Question bank — Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline

False. Constant sirf ek streamline ke saath equal guaranteed hai; alag streamlines alag constants carry kar sakti hain jab tak flow irrotational bhi na ho, jo ise global constant bana deta hai.

False. Steady sirf $\partial v/\partial t$ term ko khatam karta hai; tumhe phir bhi inviscid, incompressible aur same-streamline chahiye. Ek steady viscous flow rough pipe mein phir bhi pressure lose karta hai.

False. $P$–$v$ trade-off Bernoulli ka consequence hai, isliye ye sirf tab hold karta hai jab charo assumptions lagoo hon. Ek Boundary Layer ke andar ya pulsing pipe mein, pressure bilkul alag behave kar sakta hai.

True. Ye Newton ka 2nd law hai streamline ke saath integrate kiya gaya, jo work–energy theorem ka disguised roop hai — har assumption ek dropped energy term ki kimat hai.

True — ideal flow ke liye. $v$ aur $h$ ko continuity aur geometry se fix karke, Bernoulli $P_1=P_2$ force karta hai. Koi bhi measured drop viscosity signal karta hai, yaani inviscid assumption fail ho rahi hai.

False. Ek pump energy inject karta hai, isliye "constant" uske across upar jump karta hai. Tumhe pump-head term $h_{\text{pump}}$ ke saath extended energy equation use karni hogi.

False. Mach $\approx 0.3$ se neeche (yaani $M=v/c<0.3$) density changes negligible hain, isliye air incompressibly behave karti hai aur Bernoulli legitimate hai — ye exactly low-speed-wing case hai. Dekho Mach Number and Compressibility.

Practice mein False. Iska matlab hai viscous forces is flow region ke liye negligible hain — walls se door sach, lekin thin boundary layer ke andar jhooth jahan shear dominate karta hai.

False. $\rho gh$ term sirf isliye exist karta hai kyunki gravity ek height potential se aati hai; ek non-conservative body force ko clean potential ke roop mein nahi likha ja sakta, isliye koi single "constant" ise absorb nahi kar sakta.

Ek free vortex (idealized drain) mein fluid centre ke around speed $v\propto 1/r$ se circle karta hai — andar ki taraf tez — isliye do points alag circular streamlines par hain; same-streamline license apply nahi hota, aur jab tak flow irrotational na ho constants differ karte hain. (Neeche ki figure mein free-vortex sketch dekho.)

Start-up ke dauran flow unsteady hoti hai, $\partial v/\partial t\ne 0$ har jagah, isliye dropped inertial term nonzero hai — steady assumption violated hai aur Bernoulli mein water-hammer term missing hai.

Kyunki high Mach par ($M=v/c$ chota nahi) $\rho$ $P$ ke saath vary karta hai, isliye ise integral se bahar nahi nikal sakte; incompressible assumption fail hoti hai aur internal-energy correction chahiye.

Gauge theek hai; assumption toot gayi hai. Ek wall tapping static pressure padhta hai, aur real pipes mein wall friction hoti hai, isliye inviscid premise fail hoti hai aur mechanical energy heat mein leak hoti hai, downstream static pressure drop karta hai.

Flow ki taraf face karta Pitot tube total (stagnation) pressure $P+\frac{1}{2}\rho v^2$ padhta hai, jabki sidewall tapping static pressure $P$ padhta hai; difference dynamic pressure hai, isliye wo jahan bhi $v\ne 0$ hai wahan agree nahi karne chahiye.

Surface par bilkul tum boundary layer ke andar hote ho, jahan viscosity dominate karti hai — wahan inviscid fail hoti hai. Lift-explaining Bernoulli un streamlines par apply hota hai jo us thin layer ke bahar hain.

Steady sirf $\partial v/\partial t$ ko khatam karta hai. Convective term survive karta hai — narrowing pipe mein paani phir bhi perfectly steady flow mein accelerate karta hai kyunki wo $s$ ke saath ek faster region mein move karta hai.

Viscosity mechanical energy ko heat mein convert karke irreversible work karti hai, isliye $P+\frac{1}{2}\rho v^2+\rho gh$ downstream steady decay karta hai fixed rehne ki jagah. Dekho Viscosity and Poiseuille Flow.

Ek parcel ya toh isliye accelerate hota hai kyunki field time mein change hoti hai ($\partial v/\partial t$) ya isliye kyunki wo $s$ ke saath ek faster region mein move karta hai ($v\partial v/\partial s$). Steady pehle wale ko delete karta hai, local term ko; dekho Material Derivative.

Newton ka law $s$ ke saath force component ke liye likha gaya tha, isliye sirf streamline ke saath motion appear hota hai; perpendicular balance ek alag equation hai, isliye constant sirf ek streamline ke saath guaranteed hai.

Irrotational Flow ke liye cross-streamline pressure balance har streamline ko same constant share karne par force karta hai, isliye equation field mein kinhi bhi do points ko relate karti hai.

Ek isentropic gas ke liye fractional density change $\frac{\Delta \rho }{\rho }\approx \frac{M^2}{2}$ (leading term) ki tarah badhti hai, isliye $M=0.3$ par ye sirf $0.3^2/2\approx 4.5\%$ hai — ignore karne ke liye kaafi choti, lekin $0.3$ se upar tezi se badhti hai. Related pressure-change estimate mein $\gamma$ (adiabatic index) ka factor hota hai kyunki gas mein pressure aur density $P\propto {\rho }^{\gamma }$ se linked hain. Dekho Mach Number and Compressibility.

Constant ek passive parcel ke liye ek closed energy balance hai; pump ek external energy source hai (uska pump-head $h_{\text{pump}}$) jo us closure ko tod deta hai, isliye uska head explicitly add karna padta hai.

Sirf ek conservative force ko potential ke roop mein likha ja sakta hai (yahan $gh$), jo iske kaam ko single clean term $\rho gh$ mein integrate hone deta hai; warna kaam path par depend karta aur koi fixed "constant" exist nahi kar sakta.

Haan — ye reduce hota hai $P+\rho gh=\text{const}$ tak, jo hydrostatic pressure law hai, same equation ki zero-velocity limit.

$P+\frac{1}{2}\rho v^2=\text{const}$ tak, isliye pressure aur speed directly trade off karte hain — classic Venturi/wing result, sirf tab valid jab charo assumptions hold karein.

Haan. Incompressibility sirf $\rho$ constant chahti hai, chota nahi; derivation $dP/\rho$ ko cleanly integrate karti hai $\rho$ ki value se regardless.

Incompressible step: $\int dP/\rho$ ab $P/\rho$ nahi hota, isliye tumhe standard wale ki jagah compressible form of Bernoulli chahiye.

Sirf ek achha approximation — density pehle hi kuch percent shift ho chuki hai ($\approx M^2/2\approx 4.5\%$), isliye incompressible result mein ek chota lekin real error hai jo $0.3$ se upar tezi se badhta hai.

Bilkul "exact" nahi — negligible boundary layer ke baad bhi, curved core streamlines cross-stream pressure gradient carry karti hain (isliye alag streamlines alag constants hold karti hain), aur entrance/exit aur area-change losses phir bhi hoti hain; Bernoulli sirf un effects se door ek single straight core streamline ke saath exact hai. Dekho Boundary Layer.