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

2.2.14 · D5 · Physics › Fluid Mechanics › Bernoulli's equation — derivation from F = ma along streamli

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Symbol & assumption refresher (pehle yeh padho)

Definition Neeche use hone wala har symbol, seedhi bhaasha mein

$P$ = static pressure — kisi point par sadharan squeeze, pascals mein measure hoti hai ( $Pa = N/m^{2} = J/m^{3}$ ).
$ρ$ (Greek "rho") = density — har cubic metre mein kitna mass packed hai ( $kg/m^{3}$ ); yahan constant maana gaya hai.
$v$ = kisi point par fluid ki flow speed ( $m/s$ ).
$g$ = gravitational acceleration, lagbhag $9.8 m/s^{2}$ neeche ki taraf.
$y$ = kisi chosen reference level se us point ki height ( $m$ ).
$h$ = ek height difference (jaise paani ki surface ke neeche kisi hole ki depth), $m$ .
$A$ = pipe ya blob ka cross-sectional area ( $m^{2}$ ).
$s$ = streamline ke saath measure ki gayi distance — tum flow ke curved path par "chalte" ho aur $s$ batata hai tum kitna aage gaye; $d s$ us path par ek chhota sa step hai. Yahi woh coordinate hai jo $v \frac{d v}{d s}$ ko meaningful banata hai.
$θ$ = woh angle jo streamline horizontal se banati hai, toh ek step $d s$ tumhari height $d y = d s sin θ$ jitni badhata hai.

Yeh items in char idealisations, force-vs-energy readings, Equation of Continuity, aur convective-acceleration trick par lean karte hain. Neeche ke do figures woh scenarios hain jo questions mein baar baar aate hain — shuru karne se pehle unhe dekh lo.

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

True ya false — justify karo

Bernoulli's constant ki value ek flowing fluid mein har jagah same hoti hai.

Aam taur par False.

P + \frac{1}{2} ρ v^{2} + ρ g y

ka sum ek streamline ke saath constant hota hai; ek alag streamline alag constant carry kar sakti hai, jab tak ki poori flow irrotational na ho.

"Steady flow" ka matlab hai ki har fluid parcel ka acceleration zero hota hai.

False. Steady ka matlab hai ki kisi fixed point par velocity time ke saath nahi badalti; ek parcel jo kisi tez region mein ja raha hai woh phir bhi convective acceleration

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

se accelerate hota hai, jahan

s

streamline ke saath distance hai.

Agar ek horizontal pipe narrow ho jaaye, toh narrow part mein pressure kam hota hai.

True. Equation of Continuity fluid ko constriction mein speed up karne par majboor karti hai (upar wali Venturi figure dekho), aur Bernoulli phir require karta hai ki static pressure

P

drop kare taaki total constant rahe.

Paani kisi hole se jitna gehri hole surface se neeche hogi utni tez nikalti hai.

True. Torricelli deta hai

v = 2 g h

jahan

h

free surface se neeche depth hai (upar wali left figure), toh zyada gehri hole (bada

h

) matlab tez jet — Torricelli's Law.

Bernoulli's equation energy conservation ki ek form hai.

True. Derived line

d P + ρ g d y + ρ v d v = 0

ko volume se multiply karo aur yeh hai Work-Energy Theorem per unit volume: pressure-work plus gravity-work equals kinetic energy mein change.

Tum Bernoulli ko ek water pump ke across apply kar sakte ho.

False. Ek pump fluid mein energy add karta hai, jo "no energy added" assumption (idealisation 3) ko violate karta hai; tumhe ek pump-head term daalna hoga, warna equation wahan simply galat hai.

Low pressure hi woh cheez hai jo fluid ko fast move karaati hai.

Yeh statement galat hai. Causation

F = ma

se hota hai: ek pressure difference woh net force hai jo speed change produce karta hai (

- d P = ρ v d v

). Speed aur low pressure saath aate hain, lekin difference kaaran hai.

\frac{1}{2} ρ v^{2}

term ki units pressure ki hain.

True. Har Bernoulli term energy per unit volume hai, yaani

J/m^{3} = Pa

— isi liye inhe static pressure

P

mein add kiya ja sakta hai.

Rest mein fluid ke liye, Bernoulli's equation

P + ρ g y = constant

mein reduce ho jaati hai.

True. Jab

v = 0

har jagah ho toh dynamic term gayab ho jaata hai aur tumhe Hydrostatic Pressure milti hai: pressure depth ke saath linearly badhta hai.

Bernoulli ek patli tube mein dheere behte honey ke liye utni hi achhi tarah kaam karti hai jitni ek wide tube mein paani ke liye.

False. Honey bahut zyada viscous hai, toh internal friction energy dissipate karta hai (idealisation 3 tooti) — woh regime Viscosity and Poiseuille Flow se govern hota hai, inviscid Bernoulli se nahi.

Error pakdo

P_{1} = P_{2}

draining-tank streamline ke dono ends par hai, toh do static pressures cancel ho jaate hain aur paani move nahi kar sakta."

Pressures cancel hote hain, haan, lekin gravity nahi hoti —

ρ g h

height difference woh bachha hua force hai, aur yahi jet ko

2 g h

tak accelerate karta hai (left figure).

"Venturi mein, area shrink hoti hai, toh

F = ma

se mass shrink hoti hai aur parcel slow ho jaata hai."

Galat link. Equation of Continuity volume flow

A v

ko constant fix karta hai, toh chhoti area

A

ka matlab hai badi speed

v

; parcel speed up hota hai, aur yahan mass conserve nahi ho raha.

"Kyunki flow steady hai,

\frac{d v}{d t} = 0

, toh koi acceleration term nahi hai aur Bernoulli mein

\frac{1}{2} ρ v^{2}

hona hi nahi chahiye."

Parcel ka

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

steady flow mein bhi nonzero hota hai (convective acceleration, jahan

s

= streamline ke saath arc length); yahi term integrate hokar

\frac{1}{2} ρ v^{2}

piece banata hai.

"Gravity

m g

neeche ki taraf hai, toh

F = ma

mein gravity term

- ρ A d s g

hai."

Tum streamline par project karna bhool gaye. Sirf flow ke saath wala component kaam karta hai, jo deta hai

- ρ A g d y

jahan

d y = d s sin θ

woh rise hai (poora step length

d s

nahi), aur

θ

streamline ka tilt hai.

"Ek wing ka upar pressure kam hota hai, aur low pressure upar ki taraf suck karta hai, isliye Bernoulli flight ke liye energy create karta hai."

Koi energy create nahi hoti. Pressure difference ek force hai jo flow ki kinetic aur potential energy mein already account ki gayi hai — dekho Aerodynamic Lift; Bernoulli bookkeeping hai, source nahi.

"Pitot tube ke mooh par fluid ruk jaata hai, toh uski energy gayab ho jaati hai."

Kinetic energy gayab nahi hoti, woh extra static pressure mein convert hoti hai:

P_{0} - P = \frac{1}{2} ρ v^{2}

, aur yahi gauge read karta hai — Pitot Tube.

Why questions

Derivation mein cross-sectional area

A

kyun cancel ho jaata hai?

Kyunki pressure force, gravity component, aur mass sabhi

A

ke proportional hain (har face ki area

A

hai, volume

A d s

hai), toh woh divide out ho jaata hai — result streamline ki property hai, blob ki motaai ki nahi.

Hume streamline ke saath integrate kyun karna chahiye, na ki kisi bhi direction mein?

Humne jo

F = ma

likha usne sirf forces aur acceleration flow ke saath use kiye (yahi woh coordinate

s

track karta hai); streamline ke across components alag balance hote hain, toh yeh saaf exact-differential form sirf streamline follow karte hue hold karta hai.

Pressure term difference

- A d P

kyun aata hai,

P A

kyun nahi?

Ek uniform pressure dono faces ko equally squeeze karta hai aur cancel ho jaata hai; sirf parcel ke across change

d P

ek net push deta hai, isliye physics pressure gradient par depend karta hai.

Horizontal pipe ke liye "fast matlab low pressure" kyun sach hai lekin rising pipe ke liye automatically nahi?

Horizontal streamline par height term

ρ g y

fixed hoti hai, toh

\frac{1}{2} ρ v^{2}

aur

P

directly trade off karte hain; agar fluid upar bhi chadh raha hai, toh

ρ g y

term bhi bargain mein shamil ho jaata hai aur pressure comparison flip kar sakta hai.

Incompressibility integration step ke liye kyun matter karti hai?

Jab

ρ

constant ho toh woh integrals se saaf bahar aa jaata hai, jo deta hai neat

\frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})

aur

ρ g (y_{2} - y_{1})

; agar

ρ

vary karta toh tum use bahar nahi nikal sakte aur equation ko compressible correction ki zarurat padti.

Shower curtain paani ki taraf andar kyun khicha jaata hai?

Curtain ke andar fast-moving air (spray se drag hoti) ki static pressure bahar ki still room air se kam hoti hai, aur bahar ki pressure curtain ko andar push karti hai — wahi fast=low-pressure trade jo Venturi Meter mein dikhi (right figure).

Edge cases

Jab

v = 0

har jagah ho (koi flow nahi) toh Bernoulli kya kehta hai?

Woh collapse hokar

P + ρ g y = constant

ban jaata hai, yaani hydrostatic pressure law — Bernoulli statics ko ek special case mein contain karta hai.

Jab hole surface tak utha jaaye (

h \to 0

) toh exit speed formula

2 g h

ka kya hota hai?

Jet speed zero ho jaati hai: hole ke level par paani ke paas koi head

h

nahi hota jo use push kare, toh kuch bhi nahi nikalti.

Agar tank ka hole chhota nahi hai (tank ke cross-section ke comparable hai), toh kya

v_{2} = 2 g h

ab bhi exact hai?

Nahi — humne Equation of Continuity use karke

v_{1} \approx 0

assume kiya tha; bade hole ke saath surface noticeably fast girti hai,

v_{1}

negligible nahi rahti, aur

2 g h

overestimate ban jaata hai.

Stationary fluid (

v = 0

) mein Pitot tube kya read karta hai?

P_{0} = P

exactly — koi flow nahi toh convert karne ke liye koi dynamic pressure nahi, toh gauge sirf ambient static pressure dikhata hai aur zero speed report karta hai.

Agar streamline seedhi upar jaaye bina koi speed change ke (

v_{1} = v_{2}

) toh kya?

Dynamic terms cancel ho jaate hain aur bachta hai

P_{1} - P_{2} = ρ g (y_{2} - y_{1})

, yaani ek vertical column ke saath pure hydrostatics.

Kya Bernoulli us moment apply hoti hai jab tum tap khol dete ho, flow settle hone se pehle?

Nahi — woh unsteady startup flow hai (

\partial / \partial t \neq = 0

), toh steadiness assumption (idealisation 1) fail ho jaati hai aur flow settle hone tak ek extra time-dependent term chahiye.

Recall Ek-line self-test

Agar tum upar ke har trap ke liye yeh bata sako ki char idealisations mein se kaun sa ya derivation ka kaun sa step woh attack karta hai, toh tumne concept map master kar liya hai — sirf algebra nahi.

Item :: koi bhi trap chuno, woh assumption batao jo woh target karta hai, phir reveal karo

Steady / incompressible / inviscid / single-streamline, ya convective-acceleration step — is page par har trap inhi mein se exactly ek par map karta hai.