Momentum Control Volumes and Turbomachinery

Energy answers head; momentum answers force

Many fluid problems contain both energy and momentum ideas, but the question wording tells you which one leads. If the target is pressure, head, pump power, or pipe loss, start with the mechanical energy equation. If the target is a support reaction, thrust, force on a bend, jet impact, or nozzle holding force, start with a control-volume momentum balance.

For steady one-inlet, one-outlet flow, the linear momentum equation says that the sum of external forces on the control volume equals mass flow rate times the change in velocity vector. The word vector matters. A 90-degree elbow may have the same speed at inlet and outlet, but the momentum direction changes, so force is still required.

Set up components

Choose axes before writing numbers. Draw the control volume, show inlet and outlet velocities, show pressure forces acting inward on the control surface, include weight only when it is material, and include support forces as unknown reactions. Then write separate x and y equations.

Mass flow rate is m_dot = rho Q = rho A V for incompressible flow. Do not multiply velocity change by Q alone unless the equation has been written per unit density or the problem has already converted terms. Momentum force units must become newtons or pounds force, not m^4/s^2.

Pressure and gauge choices

Momentum equations often use gauge pressure because atmospheric pressure cancels around a control volume exposed to atmosphere. That does not mean pressure is always zero. A pressurized pipe elbow has pressure forces at inlet and outlet. A free jet discharging to surrounding air has zero gauge pressure at the free surface. The FE trap is applying one of these facts to the wrong boundary.

If the problem asks for the force exerted by fluid on the pipe, remember Newton's third law. The force of pipe on fluid comes from the control-volume equation. The force of fluid on pipe is equal and opposite, plus any convention changes for support reactions.

Turbomachinery basics

Turbomachinery transfers energy between a rotating machine and a fluid. Pumps and fans add energy to the fluid. Turbines extract energy from the fluid. FE Mechanical usually stays at the head, power, and similarity level rather than detailed blade velocity triangles, although the handbook includes relationships for idealized analysis.

For pumps, the useful fluid power is rho g Q H; shaft input is larger after efficiency is considered. For hydraulic turbines, useful shaft output is less than the fluid power drop. Fans often use pressure rise times volume flow rate, with attention to consistent units.

Affinity laws are high-yield for geometrically similar pumps or fans operating with the same fluid. For constant impeller diameter, flow rate varies directly with rotational speed, head varies with speed squared, and power varies with speed cubed. For constant speed with diameter changes, similar proportional patterns apply with diameter powers. These are approximate similarity rules, so use them only when the problem states same pump family, same geometry, or similar operation.

Exam workflow

For force questions, write the momentum equation before doing arithmetic. For machine questions, decide whether the machine consumes or produces shaft power, then apply efficiency in the correct direction. If an answer choice differs by a sign, a factor of density, or a squared speed ratio, those are not random; they are the traps the problem is built to expose.