Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline: Worked examples

2.2.15 · D3 · Physics › Fluid Mechanics › Assumptions in Bernoulli — steady, inviscid, incompressible,

Hum sab kuch usi master statement se banate hain (pehle earn kiye bina kabhi use nahi kiya):

Formula Poori parcel equation (kuch bhi phenkne se pehle)

$unsteady ρ d s \frac{\partial v}{\partial t} + convective ρ v d v = pressure - d P gravity - ρ g d h + friction d F_{visc} + added energy d F_{pump}$ Ise Newton's 2nd law on a sliding fluid blob ki tarah padho. $P$ = pressure (force per area, unit pascal $= N/m^{2}$ ), $v$ = speed, $h$ = height, $ρ$ = density (mass per volume, $kg/m^{3}$ ), $g = 9.8 m/s^{2}$ , $d s$ = streamline par ek chhota sa step, $A$ = parcel ka cross-sectional area ( $m^{2}$ ). Bernoulli chhe mein se teen terms drop karta hai: wo demand karta hai $\partial v / \partial t = 0$ taaki unsteady term gayab ho jaye, aur assume karta hai ki koi friction nahi aur koi pump nahi to $d F_{visc} = 0$ aur $d F_{pump} = 0$ . Jo bachta hai wo convective, pressure aur gravity terms ka balance hai — yahi poora Bernoulli hai. Neeche har failure mein wohi teen dropped terms mein se ek wapas zinda ho jaati hai.

Definition Paanch off-page tools, yahan restate kiye taaki ye page apne aap mein complete ho

Material derivative — ek moving parcel ki acceleration ke do hisse hote hain: $\frac{D v}{D t} = \frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}$ . Pehla hai ek fixed spot par time-change, doosra hai ek tezi wale region mein move karke speed up hona. (Poora build Material Derivative mein hai.)
Continuity equation — incompressible flow ke liye mass in = mass out, to $A_{1} v_{1} = A_{2} v_{2}$ : ek narrow pipe tezi wala flow force karta hai. (Dekho Continuity Equation.)
Viscosity — fluid ki internal friction; wall ke paas ye flow ko drag karti hai aur mechanical energy ko heat mein badal deti hai, Poiseuille flow se describe hota hai pipes mein.
Boundary layer — fluid ki ultra-thin skin jo surface ke bilkul upar hoti hai jahan speed free-stream value se zero tak girta hai; viscosity yahan hamesha dominate karti hai chahe outer flow inviscid ho. (Dekho Boundary Layer.)
Mach number $M = v / c$ ( $c$ = speed of sound) — measure karta hai ki flow kitna compressible act karta hai; $M < 0.3$ se neeche density change ~4.5% se kam rehta hai aur hum gas ko incompressible treat karte hain. (Dekho Mach Number and Compressibility.)

The scenario matrix

Har situation jo ek Bernoulli problem mein aa sakti hai, neeche bilkul ek row mein fit hoti hai. Hamare examples (Ex1–Ex9) us cell ke saath tagged hain jise wo cover karte hain.

Cell	Situation class	Kaun sa term jaag uthta hai?	Example
A	Ideal case — chaaon assumptions hold karti hain	koi nahi (Bernoulli exact)	Ex1
A′	Ideal case with a height change ( $h_{1} \neq = h_{2}$ )	gravity term $ρ g d h$ active	Ex2
B	Friction present (rough/long/viscous)	$d F_{visc}$	Ex3
C	Unsteady / start-up / pulsing	$\partial v / \partial t$	Ex4
D	Compressible gas (high Mach)	$ρ$ not constant in $\int d P / ρ$	Ex5
E	Do alag streamlines	constant har streamline par alag hota hai	Ex6
F	Energy added by a pump/fan	$d F_{pump}$	Ex7
G	Degenerate / zero input ( $v = 0$ , flat, equal area)	terms vanish → limiting form	Ex8
H	Exam twist — ek "valid-lagta-hua" trap jo cells combine karta hai	hidden broken assumption pakdo	Ex9

Streamline ka idea (ek curve jo hamesha velocity ke tangent ho, jise koi fluid cross nahi karta) aur Material Derivative (do-hissa acceleration) wo do tools hain jine hum lean on karte hain; dono parent mein define hain.

Cell A — the ideal case (sabhi assumptions hold karti hain)

Worked example Ex1 · Paani ek smooth nozzle mein tezi se bahna (Cell A)

Water ( $ρ = 1000 kg/m^{3}$ ) ek smooth, level nozzle mein steadily flow karta hai. Wide inlet par $v_{1} = 2 m/s$ , $P_{1} = 300 kPa$ . Nozzle narrow ho jaata hai taaki outlet par $v_{2} = 8 m/s$ . $P_{2}$ nikaliye.

Forecast: Paani 4× fast ho jaata hai. Kya pressure badhega ya ghate ga, aur roughly kitna — tens of kPa, ya hundreds?

License check karo. Steady ✔, smooth ⇒ inviscid ✔, water ⇒ incompressible ✔, single tube ⇒ one streamline ✔. Ye step kyun? Cell A ka matlab hai hum plain Bernoulli tabhi use kar sakte hain jab chaaon confirm ho jaayein; ye skip karna wahi galti hai jo parent mein steel-manned hai.
Bernoulli likho, height term cancel karo. Level pipe ⇒ $h_{1} = h_{2}$ , to $ρ g h$ drop ho jaata hai: $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$ Ye step kyun? Gravity gone hone ke baad, sirf pressure ⇄ kinetic energy ka trade hota hai.
$P_{2}$ solve karo. $P_{2} = P_{1} + \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) = 300000 + \frac{1}{2} (1000) (4 - 64)$ $P_{2} = 300000 - 30000 = 270000 Pa = 270 kPa$ Ye step kyun? Tez paani ⇒ kam pressure; $30 kPa$ ka drop "kinetic tax" hai.

Verify: Units: $ρ v^{2}$ mein $kg/m^{3} \cdot m^{2} / s^{2} = kg/(m\cdots^{2}) = Pa$ ✔. Sanity: pressure gira (sahi — speed badhi), aur total $P + \frac{1}{2} ρ v^{2}$ inlet par $302 kPa$ aur outlet par bhi $302 kPa$ hai. ✔

Cell A′ — ideal case with a height change

Worked example Ex2 · Paani ek rooftop tank mein upar jaana (Cell A′)

Water ( $ρ = 1000 kg/m^{3}$ ) ek pipe mein steadily aur smoothly upar flow karta hai. Ground point par $P_{1} = 400 kPa$ , $v_{1} = 3 m/s$ , $h_{1} = 0$ . Ye $h_{2} = 12 m$ oopar ek rooftop nozzle par nikalta hai jahan speed $v_{2} = 3 m/s$ hai (same area). $P_{2}$ nikaliye.

Forecast: Yahan sirf height badal raha hai. Kya upar ka pressure ground se zyada hoga ya kam, aur kitne kPa se?

License check karo. Steady ✔, smooth ⇒ inviscid ✔, water ⇒ incompressible ✔, one pipe ⇒ one streamline ✔. Ye step kyun? Chaaon hold karti hain, to Bernoulli exact hai — lekin ab gravity term rehta hai.
$ρ g h$ term rakho. Kyunki $h_{1} \neq = h_{2}$ to hum ise drop nahi kar sakte: $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}$ Ye step kyun? Cell A′ wahi reason hai jiske liye gravity term parent ki derivation mein exist karta hai; equal speeds se kinetic terms cancel ho jaate hain.
$P_{2}$ solve karo. $v_{1} = v_{2}$ se kinetic terms cancel ho jaate hain: $P_{2} = P_{1} - ρ g (h_{2} - h_{1}) = 400000 - 1000 \cdot 9.8 \cdot 12$ $P_{2} = 400000 - 117600 = 282400 Pa \approx 282.4 kPa$ Ye step kyun? Paani ko 12 m utha ne mein pressure energy lagti hai — gravity tax, Ex1 mein kinetic tax ka mirror image.

Verify: $ρ g Δ h = 1000 \cdot 9.8 \cdot 12 = 117600 Pa$ ✔. Units: $kg/m^{3} \cdot m/s^{2} \cdot m = Pa$ ✔. Pressure upar jaane par gira — sahi, gravity rise ke against hai.

Cell B — friction jaag uthti hai

Worked example Ex3 · Rusty 50 m wali pipe (Cell B)

Paani ek 50 m rough horizontal pipe mein flow karta hai jiska area constant hai. $P_{1} = 200 kPa$ , $P_{2} = 150 kPa$ ; $v$ aur $h$ dono ends par equal hain. Bernoulli predict karta hai $P_{1} = P_{2}$ — kya galat hua, aur kitni energy per volume kho gayi?

Forecast: Bernoulli insist karta hai pressures equal hain, phir bhi wo 50 kPa se alag hain. Chaaon mein se kaunsi term hum ne illegally cross out kar di?

Naive Bernoulli apply karo. Constant area ⇒ $v_{1} = v_{2}$ (Continuity Equation se: $A_{1} v_{1} = A_{2} v_{2}$ , to equal areas equal speeds force karti hain); horizontal ⇒ $h_{1} = h_{2}$ . To $P_{1} = P_{2}$ predict hota hai. Ye step kyun? Hum prediction dikhate hain taaki contradiction undeniable ho.
Contradiction observe karo. Reality: $Δ P = 50 kPa$ lost. Ye step kyun? 50 kPa se fail hone wali prediction ek missing term signal karti hai, math ki galti nahi.
Woken term identify karo. Rough wall viscous shear exert karta hai (wo fluid friction jo wall ke paas flow ko drag karti hai); cross-out kiya hua $d F_{visc}$ wapas aata hai. Lost mechanical energy per unit volume pressure drop ke equal hai, $Δ P = 50000 Pa$ , heat mein convert hoti hai. Ye step kyun? Ek level, equal-speed pipe mein pressure ke liye aur kuch nahi ban sakta — to ye dissipation hona chahiye.

Verify: Head-loss form $P_{1} = P_{2} + Δ P_{loss}$ ⇒ $Δ P_{loss} = 200 - 150 = 50 kPa$ ✔. Ye friction wall ke paas patli Boundary Layer ke andar hai — bilkul wahan jahan "inviscid" khatam hota hai.

Cell C — unsteady start-up

Worked example Ex4 · Achanak khula valve (Cell C)

Ek horizontal pipe jis ki length $L = 10 m$ aur cross-sectional area $A$ (koi bhi value — cancel ho jaata hai), still water se bhari hai, khola jaata hai. Pehle instant mein paani ka column har jagah uniformly $a = 4 m/s^{2}$ se accelerate karta hai. Is acceleration ke liye do ends ke beech kaafi extra pressure difference chahiye — ek number jo plain Bernoulli nahi dekh sakta?

Forecast: $v$ abhi bhi har jagah near zero hai, to kya $\frac{1}{2} ρ v^{2}$ term us pressure drop ko explain karta hai jo is acceleration ko drive karta hai? Guess karo haan/nahi.

Galat terms kill karo. Pehle instant mein $v \approx 0$ , to convective term $ρ v d v \approx 0$ . Ye step kyun? Ye prove karta hai ki driving force woh kinetic term nahi hai jo Bernoulli rakhta hai.
Unsteady term rakho. Newton poore paani ke column par: iska mass $ρ A L$ hai (density × area × length) aur net force jo ise push karta hai wo pressure difference hai jo end face par area $A$ par act karta hai, yaani $Δ P \cdot A$ : $Δ P \cdot A = ρ A L a \Rightarrow Δ P = ρ L a (A cancels)$ Ye step kyun? $\partial v / \partial t = a \neq = 0$ bilkul wahi steady assumption hai jo toot rahi hai; area $A$ dono sides par aata hai aur drop ho jaata hai, to answer pipe ki width se independent hai.
Numbers plug karo. $Δ P = 1000 \cdot 10 \cdot 4 = 40000 Pa = 40 kPa$ Ye step kyun? Ye "inertial / water-hammer" pressure plain Bernoulli ko dikhai nahi deta.

Verify: Units: $kg/m^{3} \cdot m \cdot m/s^{2} = kg/(m\cdots^{2}) = Pa$ ✔. Jab flow settle ho jaata hai ( $a \to 0$ ), $Δ P \to 0$ aur Bernoulli wapas aata hai — parent ke Example 2 se match karta hai.

Cell D — compressible gas

Worked example Ex5 · Tez air jahan density change karti hai (Cell D)

Air ( $ρ = 1.2 kg/m^{3}$ , speed of sound $c = 340 m/s$ ) ek duct mein $v_{1} = 50 m/s$ se $v_{2} = 200 m/s$ tak accelerate karti hai. Kya Bernoulli (incompressible form) har end par trustworthy hai? Mach number $M = v / c$ use karo (tools box mein restate kiya: flow kitna compressible act karta hai).

Forecast: Ek end safe hai, ek nahi. Kaun sa end break karta hai, aur kis speed par rule of thumb $M < 0.3$ cross ho jaata hai?

Dono Mach numbers compute karo. $M_{1} = \frac{50}{340} \approx 0.147, M_{2} = \frac{200}{340} \approx 0.588$ Ye step kyun? $M$ measure karta hai ki flow gas ko kitna compress kar sakta hai; $\int d P / ρ$ tabhi clean hai jab $ρ$ constant ho.
Actual density change estimate karo. Ek smooth (isentropic) gas flow ke liye fractional density change approximately $\frac{Δ ρ}{ρ} \approx \frac{1}{2} M^{2}$ hota hai. $At M_{1} : \frac{1}{2} (0.147)^{2} \approx 0.0108 (\approx 1%), At M_{2} : \frac{1}{2} (0.588)^{2} \approx 0.173 (\approx 17%)$ Ye step kyun? Ye rule of thumb ke peeche ka number hai: $M = 0.3$ par, $\frac{1}{2} M^{2} \approx 0.045$ , yaani ~4.5% density change — jo sabse bada error hai jo hum usually tolerate karte hain. $M_{2}$ par ek 17% density swing itni badi hai ki ignore nahi kar sakte.
Verdict. $M_{1}$ : ~1% density change ⇒ incompressible Bernoulli theek. $M_{2}$ : ~17% ⇒ cross-out kiya hua variable- $ρ$ term jaag uthta hai; compressible Bernoulli use karo (dekho Mach Number and Compressibility). Ye step kyun? Dono ends ko ek incompressible constant se mix karna outlet par ~17% error laayega.

Verify: $50/340 = 0.1471$ , $200/340 = 0.5882$ ✔. $\frac{1}{2} (0.3)^{2} = 0.045$ ✔ (~4.5% threshold). Crossover speed $v = 0.3 \times 340 = 102 m/s$ — outlet ka 200 m/s isse kaafi aage hai.

Cell E — do streamlines, do constants

Worked example Ex6 · Free vortex, inner vs outer streamline (Cell E)

Ek free vortex (draining sink) mein speed $v = k / r$ hai jahan $k = 0.6 m^{2} / s$ . Ek inner streamline $r_{1} = 0.1 m$ par aur ek outer $r_{2} = 0.3 m$ par compare karo, same height. Naively ek Bernoulli constant equate karte hue, kaun sa pressure difference predict hota hai — aur ye number automatically legitimate kyun nahi hai?

Forecast: Inner wala paani tez hai. Kya ek Bernoulli constant in do points ko compare karne deta hai? Step 1 se pehle valid/invalid guess karo.

Neeche ka figure do circular streamlines dikhata hai. Ise poore argument ke map ki tarah padho: red inner circle ( $r_{1} = 0.1 m$ ) aur blue outer circle ( $r_{2} = 0.3 m$ ) do alag streamlines hain jo kabhi nahi miltin — yahi visual reason hai ki Bernoulli ka constant unke beech share nahi hota. Yellow arrows local velocities hain: red arrow lamba hai (fast, $6 m/s$ ), blue arrow chhota (slow, $2 m/s$ ), to arrow lengths literally $v = k / r$ law draw karti hain. Jab tum kisi bhi circle ke around poora chakkar lagao, tum ek streamline ke along move kar rahe ho — wahan Bernoulli guarantee hai; radially red se blue tak jump karna woh illegal move hai jise hum test kar rahe hain.

Do speeds nikalo. $v_{1} = \frac{0.6}{0.1} = 6 m/s, v_{2} = \frac{0.6}{0.3} = 2 m/s$ Ye step kyun? Alag radii ⇒ alag speeds ⇒ alag kinetic terms — figure mein do arrow lengths se match karta hai.
Naive single-constant Bernoulli kya kahega. Dono par ek constant set karte hue: $P_{2} - P_{1} = \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) = \frac{1}{2} (1000) (36 - 4) = 16000 Pa$ Ye step kyun? Hum ise precisely compute karte hain taaki trap expose ho sake.
Streamlines cross karna tabhi allowed hai jab flow irrotational ho — aur ye flow hai. Ye alag streamlines hain (figure mein concentric circles kabhi nahi milte), to Bernoulli ka constant sirf unme se ek ke along guarantee hai. Cross karna tabhi legal hai jab flow irrotational ho (zero curl — koi local spin nahi ek tiny paddle wheel mein). Ek 2-D flow ke liye jisme sirf tangential speed $v_{θ} (r)$ hai, curl (vorticity) hai $ω = \frac{1}{r} \frac{d ( r v _{θ} )}{d r} .$ Free vortex ke liye $v_{θ} = k / r$ , hamare paas $r v_{θ} = k = const$ hai, to $\frac{d ( r v _{θ} )}{d r} = 0$ aur isliye $ω = 0$ har jagah except $r = 0$ (core, jahan $k / r$ blow up karta hai). To number yahan survive karta hai. Ye step kyun? Ye wo condition dikhata hai jo calculation ko bachati hai — aur warn karta hai ki ek forced (rigid-body) vortex $v_{θ} = Ω r$ deta hai $r v_{θ} = Ω r^{2}$ , $ω = 2Ω \neq = 0$ , to wahan naive number galat hoga.

Verify: $\frac{1}{2} (1000) (36 - 4) = 16000 Pa$ ✔; aur $\frac{d}{d r} (r \cdot \frac{k}{r}) = \frac{d k}{d r} = 0$ ✔ zero vorticity confirm karta hai. Rule: dono streamlines ke across constants tab tak equate mat karo jab tak irrotationality check nahi kar lete (parent Example 4).

Cell F — ek pump energy add karta hai

Worked example Ex7 · Line mein pump (Cell F)

Ek pump level pipe mein hai aur paani ko $P_{1} = 100 kPa$ se $P_{2} = 250 kPa$ tak raise karta hai bina $v$ change kiye (equal areas dono sides, to continuity deta hai $v_{1} = v_{2}$ ) ya $h$ change kiye. Plain Bernoulli kehta hai constant change nahi ho sakta. Kaun sa pump-head (metres mein) is jump ko account karta hai?

Forecast: Bernoulli forbid karta hai ki pressure $v, h$ fixed hone par badhe — phir bhi ye 150 kPa badh gaya. Pump ne kaun sa term switch on kiya?

Naive Bernoulli. $v_{1} = v_{2}$ aur $h_{1} = h_{2}$ se, dono kinetic aur gravity terms cancel ho jaate hain, to plain Bernoulli predict karta hai $P_{1} = P_{2}$ . Ye step kyun? Hum prediction dikhate hain taaki contradiction undeniable ho — friction case (Ex3) ki tarah spirit mein same, lekin yahan energy add ho rahi hai, kho nahi rahi.
Pump term $d F_{pump}$ restore karo. Master parcel equation par wapas jao aur rakho wo term jo Bernoulli ne cross out kiya. Pump ke across added energy per unit volume integrate karne se head-rise $H_{pump}$ milta hai, to energy balance ban jaata hai $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} + ρ g H_{pump} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2} .$ $v_{1} = v_{2}$ aur $h_{1} = h_{2}$ se ye collapse ho jaata hai $ρ g H_{pump} = P_{2} - P_{1} = 150000 Pa .$ Ye step kyun? Bernoulli ne jo $d F_{pump}$ term delete kiya tha exactly yahi injected energy hai; "constant" pump ke across jump karna supposed hai.
Head solve karo. $H_{pump} = \frac{P _{2} - P _{1}}{ρ g} = \frac{150000}{1000 \cdot 9.8} \approx 15.31 m .$ Ye step kyun? Head ki units metres hain — equivalent height of water jitna pump ki energy utha sakti hai.

Verify: $150000/ (1000 \cdot 9.8) = 15.306 m$ ✔. Units: $Pa / (kg/m^{3} \cdot m/s^{2}) = m$ ✔. Sanity: pressure pump ke across badhi — sahi, kyunki pump energy input karta hai (parent ki fourth mistake se match karta hai).

Cell G — degenerate / zero inputs

Worked example Ex8 · Static tank (Cell G, zero-velocity limit)

Ek bada open tank of water bilkul still hai ( $v = 0$ har jagah). Ek point free surface se $d = 3 m$ neeche hai (jahan $P_{atm} = 101 kPa$ ). Bernoulli kya deta hai, aur kya ye ek jaane-pahchaane law mein collapse ho jaata hai?

Forecast: Jab har speed zero hai, $\frac{1}{2} ρ v^{2}$ term khatam ho jaata hai. Bernoulli ki limiting form kaun sa famous formula hai?

Kinetic term zero karo. Dono points par $v = 0$ ⇒ $\frac{1}{2} ρ v^{2} = 0$ . Ye step kyun? Degenerate input $v = 0$ ek poora term hataa deta hai — limiting case jo test karta hai ki Bernoulli consistent rehta hai ya nahi.
Jo bachta hai wo hydrostatics hai. Surface point lo (height $h_{surface} = d$ , pressure $P_{atm}$ ) aur deep point (height $0$ , pressure $P_{deep}$ ): $P_{atm} + ρ g d = P_{deep} + ρ g (0)$ $\Rightarrow P_{deep} = P_{atm} + ρ g d .$ Ye step kyun? Bernoulli ko zaroor $P = P_{0} + ρ g d$ tak reduce karna chahiye jab kuch nahi hilta — poori theory par ek consistency check.
Numbers plug karo. $P_{deep} = 101000 + 1000 \cdot 9.8 \cdot 3 = 130400 Pa \approx 130.4 kPa .$ Ye step kyun? Confirm karta hai ki pressure-with-depth law sirf Bernoulli ka still-water shadow hai.

Verify: $101000 + 1000 \cdot 9.8 \cdot 3 = 130400 Pa$ ✔. $v$ term hatana "degenerate cell" hai — Bernoulli gracefully hydrostatics ban jaata hai, to koi assumption violate nahi hoti; kinetic term simply zero hai. ✔

Cell H — exam trap

Worked example Ex9 · "Valid lagta hua" wing question (Cell H, hidden failure)

Air ( $ρ = 1.2 kg/m^{3}$ ) $30 m/s$ par ek wing ke upar flow karti hai; upar ye $45 m/s$ tak pahunch jaati hai. (a) Bernoulli se pressure drop predict karo. (b) Exam phir poochha surface ki skin ke bilkul andar, thin surface layer mein pressure kya hai — kya tumhara answer abhi bhi valid hai?

Forecast: Part (a) clean Cell A hai. Part (b) twist hai. Kaun si assumption metal surface par silently fail hoti hai?

Compressibility ke liye Mach check karo. $M = 45/340 \approx 0.132 < 0.3$ ⇒ incompressible OK. Ye step kyun? Cell D rule out karta hai taaki hum Mach Number and Compressibility se compressible correction waste na karein.
Part (a): Bernoulli pressure drop. $Δ P = \frac{1}{2} ρ (v_{top}^{2} - v_{\infty}^{2}) = \frac{1}{2} (1.2) (4 5^{2} - 3 0^{2})$ $= \frac{1}{2} (1.2) (2025 - 900) = 675 Pa .$ Ye step kyun? Tez top surface ⇒ kam pressure ⇒ lift; surface layer ke bahar valid.
Part (b): hidden failure pakdo. Bilkul skin par, velocity Boundary Layer ke through zero tak girta hai (tools box mein restate ki gayi ultra-thin viscous skin), aur viscous shear dominate karta hai — inviscid assumption khatam ho jaati hai. Bernoulli ka $675 Pa$ sirf layer ke bahar apply hota hai, metal par nahi. Ye step kyun? Exam trap ek valid answer ko ek invalid region mein apply karna hai — ek alag streamline family jahan ek cross-out term (friction) jaag gayi hai.

Verify: $\frac{1}{2} (1.2) (2025 - 900) = 675 Pa$ ✔. $45/340 = 0.1324 < 0.3$ ✔. Lesson: Bernoulli trust karne se pehle hamesha poochho ki flow mein tum kahan ho — dekho Bernoulli's Equation.

Recall Self-test — failing assumption batao

Rough 50 m pipe, equal $v$ aur $h$ , pressure drop ::: Inviscid (friction → heat). Cell B. Valve abhi-abhi khula, paani har jagah accelerate ho raha hai ::: Steady (unsteady term $\partial v / \partial t$ ). Cell C. Air duct jahan $M = 0.6$ ::: Incompressible ( $ρ$ varies ~17%). Cell D. Vortex ki inner aur outer streamlines compare karna ::: Along a streamline (cross karne ke liye irrotational hona zaroori). Cell E. Pump pressure raise karta hai $v, h$ fixed hone par ::: Inme se koi nahi — energy added hai, constant jump karta hai. Cell F. Paani 12 m upar jaata hai, chaaon assumptions hold karti hain ::: Koi fail nahi — gravity term rakho. Cell A′. Still tank, sabhi speeds zero ::: Koi fail nahi — Bernoulli hydrostatics mein reduce ho jaata hai. Cell G.