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. The tailwater and backwater calculation applies the energy balance with losses to find upstream water levels from a known downstream level. Combining outlet relationships with conservation of mass lets us calculate how pond storage changes with time under the level-pool approximation.

We start with the forces on a small volume of water. The viscous-stress relation gives us the Navier–Stokes equation; neglecting viscous stresses leaves Euler’s equation. We then follow the acceleration along a streamline and integrate to obtain Bernoulli’s equation, stating the assumptions at each step.

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:

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

Here m is mass in kilograms, \boldsymbol{F} is force in newtons, and \boldsymbol{a} is acceleration in metres per second squared. Bold symbols have both magnitude and direction.

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 denote time. The acceleration of that moving water is written

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

The capital D/Dt means the rate of change following the water. 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. The operator \nabla describes spatial derivatives; \boldsymbol{u}\cdot\nabla takes those derivatives in the direction of motion, weighted by the velocity components.

Which forces act on it? #

The forces on this water are pressure and viscous forces on its surface, and gravity acting throughout its mass. We exclude other body forces, such as electromagnetic forces.

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

\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} . The left side is mass per volume multiplied by acceleration. The right side collects the forces that produce that acceleration. This balance does not yet specify how the fluid’s viscous stresses depend on its motion.

Conservation of Mass #

Newton’s law tells us how the water accelerates. To determine how much water occupies a region, we also need conservation of mass. We choose a fixed control volume, a region of space through whose boundary water may flow. The material volume follows the water; the control volume stays in place.

Let \mathcal V denote the fixed volume, \partial\mathcal V its boundary, d\mathcal V a small volume within it, and \boldsymbol n the outward unit normal on that boundary. A small boundary area dA carries outward mass at rate \rho(\boldsymbol u\cdot\boldsymbol n)dA . The dot product is positive for outflow and negative for inflow. 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.

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, or accelerate through a narrowing pipe. We use this zero-divergence condition below to simplify the viscous force. The fixed-volume mass and momentum balances are developed in NPTEL’s conservation-equations lecture.

Deriving Bernoulli from Netwon’s law #

Newton’s second law gives the relationship between various forces. For constant-density water, the Newtonian stress relation with constant viscosity turns that balance into the incompressible Navier–Stokes equation. Neglecting viscous stress gives Euler’s equation. For steady flow with constant downward gravity, taking Euler’s component along a streamline and integrating gives Bernoulli’s equation.

Navier–Stokes and Euler describe the local velocity and pressure fields. Bernoulli is the resulting mechanical-energy relation along one streamline under the additional assumptions. Each arrow in the diagram identifies the step that lets us move to the next equation.

We can also obtain Euler’s equation directly from the general momentum balance by neglecting viscous stresses, without first specifying the Newtonian relation. Navier–Stokes makes the viscous part explicit before we decide whether it can be neglected.

Navier–Stokes Derivation #

Water resists deformation #

For ordinary water-flow calculations, we model water as a Newtonian fluid: its viscous stresses are proportional to its rates of deformation. Here “Newtonian” names a fluid stress relation.

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, 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).

The subscripts i and j each select one of the three coordinate directions. Thus \tau_{ij} is viscous force per area in direction i on a face normal to direction j . 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} .

The advantage is mathematical simplification: we express the viscous force through the velocity field we are already solving for, instead of carrying the separate stress components through the momentum equation. Physically, this term describes viscosity transferring momentum between neighbouring regions, tending to smooth differences in velocity. Its mathematical form is a diffusion term, which helps us analyse and calculate viscous flow.

For our route to Bernoulli, this gives the viscous term in Navier–Stokes that we later neglect to obtain Euler’s equation. We now derive this simpler form from the stress relation above.

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}.

Together with continuity and the boundary and initial conditions, this equation describes how pressure, gravity and viscosity affect the moving water.

This reduction depends on constant viscosity and incompressibility. Variable viscosity produces additional terms when we differentiate the stress; compressible flow need not have zero velocity divergence and requires a different stress relation.

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}.

The acceleration, pressure gradient and gravity remain. In particular, (\boldsymbol{u}\cdot\nabla)\boldsymbol{u} describes acceleration as water moves through the velocity field, so it stays in the equation. 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 water that we follow as it moves, keeping track of the same water. The diagram shows the velocity field at one instant. In steady flow, that field stays fixed, so the parcel follows the streamline. In unsteady flow, its path need not follow a streamline drawn at an earlier instant.

We consider a smooth portion of the flow where these derivatives exist. Then ds/dt=v . Applying the chain rule gives the tangential acceleration:

\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.

The left panel shows a short uphill piece of streamline. Its length is ds and the elevation gain is dz . The ratio 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.

With constant downward gravity, the gravitational potential energy per unit mass is gz .

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.

Thus, the pressure, speed and elevation changes balance along the chosen streamline under the assumptions made so far.

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}.

Or, using specific weight \gamma=\rho g in newtons per cubic metre,

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

Here H_B is the Bernoulli head, measured in metres: the sum of the three head contributions.

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. Pressure may be absolute or gauge pressure; subtracting the same reference pressure leaves the pressure difference unchanged. Changing the elevation datum likewise leaves z_1-z_2 unchanged.

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 . With losses excluded in the ideal calculation, 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 units also follow: gH has units \mathrm{m^2/s^2} , and its square root is a speed. 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 for the stated opening and head:

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

Here Q_o is discharge in cubic metres per second. Bernoulli supplies the ideal velocity dependence; it does not determine C_d . For large openings, partial submergence and downstream tailwater, the head and geometry need the separate treatment developed 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. 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.

In the left panel, the flow is normal to the section, so u_n is also the local speed v . The right panel replaces these varying speeds with a uniform speed V carrying the same discharge. The factor \alpha accounts for the difference in kinetic-energy transport. The profiles are schematic, with no numerical values assigned.

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

Head losses, pumps and turbines #

Ideal Bernoulli assumes no mechanical energy is added or removed along the streamline. A pump adds energy to the water; a turbine extracts it. To include these transfers and head losses, we extend the equation to a mechanical-energy balance. For steady, incompressible flow between two suitable sections, we write

\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.

Here h_p is head supplied to the water by a pump, h_t is head removed by a turbine, and h_L is head dissipated between the sections. All are measured in metres. The section heads and friction-loss interpretation are illustrated in the US Bureau of Reclamation’s energy-balance guidance.

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.

In the diagram, the difference between the two piezometer levels is therefore h_L . The elevation and velocity heads are unchanged, so the reduction in total head appears entirely as a drop in pressure head.

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.