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Bernoulli's equation in real practice

Short answer

Bernoulli: p/ρg + V²/2g + z = constant. Assumptions: ideal, steady, incompressible, irrotational flow. Real practice modification adds head loss hL, kinetic energy factor α, pump head Hp, turbine head Ht: p1/ρg + α1V1²/2g + z1 + Hp = p2/ρg + α2V2²/2g + z2 + Ht + hL.

State Bernoulli's equation. Mention the assumptions made. How is it modified while applying in real practice?

Statement

For steady, ideal, incompressible flow along a streamline, the total energy per unit weight (sum of pressure, velocity and datum heads) is constant:

pρg+V22g+z=constant\frac{p}{\rho g} + \frac{V^2}{2g} + z = \text{constant}

Assumptions

  1. Ideal fluid (no viscosity, so no friction loss).
  2. Steady flow.
  3. Incompressible flow.
  4. Irrotational flow along a streamline, with uniform velocity over the section.

Modification for real flow

Real fluids are viscous, and velocity is not uniform across a section. So:

  1. A head loss term hLh_L (friction loss hf=fLV22gDh_f = \frac{fLV^2}{2gD} plus minor losses at bends, valves, entry and exit) is added on the downstream side.
  2. A kinetic energy correction factor α\alpha is applied to the velocity head (α≈2\alpha \approx 2 for laminar, 1.03–1.1 for turbulent pipe flow).
  3. Energy added by a pump (HpH_p) or extracted by a turbine (HtH_t) is included.
p1ρg+α1V122g+z1+Hp=p2ρg+α2V222g+z2+Ht+hL\frac{p_1}{\rho g} + \alpha_1\frac{V_1^2}{2g} + z_1 + H_p = \frac{p_2}{\rho g} + \alpha_2\frac{V_2^2}{2g} + z_2 + H_t + h_L

Example: in a hydropower penstock, Hnet=Hgross−hLH_{net} = H_{gross} - h_L.

✓ Answer checked by Claude Opus 5.5 (high effort)

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