Bernoulli equation applied

Definition and principle

The Bernoulli equation expresses the principle of conservation of energy for a steady, inviscid, and incompressible flow. Along a streamline, the sum of pressure, kinetic, and potential energies per unit mass remains constant: in a horizontal pipe, an increase in velocity implies a decrease in pressure and vice versa. The Bernoulli principle is the basis of the theorem of the same name, which is a special form of the Euler equation derived along a streamline.

Equations

Below are the most commonly used forms of the Bernoulli equation, including the energy loss terms.

Energy form
For an incompressible flow in steady state:

E = p₁ / ρ + v₁² / 2 + g h₁ = p₂ / ρ + v₂² / 2 + g h₂ – E_loss

where:

• E: energy per unit mass (J/kg or Btu/slug)
• p: fluid pressure (Pa or psf)
• ρ: fluid density (kg/m³ or slug/ft³)
• v: fluid velocity (m/s or ft/s)
• g: acceleration due to gravity (9.81 m/s² or 32.2 ft/s²)
• h: height with respect to a reference plane (m or ft)
• E_loss: energy loss per unit mass (J/kg or Btu/slug)

Head form
Dividing the above equation by g yields terms with length units:

h = p₁ / γ + v₁² / (2 g) + h₁ = p₂ / γ + v₂² / (2 g) + h₂ – E_loss / g

where γ = ρ g is the specific weight of the fluid (N/m³ or lbf/ft³). This form is called “head” because all terms have dimensions of fluid column.

Dynamic pressure
If the height variation is neglected (horizontal flow), the Bernoulli equation simplifies to:

p₁ + ½ ρ v₁² = p₂ + ½ ρ v₂² – p_loss

The term ½ ρ v² is known as dynamic pressure and relates the velocity to the recoverable pressure at a stagnation point. The total pressure is the sum of static pressure and dynamic pressure.

Dynamic pressure reference table

Values of dynamic pressure for water (ρ = 1000 kg/m³ / 1.940 slug/ft³ at 20 °C) as a function of velocity. Calculations are based on the expression p_d = ½ ρ v².

Velocity (m/s)	Velocity (ft/s)	Dynamic pressure (Pa)	Dynamic pressure (psi)
1 m/s	3.28 ft/s	500 Pa / 0.0725 psi	0.50 kPa / 0.073 psi
2 m/s	6.56 ft/s	2 000 Pa / 0.290 psi	2.0 kPa / 0.29 psi
5 m/s	16.4 ft/s	12 500 Pa / 1.81 psi	12.5 kPa / 1.81 psi
10 m/s	32.8 ft/s	50 000 Pa / 7.25 psi	50.0 kPa / 7.25 psi
15 m/s	49.2 ft/s	112 500 Pa / 16.3 psi	112.5 kPa / 16.3 psi
20 m/s	65.6 ft/s	200 000 Pa / 29.0 psi	200 kPa / 29.0 psi
25 m/s	82.0 ft/s	312 500 Pa / 45.3 psi	312.5 kPa / 45.3 psi
30 m/s	98.4 ft/s	450 000 Pa / 65.3 psi	450 kPa / 65.3 psi

Note: 1 psi = 6894.76 Pa = 144 psf. Imperial conversions use ρ = 1.940 slug/ft³.

Typical applications

The Bernoulli equation, with its adaptations, allows solving numerous practical problems:

• Discharge from an open tank (Torricelli’s Theorem): If the surface area of the tank is much larger than the discharge orifice area and the external pressure equals the internal pressure (vented tank), the exit velocity is v₂ = √(2 g h), where h is the height difference between the free surface and the center of the orifice. For a head of 10 m, v₂ = 14.0 m/s.
• Discharge from a pressurized tank: When the tank pressure (p₁) is different from the external pressure (p₂), the exit velocity is calculated with:

v₂ = c · √{ (2 / [1 – (A₂/A₁)²]) · [ (p₁ – p₂) / ρ + g h ] }

where c is the discharge coefficient of the orifice.

• Flow meters: The Venturi tube and the orifice plate use the pressure drop across a restriction to determine the flow rate, applying Bernoulli along with the continuity equation.
• Pitot tube: Measures the total pressure at the stagnation point (v = 0). The difference between total and static pressure gives the dynamic pressure, from which the flow velocity is obtained.
• Pressure reducing valves: The energy loss through a valve can be modeled as E_loss = (p₁ – p₂) / ρ, neglecting changes in velocity and height.

Discharge coefficients

To account for real losses, the theoretical velocity obtained from the Bernoulli equation is multiplied by a discharge coefficient (c), determined experimentally.

Orifice geometry	Discharge coefficient (c)
Sharp-edged orifice	0.60 – 0.64
Rounded-edged orifice	0.72 – 0.80
Short and smooth orifice	0.80 – 0.85
Smooth discharge nozzle	0.95 – 1.00

Limitations of the equation

The simple form of the Bernoulli equation is subject to the following restrictions:

1. Steady (time-independent) and inviscid (negligible shear stress effects) flow.
2. Incompressible fluid; in gases only valid for low Mach numbers (Ma < 0.3).
3. Must be applied along the same streamline; if the flow is irrotational, the Bernoulli constant is the same for all streamlines.
4. Does not consider head losses due to friction, valves, or sudden changes in cross-section; to include these effects, the loss term ΔE or p_loss is added.
5. Not suitable for flows with recirculation, intense turbulence, or significant heat exchange (non-adiabatic processes).

Calculation example

Problem: A pressurized tank contains water (ρ = 1000 kg/m³) at a gauge pressure p₁ = 0.2 MPa. The external pressure is p₂ = 0.1 MPa. The discharge orifice is 10 m below the free surface. The area ratio is A₂/A₁ = 0.01. Assuming a discharge coefficient c = 1.0 (smooth orifice), determine the exit velocity.

Solution:
We apply the equation for pressurized tank:

v₂ = √{ (2 / [1 – (A₂/A₁)²]) · [ (p₁ – p₂) / ρ + g h ] }

Substituting:
A₂/A₁ = 0.01 → (A₂/A₁)² = 0.0001 → 1 – (A₂/A₁)² ≈ 0.9999
p₁ – p₂ = 0.2×10⁶ – 0.1×10⁶ = 0.1×10⁶ Pa
g = 9.81 m/s², h = 10 m

v₂ = √{ (2 / 0.9999) [ (0.1×10⁶ Pa) / (1000 kg/m³) + (9.81 m/s²)(10 m) ] }
v₂ = √{ 2.0002 [ 100 + 98.1 ] } ≈ √{ 2.0002 × 198.1 } ≈ √{ 396.2 } ≈ 19.9 m/s

The exit velocity is 19.9 m/s (65.3 ft/s). If the height contribution had been neglected (h ≈ 0), the velocity would be 14.1 m/s, demonstrating the importance of considering all energy contributions.

Frequently Asked Questions (FAQ)

What is the exit velocity of water in an open tank with a height of 10 m relative to the orifice?

The theoretical velocity is 14.0 m/s (45.9 ft/s) according to Torricelli’s theorem, assuming ideal flow without losses.

What dynamic pressure does a water flow at 5 m/s generate?

A water flow at 5 m/s (16.4 ft/s) develops a dynamic pressure of 12,500 Pa (12.5 kPa / 1.81 psi).

What pressure difference between the tank and the outside is needed to obtain 20 m/s exit velocity without height influence?

Solving from v = √(2 Δp / ρ), Δp = ½ ρ v² = 0.5 × 1000 × 400 = 200,000 Pa = 200 kPa (29.0 psi).

What is the typical discharge coefficient for a sharp-edged orifice?

The discharge coefficient for a sharp-edged orifice varies between 0.60 and 0.64, so the actual velocity will be approximately 60% of the ideal velocity.

In what units is the head form of the Bernoulli equation measured?

The terms in the head form have units of length (m or ft) and represent the height of the fluid column in circulation.

Up to what air velocity can the flow be considered incompressible when applying Bernoulli?

For air under normal conditions, the incompressible approximation is acceptable up to velocities of approximately 100 m/s (328 ft/s), which corresponds to a Mach number close to 0.3.

References

• engineeringtoolbox.com: https://www.engineeringtoolbox.com/bernouilli-equation-d_183.html
• efunda.com: https://www.efunda.com/formulae/fluids/bernoulli.cfm
• thermopedia.com: https://www.thermopedia.com/content/579/