Motivation #

When water flows through a drain or leaves a pond, we want to calculate its speed, pressure and discharge. How are these quantities related to the water’s elevation and the shape of the structure? Bernoulli’s equation gives us one of these relationships. But how does it come about, and how is it connected to Newton’s laws of motion?

In this article, we will derive Bernoulli’s equation from Newton’s second law: force equals mass times acceleration. Bernoulli then provides the basis for orifice discharge equations, weir discharge equations and culvert equations. Combining outlet relationships with conservation of mass lets us calculate how pond storage changes with time under the level-pool approximation.

Newton’s law for a moving volume of water #

Follow the water, not a fixed point #

Newton’s second law relates the net external force on a body to its acceleration. Let m be mass in kilograms, \boldsymbol{F} force in newtons, and \boldsymbol{a} acceleration in metres per second squared. Bold symbols have both magnitude and direction:

\displaystyle \sum \boldsymbol{F}=m\boldsymbol{a}.

For water, we follow a material volume: the same water as it moves and deforms. Its mass remains fixed. We treat water as a continuum, assigning pressure, density and velocity at each point without tracking individual molecules. We describe its motion using fixed spatial coordinates in an inertial reference frame, where Newton’s law has the form written above.

Let \boldsymbol{u} denote the local velocity vector and t time. The operator D/Dt means the rate of change following the water, while \nabla describes spatial derivatives. The acceleration is

\displaystyle \boldsymbol{a}=\frac{D\boldsymbol{u}}{Dt}=\frac{\partial\boldsymbol{u}}{\partial t}+(\boldsymbol{u}\cdot\nabla)\boldsymbol{u}.

The first term is the change in velocity with time at a fixed position. The second is the change encountered as the water moves into places with different velocities; \boldsymbol{u}\cdot\nabla takes spatial derivatives in the direction of motion, weighted by the velocity components.

Which forces act on it? #

We consider pressure and viscous forces on the water’s surface, and gravity acting throughout its mass.

The pressure-force sign can be seen using a small rectangular element. Let x increase to the right, let the element length be dx , and let each face perpendicular to x have area A_f . Pressure p is force per unit area, measured in pascals. Pressure on the left face pushes right; pressure on the right face pushes left.

To first order, the pressure on the right face is p+(\partial p/\partial x)dx . Subtracting the opposing forces gives the net pressure force in the x direction, F_{p,x} :

\displaystyle F_{p,x}=pA_f-\left(p+\frac{\partial p}{\partial x}dx\right)A_f=-\frac{\partial p}{\partial x}A_fdx.

Dividing by the element volume A_fdx gives the pressure force per unit volume:

\displaystyle \frac{F_{p,x}}{A_fdx}=-\frac{\partial p}{\partial x}.

The same reasoning applies in the other two directions. Together, the pressure forces per unit volume are -\nabla p . If pressure decreases to the right, \partial p/\partial x<0 , and the net pressure force points right, towards the lower pressure.

Let \rho be density and \boldsymbol{g} the gravitational acceleration vector. The gravity force per unit volume is \rho\boldsymbol{g} . Let \boldsymbol{\tau} describe the viscous stresses; \nabla\cdot\boldsymbol{\tau} is their net surface force divided by volume. Taking the small-volume limit of Newton’s law gives the continuum momentum equation:

\displaystyle \rho\frac{D\boldsymbol{u}}{Dt}=-\nabla p+\rho\boldsymbol{g}+\nabla\cdot\boldsymbol{\tau}.

Every term has units of force per volume, \mathrm{N/m^3} . We still need a relation between viscous stress and the water’s motion.

Conservation of Mass #

We also need conservation of mass. We choose a fixed control volume, a region of space through whose boundary water may flow. Unlike the material volume, it stays in place.

Let \mathcal V denote the fixed volume, \partial\mathcal V its boundary, d\mathcal V a small volume within it, and dA a small area on its boundary. The outward unit normal \boldsymbol n points perpendicular to the boundary, away from the volume. The local water velocity is \boldsymbol u .

The orange element d\mathcal V is used to add up mass inside the volume. The green patch dA is used to add up mass crossing its boundary. Only the velocity component perpendicular to the patch, \boldsymbol u\cdot\boldsymbol n , carries water across it. A positive value means outflow; a negative value means inflow. The outward mass flow rate through the patch is \rho(\boldsymbol u\cdot\boldsymbol n)dA . With no mass created or destroyed,

\displaystyle \frac{d}{dt}\int_{\mathcal V}\rho\,d\mathcal V+\int_{\partial\mathcal V}\rho\boldsymbol u\cdot\boldsymbol n\,dA=0.

The first term is the rate at which mass accumulates inside; the second is net outward mass flow. For smooth fields within the water, the divergence theorem converts the boundary integral to \int_{\mathcal V}\nabla\cdot(\rho\boldsymbol u)d\mathcal V . On a fixed bounded region, with the density and its time derivative sufficiently regular to interchange differentiation and integration, we can take the time derivative inside the volume integral. Thus,

\displaystyle \int_{\mathcal V}\left[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol u)\right]d\mathcal V=0.

Because this must hold for any such small region and the integrand is continuous, the integrand is zero. Otherwise, continuity would give a small neighbourhood where it has one sign and a nonzero integral:

\displaystyle \frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol u)=0.

This is the local continuity equation. Expanding the product gives

\displaystyle \frac{D\rho}{Dt}+\rho\nabla\cdot\boldsymbol u=0.

Let u_x,u_y,u_z be the velocity components along coordinates x,y,z . For our constant-density water model, D\rho/Dt=0 and \rho>0 , so

\displaystyle \nabla\cdot\boldsymbol u=\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}=0.

The moving water volume does not expand overall, although it can stretch in one direction and contract in another. We use this zero-divergence condition to simplify the viscous force below.

The route from Newton to Bernoulli #

The diagram shows the assumptions and operations that take us from Newton’s force balance to Bernoulli’s equation. We derive these steps below.

Navier–Stokes Derivation #

Water resists deformation #

We model water as a Newtonian fluid: its viscous stresses are proportional to its rates of deformation.

For a simple shear flow, imagine adjacent layers moving horizontally at different speeds. Let u_x be their horizontal velocity, y the distance across the layers, and \mu the dynamic viscosity in pascal seconds. The shear stress \tau_{xy} is viscous force in the x direction per unit area on a face normal to y :

\displaystyle \tau_{xy}=\mu\frac{\partial u_x}{\partial y}.

The faster water above pulls the highlighted layer forwards, while the slower water below pulls it backwards. The orange arrows are forces on that layer; the blue arrows show how the water moves. Opposite faces have opposite outward normals. For the same shear-stress value at both faces, their forces cancel. A difference between the face stresses gives a net force and changes the layer’s momentum.

This simple form assumes that velocity varies only across the layers. In three-dimensional flow, let i and j select coordinate directions, with u_i the velocity component in direction i and x_j the coordinate in direction j . The stress \tau_{ij} is force per area in direction i on a face normal to direction j . The incompressible Newtonian relation is

\displaystyle \tau_{ij}=\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right).

Both derivatives are needed because deformation can involve motion in either direction.

We can read the stress tensor as a table: choose a column for the face normal, then a row for the force direction. Each column gives the three viscous-force components per unit area on the corresponding positive coordinate face. A normal stress acts perpendicular to the face; a shear stress acts along it. The diagram shows the x and y components on two faces, with the full three-dimensional table alongside.

For example, \tau_{xy} selects the y -normal face and the x force direction: it is the rightward orange arrow on the top face. Swapping the subscripts gives \tau_{yx} , the upward orange arrow on the right face. The Newtonian relation above makes these two stress values equal at a point, but they describe forces on different faces. The diagonal entries describe normal viscous stresses; pressure is accounted for separately in the momentum equation.

Why express the viscous force using a Laplacian? #

The Laplacian, written \nabla^2 , adds the second spatial derivatives in the three coordinate directions. For velocity, we apply it to each component. Under our Newtonian, constant-viscosity and incompressible assumptions, the viscous force per unit volume reduces to \mu\nabla^2\boldsymbol{u} .

This expresses the viscous force through the velocity field, so we do not have to carry the separate stress components through the momentum equation. Physically, it describes viscosity transferring momentum between neighbouring regions, tending to smooth differences in velocity.

We use the constant-density, zero-divergence condition just derived and take \mu as constant. For a velocity component u_i , the corresponding viscous-force component is the sum of stress gradients in all three directions:

\displaystyle (\nabla\cdot\boldsymbol\tau)_i=\sum_{j=1}^{3}\frac{\partial\tau_{ij}}{\partial x_j}.

To find the net force in the x direction, we compare \tau_{xx} across the two x faces, \tau_{xy} across the two y faces, and \tau_{xz} across the two z faces. Dividing each opposing-face force difference by the element volume gives a spatial derivative. Adding the three contributions gives the x row of the stress divergence. Stress describes force per area at a face; its divergence describes the resulting net force per volume.

Substitute the Newtonian stress relation. Constant viscosity can be taken outside the derivatives:

\displaystyle (\nabla\cdot\boldsymbol\tau)_i=\mu\sum_{j=1}^{3}\left[\frac{\partial^2u_i}{\partial x_j^2}+\frac{\partial^2u_j}{\partial x_j\partial x_i}\right].

We assume continuous second spatial derivatives in this region, so we can interchange the order of differentiation. The second sum then becomes the derivative of the velocity divergence:

\displaystyle (\nabla\cdot\boldsymbol\tau)_i=\mu\nabla^2u_i+\mu\frac{\partial}{\partial x_i}\left(\sum_{j=1}^{3}\frac{\partial u_j}{\partial x_j}\right)=\mu\nabla^2u_i+\mu\frac{\partial}{\partial x_i}(\nabla\cdot\boldsymbol u).

Continuity makes the velocity divergence zero throughout the smooth region, so its spatial gradient is also zero. For example, the remaining force in the x direction is

\displaystyle (\nabla\cdot\boldsymbol\tau)_x=\mu\left(\frac{\partial^2u_x}{\partial x^2}+\frac{\partial^2u_x}{\partial y^2}+\frac{\partial^2u_x}{\partial z^2}\right).

Combining all three components gives

\displaystyle \nabla\cdot\boldsymbol{\tau}=\mu\nabla^2\boldsymbol{u}.

Substituting this viscous force into the momentum balance gives the incompressible Navier–Stokes momentum equation:

\displaystyle \rho\left[\frac{\partial\boldsymbol{u}}{\partial t}+(\boldsymbol{u}\cdot\nabla)\boldsymbol{u}\right]=-\nabla p+\rho\boldsymbol{g}+\mu\nabla^2\boldsymbol{u}.

From Navier–Stokes to Euler’s equation #

Neglecting viscous stresses reduces Navier–Stokes to Euler’s equation. This is the inviscid approximation. We remove the viscous term \mu\nabla^2\boldsymbol{u} , leaving

\displaystyle \rho\left[\frac{\partial\boldsymbol{u}}{\partial t}+(\boldsymbol{u}\cdot\nabla)\boldsymbol{u}\right]=-\nabla p+\rho\boldsymbol{g}.

Dividing by \rho and writing the acceleration as D\boldsymbol{u}/Dt gives

\displaystyle \frac{D\boldsymbol{u}}{Dt}=-\frac{1}{\rho}\nabla p+\boldsymbol{g}.

This is Euler’s equation of motion with gravity as the body force. We keep the constant-density water model, but have not yet assumed steady flow.

From Euler to Bernoulli along a streamline #

The acceleration along the flow #

A streamline is a line whose tangent follows the local velocity direction. We now assume steady flow, so the velocity at each fixed point does not change with time.

Let s be distance in metres along that streamline, increasing in the direction of flow, and let v be the local speed in metres per second.

Each blue arrow shows the water velocity at the position where the arrow starts. The purple curve is drawn so that its direction at each point agrees with the water velocity at that point. At P, the dashed line shows the curve’s local direction, called its tangent. The red velocity arrow points along this line; its length represents the local speed v . The distance s is measured along the purple curve, increasing in the direction of flow.

A water parcel is a small volume of the same water that we follow as it moves. In steady flow, the velocity field stays fixed, so the parcel follows the streamline.

On a smooth portion of the flow, ds/dt=v . Applying the chain rule gives the acceleration along the tangent, a_s :

\displaystyle a_s=\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}.

For example, if speed increases downstream, dv/ds>0 , and the tangential acceleration is positive. If the streamline bends, the parcel also accelerates perpendicular to the flow. A unit vector along the tangent has fixed length, so its derivative is perpendicular to it. Taking the acceleration component along the tangent therefore removes the bending contribution and leaves v\,dv/ds .

Pressure and gravity along the same direction #

Let z be elevation above a fixed datum, positive upwards, and let g=9.81\ \mathrm{m/s^2} be the magnitude of gravitational acceleration. We treat gravity as constant and downward over the elevation range considered.

In the diagram, s measures distance along the streamline in the direction of flow. A short movement of length ds raises the water by dz . The right panel compares the pressures p and p+dp at the two ends of that short water element.

In the left panel, dz/ds measures how steeply the path rises. Gravity acts vertically down; its component along the uphill tangent is negative, giving -g\,dz/ds .

For an uphill streamline, dz/ds>0 , so gravity opposes the motion. For a downhill streamline it assists the motion. Along a horizontal streamline, its tangential component is zero.

The right panel shows an example where pressure falls in the direction of flow. The upstream face is at pressure p and the downstream face at p+dp , with dp<0 . Both pressure forces push into the water element, but the upstream push is larger. Dividing the net tangential pressure force by the element mass gives -(1/\rho)\,dp/ds . If pressure rises downstream instead, that contribution points against the flow.

Taking the streamline component of Euler’s equation gives

\displaystyle v\frac{dv}{ds}=-\frac{1}{\rho}\frac{dp}{ds}-g\frac{dz}{ds}.

Multiplying by ds and moving all terms to the left gives

\displaystyle \frac{dp}{\rho}+v\,dv+g\,dz=0.

Integrate to obtain Bernoulli’s equation #

We choose points 1 and 2 on the same streamline. At point 1, the pressure and speed are p_1 and v_1 ; at point 2 they are p_2 and v_2 . The elevations z_1 and z_2 are the vertical heights of those points above one common datum, measured in metres.

Here point 2 is higher, so z_2-z_1>0 . The elevation change is this vertical difference, not the distance travelled along the streamline. The derivation also applies when point 2 is lower or at the same elevation.

With constant density as assumed for our water model, and constant g , we integrate the differential balance between these two points:

\displaystyle \int_{p_1}^{p_2}\frac{dp}{\rho}+\int_{v_1}^{v_2}v\,dv+\int_{z_1}^{z_2}g\,dz=0.

Because \rho is constant, it comes outside the first integral. Since the derivative of v^2/2 is v , the three evaluated integrals are

\displaystyle \frac{p_2-p_1}{\rho}+\frac{v_2^2-v_1^2}{2}+g(z_2-z_1)=0.

Rearranging gives Bernoulli’s equation per unit mass:

\displaystyle \frac{p_1}{\rho}+\frac{v_1^2}{2}+gz_1=\frac{p_2}{\rho}+\frac{v_2^2}{2}+gz_2.

The pressure term p/\rho accounts for pressure work per unit mass; v^2/2 is kinetic energy per unit mass, and gz is gravitational potential energy per unit mass. Each has units of joules per kilogram. Dividing by g expresses the balance as energy per unit weight, measured in metres. We call these three contributions heads:

The same balance in head form is:

\displaystyle \frac{p_1}{\rho g}+z_1+\frac{v_1^2}{2g}=\frac{p_2}{\rho g}+z_2+\frac{v_2^2}{2g}.

Using specific weight \gamma=\rho g in newtons per cubic metre, we write the sum as H_B . Here H_B is the Bernoulli head, measured in metres:

\displaystyle \frac{p}{\gamma}+z+\frac{v^2}{2g}=H_B,\qquad H_B\text{ is constant along the chosen streamline}.

What stays constant along a streamline? #

The quantity that stays constant is the sum H_B=p/\gamma+z+v^2/(2g) , not each contribution separately. Under the stated steady, inviscid, constant-density and constant-gravity assumptions, an increase in velocity head must be balanced by a decrease in pressure head, elevation head, or both. This result applies along the chosen streamline; different streamlines may have different Bernoulli heads.

For example, if two points on that streamline have the same elevation, the elevation heads cancel and

\displaystyle p_2-p_1=-\frac{\rho}{2}(v_2^2-v_1^2).

An increase in speed then requires a decrease in pressure. If elevation also changes, we must retain its contribution to the same head balance.

A short application: water leaving a pond #

Consider a wide open pond discharging through a small outlet into air. We take point 1 at the pond surface and point 2 in the free jet, both at atmospheric pressure p_{atm} . Let H=z_1-z_2 be the vertical distance from the surface to the jet point. We follow an ideal streamline between them and use a quasi-steady approximation: the pond level changes slowly enough for the acceleration caused by that change to be negligible in this calculation.

Use the same pressure reference and elevation datum at both points.

Both points are at atmospheric pressure, so their pressure heads cancel. The pond’s horizontal surface area is much larger than the outlet area, so the falling surface speed is negligible compared with the jet speed: v_1\approx0 . Let v_{2,ideal} be the ideal jet speed. With losses excluded, Bernoulli reduces to

\displaystyle z_1=z_2+\frac{v_{2,ideal}^2}{2g}.

Using H=z_1-z_2 , we obtain

\displaystyle v_{2,ideal}=\sqrt{2gH}.

For H=2.00\ \mathrm{m} ,

\displaystyle v_{2,ideal}=\sqrt{2\times9.81\times2.00}=6.26\ \mathrm{m/s}.

The available elevation head supplies the ideal velocity head.

For a small opening under nearly uniform head, continuity gives ideal discharge as opening area multiplied by ideal velocity. A real opening contracts the jet and dissipates energy, so we introduce a dimensionless discharge coefficient C_d and opening area A_o in square metres. The coefficient is actual discharge divided by ideal discharge. Here Q_o is discharge in cubic metres per second:

\displaystyle Q_o=C_dA_o\sqrt{2gH}.

Bernoulli does not determine C_d . Large openings, partial submergence and downstream tailwater are treated in the orifice discharge article.

To calculate how the pond level changes with time, combine the outlet relationship with the mass balance developed in From Conservation of Mass to the Level-Pool Routing Equation.

Using Bernoulli in pipe and drainage calculations #

So far, we have followed one streamline and neglected viscous stresses. For pipes and drains, we usually calculate between whole flow sections. We therefore account for the velocity distribution across each section and head losses between them. If a pump or turbine lies between the sections, we also include the energy it adds to or removes from the water.

From a local velocity to a section-average velocity #

So far, v has denoted the local speed at a point. Pipe and drainage calculations usually use the section-average velocity V=Q/A in metres per second, where Q is discharge in cubic metres per second and A is flow area in square metres. We choose two cross-sections, labelled 1 and 2 in the diagram:

The subscripts identify the section: V_1=Q_1/A_1 at section 1 and V_2=Q_2/A_2 at section 2. For the steady, incompressible flow shown, the same discharge passes through both sections, so the smaller area at section 2 gives a larger average speed. Within either section, local speeds can still vary. They all equal V only when the velocity profile is uniform across that section; steady flow alone does not make it uniform.

We therefore need a correction factor when using V in the energy balance. The average V=Q/A gives the correct discharge, but faster parts of the flow carry more kinetic energy per kilogram and pass more water through the section each second.

To account for this, consider forward flow approximately normal to the section. Let u_n be the local velocity normal to the section and dA a small area within it.

In the left panel, flow is normal to the section, so u_n is also the local speed v . The right panel shows a uniform speed V carrying the same discharge.

The mass passing through that area each second is \rho u_n\,dA , and its kinetic energy per kilogram is u_n^2/2 . Multiplying these and adding over the section gives the kinetic-energy transfer rate, in watts:

\displaystyle \frac{\rho}{2}\int_A u_n^3\,dA.

A uniform profile at speed V would instead carry kinetic energy at the rate \rho A V^3/2 . We define the kinetic-energy correction factor as the ratio of the actual rate to this uniform-profile rate:

\displaystyle \alpha=\frac{\int_A u_n^3\,dA}{A V^3}.

The factor \alpha is dimensionless and requires positive section area and nonzero mean speed. Dividing the kinetic-energy transfer rate by the weight flow rate \rho gQ=\rho gAV gives \alpha V^2/(2g) , the kinetic-energy head we use in the section balance below.

For uniform velocity, \alpha=1 . For a nonuniform profile, taking \alpha\approx1 means neglecting the kinetic-energy correction.

Head losses, pumps and turbines #

A pump adds mechanical energy to the water; a turbine extracts it. Let h_p be head supplied by a pump, h_t head removed by a turbine, and h_L head dissipated between the sections, all in metres. Including these terms gives the section energy balance for steady, incompressible flow. Subscripts on \alpha identify each section’s kinetic-energy correction factor:

\displaystyle \frac{p_1}{\gamma}+z_1+\alpha_1\frac{V_1^2}{2g}+h_p=\frac{p_2}{\gamma}+z_2+\alpha_2\frac{V_2^2}{2g}+h_t+h_L.

If there is no pump or turbine between the sections, h_p=h_t=0 . Friction and turbulence convert mechanical energy into internal energy; h_L measures the reduction in usable mechanical energy, while total energy remains conserved.

For example, consider the full, level pipe below. It has constant diameter, equal velocity correction factors and no pump or turbine. The open tubes are piezometers: their water levels show the pressure head above the common pipe centreline. Here p_1 and p_2 are gauge pressures relative to the atmosphere, and both permit positive water-column readings.

Steady incompressible flow gives the same discharge and section-average speed at both sections. The velocity-head terms cancel, and the elevations are equal. The section balance therefore reduces to

\displaystyle \frac{p_1-p_2}{\gamma}=h_L.

The difference between the two piezometer levels is therefore h_L : the pressure-head drop equals the head loss.

Conclusion #

Starting from Newton’s second law, we obtain Navier–Stokes by applying the Newtonian viscous-stress relation. Neglecting viscous stresses leaves Euler’s balance of pressure, gravity and acceleration. For steady flow, the tangential acceleration is v\,dv/ds . With constant density and gravity, we integrate the pressure, speed and elevation changes to obtain

\displaystyle \frac{p}{\gamma}+z+\frac{v^2}{2g}=\text{constant along a streamline}.

For specific conditions, the balances developed here give simpler formulas: for a slowly draining wide pond and an ideal free jet, Bernoulli’s equation gives v=\sqrt{2gH} ; for the level, constant-diameter pipe considered above, the energy balance including losses reduces to (p_1-p_2)/\gamma=h_L .

The orifice, weir and culvert derivations introduced at the start take the simplification further. Using each structure’s geometry and flow conditions, we obtain discharge relationships short enough to print on a T-shirt—and for engineers to use with just pen and paper.