Worked examples — Dimensional analysis — Buckingham π theorem

2.2.26 · D3 · Physics › Fluid Mechanics › Dimensional analysis — Buckingham π theorem

Yeh parent topic ka "paces se guzaaro" companion hai. Parent ne aapko recipe dikhayi. Yahan hum deliberately har tarah ki situation dhundte hain jo theorem aapke saamne rakh sakta hai — clean cases, awkward ones, traps, aur exam twists — aur har ek ko last symbol tak kaam karte hain.

Hum recipes par sirf yeh operations use karenge:

• Do quantities ko Multiply karo → unke exponents add karo ($L^1{T}^{-1}\cdot L^1=L^2{T}^{-1}$).
• Kisi power $a$ tak Raise karo → har exponent ko $a$ se multiply karo.
• Dimensionless matlab har exponent $0$ ke barabar ho.

Yahi poora algebra hai. Agar tum chhote whole numbers aur fractions add aur multiply kar sako, toh tum yeh sab kar sakte ho.

The scenario matrix

Theorem jo bhi problem aapke saamne daale, woh in case classes mein se ek mein aati hai. Har row ek distinct "shape" ki situation hai; neeche ke examples un cells ke saath label kiye gaye hain jo woh cover karte hain, taaki tum dekh sako ki kuch chhoot nahi raha.

Ex 1 — Baseline clean case (Cell C1)

Forecast: abhi andaza lagao — kitne independent pure numbers? (Aage padhne se pehle likh lo.)

1. Sab kuch count karo. $n=5$ variables; base dimensions jo actually present hain woh hain $M,L,T$, toh $k=3$. Yeh step kyun? $p=n-k$ tabhi kaam karta hai jab aapko dono counts pata hon. Theorem pehle ek counting statement hai.

2. Groups ki sankhya. $p=5-3=2$. Yeh step kyun? Yeh batata hai ki physics mein exactly do knobs hain — koi aur clever variable freedom add nahi kar sakta.

3. Repeaters choose karo $\rho ,v,D$. Yeh teen kyun? Inhe (a) milke $M,L,T$ contain karna chahiye aur (b) khud mein koi group form nahi karna chahiye. Unki recipes dekho: sirf $\rho$ mein $M$ hai, sirf $v$ mein $T$ hai, aur $D$ ek "spare" $L$ supply karta hai lengths fix karne ke liye. Yahi exactly woh independence hai jo chahiye — unka koi product pure number mein collapse nahi hota.

4. ${\pi }_1$ banao $G$ ke saath: likho ${\pi }_1={\rho }^a{v}^b{D}^{c}G$ aur demand karo ki har exponent zero ho.

   • $M:a+1=0\Rightarrow a=-1$
   • $T:-b-2=0\Rightarrow b=-2$
   • $L:-3a+b+c-2=0\Rightarrow 3-2+c-2=0\Rightarrow c=1$ $\boxed{{\pi }_1=\frac{GD}{\rho v^2}}$ Yeh step kyun? Har base dimension ek equation deta hai; teen equations $a,b,c$ pin down karte hain. Isi tarah ek group forced hota hai, guess nahi.

5. ${\pi }_2$ banao $\mu$ ke saath: ${\pi }_2={\rho }^a{v}^b{D}^c\mu$ deta hai $a=-1,b=-1,c=-1$, toh $\boxed{{\pi }_2=\frac{\mu }{\rho vD}=\frac{1}{Re}}$ Reynolds number phir se.

6. Law ki shape: $\frac{GD}{\rho v^2}=\Phi (Re)$.

Verify: Check karo ki ${\pi }_1$ truly dimensionless hai. $[GD]=M{L}^{-2}{T}^{-2}\cdot L=M{L}^{-1}{T}^{-2}$. $[\rho v^2]=M{L}^{-3}\cdot L^2{T}^{-2}=M{L}^{-1}{T}^{-2}$. Ratio $=M^0{L}^0{T}^0$. ✓ Aur $[\mu ]/[\rho vD]=M{L}^{-1}{T}^{-1}/(M{L}^{-3}\cdot L{T}^{-1}\cdot L)=M{L}^{-1}{T}^{-1}/(M{L}^{-1}{T}^{-1})=1$. ✓

Ex 2 — Ek variable forced out (Cell C2)

Forecast: kya total mass $m$ answer mein appear karega? Haan ya naa guess karo.

1. Count. $n=4$; dimensions present $M,L,T$, toh $k=3$. Phir $p=4-3=1$: exactly ek group. Yeh step kyun? Ek group matlab poora law hai $\pi =\text{const}$ — ek single locked-in relation.

2. $M$ dimension dekho. $M$ appear karta hai $F_T,\lambda ,m$ mein — teen carriers, cancel karne ke liye kaafi. Lekin dekho kya hota hai jab hum actually group banate hain.

3. Repeaters $F_T,\lambda$ (woh $M,L,T$ carry karte hain unke beech: $F_T$ mein teeno hain, $\lambda$ mein $M,L$ hain). Build $\pi =F_T^a{\lambda }^{b}c$:

   • $M:a+b=0$
   • $T:-2a-1=0\Rightarrow a=-\frac{1}{2}$
   • $L:a-b+1=0$ $a=-\frac{1}{2}$ se: $b=+\frac{1}{2}$ ($M$ se), aur check $L$: $-\frac{1}{2}-\frac{1}{2}+1=0$ ✓. $\boxed{\pi =c\sqrt{\frac{\lambda }{F_T}}=\text{const}\Rightarrow c\propto \sqrt{\frac{F_T}{\lambda }}}$

4. $m$ kahan gaya? $m$ include karne ki koshish karo: ${\pi }^{\prime }=F_T^a{\lambda }^{b}m$. Phir $T:-2a=0\Rightarrow a=0$; $M:a+b+1=0\Rightarrow b=-1$; $L:a-b=0\Rightarrow b=0$ — contradiction ($b$ ek saath $-1$ aur $0$ nahi ho sakta). $m$ waala koi dimensionless group exist nahi karta. Yeh step kyun? $m=\lambda \cdot (\text{length})$, lekin list mein koi independent length variable nahi hai. Toh $m$ ek aisi length carry karta hai jise koi cancel nahi kar sakta. Woh redundant hai — physics kabhi use nahi karti.

Verify: $[c\sqrt{\lambda /F_T}]=L{T}^{-1}\cdot \sqrt{M{L}^{-1}/(ML{T}^{-2})}=L{T}^{-1}\cdot \sqrt{L^{-2}{T}^2}=L{T}^{-1}\cdot L^{-1}T=1$. ✓ Answer $c=\sqrt{F_T/\lambda }$ known string-wave speed hai. ✓

Ex 3 — Sirf length aur time bachte hain, $k=2$ (Cell C3)

Forecast: yahan $k$ kya hai — 3 ya 2? Aur isliye kitne groups?

1. Dimensions jo actually present hain unhe count karo. Har recipe scan karo: sirf $L$ aur $T$ dikhte hain. Toh $k=2$, 3 nahi. Yeh step kyun? Rule hai $k=\mathrm{rank}(D)$ = number of independent dimensions present. Blindly $k=3$ likhne par $p=3-3=0$ milega aur galat zero groups claim hoga.

2. Groups. $p=n-k=3-2=1$.

3. Build $\pi ={\ell }^a{g}^{b}c$ (repeaters $\ell ,g$; milke woh $L$ aur $T$ carry karte hain):

   • $T:-2b-1=0\Rightarrow b=-\frac{1}{2}$
   • $L:a+b+1=0\Rightarrow a-\frac{1}{2}+1=0\Rightarrow a=-\frac{1}{2}$ $\boxed{\pi =\frac{c}{\sqrt{g\ell }}=\text{const}\Rightarrow c\propto \sqrt{g\ell }}$

Verify: $[c/\sqrt{g\ell }]=L{T}^{-1}/\sqrt{L{T}^{-2}\cdot L}=L{T}^{-1}/\sqrt{L^2{T}^{-2}}=L{T}^{-1}/(L{T}^{-1})=1$. ✓ (Full theory: $c=\sqrt{g\ell /2\pi }$; woh $2\pi$ exactly woh hai jo dimensions supply nahi kar sakti.)

Ex 4 — Bad repeaters method ko khatam kar dete hain (Cell C4)

Figure dekho: har repeater ek arrow hai jiske teen coordinates uske exponents $(M,L,T)$ par hain. "Independent repeaters" ka poora point yeh hai ki yeh teen arrows exponent space mein genuinely teen alag directions mein point karein. Neeche ke red arrows woh karne se mana kar dete hain.

Forecast: student expect karta hai 2 groups banana. Woh actually kitne independently bana sakta hai?

1. Repeaters ke exponent vectors likho $(M,L,T)$ par:

   • $v$: $(0,1,-1)$
   • $d$: $(0,1,0)$
   • $t$: $(0,0,1)$ Yeh step kyun? Independence in vectors ke baare mein ek statement hai: woh ek doosre mein flattenable nahi hone chahiye.

2. Independence test karo. Koi bhi $M$ carry nahi karta — pehla coordinate teeno ke liye $0$ hai. Toh yeh teen $(L,T)$ plane mein poori tarah rehte hain. Teen vectors ek 2-dimensional plane mein thuse hue independent nahi ho sakte (figure mein red arrows dekho, sab same grey plane mein pade hain, toh ek doosron ka combination hai). Yeh step kyun? Repeater sub-matrix ka rank ab $\le 2<3$ hai. Uska determinant $0$ hai, toh woh invertible nahi hai — tum exponents uniquely solve nahi kar sakte.

3. Dikhao ki determinant zero hai explicitly. $v,d,t$ ke exponent columns rows $(M,L,T)$ par stack karo: $\det \left(\begin{array}{ccc}0& 0& 0\\ 1& 1& 0\\ -1& 0& 1\end{array}\right)=0$ kyunki poori top row zero hai. Zero determinant rank $<3$ confirm karta hai. Yeh step kyun? "Kya mere repeaters independent hain?" ka clean numerical test hai: unka exponent matrix banao aur $\det \ne 0$ check karo.

4. Consequence. Rank $2$ repeaters se tum sirf 2 constraints impose kar sakte ho, 3 nahi. Aur buri baat, woh koi $M$ carry hi nahi karte, toh $F_D$ ya $\rho$ (jo dono $M$ carry karte hain) wale koi bhi group mass mein balance karna impossible hai. Method silently ek poora dimension kho deta hai.

5. Fix. Valid choice $\rho ,v,d$ use karo (parent Ex 1). Uska exponent matrix rows $(M,L,T)$, columns $\rho ,v,d$ par, hai $\det \left(\begin{array}{ccc}1& 0& 0\\ -3& 1& 1\\ 0& -1& 0\end{array}\right)=1\cdot (1\cdot 0-1\cdot (-1))=1\ne 0,$ toh woh repeaters independent hain aur method kaam karta hai.

Verify: bad-repeater determinant $=0$ (dependent), good-repeater determinant $=1\ne 0$ (independent). Dono is page ke neeche numerically checked hain.

Ex 5 — Ek chautha base dimension aata hai (Cell C5)

Forecast: "universe mein 3 dimensions hain" reflex se aap bologe $k=3$. Kya yahan yeh sahi hai?

1. Base dimensions count karo jo present hain. Ab $M,L,T,\Theta$ sab appear karte hain → $k=4$, 3 nahi. Yeh step kyun? $\Theta$ ignore karne par groups ek se zyada count hote aur galat physics milti.

2. Variables. $n=6$. Toh $p=n-k=6-4=2$. Yeh step kyun? Do groups: exactly wo number of knobs jo convection mein hain.

3. Repeaters choose karo $\rho ,v,L_0,\kappa$ (chaar repeaters kyunki $k=4$). Check karo ki woh dimensions span karte hain: sirf $\rho$ $M$ freely carry karta hai, $v$ $T$ supply karta hai, $L_0$ ek spare $L$ supply karta hai, aur $\kappa$ $\Theta$ ka ek clean source hai. Woh independent hain. Yeh step kyun? $k=4$ ke saath hamein chaar repeaters chahiye, usual teen se ek zyada.

4. Group ${\pi }_1$ with $h$: demand karo ${\pi }_1={\rho }^a{v}^b{L}_0^c{\kappa }^{e}h$ dimensionless ho. $M,L,T,\Theta$ match karne par $a=0,b=0,c=1,e=-1$ milta hai, toh $\boxed{{\pi }_1=\frac{h{L}_0}{\kappa }=Nu}(\text{the Nusselt number}).$ Yeh step kyun? $h$ aur $\kappa$ same $M,T,\Theta$ signature carry karte hain; sirf unke length powers ek se differ karte hain, aur $L_0$ woh gap close karta hai.

5. Group ${\pi }_2$ with $c_p$: demand karo ${\pi }_2={\rho }^a{v}^b{L}_0^c{\kappa }^e{c}_p$ dimensionless ho. Matching deta hai $a=1,b=1,c=1,e=-1$, toh $\boxed{{\pi }_2=\frac{\rho v{L}_0{c}_p}{\kappa }=Re\cdot Pr}$ Reynolds number $Re=\rho v{L}_0/\mu$ aur Prandtl number $Pr=\mu c_p/\kappa$ ka product (product mein $\mu$ cancel ho jaata hai, isliye viscosity ko separately list nahi karna pada).

6. Law ki shape: $Nu=\Phi (Re\cdot Pr)$ — heat transfer sirf inhi do knobs ke through flow aur fluid par depend karta hai.

Verify: $[h{L}_0/\kappa ]=M{T}^{-3}{\Theta }^{-1}\cdot L/(ML{T}^{-3}{\Theta }^{-1})=1$ ✓ (dimensionless). Aur $[\rho v{L}_0{c}_p/\kappa ]=M{L}^{-3}\cdot L{T}^{-1}\cdot L\cdot L^2{T}^{-2}{\Theta }^{-1}/(ML{T}^{-3}{\Theta }^{-1})=1$ ✓. Dono neeche numerically checked hain.

Ex 6 — Ek group ek unique power law force karta hai (Cell C6)

Forecast: kitne free dimensionless groups hain — aur sirf ek group hone ka law ki freedom ke liye kya matlab hai?

1. Count. Play mein variables: $Q,v,\ell$ toh $n=3$. Present dimensions: sirf $L,T$, toh $k=2$. Phir $p=n-k=3-2=1$. Yeh step kyun? Ek group matlab koi free function $\Phi$ nahi ek aur argument ki — law ek single constant tak fix hai. (Drag case se compare karo, jahan do groups ek poori function experiment se determine karne ke liye chhodti hain.)

2. Group banao $\pi =v^a{\ell }^{b}Q$:

   • $T:-a-1=0\Rightarrow a=-1$
   • $L:a+b+2=0\Rightarrow -1+b+2=0\Rightarrow b=-1$ $\boxed{\pi =\frac{Q}{v\ell }=\text{const}\Rightarrow Q\propto v\ell }$ Yeh step kyun? Ek group ek constant set karne se, poori functional form fix ho jaati hai — dimensions akele dete hain $Q=(\text{number})v\ell$. Guess karne ke liye sirf woh ek pure number bachta hai.

3. "Few groups" ka lesson. Jitne kam groups, utna zyada dimensions jawab dictate karte hain. Ek group → unique power law; aur agar kabhi $p=0$ mile (koi group nahi), toh yeh signal hai ki tumhare variables dimensionlessly combine nahi kar sakte — warning ki ek variable missing hai.

Verify: $[Q/(v\ell )]=L^2{T}^{-1}/(L{T}^{-1}\cdot L)=L^2{T}^{-1}/(L^2{T}^{-1})=1$. ✓

Ex 7 — Real-world design: ek scale model (Cell C7)

Forecast: kya model real sub se slower ya faster jaayega? Compute karne se pehle guess karo.

1. Similarity condition. $C_D$ sirf $Re$ ki function hai. Prototype ka flow reproduce karne ke liye ek governing group match karo: $R{e}_m=R{e}_p$. Yeh step kyun? Equal π groups ⇒ equal dimensionless physics ⇒ model prototype ko faithfully reproduce karta hai. Yahi Model testing and similarity action mein hai.

2. $v_m$ solve karo. Same fluid ⇒ same $\rho ,\mu$, toh $Re$ equality reduce hoti hai $v_m{D}_m=v_p{D}_p$ mein: $v_m=v_p\frac{D_p}{D_m}=3\times \frac{6}{0.3}=3\times 20=60\text{m/s}.$ Yeh step kyun? $20\times$ chhota model $Re$ fixed rakhne ke liye $20\times$ speed maangta hai — chhota size tez jaane se "pura" hota hai.

3. Drag scale karo. $Re$ match hone par, model aur prototype ke liye $C_D$ equal hai, toh $F_D=C_D\rho v^2{D}^2$ same $C_D$ aur same $\rho$ ke saath. Isliye $\frac{F_{D,p}}{F_{D,m}}=\frac{v_p^2{D}_p^2}{v_m^2{D}_m^2}={\left(\frac{3}{60}\right)}^2{\left(\frac{6}{0.3}\right)}^2={\left(\frac{1}{20}\right)}^2(20{)}^2=1.$ Remarkably, $F_{D,p}=F_{D,m}$: model par measure ki gayi drag real submarine ki drag ke barabar hai. Yeh step kyun? $vD$ fixed rakhne par $v^2{D}^2$ bhi fix hota hai, toh is same-fluid matching ke under dono forces exactly coincide karte hain. Engineer model ki drag seedha prototype ki drag ke roop mein read karti hai.

Verify: $v_m=60$ m/s (checked). $R{e}_m=R{e}_p$ kyunki $v_m{D}_m=60\times 0.3=18=3\times 6=v_p{D}_p$ (checked). Force ratio $F_{D,p}/F_{D,m}=1$ (checked). Sab neeche.

Ex 8 — Exam twist: fractional & negative exponents (Cell C8)

Forecast: $R\propto t^{?}$ mein $t$ ki power guess karo solve karne se pehle.

1. Count. $n=4$; dimensions $M,L,T$ present, $k=3$. Toh $p=4-3=1$: ek group, ek locked law. Yeh step kyun? Ek group phir se matlab hai poori form ek constant tak fix hai (jise Taylor ne baad mein $\approx 1.03$ paya).

2. Build $\pi =E^a{\rho }^b{t}^{c}R$ aur har exponent zero set karo, signs carefully track karo:

   • $M:a+b=0$
   • $T:-2a+c=0$
   • $L:2a-3b+1=0$ Yeh step kyun? Teen base dimensions → teen equations → linear system solve karo $a,b,c$ ke liye.

3. Solve. $M$ se: $b=-a$. $L$ mein daalo: $2a-3(-a)+1=0\Rightarrow 5a+1=0\Rightarrow a=-\frac{1}{5}$. Phir $b=+\frac{1}{5}$, aur $T$ se: $c=2a=-\frac{2}{5}$. $\pi =E^{-1/5}{\rho }^{1/5}{t}^{-2/5}R=\text{const}$ Rearrange karo: $\boxed{R=\text{const}\times {\left(\frac{E{t}^2}{\rho }\right)}^{1/5}\Rightarrow R\propto t^{2/5}}$ Yeh step kyun? $R$ isolate karne par fractional powers invert hote hain; $t$ par ka negative $-\frac{2}{5}$ upar aa jaata hai $+\frac{2}{5}$ ban ke jab hum $\pi =\text{const}$ solve karte hain $R$ ke liye.

4. Signs ki sanity. Badi energy $E$ (positive power $+\frac{1}{5}$) → bada blast: correct. Dense air $\rho$ (negative power $-\frac{1}{5}$) → chhota blast: correct, bhaari air resist karti hai.

Verify: $[(E{t}^2/\rho {)}^{1/5}]=(M{L}^2{T}^{-2}\cdot T^2/(M{L}^{-3}){)}^{1/5}=(L^2\cdot L^3{)}^{1/5}=(L^5{)}^{1/5}=L$ ✓ (matches $[R]=L$). Exponents $a=-1/5,b=1/5,c=-2/5$ neeche checked hain.

Recall

Connections

• Reynolds number — Ex 1, 4, 7 mein master group ke roop mein phir aata hai
• Drag force and drag coefficient — Ex 7 use karta hai $C_D=\Phi (Re)$
• Model testing and similarity — Ex 7 is ka worked instance hai
• Dimensional homogeneity — har "Verify" line ek homogeneity check hai
• Fundamental and derived units — $M,L,T,\Theta$ basis har recipe ke peeche
• Navier–Stokes equations — jahan se $Re$ rigorously aata hai