Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline: Visual walkthrough

2.2.15 · D2 · Physics › Fluid Mechanics › Assumptions in Bernoulli — steady, inviscid, incompressible,

Step 1 — Road banao: ek streamline

KYA HAI. Hum ek aisi curve chunte hain aur us pe naapee gayi doori ko $s$ (metres) kehte hain. Jab blob travel karta hai, $s$ badhta hai. Uski zameen se upar ki height $h$ hai (metres, seedha upar naapa hua). Road pe uski speed $v$ hai (metres per second).

YEH TEEN HI KYUN? Kyunki ek moving blob sirf teen energy-related kaam kar sakta hai: push karna (pressure), move karna (speed $v$ ), climb karna (height $h$ ). Hum exactly yahi track karte hain, kuch aur nahi.

PICTURE. Laal curve streamline hai. Arrow blob ki velocity dikhata hai, jo hamesha road ki tangent hoti hai.

$s$ ::: curved streamline ke saath along travel ki gayi doori (seedhi-line distance nahi).
$v$ ::: blob ki speed, jo hamesha road ke along point karti hai.
$h$ ::: ek chosen floor ke upar blob ki height.

Step 2 — Ek tiny blob kaato aur uska size naam karo

KYA HAI. Zoom in karo aur road pe baitha ek tiny cylinder fluid ka kaato. Uske paas hai:

road ke along length $d s$ (ek tiny step),
cross-sectional area $A$ (uske aage aur peeche ki faces ka size),
density $ρ$ (kilograms of water per cubic metre).

CYLINDER HI KYUN? Kyunki pressure flat faces pe push karta hai, aur ek cylinder humein road ke along point karta ek clean back face aur ek clean front face deta hai — yahi ek direction hai jis ki humein parwah hai.

Uski mass hai: $mass = kg per m^{3} ρ \times volume A d s$

jahan $A d s$ hai (face area) × (length) = blob ka volume, aur density $ρ$ se multiply karne pe volume kilograms ban jaata hai.

PICTURE. Laal cylinder humara parcel hai; kaale arrows un do faces ko mark karte hain jahan pressure act karega.

Step 3 — Road ke along do forces

Hum sirf road ke along forces ki parwah karte hain ( $s$ direction), kyunki sideways forces sirf road ko modte hain; woh us ke along speed nahi badlaate.

Force A — Pressure. Pressure $P$ (pascals mein = newtons per m²) ek face pe $P \times A$ force se push karta hai.

Back face blob ko aage push karta hai: $+ P A$ .
Front face blob ko peeche push karta hai, lekin wahan pressure thoda alag hai, $P + d P$ : force $- (P + d P) A$ .

Jodne pe: $peeche se push P A - aage se push (P + d P) A = - A d P$

Dhyan do ki poora $P A$ cancel ho jaata hai — sirf change $d P$ bachta hai. Toh net pressure force hai $- A d P$ : agar pressure aage jaate waqt badhta hai ( $d P > 0$ ), toh net push peeche ki taraf hai (minus sign).

Force B — Gravity. Blob ka weight hai $mass \times g = ρ A d s g$ , seedha neeche ki taraf. Sirf woh part jo road ke along hai woh ise slow ya speed up karta hai. Woh part hai: $- ρ A d s g \frac{d h}{d s}$ Yahan $\frac{d h}{d s}$ hai "road ke ek step mein kitni height gain hoti hai" — steepness. Flat stretch pe $d h / d s = 0$ (gravity road ke along kuch nahi karta); vertical climb pe $d h / d s = 1$ (poori gravity tumse ladti hai). Minus sign kehta hai: climb karna ( $d h > 0$ ) tumhe peeche kheenchta hai.

PICTURE. Do faces pe laal pressure arrows; kaala weight arrow "road ke along" aur "sideways" pieces mein split kiya hua.

Step 4 — Newton's law ko ek acceleration chahiye: blob ke speed up hone ke do tarike

KYA HAI. Newton kehta hai $mass \times acceleration = force$ . Hamare paas mass aur force hai; humein blob ka acceleration $a$ chahiye. Lekin ek moving blob ka acceleration subtle hai — yeh material derivative hai: $a = \frac{D v}{D t} = clock term \frac{\partial v}{\partial t} + travel term v \frac{\partial v}{\partial s}$

DO PIECES KYUN? Ek blob do bilkul alag reasons se speed gain kar sakta hai:

Clock term $\frac{\partial v}{\partial t}$ — tum ek jagah khade rehte ho aur poora flow time ke saath badh jaata dekho (koi valve kholta hai). Ek stationary blob bhi speed up ho jaata.
Travel term $v \frac{\partial v}{\partial s}$ — time ke saath kuch nahi badlta, lekin blob ek narrow section mein drive karta hai jahan paani naturally faster hai. Aage wala $v$ hai "tum road kitni tezi se cover karte ho," aur $\partial v / \partial s$ hai "road ke har metre pe speed kitni badhhti hai." Dono multiply karke = travel karne se second per second speed gain.

PICTURE. Left panel: same spot, flow time ke saath upar uthta hai (clock). Right panel: time mein frozen pipe jo narrow hoti hai, toh daayein jaana matlab faster jaana (travel).

Step 5 — Blob ke liye Newton's law assemble karo

KYA HAI. Mass × acceleration = net force rakho: $mass ρ A d s (\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}) = Step 3 se net force - A d P - ρ A g d h$

$A$ se divide kyun karein? Area har term mein hai, toh yeh koi physics carry nahi karta — ise khatam karo. $A$ se divide karo: $ρ d s (\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}) = - d P - ρ g d h$

PICTURE. Blob pe poora force balance draw kiya: inertia left pe, pressure + gravity right pe, area cross out kiya hua.

Yeh equation abhi bhi fully general hai — yeh bas ek streamline pe fluid blob ke liye Newton's law hai. Abhi tak koi assumption nahi hui. Ab dekho kaise har assumption ek piece delete karta hai.

Step 6 — Assumption 1: STEADY clock term ko maar deta hai

KYA HAI. Steady matlab: kisi bhi fixed point pe, kuch bhi time ke saath nahi badlta. Toh $\frac{\partial v}{\partial t} = 0$ . Clock term gayab ho jaata hai: $ρ d s (\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}) = - d P - ρ g d h$

Dhyan do $d s \cdot \frac{\partial v}{\partial s} = d v$ (us chote step mein speed ka change), toh left side $ρ v d v$ pe saaf ho jaata hai: $ρ v d v = - d P - ρ g d h$

YEH KYUN MATTER KARTA HAI. Agar koi valve kholta hai (Step 4 left panel), toh clock term zero nahi hai, aur yeh line ek jhooth hai — blob ko time-changes shove kar rahi hain jo $P$ , $v$ , $h$ ek point pe bookkeep nahi kar sakte. Yahi "tap suddenly khula" case hai.

PICTURE. General equation jisme clock term laal mein struck out hai, aur "before/after" flow snapshots identical dikhaye hue hain (steady aisa hi lagta hai).

Step 7 — Assumption 2: INVISCID (woh force jo humne kabhi draw nahi ki)

KYA HAI. Step 3 wapas dekho. Humne pressure aur gravity draw ki — lekin koi friction nahi. Real paani walls aur khud se ragdata hai (viscosity), blob ke side pe ek shear force produce karta hai, roughly $- τ (side area)$ .

YEH ASSUMPTION KYUN HAI. Woh term kabhi na likhke, humne silently assume kar liya ki yeh zero hai — ki fluid inviscid hai. Agar yeh nahi hai, toh woh missing force mechanical energy churake heat mein badal deti hai, toh humara "constant" aage downstream quietly leak karta rehta hai. Yahi rough-pipe example hai: same $v$ , same $h$ , phir bhi pressure girta hai.

PICTURE. Blob uski side surface pe ek laal shear-drag arrow ke saath — woh arrow jo humne chod di — "energy → heat" label ke saath.

Step 8 — Assumption 3: INCOMPRESSIBLE humein cleanly integrate karne deta hai

KYA HAI. Hamare paas ab $ρ v d v = - d P - ρ g d h$ hai. Is "tiny-step" statement ko poori journey ke statement mein badalne ke liye hum road ke along integrate karte hain (point 1 se point 2 tak har step jodte hain): $\int ρ v d v + \int d P + \int ρ g d h = 0$

$\int d P = P$ — yeh hamesha theek hai, chahe density kuch bhi kare.
$\int ρ v d v = \frac{1}{2} ρ v^{2}$ — kinetic energy per volume, lekin sirf tab agar hum $ρ$ ko integral ke baahir slide kar sakein.
$\int ρ g d h = ρ g h$ — potential energy per volume, phir se sirf tab agar $ρ$ integral se bahar aaye.

INCOMPRESSIBLE KYUN — ASLI REASON. Catch $\int d P$ term mein nahi hai (woh hamesha $P$ deta hai). Catch un do terms mein hai jinmein $ρ$ andar hai. Agar density road ke along badhlti hai, toh tum $ρ$ ko baahir nahi nikal sakte — $\int ρ v d v$ aur $\int ρ g d h$ ab saaf $\frac{1}{2} ρ v^{2}$ aur $ρ g h$ pe collapse nahi hote, aur ek extra internal-energy term aa jaata hai. Incompressible assume karna ( $ρ$ constant) exactly woh permission hai jo $ρ$ ko dono integrals se baahir factor karne deti hai.

Gas ke liye, $ρ$ pressure ke saath upar-neeche hota hai, toh yeh toot jaata hai. Rule of thumb: OK jab tak Mach number $< 0.3$ .

$ρ$ constant assume karke aur sab jodke: $P + \frac{1}{2} ρ v^{2} + ρ g h = const$

PICTURE. Left: liquid blob road ke along apna size maintain karta hai (constant $ρ$ ) → $ρ$ dono energy integrals se baahir factor ho jaata hai. Right: gas blob high-pressure region mein chhota squeeze ho jaata hai (variable $ρ$ ) → $ρ$ integral ke andar phansa rehta hai.

Step 9 — Assumption 4: EK STREAMLINE KE ALONG (aur degenerate multi-road case)

KYA HAI. Step 8 mein har ek integral ek hi road $s$ ke along liya gaya tha. Toh jo constant mila woh guarantee hai sirf un points ke liye same hai jo usi streamline pe hain. Ek alag road ek alag constant carry kar sakti hai.

KYUN / degenerate case. Ek draining vortex mein, andar ki streamlines outer se tezi se spin karti hain. Ek inner point ko ek outer point se ek constant ke saath compare karna bakwaas hai — jab tak flow irrotational na ho, ek aisi special condition jo constant ko har jagah same kardi hai.

PICTURE. Do alag laal streamlines, har ek apne constant $C_{1} \neq = C_{2}$ ke saath label ki huyi; unke beech ek dashed arrow "not allowed unless irrotational" mark kiya hua.

Recall Quick self-test

Inme se kaunse pe tum Bernoulli seedha equate kar sakte ho? ::: Sirf do aisa points jo same streamline pe hain (ya kahin bhi, agar flow irrotational ho).

Ek-picture summary

Upar ki single figure poori story stack karti hai: blob → forces → Newton with material derivative → clock term strike karo (steady) → missing shear note karo (inviscid) → $ρ$ ko energy integrals se baahir factor karo (incompressible) → ek hi road pe rakho (streamline ke along) → Bernoulli.

Recall Feynman retelling — poori walk simple words mein

Socho ek marble-sized paani ka blob ek curved slide pe neeche slide kar raha hai. Do cheezein use slide ke along push karti hain: pressure (peeche ka paani zyada tezi se shove karta hai front ke paani se, toh sirf pressure ka difference count hota hai) aur gravity (sirf woh part jo neeche-slope point karta hai help ya fight karta hai, is se set hota hai ki slide kitni steep hai).

Newton kehta hai push equals mass times acceleration. Lekin ek moving blob do tareekon se accelerate karta hai: kyunki poori nadi time ke saath speed up ho rahi hai (clock), ya kyunki woh ek faster stretch mein slide karta hai (ek narrow part mein travel karna). Dono milke "material derivative" hain.

Ab hum chaar promises karte hain. Steady: nadi time ke saath nahi badal rahi — clock term mar jaata hai. Inviscid: hum pretend karte hain friction nahi hai — humne woh drag arrow simply kabhi draw nahi ki, toh hum assume kar rahe hain yeh zero hai. Incompressible: blob kabhi squish nahi hota, toh uski density wahi rehti hai — yahi humein $ρ$ ko kinetic aur potential integrals se baahir nikalne deta hai aur sab kuch cleanly add karne deta hai. Along one streamline: humne sab kuch sirf ek road ke along add kiya jis pe blob tha, toh humara final answer sirf us road ke baare mein ek promise hai.

Kinetic energy ( $\frac{1}{2} ρ v^{2}$ ), pushing energy ( $P$ ), aur climbing energy ( $ρ g h$ ) jodo — unka sum constant rehta hai. Chaar promises mein se koi ek todo aur sum drift hone lagta hai: friction ise heat mein bleed karta hai, unsteadiness ek hidden shove ghus aati hai, compression ise gas ke andar store karta hai, aur doosri road pe jump karna tumhe bilkul alag number pe le jaata hai.

Visual walkthrough — Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline