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

WHY do we even need this?

WHAT we assume (so the derivation is honest)

HOW we derive it — $F=ma$ on a fluid blob

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

Step 1 — Pick a tiny cylindrical parcel

Take a small fluid element on the streamline: cross-sectional area $A$ , length $ds$ along the streamline.

Volume: $dV = A\,ds$
Mass: $dm = \rho\,A\,ds$

Why this step? Newton's law is about a definite mass. A small cylinder aligned with the flow lets us add up only the forces along the direction of motion.

Step 2 — Forces along the streamline

Let $s$ measure distance along the streamline (increasing in the flow direction).

(a) Pressure force. Pressure $P$ pushes on the back face (area $A$ ) forward; pressure $P+dP$ pushes on the front face backward.

dF_{\text{pressure}} = PA - (P+dP)A = -A\,dP

Why this step? Only a difference in pressure across the parcel produces a net force; a uniform pressure squeezes equally on both ends and cancels.

(b) Gravity component. Gravity is $-dm\,g\,\hat{y}$ . The component along the streamline depends on how steeply the streamline rises. If the parcel rises by $dy$ over length $ds$ , the slope is $\sin\theta = dy/ds$ :

dF_{\text{gravity}} = -dm\,g\sin\theta = -\rho A\,ds\, g\,\frac{dy}{ds} = -\rho A\, g\, dy

Why this step? Only the part of gravity pointing along the streamline can speed up or slow the parcel; the sideways part is balanced by pressure across the streamline.

Step 3 — Acceleration along the streamline (the clever bit)

In steady flow, the speed $v$ at a fixed point is constant in time, but a moving parcel still accelerates because it travels to a place with a different $v$ . Using the chain rule:

a = \frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = v\frac{dv}{ds}

Why this step? This is convective acceleration: the parcel "carries itself" into a faster region. This is why a steady flow can still accelerate — a subtle point people forget.

Step 4 — Apply $F=ma$

dF_{\text{pressure}} + dF_{\text{gravity}} = dm\, a

-A\,dP - \rho A g\,dy = (\rho A\,ds)\left(v\frac{dv}{ds}\right)

Notice $A$ cancels and $ds$ cancels on the right ( $ds\cdot \frac{1}{ds}$ ):

-dP - \rho g\,dy = \rho v\,dv

Rearrange so everything is a differential:

dP + \rho g\,dy + \rho v\,dv = 0

Why this step? Each term is now an exact differential — we can integrate term by term along the streamline.

Step 5 — Integrate along the streamline

With $\rho$ = const, integrate from point 1 to point 2:

\int dP + \rho g \int dy + \rho \int v\,dv = 0

(P_2-P_1) + \rho g(y_2-y_1) + \tfrac{1}{2}\rho(v_2^2 - v_1^2) = 0

Group by point:

Worked Examples

Worked example 2 — Venturi / horizontal pipe constriction

Horizontal pipe ( $y_1=y_2$ ) narrows from area $A_1$ to $A_2$ . Show pressure drops where it's narrow.

Continuity: $A_1 v_1 = A_2 v_2 \Rightarrow v_2 = v_1 A_1/A_2 > v_1$ . Why? Same volume per second must pass through smaller area, so it speeds up.

Bernoulli (height term cancels):

P_1 + \tfrac12\rho v_1^2 = P_2 + \tfrac12\rho v_2^2

P_1 - P_2 = \tfrac12\rho(v_2^2 - v_1^2) > 0

So $P_2 < P_1$ : fast region = low pressure. Why this step? $v_2>v_1$ makes the bracket positive, forcing $P_1>P_2$ .

Common Mistakes (Steel-manned)

Recall Feynman: explain it to a 12-year-old

Imagine a crowd walking down a hallway. When the hallway gets narrow, people have to walk faster to keep moving (no one disappears). To walk faster they need a push from behind — so the people behind are squished together (high "push"/pressure) and the fast people in the narrow part feel less squished (low pressure). Bernoulli's equation is just bookkeeping: squish (pressure) + speed-energy + height-energy always add up to the same total as you follow one person down the hall. If one goes up, another must go down.

Forecast-then-Verify

Active Recall

What law is Bernoulli's equation a disguised form of?

Newton's second law

F=ma

, integrated along a streamline (equivalently, work–energy theorem per unit volume).

List the four assumptions behind Bernoulli's equation.

Steady, incompressible (constant

\rho

), non-viscous (inviscid), and along a single streamline.

Why does a fluid parcel accelerate in steady flow?

Convective acceleration

a=v\,dv/ds

: it moves into a region of different velocity, even though the field is time-independent.

Write the differential form before integrating.

dP + \rho g\,dy + \rho v\,dv = 0

What are the units of every term in Bernoulli's equation?

Pressure, i.e.

\text{Pa}=\text{J/m}^3

(energy per unit volume).

Derive Torricelli's law from Bernoulli.

Open tank,

P_1=P_2

v_1\approx0

, drop

h

⇒

\rho g h = \frac12\rho v_2^2

⇒

v_2=\sqrt{2gh}

In a constriction, where is pressure lowest?

Where the area is smallest, because speed is highest there and

P+\frac12\rho v^2=

const.

What net force did we balance against

ma

in the derivation?

Net pressure force

-A\,dP

plus the streamline-component of gravity

-\rho A g\,dy

Why is

A

and

ds

allowed to cancel in the derivation?

A

multiplies every force/mass term;

ds

from the mass cancels the

1/ds

a=v\,dv/ds

, leaving an equation in pure differentials.

A Pitot tube reads

P_0-P=200

Pa in air (

\rho=1.2

). What's the airspeed?

v=\sqrt{2(200)/1.2}\approx 18.3\ \text{m/s}

Can Bernoulli compare points on two different streamlines?

Generally no — the constant differs per streamline, unless the flow is irrotational.

Connections

Equation of Continuity — supplies the $A_1v_1=A_2v_2$ relation used in every example.
Newton's Second Law — the parent law we integrated.
Work-Energy Theorem — the alternative derivation route (energy reading).
Hydrostatic Pressure — the $\rho g y$ term reduces to $P=P_0+\rho g h$ when $v=0$ .
Viscosity and Poiseuille Flow — what to add when "inviscid" fails.
Torricelli's Law · Venturi Meter · Pitot Tube · Aerodynamic Lift — applications.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho yaar, Bernoulli's equation koi alag se yaad karne wali cheez nahi hai — ye sirf Newton ka $F=ma$ hai, lekin ek fluid ke chhote se tukde (parcel) par streamline ke along likha hua. Fluid ka ek chhota cylinder lo jiska area $A$ aur lambai $ds$ hai. Uspar do tarah ke forces lagte hain: ek pressure difference (peeche $P$ aage $P+dP$ , to net force $-A\,dP$ ), aur doosra gravity ka streamline-wala component ( $-\rho A g\,dy$ ). Inko $dm \cdot a$ ke barabar rakh do.

Ab ek important twist: steady flow mein bhi parcel accelerate karta hai! Kyunki jab wo aage badhta hai to wo ek alag speed wali jagah pahunch jata hai. Isliye $a = v\,\frac{dv}{ds}$ (isko convective acceleration kehte hain). Sab milake equation banta hai $dP + \rho g\,dy + \rho v\,dv = 0$ . Isko streamline ke along integrate karo, $\rho$ constant maan ke, aur mil jata hai: $P + \tfrac12 \rho v^2 + \rho g y = \text{constant}$ .

Iska matlab kya hai? Teen tarah ki "energy per unit volume" — pressure energy, speed (kinetic) energy, aur height (potential) energy — ka total constant rehta hai. Jahan fluid tej chalta hai (narrow pipe), wahan pressure kam ho jata hai. Yahi reason hai venturi meter, aeroplane ke wing ka lift, aur shower curtain andar khinchne ka. Yaad rakhna: ye sirf ek streamline par sach hai, har jagah nahi; aur agar pump ya viscosity ho to extra terms add karne padenge.

Exam tip: agar dono points atmosphere mein hain to $P$ cancel ho jata hai (Torricelli, $v=\sqrt{2gh}$ ). Agar pipe horizontal hai to $\rho g y$ cancel ho jata hai (venturi). Bas streamline trace karo, har term ko samajh ke daalo — formula automatically simple ho jayega.

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

WHY do we even need this?

WHAT we assume (so the derivation is honest)

HOW we derive it — $F=ma$ on a fluid blob

Step 1 — Pick a tiny cylindrical parcel

Step 2 — Forces along the streamline

Step 3 — Acceleration along the streamline (the clever bit)

Step 4 — Apply $F=ma$

Step 5 — Integrate along the streamline

Worked Examples

Common Mistakes (Steel-manned)

Forecast-then-Verify

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Fluid Mechanics

Connections