Account for pressure work in ideal flow

Identify the work that accelerates the liquid

Continuity predicts a higher speed in a narrower section at fixed incompressible flow. It does not by itself explain where the increased kinetic energy comes from. A pressure difference can do net work on the moving liquid.

The inlet in this model is held at 120 kPa absolute pressure. A supply maintains the selected flow rate. The outlet is a downstream section of a pressurized pipe, not necessarily an opening to the atmosphere. Its pressure is calculated from the ideal energy balance and must be compatible with the downstream system. The model does not simultaneously prescribe an unrelated outlet pressure.

Write an energy account per volume

For steady incompressible ideal flow along the modeled stream, p + ½ρv² + ρgz remains constant when there is no intervening pump, turbine or dissipative loss. p is pressure in Pa, ρ is density in kg/m³, v is speed in m/s, g is gravitational acceleration and z is elevation in metres. Each term has units of energy per volume, J/m³, equivalent to Pa.

The two sections here have equal elevation, so their gravitational terms cancel. The relation becomes p₂ = p₁ + ½ρ(v₁² − v₂²). If the downstream speed is greater under these conditions, its pressure is smaller. The OpenStax Bernoulli reference connects this relation to pressure work and mechanical-energy change.

This is a comparison along one ideal flow with stated boundary conditions. It is not a universal rule that any faster-moving fluid anywhere must have lower pressure than any slower fluid. Different streamlines, pumps, heights, losses or transient conditions can invalidate that comparison.

Work through the default case

At flow 4 L/s, the 40 cm² inlet speed is 1 m/s. For a 20 cm² outlet, speed is 2 m/s. The increase in kinetic energy per volume is ½ × 1000 × (2² − 1²) = 1500 Pa, or 1.5 kPa.

Outlet absolute pressure is therefore 120 − 1.5 = 118.5 kPa. Relative to the 100 kPa gauge reference, this is 18.5 kPa gauge. The inlet gauge pressure is 20 kPa. Either consistent pressure reference gives the same 1.5 kPa difference.

Capture this case and compare it with an outlet area of 10 cm² at the same 4 L/s. Outlet speed becomes 4 m/s and the pressure drop becomes ½ × 1000 × (16 − 1) = 7500 Pa, or 7.5 kPa. Outlet absolute pressure is 112.5 kPa. The squared-speed term makes the pressure change nonlinear in flow speed.

Check the power ledger independently

A pressure p acting through volume flow Q has pressure-work rate pQ, measured in watts. Kinetic energy transported per second is ½ρQv² under the model’s uniform-speed assumption. A nonuniform real velocity profile requires a corresponding energy-flow correction.

For the default case, inlet pressure work is 120000 × 0.004 = 480 W. Outlet pressure work leaving is 118500 × 0.004 = 474 W. Their difference is 6 W of net pressure work on the fluid between sections.

Kinetic-energy flow is ½ × 1000 × 0.004 × 1² = 2 W at the inlet and 8 W at the outlet. Its increase is also 6 W. The sum of pressure-work and kinetic-energy rates is 482 W at each section. The equal totals check the ideal mechanical account; they are not a measurement of electrical pump consumption.

Changing both pressure references by the same constant shifts the two pressure-work entries equally because Q is the same. Their difference remains unchanged. Keep one reference throughout rather than mixing an absolute inlet reading with a gauge outlet reading.

Interpret negative gauge pressure correctly

Set flow to 8 L/s and outlet area to 10 cm². Speeds are 2 and 8 m/s. The pressure drop is ½ × 1000 × (64 − 4) = 30 kPa. Outlet pressure is 90 kPa absolute, or −10 kPa gauge relative to 100 kPa.

Negative gauge pressure means below the reference pressure, not negative absolute pressure. The modeled absolute pressure remains positive. Applying this result to a real liquid would also require checking its vapor pressure, the downstream boundary and whether the ideal assumptions remain credible. The widget does not model cavitation or select a real pipe design.

At equal areas the speeds match and the ideal pressure drop is zero. A real pipe can still require a pressure drop to overcome viscous losses. At zero flow, both pressure-work rates and kinetic-energy rates vanish, while the equal absolute pressures remain 120 kPa.

Transfer to changed speeds

An ideal horizontal stream of density 800 kg/m³ accelerates from 2 to 5 m/s. Inlet absolute pressure is 150 kPa. Find outlet absolute pressure. If volume flow is 0.003 m³/s, calculate net pressure-work rate and check it against the change in kinetic-energy flow.

Check the Bernoulli transfer

The kinetic-energy increase per volume is ½ × 800 × (25 − 4) = 8400 Pa = 8.4 kPa. Outlet absolute pressure is 141.6 kPa. Net pressure-work rate is 8400 × 0.003 = 25.2 W. Kinetic-energy flow rises from ½ × 800 × 0.003 × 4 = 4.8 W to 30 W, also an increase of 25.2 W. Equal elevation and no losses or shaft work are essential to this result.