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2.2.15Fluid Mechanics

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

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The four assumptions, derived from where they come from

We derive Bernoulli from Newton's 2nd law on a small fluid parcel sliding along a streamline. Each assumption is the price of a term we throw away.

Setting up the derivation (first principles)

Take a tiny cylindrical fluid element of length dsds, cross-section AA, lying along a streamline ss. Forces along ss:

  • Pressure pushes on the back face PAP A, on the front face −(P+dP)A-(P+dP)A. Net pressure force =−A dP= -A\,dP.
  • Gravity component along ss: weight =ρA ds g= \rho A\, ds\, g, component along streamline =−ρA ds g dhds= -\rho A\, ds\, g\,\dfrac{dh}{ds} where hh is height.

Newton's 2nd law: mass×acceleration=net force\text{mass}\times\text{acceleration} = \text{net force}. Mass =ρA ds=\rho A\,ds.

The acceleration of the parcel is the material derivative: a=DvDt=∂v∂t⏟unsteady+v∂v∂sa = \frac{Dv}{Dt} = \underbrace{\frac{\partial v}{\partial t}}_{\text{unsteady}} + v\frac{\partial v}{\partial s}

Putting it together (divide by AA): ρ ds(∂v∂t+v∂v∂s)=−dP−ρg dh\rho\,ds\left(\frac{\partial v}{\partial t} + v\frac{\partial v}{\partial s}\right) = -dP - \rho g\,dh

Now apply the assumptions one by one

(1) Steady ⇒∂v∂t=0\Rightarrow \dfrac{\partial v}{\partial t}=0. We drop the unsteady term: ρv dv=−dP−ρg dh\rho v\,dv = -dP - \rho g\,dh

(2) Inviscid — we never wrote a friction force. By assuming no viscosity we silently dropped a viscous shear term that would appear as −τ dAside-\tau\,dA_{\text{side}}. With viscosity, energy is converted to heat and the "constant" decays downstream.

(3) Incompressible ⇒ρ=const\Rightarrow \rho=\text{const}, so we can integrate dP/ρdP/\rho cleanly. Integrate along the streamline: ∫ρv dv+∫dP+∫ρg dh=0\int \rho v\,dv + \int dP + \int \rho g\,dh = 0 12ρv2+P+ρgh=const\tfrac12\rho v^2 + P + \rho g h = \text{const}

(4) Along a streamline — every step integrated along ss. The constant can be different on different streamlines unless the flow is also irrotational.

Figure — Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline

Worked examples

Common mistakes (steel-manned)

Flashcards

What are the four assumptions of Bernoulli's equation?
Steady, inviscid, incompressible, and applied along a streamline.
Which acceleration term does the "steady" assumption kill?
The local/unsteady term ∂v/∂t\partial v/\partial t in the material derivative.
Why must Bernoulli be applied along a streamline?
Because the constant is only guaranteed along one streamline; different streamlines may have different constants unless the flow is irrotational.
What physical quantity does viscosity remove from the Bernoulli balance, and into what?
Mechanical (pressure+kinetic+potential) energy, converted irreversibly into heat — so the "inviscid" assumption is needed for the sum to stay constant.
What integral becomes messy if the fluid is compressible?
∫dP/ρ\int dP/\rho — with variable ρ\rho it no longer equals P/ρP/\rho, so the incompressible assumption is needed to integrate cleanly.
Rule of thumb for treating gas flow as incompressible?
Mach number <0.3< 0.3.
In a long rough horizontal constant-area pipe, which assumption fails and what's the symptom?
Inviscid fails; pressure drops along the pipe even though vv and hh are unchanged.
What is the material (total) acceleration of a fluid parcel along a streamline?
DvDt=∂v∂t+v∂v∂s\dfrac{Dv}{Dt}=\dfrac{\partial v}{\partial t}+v\dfrac{\partial v}{\partial s}.
Recall Feynman: explain to a 12-year-old

Imagine sliding down a smooth waterslide. A tube of water keeps the same "energy budget" made of three coins: push (pressure), speed, and height. It can swap coins — trade height for speed — but the total stays the same. BUT this only works if: the slide isn't changing while you ride (steady), it's perfectly slippery so no energy is rubbed off as heat (inviscid), the water doesn't squish (incompressible), and you stay on your own slide lane (along a streamline). Break any rule and your three coins don't add up to the same total anymore.

Connections

  • Bernoulli's Equation — the result these assumptions license.
  • Continuity Equation — pairs with Bernoulli; gives the vv when area changes.
  • Viscosity and Poiseuille Flow — what happens when the inviscid assumption is dropped.
  • Boundary Layer — region where inviscid fails right next to walls.
  • Material Derivative — source of the steady vs unsteady distinction.
  • Mach Number and Compressibility — sets the limit of the incompressible assumption.
  • Irrotational Flow — the extra condition that makes Bernoulli's constant global across streamlines.

Concept Map

derived from

has

has

pressure + gravity

drops

drops viscous shear

rho constant

integration path

integrate

is

fails when

Newton 2nd law on fluid parcel

Material derivative acceleration

Unsteady term dv/dt

Convective term v dv/ds

Forces along streamline

Steady assumption

Inviscid assumption

No energy leak to heat

Incompressible assumption

Integrate dP/rho cleanly

Along streamline

Bernoulli equation P + half rho v squared + rho g h = const

Energy conservation for moving parcel

Walls, shocks, pulsing pipes, fans

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Bernoulli equation P+12ρv2+ρgh=P + \tfrac12\rho v^2 + \rho g h = constant basically energy conservation hai ek chhote se fluid parcel ke liye jo streamline ke along chal raha hai. Lekin yeh "constant" tabhi sach hai jab chaar shartein poori hon — isiliye assumptions itne important hain. Hum Newton ka second law ek tiny fluid element par lagate hain, aur har assumption ek term ko hatane ki "keemat" hoti hai.

Char shartein: (1) Steady — flow time ke saath change na ho, warna ∂v/∂t\partial v/\partial t term zinda rehta hai (jaise nal abhi-abhi khola toh paani accelerate ho raha hota hai). (2) Inviscid — koi viscosity/friction nahi, warna energy heat mein nikal jaati hai aur lambe rough pipe mein pressure girta jaata hai. (3) Incompressible — density constant, tabhi hi ∫dP/ρ\int dP/\rho clean integrate hota hai; gas ke liye yeh tab theek hai jab Mach <0.3<0.3. (4) Along a streamline — constant sirf ek hi streamline par guaranteed hai, alag-alag lanes par alag constant ho sakta hai (jab tak flow irrotational na ho).

Yeh matter kyun karta hai? Kyunki Bernoulli ka famous result "fast water = low pressure" tabhi valid hai jab yeh chaaron sahi ho. Boundary layer mein (wall ke paas) inviscid fail ho jaata hai, shock wave mein incompressible fail, pulsing flow mein steady fail. Exam aur real life dono mein, pehle assumptions check karo, phir equation lagao — warna galat answer pakka.

Yaad rakhne ka shortcut: SIIS — Steady, Inviscid, Incompressible, Streamline. "A Steady Ice Is Slippery."

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Connections

Bernoulli's equation — derivation from F = ma along streamlinePhysics · 2.2.14Continuity equation — derivation (conservation of mass), ρAv = constPhysics · 2.2.12Boundary layer — Prandtl's concept, growth along flat platePhysics · 2.2.20Mach number M = V - a — subsonic ( - 1), transonic (~1), supersonic ( - 1), hypersonic ( - 5)Physics · 3.1.4