The Mechanical Energy Balance (Bernoulli with Friction)
From ideal energy conservation to real pipes with pumps and losses
If the pressure drops along a pipe, where did that mechanical energy go? The mechanical energy balance answers that—quantitatively.
Ideal Bernoulli (no friction, no machines)
Bernoulli’s equation is a mechanical energy balance for steady, incompressible flow along a streamline. In its common form per unit mass, it states that pressure energy, kinetic energy, and potential energy trade off but sum to a constant.
Bernoulli’s ideal form neglects viscous dissipation (friction), assumes steady incompressible flow, and is best applied between points in the same flow path. Real pipes need loss terms.
Head form (per unit weight) for engineering work
Engineers often divide by g to write terms as a head (meters of fluid):
- Pressure head: P/(ρg)
- Velocity head: v²/(2g)
- Elevation head: z
Losses (like friction) also appear naturally as head terms.
hf (friction loss) is the irreversible conversion of mechanical energy into internal energy due to viscosity. hm (minor losses) captures fittings/entrances/exits. A pump adds head (positive), a turbine removes head (negative).
When velocities are the same
Many practical pipe problems use the same diameter at points 1 and 2, so v1=v2. Then the velocity-head terms cancel, simplifying the balance to mainly pressure, elevation, and losses.
- Write head balance with z1=z2 and v1=v2: P1/(ρg) + hp = P2/(ρg) + h_L.
- If the pump just needs to overcome losses with no net pressure change desired between endpoints, set P1≈P2, so hp = h_L.
- Thus hp = 12.0 m (of water).
- With no pump and z1=z2 and v1=v2: P1/(ρg) = P2/(ρg) + h_L → (P1−P2)/(ρg)=h_L.
- Compute ΔP = ρ g h_L = (998)(9.81)(5.0) = 48951.9 Pa.
Check your understanding
- Ideal Bernoulli ignores friction and machines
- Mechanical energy balance adds pump/turbine head and loss terms
- Head form uses P/(ρg), v²/(2g), and z, all in meters
- Many pipe problems simplify when v1=v2 and/or z1=z2