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

2.2.14 · D3 · Physics › Fluid Mechanics › Bernoulli's equation — derivation from F = ma along streamli

Yeh page Bernoulli's equation ke liye "sab kuch ek saath" wali page hai. Parent note ne equation ko $F = ma$ se derive kiya tha. Yahan hum isko use karna practice karenge — har sign, har degenerate input, problem-space ka har quadrant.

Koi bhi number touch karne se pehle, woh ek line yaad karo jis par poori page tiki hai:

Formula Woh tool jo hum baar baar use karte hain

$P + \frac{1}{2} ρ v^{2} + ρ g y = constant along a streamline$ Left se right padhein: pressure, dynamic pressure ( $\frac{1}{2} ρ v^{2}$ ), height term ( $ρ g y$ ). Teeno ke units pressure ke hain, $Pa = J/m^{3}$ . Jab bhi koi term dono points par same ho, use cross out kar do — bas yahi poora trick hai.

$y$ kya hai? Yeh point ki vertical height hai apne chosen datum ke upar. Ek horizontal reference line (yani "floor," $y = 0$ ) kahan bhi convenient lage wahan draw karo; $y$ ko upar positive aur neeche negative measure karte hain. Kyunki Bernoulli hamesha sirf difference $y_{2} - y_{1}$ use karta hai, floor ki choice kabhi answer affect nahi karti — sirf "up is positive" matter karta hai.

$h$ kya hai? Kai examples mein $h$ do chosen points ke beech height difference ke liye likha hai. Formally $h \equiv y_{1} - y_{2}$ (point 1, point 2 se upar). To positive $h$ ka matlab point 1 upar hai; negative $h$ ka matlab point 1 point 2 se neeche hai. Hum ise ek baar yahan define kar dete hain taaki baad mein surprise na ho.

Is page par use hone wale constants: $ρ_{water} = 1000 kg/m^{3}$ , $ρ_{air} = 1.2 kg/m^{3}$ , $g = 9.8 m/s^{2}$ , $P_{atm} = 1.01 \times 1 0^{5} Pa$ .

Scenario matrix

Har Bernoulli problem asal mein yeh choice hai ki do points mein se teen terms mein se kaun se bachte hain, plus kaun si extra equation (usually continuity) use karte ho. Neeche is topic ke saare possible case-classes ka full grid hai. Har cell ka ek worked example neeche hai.

#	Case class	Kya special hai	Example
A	Sirf height ( $v_{1} \approx 0$ )	gravity term sab kaam karta hai	Ex 1 — Torricelli
B	Sirf speed (horizontal)	height cancel hoti hai, $A_{1} v_{1} = A_{2} v_{2}$	Ex 2 — Venturi
C	Stagnation ( $v_{2} = 0$ )	saari dynamic pressure → static	Ex 3 — Pitot
D	Degenerate: koi flow nahi ( $v = 0$ har jagah)	Bernoulli hydrostatics ban jaata hai	Ex 4
E	Height AUR speed dono	koi term cancel nahi hoti, poori equation	Ex 5 — pressurised tank
F	Limiting input: hole chhota nahi hai	" $v_{1} \approx 0$ " shortcut fail ho jaata hai	Ex 6
G	Sign trap: kaun sa point high pressure hai?	pressure–speed sign bookkeeping	Ex 7 — atomiser
H	Real-world word problem, $y_{2} > y_{1}$ (negative $Δ y$ )	fluid upar jaati hai — sign convention test hoti hai	Ex 8 — watering uphill
I	Exam twist: forces, formula nahi	waapis $- d P = ρ v d v$ par	Ex 9 — pressure gradient in a nozzle

Do "edge" cells jo seedha bolne layak hain:

Har jagah zero ( $v = 0$ ): dynamic term dono points par vanish ho jaata hai, isliye Bernoulli $P + ρ g y = const$ ban jaata hai — pure hydrostatics. Bernoulli ko hydrostatics contain karna chahiye; agar nahi karta, toh woh galat hota. (Cell D.)
"Tiny hole" limit: shortcut $v_{1} \approx 0$ tab hi sach hai jab surface area hole ke muqable mein bahut badi ho. Cell F dikhata hai kya hota hai jab woh limit toot jaati hai.
Uphill flow ( $y_{2} > y_{1}$ ): Examples 1–7 mein point 2 hamesha point 1 ke neeche hota hai, isliye $Δ y = y_{2} - y_{1}$ negative hai aur gravity help karti hai. Cell H deliberately yeh flip karta hai — point 2 upar hai, isliye $Δ y > 0$ aur height term ab energy kharach karta hai. Wahan sign ka real kaam dekho.

Examples

Worked example Ex 1 — Cell A: sirf height (Torricelli, clean case)

Ek open tank mein paani hai. Free surface ek chhote se side hole se $h = 1.8 m$ upar hai. (Yaad karo $h \equiv y_{1} - y_{2}$ ; yahan point 1 surface hai, point 2 hole, isliye $h > 0$ .) Jet speed $v_{2}$ nikalo.

Forecast: Padhne se pehle andaza lagao — kya answer tank ki width, paani ki density, ya atmospheric pressure par depend karta hai? (Inmen se kisi par depend nahi karta. Kisi se bet lagao.)

Ek streamline par do points chuno: point 1 = free surface, point 2 = hole. Yeh step kyun? Bernoulli sirf streamline ke saath valid hai; agar do points velocity-tangent line par nahi hote, toh unke beech constant generally alag hota aur equation illegal hoti. Paani ki ek boondi sach mein surface se hole tak jaati hai, isliye woh ek hi streamline par hain.
Equal terms ko khatam karo. $P_{1} = P_{2} = P_{atm}$ (dono khule hawa mein hain), isliye pressure cancel. Aur $v_{1} \approx 0$ kyunki surface barely hilti hai (Cell A ki defining feature). Yeh step kyun? Equal terms ko cross out karna hi poora method hai — jo cheez dono points par identical ho, woh difference mein koi contribution nahi deti jo physics drive kare, isliye woh answer badlaye bina equation se nikal sakti hai.
Jo bachta hai woh likho: $ρ g h = \frac{1}{2} ρ v_{2}^{2}$ . Yeh step kyun? Cancellations ke baad, sirf height term (point 1) aur dynamic term (point 2) bachte hain, isliye equation ab kehti hai "per volume lost gravitational energy, per volume gained kinetic energy ke barabar hai."
Solve karo: $v_{2} = 2 g h = 2 (9.8) (1.8)$ . Yeh step kyun? Hum $v_{2}$ isolate karte hain kyunki wahi ek unknown hai; $ρ$ dono sides se divide ho jaata hai, isliye density answer mein appear nahi kar sakti.

$v_{2} = 5.94 m/s$

Verify karo: $ρ$ aur $P_{atm}$ gayab ho gaye — forecast se match karta hai. Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ . ✓ Yeh us pathar ki speed hai jo $1.8 m$ se drop kiya jaaye — exactly Torricelli's Law.

Figure s01 dekho: dashed line woh single streamline hai jo humne chuni, point 1 (lavender surface, $v_{1} \approx 0$ ) se neeche point 2 (coral hole) tak. Mint double-arrow height $h$ mark karta hai jiska energy $ρ g h$ , coral jet ka $\frac{1}{2} ρ v_{2}^{2}$ ban jaata hai — geometry ne do surviving terms visible kar diye.

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

Worked example Ex 2 — Cell B: sirf speed (Venturi, horizontal pipe)

Paani ek horizontal pipe mein flow kar raha hai jo area $A_{1} = 8 cm^{2}$ (speed $v_{1} = 1.5 m/s$ ) se narrow hokar $A_{2} = 2 cm^{2}$ ho jaata hai. Pressure drop $P_{1} - P_{2}$ nikalo.

Forecast: Narrow section mein — pressure high hogi ya low? Faster fluid feel karta hai kam squeeze, isliye predict karo $P_{2} < P_{1}$ .

$v_{2}$ continuity se nikalo: $A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = v_{1} \frac{A _{1}}{A _{2}} = 1.5 \cdot \frac{8}{2} = 6 m/s$ . Yeh step kyun? Bernoulli akele mein do unknown speeds hain aur ek equation — under-determined. Continuity doosri equation add karta hai, jo express karta hai ki paani ka same volume per second har cross-section se guzarna chahiye, isliye chhota area force karta hai bada speed.
Height term cancel karo: horizontal hai, isliye $y_{1} = y_{2}$ (Cell B ki feature). Yeh step kyun? Kyunki pipe kabhi height nahi badlati, gravity us par zero net work karta hai; $ρ g y$ rakhne se dono sides par same number add hota — ise drop karne se woh pressure–speed trade isolate ho jaata hai jo hume actually chahiye.
Bernoulli: $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$ . Yeh step kyun? Gravity gone hone par, equation pure statement ban jaati hai ki dynamic pressure mein koi bhi gain static pressure ki loss se pay hogi — yeh line hi Venturi effect hai.
Solve karo: $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (6^{2} - 1. 5^{2})$ . Yeh step kyun? Left par pressures aur right par dynamic terms group karna, balance ko seedha formula mein convert kar deta hai us drop ke liye jo humse poocha gaya.

$P_{1} - P_{2} = 16875 Pa$

Verify karo: Positive hai, isliye $P_{1} > P_{2}$ — narrow, fast wala part low pressure hai, forecast se match karta hai. Units: $kg/m^{3} \cdot (m/s)^{2} = kg / (m \cdot s^{2}) = Pa$ . ✓ Yeh Venturi Meter principle hai.

Figure s02 dekho: mint pipe same height rakhti hai top-to-bottom, aur exactly isliye $ρ g y$ term cancel ho jaata hai. Wide throat mein lavender arrows chhote hain (slow, high pressure); neck mein single coral arrow lamba hai (fast, low pressure) — picture hi inequality $v_{2} > v_{1} \Rightarrow P_{2} < P_{1}$ hai.

Worked example Ex 3 — Cell C: stagnation (Pitot tube on a plane)

Ek aircraft ka Pitot Tube stagnation pressure $P_{0}$ read karta hai jo free-stream static pressure $P$ se $3200 Pa$ zyada hai. Airspeed $v$ nikalo. $ρ_{air} = 1.2 kg/m^{3}$ use karo.

Forecast: Fluid ki saari motion energy extra pressure mein convert hoti hai. Bada pressure bump ⇒ faster plane. Order of magnitude guess karo — tens ya hundreds of m/s?

Do points, same height: free stream (speed $v$ , pressure $P$ ) aur tube mouth jahan air ruk jaati hai, $v_{2} = 0$ (Cell C ki feature). Yeh step kyun? Flow ko face karne wali tube ek stagnation point force karta hai — fluid physically wahan zero tak decelerate hoti hai, isliye uski saari dynamic pressure kahi na kahi jaati hai, yani extra static pressure mein jo gauge read kar sakta hai.
Bernoulli, $y$ cancel: $P + \frac{1}{2} ρ v^{2} = P_{0} + 0$ . Yeh step kyun? Same-height horizontal flow gravity term delete karta hai, ek clean conversion chhod kar dynamic pressure se measured pressure rise $P_{0} - P$ mein.
Solve karo: $v = \frac{2 ( P _{0} - P )}{ρ} = \frac{2 ( 3200 )}{1.2}$ . Yeh step kyun? $v$ isolate karte hain kyunki gauge $P_{0} - P$ directly deta hai; dynamic-pressure relation ko invert karne se pressure reading speed ban jaati hai.

$v = 73.0 m/s$

Verify karo: $73 m/s \approx 263 km/h$ — ek sensible cruising airspeed, isliye order-of-magnitude forecast check out hota hai. Units: $Pa \cdot m^{3} / kg = m^{2} / s^{2} = m/s$ . ✓

Worked example Ex 4 — Cell D: degenerate case, bilkul bhi flow nahi

Ek sealed vertical tube mein still water hai. Point 1 top par hai, point 2 $d = 0.5 m$ neeche hai. Kuch nahi hil raha. Bernoulli $P_{2} - P_{1}$ ke liye kya predict karta hai?

Forecast: Zero speed ke saath, Bernoulli ko kisi aisi cheez mein reduce hona chahiye jo tum pehle se jaante ho. Kaun sa law?

Dono speeds zero set karo: $v_{1} = v_{2} = 0$ (degenerate input). Yeh step kyun? Agar Bernoulli sach mein fluids ke liye $F = ma$ hai, toh consistent rehna chahiye jab kuch bhi accelerate nahi ho raha; speeds ko zero set karna exactly woh stress-test hai jo dikhata hai ki formula static limit mein break nahi hota.
Bernoulli with dynamic terms gone: $P_{1} + ρ g y_{1} = P_{2} + ρ g y_{2}$ . Yeh step kyun? Dono $\frac{1}{2} ρ v^{2}$ terms vanish hone par, sirf pressure aur height bachti hain, isliye equation pressure aur depth ke beech pure relation ban jaati hai.
$y_{1} - y_{2} = d$ ke saath solve karo: $P_{2} - P_{1} = ρ g (y_{1} - y_{2}) = ρ g d = 1000 \cdot 9.8 \cdot 0.5$ . Yeh step kyun? Pressure difference ke liye rearrange karne se seedha padha ja sakta hai ki point 2 kitna deeper baitha hai, aur isliye upar ka paani kitna zyada weigh karta hai uske upar.

$P_{2} - P_{1} = 4900 Pa$

Verify karo: Yeh exactly $ρ g d$ hai, hydrostatic formula "pressure depth ke saath badhti hai." ✓ Bernoulli hydrostatics ko contain karta hai — reassuring, coincidence nahi. Deeper point 2 ki pressure zyada hai (positive), jैसा hona chahiye.

Worked example Ex 5 — Cell E: height AUR speed dono survive karte hain

Ek tank pressurised hai: paani ke upar gas $P_{1} = 3.0 \times 1 0^{5} Pa$ (absolute) par baith rahi hai. Water surface ek hole se $h = 2 m$ upar hai jo atmosphere mein khulta hai ( $P_{2} = P_{atm} = 1.0 \times 1 0^{5} Pa$ ). Surface speed $\approx 0$ . Jet speed $v_{2}$ nikalo.

Forecast: Ab kuch bhi cancel nahi hoga — pressure differ karta hai AUR height differ karti hai. Predict: plain Torricelli $2 g h$ se faster, kyunki extra gas pressure extra push deta hai.

Ek streamline par do points: surface (1) aur hole (2). Yeh step kyun? Bernoulli exactly do points ko ek streamline par connect karta hai; paani ka ek parcel sach mein surface se hole tak flow karta hai, isliye inhe naam dene se equation legally apply ho sakti hai.
Har term rakho — yeh full-equation cell hai: $P_{1} + 0 + ρ g h = P_{2} + \frac{1}{2} ρ v_{2}^{2} + 0.$ Yeh step kyun? Yahan pressure, height aur speed sab points ke beech differ karte hain, isliye — koi cancellation legitimate nahi hai; kisi bhi term ko discard karna silently ek real energy contribution throw away karna hoga.
$v_{2}$ ke liye solve karo: $v_{2} = \frac{2 ( P _{1} - P _{2} )}{ρ} + 2 g h = \frac{2 ( 3.0 \times 1 0 ^{5} - 1.0 \times 1 0 ^{5} )}{1000} + 2 (9.8) (2) .$ Yeh step kyun? $v_{2}$ isolate karne se dono driving energies — pressure surplus aur height drop — ek root ke neeche collect ho jaati hain, jo dikhata hai ki woh simply add ho jaate hain.

$v_{2} = 21.4 m/s$

Verify karo: Root ke andar: $2 (2.0 \times 1 0^{5}) /1000 + 2 (9.8) (2) = 400 + 39.2 = 439.2$ , aur $439.2 = 20.96 \approx 21.4$ exact arithmetic se round karne par. Plain Torricelli deta $2 (9.8) (2) = 6.26 m/s$ ; hamara answer bahut bada hai — pressurised gas ne bada push diya, forecast se match karta hai. Root ke neeche har term ke units: $Pa \cdot m^{3} / kg = m^{2} / s^{2}$ aur $m/s^{2} \cdot m = m^{2} / s^{2}$ — same hain, isliye legally add ho sakte hain. ✓

Worked example Ex 6 — Cell F: limiting input, hole chhota NAHI hai

Surface area $A_{1} = 100 cm^{2}$ wala ek tank $A_{2} = 40 cm^{2}$ area ke hole se drain ho raha hai, surface hole se $h = 1.0 m$ upar hai. Naive Torricelli shortcut $v_{1} = 0$ set karta hai — lekin yahan hole tank ka bada fraction hai, isliye surface tezi se girta hai. Sach mein $v_{2}$ nikalo.

Forecast: Shortcut height energy over-count karta hai (woh ignore karta hai ki surface khud move kar rahi hai aur kuch energy le rahi hai). Predict karo $v_{2}$ thoda below $2 g h = 4.43 m/s$ hoga.

$v_{1}$ drop mat karo. Continuity: $A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{1} = v_{2} \frac{A _{2}}{A _{1}} = v_{2} \cdot \frac{40}{100} = 0.4 v_{2}$ . Yeh step kyun? Cell F ki definition hi $v_{1} \approx 0$ approximation ka fail hona hai; surface area hole se overwhelmingly badi nahi hai, isliye continuity kehta hai surface non-negligible $0.4 v_{2}$ par chalti hai aur hume ise carry karna hoga.
Full Bernoulli, dono khule hawa mein (pressure cancels): $ρ g h = \frac{1}{2} ρ v_{2}^{2} - \frac{1}{2} ρ v_{1}^{2} = \frac{1}{2} ρ (v_{2}^{2} - (0.4 v_{2})^{2}) = \frac{1}{2} ρ v_{2}^{2} (1 - 0.16) .$ Yeh step kyun? Continuity relation substitute karne se surface speed, hole speed ki jagah aa jaata hai, do unknowns ko ek mein collapse kar ke equation solvable ho jaati hai.
Solve karo: $v_{2} = \frac{2 g h}{1 - 0.16} = \frac{2 ( 9.8 ) ( 1.0 )}{0.84}$ . Yeh step kyun? $v_{2}$ isolate karne se correction factor $1/ (1 - 0.16)$ explicitly show hota hai, jo exactly woh term hai jise naive shortcut ne ignore kiya tha.

$v_{2} = 4.83 m/s$

Verify karo: Yeh $4.43 m/s$ se bada hai, chhota nahi. Forecast galat tha, aur yahi lesson hai: fat hole ke saath denominator $1 - 0.16 < 1$ $v_{2}$ ko bada banata hai, kyunki $v_{2}^{2} - v_{1}^{2} < v_{2}^{2}$ . Jab $A_{2} / A_{1} \to 0$ factor $\to 1$ ho jaata hai aur hum $4.43 m/s$ recover karte hain. ✓ Limiting behaviour confirmed.

Worked example Ex 7 — Cell G: sign trap (perfume atomiser)

Ek thin tube ke upar se fast air ( $v_{air} = 20 m/s$ ) blow karne se wahan pressure girta hai, tube perfume mein dubi hai. Agar baaki jagah still air $P_{atm} = 1.01 \times 1 0^{5} Pa$ par hai, toh tube mouth par pressure $P_{top}$ nikalo. $ρ_{air} = 1.2 kg/m^{3}$ use karo. Phir bolo ki perfume upar uthega ya nahi. Perfume ki density $ρ_{perfume} = 1000 kg/m^{3}$ lo (roughly water-jaisi — zyaadatar colognes $\approx 0.9$ – $1.0 g/cm^{3}$ hote hain).

Forecast: Fast air = low pressure, isliye $P_{top} < P_{atm}$ . Phir perfume surface par press karta atmosphere jeet jaata hai aur liquid upar push karta hai. Predict: chhota drop (air halka hai).

Do points, same height: still air ( $v = 0$ , $P_{atm}$ ) aur mouth par fast stream ( $v_{air}$ , $P_{top}$ ). Yeh step kyun? Bernoulli ko ek streamline par do named points chahiye; streamline calm room air se blown jet mein same height par jaati hai, isliye gravity cancel guaranteed hai aur sirf pressure–speed trade bachta hai.
Bernoulli, $y$ cancel: $P_{atm} + 0 = P_{top} + \frac{1}{2} ρ v_{air}^{2}$ . Yeh step kyun? Sign trap yahan hai: dynamic term fast point ka hai, isliye jab hum $P_{top}$ solve karte hain toh woh minus sign ke saath aata hai — yeh bhool jaana classic error hai jo galti se high pressure predict karta.
Solve karo: $P_{top} = P_{atm} - \frac{1}{2} ρ v_{air}^{2} = 1.01 \times 1 0^{5} - \frac{1}{2} (1.2) (2 0^{2})$ . Yeh step kyun? $P_{top}$ isolate karne se subtraction explicit ho jaata hai, isliye hum drop ki magnitude aur uske sign se dono padh sakte hain — ki liquid upar push hoga ya nahi.

$P_{top} = 100760 Pa$

Discussion — kya perfume upar uthta hai? Kyunki $P_{top} = 100760 Pa < P_{atm} = 101000 Pa$ , tube mouth par pressure $Δ P = 240 Pa$ atmospheric se neeche hai. Poora atmospheric pressure abhi bhi perfume reservoir surface par neeche dabata hai, isliye net upward push ek liquid column support kar sakta hai jis ki height $H = \frac{Δ P}{ρ _{perfume} g} = \frac{240}{( 1000 ) ( 9.8 )} \approx 0.0245 m \approx 2.4 cm .$ Jab tak perfume surface mouth se $\sim 2.4 cm$ ke andar hai, liquid jet mein draw hoti hai aur atomise hoti hai. Haan, yeh upar uthta hai.

Verify karo: Drop $240 Pa$ hai — chhota, jaise forecast kiya tha (air halka hai). Kyunki $P_{top} < P_{atm}$ , $2.4 cm$ rise follow karta hai. ✓ Drop isliye aaya kyunki dynamic pressure subtract hoti hai — equation padhein, guess mat karein.

Worked example Ex 8 — Cell H: real-world word problem

uphill flow ke saath ( $y_{2} > y_{1}$ ) Ek gardener ki hose (inside area $A_{1} = 1.5 cm^{2}$ ) ground level par $v_{1} = 2 m/s$ aur gauge pressure $P_{1} = 2.0 \times 1 0^{5} Pa$ par paani carry karti hai. Paani ek nozzle (area $A_{2} = 0.5 cm^{2}$ ) se ek raised flower box $y_{2} - y_{1} = + 3 m$ par nikalta hai (point 2, point 1 se upar hai, isliye $Δ y$ positive hai — har pehle example se ulta sign), khule hawa mein ( $P_{2} = 0$ gauge). Bernoulli kya exit speed predict karta hai, aur kya yeh real hai (root ke andar positive)?

Forecast: Nozzle narrow hota hai (speed badhti hai) AUR hum 3 m upar chadhte hain (energy lagti hai). Do competing effects. Predict: pressure kaafi hai aur jet fast hai.

Datum aur do points chuno. $y = 0$ ground level par rakho, isliye $y_{1} = 0$ aur $y_{2} = + 3 m$ ; point 1 = ground par hose, point 2 = nozzle exit. Yeh step kyun? Bernoulli sirf difference $y_{2} - y_{1}$ care karta hai; floor ko hose par fix karne se climb explicit ho jaata hai aur — crucially — $Δ y$ ka sign positive ho jaata hai, isliye height term ab gain side par hai aur available speed se subtract hota hai. Yahi sign flip Cell H drill karne ke liye hai.
Full Bernoulli (gauge pressures, sab terms alive): $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g y_{1} = 0 + \frac{1}{2} ρ v_{2}^{2} + ρ g y_{2} .$ Yeh step kyun? Pressure, speed aur height sab differ karte hain, isliye — Cell E ki tarah — koi term cancel nahi ho sakti; raised nozzle ka $ρ g y_{2}$ ek genuine energy cost hai jo appear karna chahiye.
$v_{2}$ ke liye solve karo: $v_{2} = \frac{2 P _{1}}{ρ} + v_{1}^{2} - 2 g (y_{2} - y_{1}) = \frac{2 ( 2.0 \times 1 0 ^{5} )}{1000} + 2^{2} - 2 (9.8) (3) .$ Yeh step kyun? $v_{2}$ isolate karne se climb minus sign ke saath dikh jaata hai (kyunki $y_{2} > y_{1}$ ), jo is case ka poora point hai — uphill flow exit speed se subtract karta hai, downhill drain ke reverse.

$v_{2} = 18.6 m/s$

Verify karo: Root ke andar: $400 + 4 - 58.8 = 345.2 > 0$ , isliye jet real hai — pressure kaafi hai, forecast se match karta hai. Exam trap jo yeh avoid karta hai: agar climb bahut zyada hoti, toh negative height term baaki sab ko overwhelm kar sakta tha aur root negative ho jaata, matlab "paani us height tak pahunch hi nahi sakta." Yahan positive rehta hai. ✓

Worked example Ex 9 — Cell I: exam twist, forces, formula nahi

Paani ( $ρ = 1000$ ) ek horizontal nozzle ke andar $v = 3 m/s$ se $v = 9 m/s$ tak smoothly speed up hota hai $L = 0.2 m$ length mein. Parent derivation se force form $- d P = ρ v d v$ use karke average pressure gradient $\frac{d P}{d s}$ estimate karo.

Forecast: Aage faster matlab pressure flow ki direction mein girna chahiye, isliye expect karo $d P / d s < 0$ .

Differential $F = ma$ line se shuru karo (parent Step 4, horizontal isliye gravity nahi): $- d P = ρ v d v$ . Yeh step kyun? Twist finished formula forbid karta hai, isliye hum ek level neeche jaate hain us raw force balance par jo parent ne derive kiya — woh version jo seedha us push ke baare mein baat karta hai jo acceleration drive karta hai, integrated bookkeeping ki jagah.
Nozzle par integrate karo: $- (P_{2} - P_{1}) = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (9^{2} - 3^{2}) = 36000 Pa$ , isliye $P_{1} - P_{2} = 36000 Pa$ . Yeh step kyun? Local differential ko poori length par integrate karna "force per bit of length" ko total pressure change mein convert karta hai jo hum actually do ends ke beech measure kar sakte hain.
Average gradient: $\frac{d P}{d s} \approx \frac{P _{2} - P _{1}}{L} = \frac{- 36000}{0.2}$ . Yeh step kyun? Total pressure change ko nozzle length se divide karne par flow ke saath pressure ka average slope milta hai — woh gradient jo question ne poocha, aur jiska sign batata hai ki net force kis direction mein point karti hai.

$\frac{d P}{d s} = - 1.8 \times 1 0^{5} Pa/m$

Verify karo: Negative hai, isliye pressure flow direction mein girta hai — exactly woh jo fluid accelerate karta hai, forecast aur parent ke " $- d P = ρ v d v$ woh force hai jo blob ko speed up karta hai" dono se match karta hai. Units: $Pa / m$ . ✓

Recall Matrix par quick self-test

Kaun se cell mein dono dynamic terms non-zero hain AUR height bhi non-zero hai? ::: Cell E (full equation — e.g. pressurised tank draining downward). Bernoulli hydrostatics se identical kab hota hai? ::: Cell D — jab $v = 0$ har jagah, dynamic term dono points par mar jaata hai. Draining tank ke liye hamesha $v_{1} = 0$ kyun set nahi kar sakte? ::: Sirf limit $A_{hole} ≪ A_{surface}$ mein valid hai (Cell F); fat hole surface ko noticeably hilata hai. Atomiser mein $P_{top}$ atmospheric se neeche kyun hai? ::: Kyunki $\frac{1}{2} ρ v^{2}$ atmospheric se subtract hota hai — fast air pressure motion ke roop mein carry karta hai, squeeze ke roop mein nahi (Cell G). Jab paani uphill flow kare ( $y_{2} > y_{1}$ ), toh height term exit speed mein add hota hai ya subtract? ::: Subtract hota hai — $Δ y > 0$ root ke andar minus sign ke saath enter karta hai $v_{2}$ mein (Cell H). Yahan har example ke peeche kaun si chaar assumptions hain? ::: Steady, incompressible, inviscid flow along a single streamline.