Interactive Lecture
Energy Conservation in Fluid Flow and Engineering Applications
by Dr Arnaldo Delli Carri
Bernoulli's equation is one of the most important and widely used equations in fluid mechanics. It describes the conservation of energy for flowing fluids and relates pressure, velocity, and elevation along a streamline.
For an ideal fluid flowing along a streamline, the total mechanical energy per unit volume remains constant. An increase in fluid velocity must be accompanied by a decrease in pressure or potential energy.
Daniel Bernoulli, a Swiss mathematician and physicist, published this principle in 1738 in his work "Hydrodynamica." The equation applies to incompressible, inviscid (frictionless) flow along a streamline, and it is the starting point for most pipe-flow and aerodynamics calculations.
Bernoulli's equation can be derived from Newton's second law applied to a fluid element or from the principle of conservation of energy. It states that the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.
Bernoulli's Equation (Energy Form):
\[P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}\]
Where:
Alternative Form (Head Form):
\[\frac{P}{\rho g} + \frac{v^2}{2g} + h = \text{constant}\]
Where each term represents:
Each term in Bernoulli's equation represents a different form of energy per unit volume:
Bernoulli's equation is valid under the following conditions:
Explore how fluid velocity and pressure change through a converging-diverging tube (Venturi effect). Adjust the inlet velocity and tube geometry to see how Bernoulli's equation describes the flow behaviour.
Note: At high velocities or small throat diameters, you may observe cavitation — the formation of vapour bubbles when local pressure drops below the vapour pressure limit.
Explore how Bernoulli's equation applies when elevation changes and pipe losses are significant. A water tower at height \(H\) feeds a series of sprinkler heads through a common pipe. In a frictionless system every nozzle would spray to the same height \(v^2/2g = H\); in reality, friction and fitting losses accumulate along the pipe, progressively reducing the available head — and spray height — for each downstream sprinkler. Drag the pipe diameter slider to see how this effect scales.
A large tank maintains a constant head \(H\): opening more nozzles draws more total flow, not less pressure. Without losses every sprinkler therefore sprays to the same height — series and parallel arrangements are physically identical. The progressive reduction seen downstream is caused entirely by friction and minor losses accumulating along the pipe; the series layout is worse than parallel precisely because its upstream segments must carry the combined flow of all downstream nozzles. Use the checkboxes above to explore this.
| Concept | Equation | Description |
|---|---|---|
| Bernoulli's Equation (Energy) | \(P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}\) | Conservation of energy along streamline |
| Bernoulli's Equation (Head) | \(\frac{P}{\rho g} + \frac{v^2}{2g} + h = \text{const}\) | Head form for hydraulic calculations |
| Static Pressure | \(P\) | Pressure energy per unit volume |
| Dynamic Pressure | \(\frac{1}{2}\rho v^2\) | Kinetic energy per unit volume |
| Hydrostatic Pressure | \(\rho g h\) | Potential energy per unit volume |
| Continuity Equation | \(A_1 v_1 = A_2 v_2\) | Mass conservation for incompressible flow |
| Velocity from Pressure | \(v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}\) | Velocity from pressure difference (same height) |
| Pressure Head | \(h_p = \frac{P}{\rho g}\) | Equivalent height of fluid column |
| Velocity Head | \(h_v = \frac{v^2}{2g}\) | Height equivalent to kinetic energy |
| Stagnation Pressure | \(P_0 = P + \frac{1}{2}\rho v^2\) | Total pressure when flow is brought to rest |
A large open water tank has a small hole at a depth of 4 metres below the water surface. Calculate the velocity of water exiting the hole. Assume atmospheric pressure at both the surface and the hole exit.
Show Solution
Solution:
Given:
Apply Bernoulli's Equation:
\[P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2\]
Since P₁ = P₂ and v₁ ≈ 0:
\[\rho g h_1 = \frac{1}{2}\rho v_2^2 + \rho g h_2\]
\[g(h_1 - h_2) = \frac{1}{2}v_2^2\]
\[v_2 = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 4} = \sqrt{78.48} = 8.86 \text{ m/s}\]
Answer: The exit velocity is approximately 8.86 m/s (Torricelli's theorem)
Water flows through a horizontal Venturi meter with an inlet diameter of 200 mm and a throat diameter of 100 mm. If the pressure at the inlet is 300 kPa and at the throat is 200 kPa, calculate the flow velocity at the throat and the volumetric flow rate. (Density of water = 1000 kg/m³)
Show Solution
Solution:
Given:
Step 1: Apply Continuity Equation:
\[A_1 v_1 = A_2 v_2\]
\[v_1 = v_2 \frac{A_2}{A_1} = v_2 \frac{0.00785}{0.0314} = 0.25 v_2\]
Step 2: Apply Bernoulli's Equation (horizontal, so h₁ = h₂):
\[P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\]
\[300000 + \frac{1}{2}(1000)(0.25v_2)^2 = 200000 + \frac{1}{2}(1000)v_2^2\]
\[100000 = 500v_2^2 - 500(0.0625)v_2^2\]
\[100000 = 500v_2^2(1 - 0.0625) = 468.75v_2^2\]
\[v_2 = \sqrt{\frac{100000}{468.75}} = 14.61 \text{ m/s}\]
Step 3: Calculate volumetric flow rate:
\[Q = A_2 v_2 = 0.00785 \times 14.61 = 0.115 \text{ m}^3\text{/s} = 115 \text{ L/s}\]
Answer: Throat velocity = 14.61 m/s, Flow rate = 115 L/s
A pitot tube on an aircraft measures a stagnation pressure of 105 kPa and a static pressure of 100 kPa. If the air density is 1.2 kg/m³, calculate the aircraft's airspeed.
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Solution:
Given:
Concept: At the stagnation point, velocity = 0, so:
\[P_0 = P + \frac{1}{2}\rho v^2\]
Solve for velocity:
\[\frac{1}{2}\rho v^2 = P_0 - P = 105000 - 100000 = 5000 \text{ Pa}\]
\[v^2 = \frac{2 \times 5000}{1.2} = \frac{10000}{1.2} = 8333.33\]
\[v = \sqrt{8333.33} = 91.3 \text{ m/s}\]
Answer: The aircraft's airspeed is approximately 91.3 m/s (or 328.7 km/h)