Mass, momentum and energy balances for flows

Draw the boundary before simplifying

A control volume is a chosen region whose boundary fluid may cross. At a fixed boundary, mass accumulation per second equals total mass inflow minus total mass outflow. A filling tank stores mass; its inlet and outlet rates need not match. Steady flow means properties at each fixed location do not change with time, so storage terms vanish even though fluid particles move.

For a uniform-density port, volume flow Q = AU, where A is area and U is cross-section mean normal velocity. Mass flow ṁ = ρQ, in kg/s. If velocity or density varies across a port, add the local contributions over its area; a centreline reading cannot automatically replace the mean. Equal mass flows imply equal volume flows only when the corresponding densities are equal.

Momentum connects changing motion to force

For steady, uniform inlet and outlet profiles in a fixed inertial control volume, the net external force vector on the fluid equals outgoing momentum flow minus incoming momentum flow. For one inlet and one outlet, ΣF = ṁ(v₂ − v₁), with velocities treated as vectors. Unsteady cases also require the rate of change of momentum stored inside.

The force sum includes pressure forces at the ports, wall forces and body forces such as gravity. A bend can change momentum direction even at unchanged speed. The force the fluid exerts on a wall is opposite the wall force on the fluid; it is not necessarily opposite the entire external-force sum.

Energy includes work, heat and transported states

The general flow energy balance also includes storage, heat transfer, shaft work and transport of enthalpy, kinetic energy and gravitational energy. For steady, incompressible flow with negligible viscous dissipation and no machine work, Bernoulli’s equation gives p/(ρg) + U²/(2g) + z constant along a streamline. These terms are pressure, velocity and elevation heads, in metres; U here represents the uniform-flow approximation.

A system balance can add pump head and subtract turbine head and irreversible head loss. Real nonuniform profiles may require kinetic-energy correction factors. Faster flow does not universally mean lower pressure: elevation, machines, losses, unsteadiness and the selected streamline matter. A thermodynamic energy balance remains necessary when temperature or density changes are significant.

Constructed ideal contraction checked three ways

Take steady horizontal flow of density 1,000 kg/m³ through a hypothetical smooth contraction. Assume uniform ports, negligible loss and no shaft work. Q = 0.002 m³/s through inlet area 0.001 m² gives U₁ = 2 m/s. Outlet area 0.0005 m² gives U₂ = 4 m/s. Both mass flows equal 2 kg/s, so there is no mass accumulation.

Bernoulli gives p₁ − p₂ = ½ρ(U₂² − U₁²) = ½ × 1,000 × (16 − 4) = 6,000 Pa. The corresponding kinetic-energy increase is ṁ(U₂² − U₁²)/2 = 12 W, matching pressure-work input Q(p₁ − p₂) = 12 W. Momentum requires net external force 2(4 − 2) = 4 N downstream. Finding the wall reaction additionally requires the separate port pressures and pressure-force areas.

Try the reasoning

1. A tank receives 3 kg/s and discharges 2 kg/s. Is mass conserved?

   Show the answer and reasoning

   Yes. Its stored mass increases at 1 kg/s. Setting inflow equal to outflow would wrongly impose steady storage.

2. If a flow turns 90 degrees at constant speed, is the net momentum force zero?

   Show the answer and reasoning

   No. Velocity is a vector, so changing direction changes momentum flow. The required force has components determined by both inlet and outlet vectors.

3. Can the calculated 4 N be assigned directly to the contraction wall?

   Show the answer and reasoning

   No. It is the sum of all external forces on the chosen fluid. Pressure forces and any other contributions must be separated before obtaining the wall force.

What to carry forward

Use one clear boundary, retain storage when needed, and keep force, mass and energy balances distinct.