The Bernoulli equation
An incompressible, non-viscous fluid flows steadily through a pipe of varying cross-section that also varies in height. Its speed must change as the area changes, by the equation of continuity. A change in speed needs a net force, and that force comes from a difference in pressure between regions of the pipe. Bernoulli's equation is the general relation that ties these three quantities — pressure, speed and height — together for any two points on a streamline.
Bernoulli's principle (NCERT §9.4): As we move along a streamline, the sum of the pressure $P$, the kinetic energy per unit volume $\tfrac{1}{2}\rho v^2$, and the potential energy per unit volume $\rho g h$ remains a constant.
Written between two points 1 and 2 on the same streamline, and in its general form:
$$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2$$ $$P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$$Each term has the dimensions of pressure (energy per unit volume). $P$ is the pressure energy density, $\tfrac{1}{2}\rho v^2$ the kinetic energy density, and $\rho g h$ the gravitational potential energy density. The Swiss physicist Daniel Bernoulli established this relation in 1738.
Derivation from energy conservation
Take the fluid initially between sections B and D in Figure 1. In a small interval $\Delta t$ the fluid at B (speed $v_1$, area $A_1$, pressure $P_1$) moves a distance $v_1\Delta t$ to C; the fluid at D (speed $v_2$, area $A_2$, pressure $P_2$) moves $v_2\Delta t$ to E. By continuity the same volume $\Delta V$ crosses each section. The work done on the fluid at the left face is $W_1 = P_1 A_1 (v_1\Delta t) = P_1\Delta V$, and the work done by the fluid against the pressure at the right face is $W_2 = P_2 A_2 (v_2\Delta t) = P_2\Delta V$. The net work done on the fluid is therefore $(P_1 - P_2)\Delta V$.
This work changes the kinetic and potential energies of the displaced volume. With $\Delta m = \rho\,\Delta V$, the changes are $\Delta K = \tfrac{1}{2}\rho\,\Delta V\,(v_2^2 - v_1^2)$ and $\Delta U = \rho g\,\Delta V\,(h_2 - h_1)$. The work–energy theorem gives:
$$(P_1 - P_2)\,\Delta V = \tfrac{1}{2}\rho\,\Delta V\,(v_2^2 - v_1^2) + \rho g\,\Delta V\,(h_2 - h_1)$$Dividing every term by $\Delta V$ and rearranging recovers the Bernoulli equation directly:
$$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2$$For a fluid at rest, $v = 0$ everywhere and the equation collapses to $P_1 + \rho g h_1 = P_2 + \rho g h_2$, i.e. $P_1 - P_2 = \rho g (h_2 - h_1)$ — the hydrostatic pressure relation. Bernoulli's equation is thus the dynamic generalisation of the static pressure law you met under pressure in fluids.
Assumptions and limitations
Because the derivation used the work–energy theorem with no dissipative term, the equation holds only for an idealised fluid. NEET routinely tests whether you know the four conditions and where they break.
Incompressible
Density $\rho$ is constant. The elastic energy stored in compression is not accounted for, so the relation fails for highly compressible gases at large pressure changes.
Non-viscous
Zero viscosity. Real fluids lose kinetic energy to internal friction between layers, which becomes heat. So $P_2$ is in practice lower than Bernoulli predicts.
Steady (streamline)
Velocity and pressure at each point are constant in time. The equation does not hold for turbulent or non-steady flow, where both fluctuate.
Along one streamline
Points 1 and 2 must lie on the same streamline. The constant generally differs from one streamline to another.
In short, Bernoulli's equation is exact only for an ideal fluid and a good approximation for low-viscosity, incompressible, smoothly flowing real fluids. For a turbulent rapid in a river, or for blood squeezing through a constricted artery where viscosity dominates, it cannot be applied without correction.
The pressure–velocity inverse relation
Restrict the streamline to a horizontal pipe, so $h_1 = h_2$ and the $\rho g h$ term drops out. Bernoulli's equation reduces to its most-tested form:
$$P + \tfrac{1}{2}\rho v^2 = \text{constant}$$The two terms trade against each other. Where the fluid moves faster, the kinetic-energy density $\tfrac{1}{2}\rho v^2$ is larger, so the pressure $P$ must be smaller. This is the single idea behind every Bernoulli application: high speed means low pressure, and low speed means high pressure. Combine it with continuity ($Av = \text{constant}$, so the fluid speeds up where the pipe narrows) and you can predict the pressure everywhere in a flow at a glance.
Continuity sets the speed; Bernoulli sets the pressure. Revise the link first in streamline flow & the equation of continuity.
Torricelli's law of efflux
The word efflux means fluid outflow. Torricelli discovered that the speed at which liquid spurts from an open tank equals the speed of a freely falling body. Consider a tank of liquid of density $\rho$ with a small hole in its side at height $y_1$, the free surface being at height $y_2$, and air above the surface at pressure $P$. Let $A_1$ be the hole area and $A_2$ the tank's cross-section.
By the equation of continuity $v_1 A_1 = v_2 A_2$. If the tank's area is much larger than the hole, $A_2 \gg A_1$, the surface descends so slowly that we take $v_2 \approx 0$. Applying Bernoulli between the surface (point 2) and the hole (point 1), where the pressure at the hole is atmospheric $P_a$:
$$P_a + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 = P + \rho g y_2$$Writing $h = y_2 - y_1$ and solving for the efflux speed:
$$v_1 = \sqrt{\,\dfrac{2(P - P_a)}{\rho} + 2gh\,}$$Two limiting cases follow. When the tank is sealed and the air pressure is very large, $P \gg P_a$ and $2gh$ is negligible, so $v_1 \approx \sqrt{2(P-P_a)/\rho}$ — the efflux is set by the container pressure, as in rocket propulsion. When the tank is open to the atmosphere, $P = P_a$ and the pressure terms cancel, leaving Torricelli's law:
$$\boxed{\,v_1 = \sqrt{2gh}\,}$$This is exactly the speed a body acquires falling freely through height $h$. Notice that for an open tank the density $\rho$ has cancelled out entirely — the efflux speed is independent of which liquid is in the tank.
The Venturi-meter
A Venturi-meter measures the rate of flow of a liquid through a pipe. It is a horizontal tube with a deliberate constriction; a manometer connects the wide section (area $A_1$, speed $v_1$, pressure $P_1$) to the narrow throat (area $A_2$, speed $v_2$, pressure $P_2$). Because the pipe is horizontal, the height term vanishes and only pressure and kinetic energy trade.
Continuity gives $A_1 v_1 = A_2 v_2$, so $v_2 > v_1$ at the throat. Bernoulli for the horizontal tube gives:
$$P_1 - P_2 = \tfrac{1}{2}\rho\,(v_2^2 - v_1^2)$$Eliminating $v_2$ using continuity and reading $P_1 - P_2 = \rho_m g h$ from the manometer (level difference $h$) lets us solve for the inlet speed:
$$v_1 = \sqrt{\dfrac{2(P_1 - P_2)}{\rho\left[(A_1/A_2)^2 - 1\right]}}$$Since every term except $h$ is fixed for a given meter, $v_1 \propto \sqrt{h}$, and the volume flow rate is $Q = A_1 v_1 = K'\sqrt{h}$. The pressure is least where the bore is narrowest — Venturi's principle — which is precisely the pressure–velocity inverse relation in instrument form. NEET 2023 asked the bare fact that the Venturi-meter works on Bernoulli's principle.
Dynamic lift — aerofoil and Magnus effect
Dynamic lift is the force on a body — an aircraft wing, a hydrofoil, a spinning ball — arising from its motion through a fluid. It is the most spectacular consequence of the pressure–velocity relation, and NCERT §9.4.2 treats three cases.
Ball without spin
Streamlines around a non-spinning ball are symmetric top and bottom. The air speed at corresponding points is equal, so there is zero pressure difference and no upward or downward force.
Spinning ball — Magnus effect
A spinning ball drags air with it. Air speed becomes larger on one side and smaller on the other; streamlines crowd on the fast side and rarefy on the slow side. The pressure difference produces a net sideways force that deflects the ball — the Magnus effect seen in cricket and tennis swing.
Aerofoil — wing lift
A wing is shaped and angled so streamlines crowd more above than below. Flow speed on top is higher, so pressure on top is lower. The excess pressure underneath gives an upward lift that balances the aircraft's weight.
A fully loaded Boeing aircraft has a mass of $3.3\times10^5$ kg and total wing area $500~\text{m}^2$. It is in level flight at $960~\text{km/h}$. Estimate (a) the pressure difference between the lower and upper wing surfaces, and (b) the fractional increase in air speed over the upper surface. Take $\rho_{\text{air}} = 1.2~\text{kg m}^{-3}$.
(a) The weight is balanced by the lift force $\Delta P \times A$: $\;\Delta P = \dfrac{(3.3\times10^5)(9.8)}{500} = 6.5\times10^3~\text{N m}^{-2}$.
(b) Ignoring the height difference, $\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\rho(v_2 - v_1)(v_2 + v_1)$. With average speed $v_{av} = (v_1+v_2)/2 = 960~\text{km/h} = 267~\text{m s}^{-1}$, the fractional increase $\dfrac{v_2 - v_1}{v_{av}} = \dfrac{\Delta P}{\rho\,v_{av}^2} \approx 0.08$. The air over the wing need be only about 8% faster than the air beneath.
Atomiser, sprayer and blood flow
The same low-pressure-where-fast logic runs through a family of everyday devices and one important physiological caution.
Scent / paint sprayer
Squeezing the bulb blows air fast through a narrow orifice, dropping the pressure there. Atmospheric pressure pushes liquid up the tube into the low-pressure stream, which shatters it into a fine spray.
Spray gun / Bunsen burner
A piston (or gas jet) forces air through a small hole at high speed and low pressure, sucking insecticide — or, in a Bunsen burner, air through a side hole to mix with the gas before combustion.
Arterial constriction
Where an artery narrows (a plaque), blood speeds up and the lateral pressure falls. The vessel can be squeezed shut and reopen repeatedly, fluttering. Note blood is viscous, so Bernoulli applies only qualitatively here.
Bernoulli's principle in one breath
- $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$ along a streamline — energy conservation per unit volume.
- Assumptions: incompressible, non-viscous, steady flow, single streamline. Fails for viscous, turbulent or compressible flow.
- Horizontal pipe: $P + \tfrac{1}{2}\rho v^2 = \text{constant}$ → high speed means low pressure.
- Torricelli's efflux (open tank): $v = \sqrt{2gh}$, equal to free-fall speed and independent of density.
- Venturi-meter measures flow rate from the throat pressure drop; $Q \propto \sqrt{h}$.
- Dynamic lift: faster, lower-pressure flow above an aerofoil (or one side of a spinning ball) gives lift / Magnus deflection.