Bernoulli's equation — derivation from F = ma along streamline

2.2.14 · Physics › Fluid Mechanics

YEH kyun zaroori hai?

HUM kya assume karte hain (taaki derivation honest rahe)

HUM ise kaise derive karte hain — fluid blob par $F = ma$

Figure — Bernoulli's equation — derivation from F = ma along streamline

Step 1 — Ek tiny cylindrical parcel lo

Streamline par ek chhota sa fluid element lo: cross-sectional area $A$ , streamline ke saath length $d s$ .

Volume: $d V = A d s$
Mass: $d m = ρ A d s$

Yeh step kyun? Newton ka law ek definite mass ke baare mein hai. Flow ke saath aligned ek chhota cylinder humein sirf woh forces add karne deta hai jo motion ke saath direction mein hain.

Step 2 — Streamline ke saath forces

Maano $s$ streamline ke saath distance measure karta hai (flow direction mein badhta hua).

(a) Pressure force. Pressure $P$ back face (area $A$ ) ko aage push karta hai; pressure $P + d P$ front face ko peeche push karta hai.

d F_{pressure} = P A - (P + d P) A = - A d P

Yeh step kyun? Parcel ke across pressure ka sirf antar hi ek net force produce karta hai; ek uniform pressure dono ends par equally squeeze karta hai aur cancel ho jaata hai.

(b) Gravity component. Gravity $- d m g \overset{y}{^}$ hai. Streamline ke saath component is baat par depend karta hai ki streamline kitni steeply upar jaati hai. Agar parcel length $d s$ par $d y$ upar jaaye, toh slope $sin θ = d y / d s$ hai:

d F_{gravity} = - d m g sin θ = - ρ A d s g \frac{d y}{d s} = - ρ A g d y

Yeh step kyun? Gravity ka sirf woh hissa jo streamline ke saath direction mein point kare, parcel ko speed up ya slow kar sakta hai; sideways wala hissa streamline ke across pressure se balance ho jaata hai.

Step 3 — Streamline ke saath acceleration (smart wala part)

Steady flow mein, kisi fixed point par speed $v$ time ke saath constant hoti hai, lekin ek moving parcel phir bhi accelerate karta hai kyunki woh ek aisi jagah jaata hai jahan alag $v$ hoti hai. Chain rule use karke:

a = \frac{d v}{d t} = \frac{d v}{d s} \frac{d s}{d t} = v \frac{d v}{d s}

Yeh step kyun? Yeh convective acceleration hai: parcel "khud ko" ek faster region mein le jaata hai. Isliye ek steady flow phir bhi accelerate kar sakti hai — ek subtle point jise log bhool jaate hain.

Step 4 — $F = ma$ apply karo

d F_{pressure} + d F_{gravity} = d m a

- A d P - ρ A g d y = (ρ A d s) (v \frac{d v}{d s})

Notice karo ki $A$ cancel ho jaata hai aur right side par $d s$ cancel ho jaata hai ( $d s \cdot \frac{1}{d s}$ ):

- d P - ρ g d y = ρ v d v

Rearrange karo taaki sab kuch ek differential ho:

d P + ρ g d y + ρ v d v = 0

Yeh step kyun? Har term ab ek exact differential hai — hum streamline ke saath term by term integrate kar sakte hain.

Step 5 — Streamline ke saath integrate karo

$ρ$ = const ke saath, point 1 se point 2 tak integrate karo:

\int d P + ρ g \int d y + ρ \int v d v = 0

(P_{2} - P_{1}) + ρ g (y_{2} - y_{1}) + \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = 0

Point ke hisaab se group karo:

Worked Examples

Common Mistakes (Steel-manned)

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek crowd ek hallway mein chal rahi hai. Jab hallway narrow ho jaata hai, logon ko tez chalna padta hai taaki woh aage badhte rahein (koi gayab nahi hota). Tez chalne ke liye unhe peeche se ek push chahiye — isliye peeche wale log ekdusre mein dabe hue hain (high "push"/pressure) aur narrow part mein fast chal rahe log kam dabe hue feel karte hain (low pressure). Bernoulli's equation bas ek bookkeeping hai: dabav (pressure) + speed-energy + height-energy hamesha same total hoti hai jab tum hallway mein ek insaan ko follow karo. Agar ek badh jaaye, doosra kam hona chahiye.

Forecast-then-Verify

Active Recall

Bernoulli's equation kiski disguised form hai?

Newton's second law

F = ma

ki, jo streamline ke saath integrate kiya gaya hai (equivalently, work–energy theorem per unit volume).

Bernoulli's equation ke peeche chaar assumptions kya hain?

Steady, incompressible (constant

ρ

), non-viscous (inviscid), aur ek single streamline ke saath.

Steady flow mein ek fluid parcel accelerate kyun karta hai?

Convective acceleration

a = v d v / d s

: woh alag velocity ke ek region mein move karta hai, chahe field time-independent ho.

Integrate karne se pehle differential form likhо.

d P + ρ g d y + ρ v d v = 0

Bernoulli's equation mein har term ki units kya hain?

Pressure, yaani

Pa = J/m^{3}

(energy per unit volume).

Bernoulli se Torricelli's law derive karo.

Open tank,

P_{1} = P_{2}

v_{1} \approx 0

, drop

h

⇒

ρ g h = \frac{1}{2} ρ v_{2}^{2}

⇒

v_{2} = 2 g h

Constriction mein pressure sabse kam kahan hoti hai?

Jahan area sabse chhota hota hai, kyunki wahan speed sabse zyada hoti hai aur

P + \frac{1}{2} ρ v^{2} =

const.

Derivation mein humne

ma

ke against kaun sa net force balance kiya?

Net pressure force

- A d P

aur gravity ka streamline-component

- ρ A g d y

Derivation mein

A

aur

d s

cancel kyun ho jaate hain?

A

har force/mass term ko multiply karta hai; mass ka

d s

a = v d v / d s

mein

1/ d s

ko cancel kar deta hai, aur ek pure differentials mein equation bacha leta hai.

Ek Pitot tube air mein

P_{0} - P = 200

Pa read karta hai (

ρ = 1.2

). Airspeed kya hai?

v = 2 (200) /1.2 \approx 18.3 m/s

Kya Bernoulli do alag streamlines par points compare kar sakta hai?

Generally nahi — constant har streamline par alag hota hai, jab tak flow irrotational na ho.

Connections

Equation of Continuity — har example mein use ki jaane wali $A_{1} v_{1} = A_{2} v_{2}$ relation supply karta hai.
Newton's Second Law — woh parent law jise humne integrate kiya.
Work-Energy Theorem — alternative derivation route (energy reading).
Hydrostatic Pressure — $ρ g y$ term $P = P_{0} + ρ g h$ mein reduce ho jaata hai jab $v = 0$ ho.
Viscosity and Poiseuille Flow — jab "inviscid" fail ho tab kya add karna hai.
Torricelli's Law · Venturi Meter · Pitot Tube · Aerodynamic Lift — applications.