Dimensional analysis — Buckingham π theorem: Exercises

2.2.26 · D4 · Physics › Fluid Mechanics › Dimensional analysis — Buckingham π theorem

Shuru karne se pehle, ek reminder un teenon "grammar" pieces ki jo hum har problem mein use karte hain, simple words mein:

Neeche wali picture har exercise ke peeche ka mental model hai: variables andar jaate hain, apne units cancel karte hain, aur pure numbers ki tarah bahar aate hain.

Figure — Dimensional analysis — Buckingham π theorem

Level 1 — Recognition

Exercise 1.1

Har ek ke dimensions $M, L, T$ mein batao: force $F$ , pressure $P$ , dynamic viscosity $μ$ , kinematic viscosity $ν$ , power $W$ .

Recall Solution 1.1

HUM KYA KARTE HAIN: har ek ko definitions se build karo, phir exponents padho.

Force $=$ mass $\times$ acceleration: $[F] = M \cdot L T^{- 2} = M L T^{- 2}$ .
Pressure $=$ force / area: $[P] = \frac{M L T ^{- 2}}{L ^{2}} = M L^{- 1} T^{- 2}$ .
Dynamic viscosity: drag/stress law $τ = μ \frac{d v}{d y}$ se $[μ]$ nikaalte hain, $[μ] = \frac{M L ^{- 1} T ^{- 2}}{T ^{- 1}} = M L^{- 1} T^{- 1}$ .
Kinematic viscosity $ν = μ / ρ$ : $[ν] = \frac{M L ^{- 1} T ^{- 1}}{M L ^{- 3}} = L^{2} T^{- 1}$ .
Power $=$ energy / time $=$ force $\times$ speed: $[W] = M L T^{- 2} \cdot L T^{- 1} = M L^{2} T^{- 3}$ .

Exercise 1.2

Sphere-drag problem ${F, d, v, ρ, μ}$ ke liye kitne independent dimensionless groups exist karte hain? Pendulum ${T, L, g, m}$ ke liye?

Recall Solution 1.2

COUNT KYUN MATTER KARTA HAI: $p$ batata hai ki physics mein kitne "knobs" sach mein hain.

Sphere drag: $n = 5$ variables, dimensions present hain $M, L, T$ toh $k = 3$ , giving $p = 5 - 3 = 2$ .
Pendulum: $n = 4$ , dimensions present $M, L, T$ toh $k = 3$ , giving $p = 4 - 3 = 1$ .

Exercise 1.3

Inn mein se kaun se already dimensionless hain? (a) $R e = ρ v d / μ$ , (b) $F / (ρ v^{2} d^{2})$ , (c) $v / g L$ , (d) $ρ v d$ .

Recall Solution 1.3

HUM KYA CHECK KARTE HAIN: kya $M, L, T$ har ek power zero tak cancel hota hai?

(a) $\frac{( M L ^{- 3} ) ( L T ^{- 1} ) ( L )}{M L ^{- 1} T ^{- 1}} = \frac{M L ^{- 1} T ^{- 1}}{M L ^{- 1} T ^{- 1}} = 1$ . ✅ dimensionless — yeh hai Reynolds number.
(b) $\frac{M L T ^{- 2}}{( M L ^{- 3} ) ( L ^{2} T ^{- 2} ) ( L ^{2} )} = \frac{M L T ^{- 2}}{M L T ^{- 2}} = 1$ . ✅ dimensionless.
(c) $\frac{L T ^{- 1}}{( L T ^{- 2} ) ( L )} = \frac{L T ^{- 1}}{L T ^{- 1}} = 1$ . ✅ dimensionless (Froude number).
(d) $[M L^{- 3} \cdot L T^{- 1} \cdot L] = M L^{- 1} T^{- 1}$ . ❌ dimensionless nahi (yeh $μ$ ke dimensions hain).

Level 2 — Application

Exercise 2.1

Pendulum ${T, L, g}$ ke liye single π group nikalo (mass pehle se exclude hai). Confirm karo $T \propto L / g$ .

Recall Solution 2.1

HUM KYA KARTE HAIN: group ko $π = T g^{a} L^{b}$ likhte hain aur har dimension ko cancel karte hain. Dimensions: $[T] = T$ , $[g] = L T^{- 2}$ , $[L] = L$ . $[π] = T \cdot (L T^{- 2})^{a} \cdot L^{b} = L^{a + b} T^{1 - 2 a} .$

$T$ : $1 - 2 a = 0 \Rightarrow a = \frac{1}{2}$ .
$L$ : $a + b = 0 \Rightarrow b = - \frac{1}{2}$ . Toh $π = T g^{1/2} L^{- 1/2} = T g / L = const$ , isliye $T \propto L / g$ . ✅ Constant ( $2 π$ ) woh hai jo dimensions nahi de sakte.

Exercise 2.2

Deep water par ek wave ki speed $c$ , wavelength $λ$ aur gravity $g$ par depend karti hai. $c$ ki form nikalo.

Recall Solution 2.2

DIMENSIONAL ANALYSIS KYUN: full water-wave PDE solve karna mushkil hai, lekin sirf ${c, λ, g}$ aur dimensions $L, T$ ke saath hame $n = 3$ , $k = 2$ milta hai, toh $p = 1$ — ek group ka matlab ek fixed form hai. $π = c g^{a} λ^{b}$ likhte hain jahan $[c] = L T^{- 1}$ , $[g] = L T^{- 2}$ , $[λ] = L$ : $[π] = L T^{- 1} (L T^{- 2})^{a} L^{b} = L^{1 + a + b} T^{- 1 - 2 a} .$

$T$ : $- 1 - 2 a = 0 \Rightarrow a = - \frac{1}{2}$ .
$L$ : $1 + a + b = 0 \Rightarrow b = - \frac{1}{2}$ . $π = \frac{c}{g λ} = const \Rightarrow c \propto g λ .$

Exercise 2.3

Sphere drag dobara karo lekin repeating variables $μ, v, d$ chuno $ρ, v, d$ ki jagah. $F$ wala group banao.

Recall Solution 2.3

HUM KYA KARTE HAIN: $π = μ^{a} v^{b} d^{c} F$ . $[μ] = M L^{- 1} T^{- 1}$ , $[v] = L T^{- 1}$ , $[d] = L$ , $[F] = M L T^{- 2}$ ke saath: $[π] = (M L^{- 1} T^{- 1})^{a} (L T^{- 1})^{b} L^{c} (M L T^{- 2}) .$

$M$ : $a + 1 = 0 \Rightarrow a = - 1$ .
$T$ : $- a - b - 2 = 0 \Rightarrow 1 - b - 2 = 0 \Rightarrow b = - 1$ .
$L$ : $- a + b + c + 1 = 0 \Rightarrow 1 - 1 + c + 1 = 0 \Rightarrow c = - 1$ . $π = \frac{F}{μv d} .$ YEH THEEK KYUN HAI: yeh ek valid (lekin kam famous) drag group hai. Note karo $\frac{F}{μv d} = \frac{F}{ρ v ^{2} d ^{2}} \cdot R e$ — do standard groups ka product. Repeaters ke dono choices valid hain; ye sirf unhi do knobs ko relabel karte hain. Dekho Drag force and drag coefficient.

Level 3 — Analysis

Exercise 3.1

Pipe mein pressure drop $Δ P$ per unit length $Δ P / ℓ$ , diameter $D$ , mean speed $v$ , density $ρ$ , viscosity $μ$ par depend karta hai. Kitne groups hain, aur wo kya hain?

Recall Solution 3.1

HUM KYA KARTE HAIN: variables ${Δ P / ℓ, D, v, ρ, μ}$ , toh $n = 5$ . Dimensions $M, L, T$ toh $k = 3$ , $p = 2$ . $[Δ P / ℓ] = [P] / L = M L^{- 1} T^{- 2} / L = M L^{- 2} T^{- 2}$ . Repeaters $ρ, v, D$ . $Δ P / ℓ$ wala group: $π_{1} = ρ^{a} v^{b} D^{c} (Δ P / ℓ)$ :

$M$ : $a + 1 = 0 \Rightarrow a = - 1$ .
$T$ : $- b - 2 = 0 \Rightarrow b = - 2$ .
$L$ : $- 3 a + b + c - 2 = 0 \Rightarrow 3 - 2 + c - 2 = 0 \Rightarrow c = 1$ . $π_{1} = \frac{( Δ P / ℓ ) D}{ρ v ^{2}} .$ $μ$ wala group jaana-pehchana $π_{2} = \frac{μ}{ρ v D} = 1/ R e$ deta hai. Toh $\frac{( Δ P / ℓ ) D}{ρ v ^{2}} = Φ (R e)$ — Darcy friction-factor chart ka basis.

Exercise 3.2

Dikhao ki repeaters ${v, D, ν}$ (kinematic viscosity $ν = L^{2} T^{- 1}$ ke saath) chunnaa drag problem ke liye saari zaroorat ki dimensions span karne mein fail karta hai. Kaun si dimension missing hai?

Recall Solution 3.2

YEH POINT KYUN HAI: repeaters ko (a) milkar saare present dimensions contain karne chahiye aur (b) independent hone chahiye. Trio ki dimensions: $[v] = L T^{- 1}$ , $[D] = L$ , $[ν] = L^{2} T^{- 1}$ . Har ek mein sirf $L$ aur $T$ hain — kahin bhi $M$ nahi hai. Lekin drag problem mein $M$ hai (in $F$ aur $ρ$ ). Toh ye teen repeaters $π_{1} = v^{a} D^{b} ν^{c} F$ se mass cancel nahi kar sakte — $M$ -equation hai $0 + 1 = 0$ , jo impossible hai. Missing dimension: $M$ . Fix: repeaters mein ek aisa variable shamil karo jo $M$ carry kare (jaise $ρ$ ).

Exercise 3.3

${F, d, v, ρ, μ}$ ke liye $(M, L, T)$ par dimension matrix $D$ hai

D = 11 - 2 010 01 - 1 1 - 3 0 1 - 1 - 1 .

Verify karo $rank (D) = 3$ aur isliye $p = n - rank (D) = 2$ .

Recall Solution 3.3

RANK KA MATLAB KYA HAI: columns ke beech dimensionally independent directions ki sankhya. Pehle teen columns lo ( $F, d, v$ ke exponents):

11 - 2 010 01 - 1 .

Iska determinant: top row ke saath expand karo $= 1 \cdot (1 \cdot (- 1) - 1 \cdot 0) = - 1 \neq = 0$ . Ek non-zero $3 \times 3$ minor ka matlab hai $rank (D) = 3$ . Isliye $p = 5 - 3 = 2$ , jo Exercise 1.2 se match karta hai. KYUN: ek $k \times n$ matrix ka null-space dimension $n - rank$ hota hai, aur har null-space vector ek independent π hai. Parent note mein $p = n - k$ ki derivation dekho.

Level 4 — Synthesis

Exercise 4.1

Radius $r$ ka ek liquid drop surface tension $σ$ (dimensions $[σ] = M T^{- 2}$ , force per length) aur density $ρ$ ke under frequency $f$ se oscillate karta hai. $f$ ki form derive karo.

Recall Solution 4.1

HUM KYA KARTE HAIN: variables ${f, r, σ, ρ}$ , $n = 4$ . Dimensions present: $M, L, T$ ? Check karo — $[f] = T^{- 1}$ , $[r] = L$ , $[σ] = M T^{- 2}$ , $[ρ] = M L^{- 3}$ . $M, L, T$ sab aate hain, $k = 3$ , toh $p = 1$ : ek group, ek fixed form. $π = f ρ^{a} r^{b} σ^{c}$ form karo: $[π] = T^{- 1} (M L^{- 3})^{a} L^{b} (M T^{- 2})^{c} = M^{a + c} L^{- 3 a + b} T^{- 1 - 2 c} .$

$M$ : $a + c = 0$ .
$T$ : $- 1 - 2 c = 0 \Rightarrow c = - \frac{1}{2} \Rightarrow a = \frac{1}{2}$ .
$L$ : $- 3 a + b = 0 \Rightarrow b = 3 a = \frac{3}{2}$ . $π = f ρ^{1/2} r^{3/2} σ^{- 1/2} = f \frac{ρ r ^{3}}{σ} = const \Rightarrow f \propto \frac{σ}{ρ r ^{3}} .$ Feel se check: zyaada surface tension → stiffer → faster wobble ( $f ↑$ ), bada/bhaari drop → slower ( $f ↓$ ). ✅

Exercise 4.2

Non-dimensionalise karo: ek ship ka scale model test kiya jaata hai. Real ship ke flow se match karne ke liye, model aur prototype par kaun sa π group equal hona chahiye agar gravity waves dominate karti hain? (Hint: ${v, L, g}$ use karo.)

Recall Solution 4.2

KYUN: do flows dynamically similar hote hain jab unke governing π groups match karte hain — Model testing and similarity ka core idea. Exercise 1.3(c) / 2.2 se, ${v, L, g}$ ka gravity group hai Froude number $F r = \frac{v}{g L}$ . Match karne ka matlab hai $\frac{v _{m}}{g L _{m}} = \frac{v _{p}}{g L _{p}}$ . $g$ same hone par, model speed scale karti hai as $v_{m} = v_{p} \frac{L _{m}}{L _{p}} .$ Numeric: $v_{p} = 10 m/s$ par chal rahe ship ka $1 : 100$ model ( $L_{m} / L_{p} = 1/100$ ) ko $v_{m} = 10 1/100 = 1 m/s$ par tow karna hoga.

Exercise 4.3

Heat transfer: convective coefficient $h$ (dimensions $M T^{- 3} Θ^{- 1}$ , jahan $Θ$ = temperature) velocity $v$ , pipe diameter $D$ , fluid density $ρ$ , viscosity $μ$ , specific heat $c_{p}$ ( $L^{2} T^{- 2} Θ^{- 1}$ ), aur thermal conductivity $κ$ ( $M L T^{- 3} Θ^{- 1}$ ) par depend karta hai. Kitne π groups hain?

Recall Solution 4.3

HUM KYA COUNT KARTE HAIN: $n = 7$ variables ${h, v, D, ρ, μ, c_{p}, κ}$ . Ab chaar dimensions appear karte hain: $M, L, T, Θ$ (temperature $h, c_{p}, κ$ ke zariye aata hai). Toh $k = 4$ . $p = n - k = 7 - 4 = 3.$ Ye teen Nusselt, Reynolds aur Prandtl numbers nikle hain — standard heat-transfer trio. $k = 4$ kyun, 3 nahi: temperature $Θ$ yahan ek genuine, independent fundamental dimension hai — Level 1 trap warning aur Fundamental and derived units dekho.

Level 5 — Mastery

Exercise 5.1

Ek wing par lift force $F_{L}$ , air density $ρ$ , speed $v$ , wing area $S$ (dimensions $L^{2}$ ), viscosity $μ$ , aur speed of sound $a$ ( $L T^{- 1}$ ) par depend karta hai. $p$ nikalo aur har group ko physically identify karo.

Recall Solution 5.1

HUM KYA KARTE HAIN: $n = 6$ variables ${F_{L}, ρ, v, S, μ, a}$ , dimensions $M, L, T$ toh $k = 3$ , giving $p = 6 - 3 = 3$ . Repeaters $ρ, v, S$ (note karo $[S] = L^{2}$ , phir bhi $L$ span karta hai aur $ρ, v$ ke saath milkar saare $M, L, T$ ).

$F_{L}$ wala group: $π_{1} = ρ^{a} v^{b} S^{c} F_{L}$ . $[S^{c}] = L^{2 c}$ ke saath:
- $M$ : $a + 1 = 0 \Rightarrow a = - 1$ .
- $T$ : $- b - 2 = 0 \Rightarrow b = - 2$ .
- $L$ : $- 3 a + b + 2 c + 1 = 0 \Rightarrow 3 - 2 + 2 c + 1 = 0 \Rightarrow c = - 1$ . $π_{1} = \frac{F _{L}}{ρ v ^{2} S} (lift coefficient C_{L}) .$
$μ$ wala group $π_{2} = \frac{μ}{ρ v S}$ deta hai ( $S$ as length scale wala Reynolds number).
$a$ wala group: $π_{3} = ρ^{a} v^{b} S^{c} a$ → dimensionally $a$ aur $v$ dono $L T^{- 1}$ hain, toh sirf cancellation deta hai $π_{3} = \frac{a}{v}$ — yani $π_{3} = \frac{1}{M a}$ , inverse Mach number. Toh $C_{L} = Φ (R e, M a)$ — teen knobs, exactly wahi jo wind-tunnel aerodynamics use karta hai.

Exercise 5.2

Degenerate case. Vacuum mein swing karta pendulum: ${T, L, g}$ lekin ab saath mein (irrelevant) air density $ρ$ bhi list karo. Naively $n = 4$ , dimensions $M, L, T$ → $p = 1$ ? Dikhao ki actually kya hota hai aur kyun.

Recall Solution 5.2

YEH SUBTLE KYUN HAI: $ρ$ add karna dimension $M$ introduce karta hai, lekin koi aur cheez $M$ carry nahi karti. Toh $M$ exactly ek variable mein appear karta hai. Koi bhi group set up karo $π = T^{a} L^{b} g^{c} ρ^{d}$ . $M$ -equation hai $d = 0$ (sirf $ρ$ mein $M$ hai). Yeh $ρ$ ko har group se bahar force karta hai — bilkul waise jaise plain pendulum mein mass hota hai. Ab $(M, L, T)$ par dimension matrix mein ek $M$ -row hai jo zero hai sirf $ρ$ ke column ko chhodkar, aur $ρ$ ke column mein $L, T$ entries $- 3, 0$ hain. Effectively $rank (D) = 3$ hai lekin ek variable ( $ρ$ ) dead hai. Usable groups ki real count abhi bhi $1$ hai (wahi $T g / L$ ). Lesson: ek aisa variable add karna jo ek aisi dimension carry kare jise koi aur share nahi karta ek knob add nahi karta — yeh ek aisa variable add karta hai jiska exponent zero hona zaroori hai. Fix: un variables ko drop karo jiske dimension ko balance nahi kiya ja sakta (parent note ka Mistake A).

Exercise 5.3

Full synthesis. Bahut viscous fluid mein settle ho rahe ek chhote sphere (radius $r$ ) ki terminal velocity $v_{t}$ , $r$ , $μ$ , aur effective weight per volume $Δ ρ g$ (buoyancy-corrected, dimensions $M L^{- 2} T^{- 2}$ ) par depend karti hai. $v_{t}$ ki form predict karo, phir Stokes' law se compare karo.

Recall Solution 5.3

HUM KYA KARTE HAIN: variables ${v_{t}, r, μ, Δ ρ g}$ , $n = 4$ . Dimensions $M, L, T$ toh $k = 3$ , $p = 1$ — ek fixed form. Group $π = v_{t} μ^{a} r^{b} (Δ ρ g)^{c}$ jahan $[v_{t}] = L T^{- 1}$ , $[μ] = M L^{- 1} T^{- 1}$ , $[r] = L$ , $[Δ ρ g] = M L^{- 2} T^{- 2}$ : $[π] = L T^{- 1} (M L^{- 1} T^{- 1})^{a} L^{b} (M L^{- 2} T^{- 2})^{c} = M^{a + c} L^{1 - a + b - 2 c} T^{- 1 - a - 2 c} .$

$M$ : $a + c = 0$ .
$T$ : $- 1 - a - 2 c = 0$ . $a = - c$ substitute karo: $- 1 + c - 2 c = 0 \Rightarrow - 1 - c = 0 \Rightarrow c = - 1$ , toh $a = 1$ .
$L$ : $1 - a + b - 2 c = 0 \Rightarrow 1 - 1 + b + 2 = 0 \Rightarrow b = - 2$ . $π = \frac{v _{t} μ}{r ^{2} Δ ρ g} = const \Rightarrow v_{t} \propto \frac{Δ ρ g r ^{2}}{μ} .$ Compare karo: Stokes' law deta hai $v_{t} = \frac{2}{9} \frac{Δ ρ g r ^{2}}{μ}$ — bilkul same shape, constant $\frac{2}{9}$ ke saath jo dimensions akele supply nahi kar sakte (parent note ka Mistake C). ✅

Common mistake Level 5 trap — physics ke upar naive

$n - k$ par trust karna Exercise 5.2 mein naive count kehta hai " $p = 1$ , theek hai" aur student kabhi notice nahi karta ki $ρ$ dead weight tha; richer problems mein (ek variable jo ek unique dimension carry karta hai) yeh $p$ ko inflate karta hai aur ek phantom group produce karta hai. Kyun sahi lagta hai: formula $p = n - k$ ko gospel ki tarah drill kiya jaata hai. Kyun galat hai: formula tabhi valid hai jab $rank (D) = k$ aur har listed variable physically relevant ho. Ek aisa variable jiske dimension ko koi aur variable balance nahi kar sakta, wo exponent zero par force hota hai — na toh koi group add karta hai na list mein belong karta hai. Fix: count par trust karne se pehle hamesha sanity-check karo yeh poochhkar "kya is dimension ko koi aur variable actually cancel kar sakta hai?" Yeh Navier–Stokes equations non-dimensionalisation ke peeche ka analysis hai, jahan har retained term apni jagah earn karta hai.

Recall Self-test summary (cloze)

Independent π groups ki count hai == $p = n - rank (D)$ ==, jo $n - k$ ke barabar tabhi hoti hai jab ==dimension matrix full rank $k$ ho. Repeaters ko milkar saare present dimensions span karne chahiye aur dimensionally independent (koi π na banayein) hone chahiye. Ek aisa variable jो ek aisi dimension carry kare jo koi aur variable share nahi karta== exponent zero par force hota hai aur drop ho jaata hai.

Connections

Reynolds number — Exercises 2.3, 3.1, 5.1 mein recurring $μ$ -based group
Drag force and drag coefficient — Exercises 2.3 aur 5.1 mein $C_{D}$ , $C_{L}$ build karte hain
Dimensional homogeneity — woh principle jis par har solution tika hai
Model testing and similarity — Exercise 4.2 ka Froude matching
Navier–Stokes equations — mastery-level non-dimensionalisation
Fundamental and derived units — Exercise 4.3 mein use hua $M, L, T, Θ$ basis