Mechanics

A fountain pump does not create height directly. It transfers shaft work to water, and that energy appears as pressure, speed, elevation, heat, and sound. The useful mechanical question is therefore: where does each joule go?

From force to hydraulic power

For a small displacement, work is dW = F · dr. Divide by time and use dr/dt = v:

P = dW/dt = F · v

For a steady stream, the mass lifted each second is ṁ = ρQ, where ρ is water density in kg/m³ and Q is volume flow rate in m³/s. Lifting that mass through height H increases gravitational energy at rate

P_lift = ṁgH = ρgQH.

The unit check is kg/m³ · m/s² · m³/s · m = kg·m²/s³ = W.

A number worth checking

If Q = 0.004 m³/s, H = 3 m, and ρ = 998 kg/m³, then

P_lift = 998 · 9.81 · 0.004 · 3 = 117.4 W.

That is only the elevation term. If pipe and nozzle losses consume another 2 m of head, the pump must supply ρgQ(3 + 2) = 195.9 W to the water. At 55% total efficiency, electrical input is 195.9/0.55 = 356 W.

Why the support feels a force

A nozzle changes water momentum. For steady flow, the control-volume balance is

ΣF_ext = ṁ(v_out − v_in).

The pressure forces at the cut sections, the water’s weight, and the wall reaction all belong in ΣF_ext. If a 2 kg/s stream turns from (3, 0) to (0, 3) m/s, its momentum change is (-6, +6) N; the water pushes the bend back with (6, -6) N before pressure corrections.

Test the model

Measure Q, elevation change, pressure before the nozzle, pipe diameter, and electrical real power at one operating point. Calculate hydraulic power two ways—ρgQH for elevation alone and ΔpQ for a measured pressure rise—and explain the difference as velocity and loss terms. If the inferred efficiency exceeds one, the boundary or units are wrong.