Stream function, velocity potential: Question bank

2.2.11 · D5 · Physics › Fluid Mechanics › Stream function, velocity potential

Shuru karne se pehle, teen symbols ka ek quick refresher taaki neeche ki har line zero se bhi readable ho:

True ya false — justify karo

ψ

ka exist karna incompressibility par depend karta hai, spin par nahi

True —

ψ

ko sirf

\nabla \cdot V = 0

chahiye (2D). Ek flow whirlpool ki tarah spin kar sakta hai aur phir bhi ek perfectly good stream function rakh sakta hai.

ϕ

ka exist karna irrotationality par depend karta hai, incompressibility par nahi

True —

ϕ

ko sirf

\nabla \times V = 0

chahiye. Ek compressible-but-irrotational flow mein phir bhi

ϕ

hota hai; bas uska

ϕ

Laplace obey nahi karega.

Har woh flow jiske paas

ψ

hai, uske paas

ϕ

bhi hota hai

False —

ψ

ko incompressible chahiye,

ϕ

ko irrotational. Ek rotating incompressible flow (solid-body rotation) mein

ψ

hoga lekin single-valued

ϕ

nahi.

Agar dono

ϕ

aur

ψ

exist karte hain, to dono ko Laplace's equation satisfy karni chahiye

True — dono ka exist karna matlab hai incompressible AUR irrotational, aur yahi do conditions hain jo [[Laplace Equation|

\nabla^{2} ϕ = 0

aur

\nabla^{2} ψ = 0

]] force karti hain.

ψ

Laplace's equation satisfy karta hai kyunki flow incompressible hai

False —

ψ

Laplace obey karta hai kyunki flow irrotational hai; incompressibility woh cheez hai jo

ψ

ko exist karne deti hai. Roles

ϕ

se swap hain.

ϕ

Laplace's equation satisfy karta hai kyunki flow incompressible hai

True —

V = \nabla ϕ

\nabla \cdot V = 0

mein plug karo aur tumhe

\nabla^{2} ϕ = 0

milta hai. Yahan incompressibility Laplace ka source hai.

Streamlines aur equipotential lines hamesha right angles par cross karti hain

Jaisa kaha gaya waise False — sirf jab DONO

ψ

aur

ϕ

exist karte hain (ideal flow) tab

\nabla ψ \cdot \nabla ϕ = 0

hota hai. Purely incompressible-but-rotational flow mein koi

ϕ

hi nahi hota jo perpendicular ho.

ψ_{2} - ψ_{1}

ka difference un do streamlines ke beech volume flow rate ke barabar hota hai

True — flux

= \int (u d y - v d x) = \int d ψ = ψ_{2} - ψ_{1}

per unit depth. Yahi

ψ

ka defining physical meaning hai.

ψ

mein constant add karne se flow change hota hai

False — velocities

ψ

ke derivatives hain, aur constant ka derivative zero hota hai. Sirf

ψ

ke differences (fluxes) physical hain.

ϕ

mein constant add karne se velocity change hoti hai

False —

V = \nabla ϕ

, aur

\nabla (ϕ + c) = \nabla ϕ

. Potential ka zero-level hamesha arbitrary hota hai, bilkul Electrostatic Potential ki tarah.

Error dhundo

"Continuity se

u_{x} + v_{y} = 0

milta hai, to maine

u = ψ_{x}, v = ψ_{y}

set kiya."

Galat sign/index pattern: isse

ψ_{xx} + ψ_{y y}

milta hai, zero nahi. Sahi choice

u = ψ_{y}, v = - ψ_{x}

continuity ko automatically

ψ_{y x} - ψ_{x y} = 0

padh deti hai.

v = - ψ_{x}

mein minus sign sirf ek convention hai jise tum freely flip kar sakte ho."

Yahi sign hai jo mixed partials ko cancel karta hai taaki continuity identically hold ho. Isko flip karo aur

ψ

ab incompressibility guarantee nahi karta — yeh valid stream function rehna band kar deta hai.

ϕ =

const streamlines hain kyunki fluid scalar ki level curves follow karta hai."

ϕ =

const equipotentials hain, flow ke perpendicular; fluid inhe cross karta hai.

ψ =

const woh streamlines hain jinpe fluid actually travel karta hai.

\nabla \times (\nabla ϕ) = 0

prove karta hai ki har flow irrotational hai."

Yeh sirf prove karta hai ki

\nabla ϕ

ke roop mein likhe gaye flows irrotational hain. Physics converse hai: tum

V = \nabla ϕ

tabhi likh sakte ho jab check kar lo ki flow irrotational hai.

"Source ke liye

ϕ = \frac{m}{2 π} ln r

, aur kyunki

ln

r = 0

par blow up karta hai, model galat hai."

r = 0

par singularity source point itself hai — fluid ka ek idealised injection. Model wahan ke alawa har jagah valid hai, jo deliberate hai. Aise point singularities ka use dekho Bernoulli Equation mein.

"Cauchy–Riemann relations sirf complex analysis mein matter karti hain, fluids mein nahi."

Fluid pair

ϕ_{x} = ψ_{y}, ϕ_{y} = - ψ_{x}

wahi hain Cauchy-Riemann Equations; isliye

ϕ + i ψ

ek analytic function hai aur complex methods ideal flows solve karte hain.

"Ek source mein

ψ = \frac{m}{2 π} θ

hai, jo single-valued hai."

Yeh multi-valued hai: ek baar ghoomne par (

θ \to θ + 2 π

)

ψ

m

se badh jaata hai. Yeh jump exactly woh flux

m

hai jo source se nikal raha hai, isliye multivaluedness physical hai, bug nahi.

Why questions

Incompressible flow ke liye do velocity components ko ek scalar se kyun replace kiya ja sakta hai?

Incompressibility ek equation hai jo

u

aur

v

ko aapas mein baandh deti hai, ek degree of freedom hata deti hai — isliye ek akela function (

ψ

) dono ko encode kar sakta hai aur construction se woh constraint satisfy karta hai.

Velocity

\nabla ϕ

ke saath kyun point karti hai lekin

ψ

-lines ke tangent ke

\nabla

's ke across?

\nabla ϕ = (u, v)

velocity hai, isliye woh jahan

ϕ

sabse tezi se badhta hai wahan point karta hai. Streamline direction bhi

(u, v)

hai, isliye

\nabla ψ = (- v, u)

uske perpendicular hai.

Ideal flow mein equipotentials aur streamlines perpendicular kyun hote hain?

\nabla ϕ = (u, v)

aur

\nabla ψ = (- v, u)

; unka dot product

- uv + uv = 0

hai, isliye unki level curves right angles par cross karti hain, ek flow net banati hain.

Ek irrotational flow mein vortex ke around single-valued

ϕ

kyun nahi hota?

Vortex ke around circulation nonzero hai, isliye

V \cdot d r

integrate karne se ek fixed amount milta hai ek chakkar mein —

ϕ

har loop mein jump karta hai, isliye woh single-valued nahi ho sakta chahe flow center ke alawa irrotational ho.

ψ

streamline ke saath constant kyun rehta hai?

Flow ke saath move karte hue,

d ψ = ψ_{x} d x + ψ_{y} d y = - v d x + u d y

, aur streamline par

u d y = v d x

, isliye

d ψ = 0

. Koi fluid streamline cross nahi karta, isliye "lane number" nahi badlta.

Streamlines ke beech flux chosen path par kyun depend nahi karta?

Kyunki

\int d ψ

sirf endpoints ke

ψ

values dekhta hai,

Q = ψ_{2} - ψ_{1}

connecting curve chahe koi bhi ho — bilkul wahi reason jaise potential field mein work path-independent hota hai.

Edge cases

Solid-body rotation

V = (- Ω y, Ω x)

: kya

ψ

exist karta hai? Kya

ϕ

\nabla \cdot V = 0

isliye

ψ = - \frac{Ω}{2} (x^{2} + y^{2})

exist karta hai; lekin

\nabla \times V = 2Ω \neq = 0

isliye yeh rotational hai aur

ϕ

exist nahi karta.

Free vortex

u_{θ} = \frac{Γ}{2 π r}

: kya

ϕ

exist karta hai? Kya yeh single-valued hai?

Yeh

r = 0

ke alawa har jagah irrotational hai, isliye locally

ϕ = \frac{Γ}{2 π} θ

exist karta hai — lekin yeh har loop mein

Γ

se badhta hai, isliye globally multi-valued hai. Classic irrotational-yet-no-single-

ϕ

case.

Uniform flow at rest (

V = 0

ϕ

aur

ψ

kya hain?

Dono constants hain (koi bhi constants), kyunki unke saare derivatives vanish ho jaate hain. Degenerate lekin consistent — koi velocity nahi matlab koi meaningful level curves nahi.

Compressible steady flow: kya phir bhi stream function define kar sakte hain?

Ordinary

ψ

nahi; tumhe density-weighted version chahiye

ρ u = ψ_{y}, ρ v = - ψ_{x}

taaki

(ρ u)_{x} + (ρ v)_{y} = 0

ho (mass, volume nahi, conserved hota hai).

Stagnation point par jahan

u = v = 0

, flow net ka kya hota hai?

Wahan

\nabla ϕ = 0

aur

\nabla ψ = 0

hote hain, isliye orthogonality argument degenerate ho jaata hai — streamlines us ek point par cross kar sakti hain (jaise cylinder ka front), woh ek jagah jahan right angles zaroori nahi.

ψ

ek source ke exact centre par kya matlab rakhta hai?

Wahan undefined hai (

θ

ki koi value nahi

r = 0

par); woh point ek singularity hai jahan fluid create hota hai, isliye koi finite lane number apply nahi hoti.

Agar flow irrotational hai lekin compressible hai, kya

ϕ

Laplace's equation satisfy karta hai?

Nahi —

ϕ

exist karta hai (irrotational) lekin

\nabla \cdot V \neq = 0

, isliye

\nabla^{2} ϕ = \nabla \cdot V \neq = 0

. Laplace ko incompressibility bhi chahiye.

Recall Ek-line self-test

Upar sab cover karo aur jawab do: kaun si condition $ψ$ deti hai, kaun si $ϕ$ , aur dono ke liye Laplace kaun deta hai? $ψ$ ← incompressible; $ϕ$ ← irrotational; Laplace for $ϕ$ ← incompressible; Laplace for $ψ$ ← irrotational. Laplace conditions cross-over hain — yahi poora trap hai.