Engineering & Technology · Fluid Mechanics, Flow Systems & Machines

Pumps, turbines, fans and system matching

A pump label cannot determine the flow through every connected system. The same machine can produce different flow rates when pipe resistance or elevation changes. Its behaviour emerges from an interaction between the machine, the fluid and the surrounding network.

Review force vectors, energy and power, control volumes and steady-flow thermodynamics, SI units and measurement uncertainty. Be comfortable with ratios, areas, square roots and reading graphs. Symbols are defined where used: ρ is density, μ dynamic viscosity, Q volume flow and ṁ mass flow. For the constructed liquid cases, use g = 9.81 m/s² where stated. Prior physics study of fluids is helpful but not required.

Back to the chapter

After reading, you can

Relate fluid-machine power to head and efficiency, and find a fictional operating point from a matched system curve.

Machines exchange mechanical energy with a flow

A pump receives shaft work and raises a liquid’s mechanical energy. A hydraulic turbine extracts part of the fluid’s available mechanical energy as shaft work. Rotating blades exchange momentum with fluid, producing torque; positive-displacement machines move enclosed portions of fluid instead. These mechanisms have different characteristics and limits.

For an incompressible flow, useful hydraulic power is Ph = ρgQH, where H is the relevant total-head rise across a pump or total-head drop through a turbine. The units are kg/m³ × m/s² × m³/s × m = W. Pump efficiency ηp = Ph/Pshaft; turbine efficiency ηt = Pshaft/Ph. A motor’s electrical losses are separate from pump shaft-to-fluid efficiency. State which boundaries a quoted efficiency covers.

Match the whole system

A pump characteristic relates developed head to flow at stated speed, geometry and fluid conditions. A system characteristic gives the total head needed at each flow, including elevation or boundary-pressure differences and flow-dependent losses. Where the two curves intersect, the pump’s head matches the system’s requirement. A consistent steady operating point must also satisfy the rest of the flow model.

The intersection need not coincide with maximum efficiency, and real curves may have restricted or unstable regions. A system’s approximately quadratic loss term assumes nearly fixed loss coefficients; its static head need not be zero. Changing the system can move the intersection even when the machine itself is unchanged. Model curves and measured curves should be distinguished, including the conditions and uncertainty of measured data.

Fans and cavitation add further limits

A fan transfers work to a gas. At a small pressure rise with approximately constant density, fluid power can be estimated from Q times the total-pressure rise. Static-pressure and total-pressure efficiency conventions give different numbers. Larger density or temperature changes require a compressible energy treatment rather than simply borrowing a liquid head calculation.

In liquid machines, local absolute pressure can approach the liquid’s vapor pressure at its temperature. Vapor cavities may form and subsequently collapse as pressure rises, changing performance and potentially damaging surfaces. A satisfactory head-flow intersection alone establishes neither cavitation margin nor an acceptable operating range. Those require additional machine and system evidence; the learning curves here are not equipment-setting instructions.

Horizontal axis: volume flow Q, 0 to 2.2 L/s. Vertical axis: head H, 0 to 12 m. Solid blue pump head decreases from 10 m; dashed orange system head rises from a static 2 m. Their marked intersection is approximately 1.633 L/s and 4.667 m. A real operating range also needs efficiency, cavitation and stability evidence; none is certified by this fictional curve.

Constructed pump-system intersection and power

Suppose fictional pump head is H = 10 − 2Q² and system head is H = 2 + Q², with H in metres and Q in litres per second. Coefficients carry the units required by that convention. Set the expressions equal: 8 = 3Q². Thus Q = √(8/3) = 1.633 L/s and H = 4.667 m. A negative algebraic root lies outside the positive-flow model being considered.

Using ρ = 1,000 kg/m³, g = 9.81 m/s² and Q = 0.001633 m³/s, hydraulic power is approximately 74.8 W. At an assumed pump efficiency of 0.65 at that point, required shaft power is 74.8/0.65 ≈ 115 W. Electrical input would be higher if the drive has losses. The assumed efficiency does not locate a best-efficiency point, and the two fictional curves do not certify a real pump.

Try the reasoning

If the system needs greater head at every positive flow, what happens to this intersection?
Show the answer and reasoning
For these descending pump and rising system curves, it moves toward lower flow. The exact change requires the new system relation; real characteristic shapes can be more complicated.
Can turbine shaft output equal ρgQH/η when η is below one?
Show the answer and reasoning
No. With η defined as shaft output divided by available hydraulic power, output is ηρgQH. Dividing would predict more output than the stated available input.
Why is zero gauge pressure insufficient for assessing cavitation?
Show the answer and reasoning
Vapor formation depends on local absolute pressure and temperature-dependent vapor pressure. The gauge reference atmosphere must be included, and pressure can vary locally inside the machine.

What to carry forward

Match the machine and system curves, then assess power, efficiency definitions and operating limits separately.

References

Teaching resources and research supporting this lesson’s explanations and examples.