03 / CHAPTER 12
Energy audit studio
Build an honest power and loss budget for a pump-driven water feature.
ANCHOR EQUATION
Pump efficiency
In symbols: η = Pout/Pin
Efficiency compares useful hydraulic output with supplied power.
A pump support feels the water’s reaction
The impeller and pipe exert forces that change the water’s momentum; the water pushes back on the hardware. Estimating the load requires a control volume around the water region, not just the pump’s electrical rating.
Separate pressure, momentum, and weight
For steady water in a control volume, R + F_pressure + W = ṁ(v_out − v_in), where R is the wall/support force on the fluid. Solve for R, then reverse it to obtain the fluid’s force on the support. Each term has units of newtons and a clear physical origin.
A 3 kg/s stream turns from (4,0) to (0,4) m/s. Momentum change is (−12,12) N. If pressure and weight terms cancel, the pipe pushes the fluid with that vector, so water loads the pipe with (12,−12) N, magnitude 17 N.
Hydraulic power is not force
A large flow at modest pressure can have the same power as a small flow at high pressure, but the momentum reaction depends on mass flow and velocity change. Do not infer structural load from watts alone.
Work it: Choose a pipe bend geometry, specify Q and the inlet/outlet velocity vectors, and calculate the ideal support reaction. List the pressure and gravity data still needed for a complete model.