab body slow hoti hai.

Step 2 — Case E, subtle wala. $μ = 0$ ke saath koi boundary layer nahi hai aur koi flow separation nahi hai. Pressure front-to-back symmetrically wrap karta hai, toh $p_{front}$ equal $p_{back}$ tak recover karta hai aur: $D_{form} = \oint (p_{front} - p_{back}) d A_{⊥} = 0$ aur $D_{friction} = 0$ kyunki rub karne ke liye koi viscosity nahi hai. Total $D = 0 N$ . Yeh step kyun? Yeh d'Alembert's paradox hai: viscosity dono drags ka hidden cause hai. $μ$ kill karo aur dono vanish — chahe body move kar rahi ho.

Example 5 — Cell F: skydiver terminal velocity (word problem)

Step 1 — Force balance. $D = m g \Rightarrow \frac{1}{2} ρ v^{2} A C_{D} = m g$ Yeh step kyun? Constant velocity matlab net force $= 0$ ; sirf do vertical forces hain — weight down aur drag up.

Step 2 — $v$ ke liye solve karo. $v = \frac{2 m g}{ρ A C _{D}} = \frac{2 ( 90 ) ( 9.8 )}{( 1.2 ) ( 25 ) ( 1.4 )}$ Yeh step kyun? Hum $v^{2}$ ko square root lekar invert karte hain — wahi $v^{2}$ jo Example 1 mein humein punish kiya tha ab help karta hai: thodi si speed bahut zyada drag buy karti hai isliye terminal speed chhoti rehti hai.

Step 3 — Number. $v = \frac{1764}{42} = 42 \approx 6.48 m/s$ Yeh step kyun? Hum root ke under arithmetic finish karte hain ( $1764/42 = 42$ ) aur square root lete hain symbolic answer ko actual landing speed mein convert karne ke liye jo ek designer reality ke against check kar sake.

Example 6 — Cell G: same car, sea level vs altitude

Step 1 — Ratio, $v, A, C_{D}$ fixed hold karke. $\frac{D _{1}}{D _{0}} = \frac{ρ _{1}}{ρ _{0}} = \frac{0.9}{1.2} = 0.75$ Yeh step kyun? $ρ$ ke alawa har factor cancel ho jaata hai, toh yeh ek clean proportion hai — $C_{D}$ ki actual value ki zaroorat nahi.

Step 2 — Scale karo. $D_{1} = 0.75 \times 400 = 300 N$ Yeh step kyun? Ratio $0.75$ batata hai ki naya drag purane ka three-quarters hai; known sea-level drag ( $400$ N) ko us fraction se multiply karna proportion ko actual altitude drag mein convert karta hai.

Example 7 — Cell H: curved velocity profile

Step 1 — Poora profile differentiate karo. $\frac{d u}{d y} = 6000 y + 150$ Yeh step kyun? Humein rate of shear chahiye, jo ki derivative hai — Newton's law sirf slopes read karta hai, $u$ ki value kabhi nahi.

Step 2 — Wall $y = 0$ par evaluate karo. $\frac{d u}{d y}_{y = 0} = 6000 (0) + 150 = 150 s^{- 1}$ Yeh step kyun? Skin friction surface par generate hoti hai; curved $y^{2}$ term wahan kuch contribute nahi karta kyunki uska slope zero se start hota hai (no-slip condition curve ko anchor karti hai).

Step 3 — Stress aur force. $τ_{w} = (1.8 \times 1 0^{- 5}) (150) = 2.7 \times 1 0^{- 3} Pa$ $D_{friction} = τ_{w} A = (2.7 \times 1 0^{- 3}) (4) = 1.08 \times 1 0^{- 2} N$

Example 8 — Cell I: exam twist (formula invert karo)

Step 1 — Master formula rearrange karo. $D = \frac{1}{2} ρ v^{2} A C_{D} \Rightarrow C_{D} = \frac{D}{\frac{1}{2} ρ v ^{2} A}$ Yeh step kyun? $C_{D}$ ek matra unknown hai, toh isko isolate karo dynamic-pressure block se dono sides divide karke. Yeh exactly waise hai jaise drag coefficients experimentally measure kiye jaate hain.

Step 2 — Denominator compute karo. $\frac{1}{2} (1.2) (2 0^{2}) (0.3) = \frac{1}{2} (1.2) (400) (0.3) = 72 N$ Yeh step kyun? Denominator dynamic-pressure block hai (ek force in N); isko pehle work out karna ek clean number deta hai divide karne ke liye, aur uski unit (N) guarantee karta hai ki final answer dimensionless hoga.

Step 3 — Divide karo. $C_{D} = \frac{54}{72} = 0.75$ Yeh step kyun? Measured drag (N) ko block (N) se divide karne se units cancel ho jaate hain, pure shape number $C_{D}$ bachta hai jo experiment extract karne ke liye design kiya gaya tha.

Example 9 — Cell J: creeping (Stokes) flow, drag linear in speed

Step 1 — Regime recognize karo aur sahi law pick karo. $D = 6 π μ R v$ Yeh step kyun? Low Reynolds number par viscous forces dominate karte hain aur inertia ( $ρ v^{2}$ momentum-flux argument) negligible hai, toh dynamic-pressure formula ab apply nahi hota — Stokes ne yeh viscosity-only result derive kiya tha. Yeh essentially saari friction-type drag hai, $v$ , $μ$ , aur $R$ mein linear.

Step 2 — Numbers plug karo. $D = 6 π (0.1) (1 \times 1 0^{- 4}) (2 \times 1 0^{- 3})$ Yeh step kyun? Har factor first power mein enter karta hai, toh master formula ke unlike square karne ki koi baat nahi — yeh practical mein "linear in velocity" aise dikhta hai.

Step 3 — Number. $D = 6 π \times 2 \times 1 0^{- 8} = 12 π \times 1 0^{- 8} \approx 3.77 \times 1 0^{- 7} N$ Yeh step kyun? Constants collect karna ( $0.1 \times 1 0^{- 4} \times 2 \times 1 0^{- 3} = 2 \times 1 0^{- 8}$ ) aur $6 π$ se multiply karna actual settling drag deta hai — ek micronewton ka fraction, jaise forecast kiya tha.