Bernoulli's equation describes ideal steady incompressible flow with negligible viscous dissipation. Real pipe systems can also contain pumps, turbines, and irreversible losses. A practical mechanical energy equation adds those effects explicitly.

Between locations 1 and 2, divide mechanical energy per unit mass by $g$ to express each term as a head, with units of length:

$$\boxed{ \frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h_p-h_t

\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_L }.$$

Here

• $p/(\rho g)$ is pressure head;
• $V^2/(2g)$ is velocity head;
• $z$ is elevation head;
• $h_p$ is mechanical energy per unit weight added by pumps;
• $h_t$ is mechanical energy per unit weight removed by turbines;
• $h_L\ge0$ is irreversible head loss.

If $h_p=h_t=h_L=0$, the equation reduces to Bernoulli's equation.

Major losses

For a straight pipe, Darcy-Weisbach gives the distributed or major loss

$$\boxed{h_f=f\frac{L}{D}\frac{V^2}{2g}}.$$

A system can contain several pipe segments, each with its own $f$, $L$, $D$, and mean velocity. Their losses are added.

Minor losses

Localized components such as entrances, exits, elbows, valves, tees, contractions, and expansions are often modeled with a dimensionless loss coefficient $K_L$:

$$\boxed{h_m=K_L\frac{V^2}{2g}}.$$

The reference velocity must match the definition used for the coefficient. Several local losses can be summed:

$$h_{minor}=\sum_i K_{L,i}\frac{V_i^2}{2g}.$$

Despite the traditional name minor loss, these losses need not be small; in a short system with many fittings they can dominate the distributed pipe loss.

The total irreversible loss is

$$\boxed{h_L=h_f+h_{minor}}$$

for a simple one-pipe system, or the sum of all relevant major and local losses in a more complex path.

Pump and turbine head

If a pump transfers shaft work to the fluid, the ideal mechanical energy added per unit mass is

$$w_p=gh_p.$$

Thus

$$\boxed{h_p=\frac{w_p}{g}}.$$

Similarly, a turbine that extracts mechanical energy from the fluid removes head

$$\boxed{h_t=\frac{w_t}{g}}.$$

The equation tracks mechanical energy of the flowing fluid. Real pump and turbine efficiencies determine how shaft power relates to the ideal head transferred to or from the fluid.

Worked example: pumping between reservoirs

Water is pumped from a large lower reservoir to a large upper reservoir whose free surface is

$$\Delta z=20,\mathrm m$$

higher. Both reservoirs are open to the atmosphere, so

$$p_1=p_2=p_{atm}.$$

Because the reservoirs are large, free-surface speeds are negligible:

$$V_1\approx V_2\approx0.$$

Suppose the pipe and fittings produce a total head loss

$$h_L=6.0,\mathrm m.$$

There is no turbine. The mechanical energy equation becomes

$$z_1+h_p=z_2+h_L.$$

Therefore

$$h_p=(z_2-z_1)+h_L =20+6 =\boxed{26,\mathrm m}.$$

The pump must supply enough mechanical energy to raise the water by $20,\mathrm m$ and replace the $6,\mathrm m$ dissipated by hydraulic losses.

The ideal hydraulic power delivered to a volume flow rate

$$Q=0.010,\mathrm{m^3/s}$$

is

$$\dot W_{hyd}=\rho gQh_p.$$

For water,

$$\dot W_{hyd} =(1000)(9.81)(0.010)(26) \approx2.55\times10^3,\mathrm W.$$

Thus

$$\boxed{\dot W_{hyd}\approx2.55,\mathrm{kW}}.$$

If the pump is less than perfectly efficient, the required shaft input is larger than this hydraulic power.

Modeling workflow

For an incompressible pipe-system problem:

1. choose two useful sections or reservoir surfaces;
2. write pressure, velocity, and elevation heads at both locations;
3. add pump head and subtract turbine head according to energy-transfer direction;
4. calculate major losses with Darcy-Weisbach;
5. add relevant local-loss coefficients;
6. solve the resulting mechanical energy balance.

The equation does not replace detailed viscous-flow theory. It packages irreversible mechanical-energy loss into head-loss models so complete hydraulic systems can be analyzed without resolving the full velocity and stress field everywhere.