Pump operating point

Where does this pump actually run against this loop?

h = h0 \cdot \left( 1 - \left( \frac{Q}{Q0} \right)^2 \right) = \frac{Δp}{ρ \cdot g}

Method

1. Convert the rating to SI: 500 gal/h is 5.26 × 10⁻⁴ m³/s, one foot is 0.3048 m.
2. Draw the pump curve as a parabola through the rated point: shutoff head h₀ = 2·h_r at zero flow, free delivery Q₀ = √2·Q_r at zero head, h(Q) = h₀·(1 − (Q ÷ Q₀)²). It passes through (Q_r, h_r) by construction.
3. Draw the loop curve: at each flow the velocity is Q over the bore's area, the pressure drop follows from Darcy–Weisbach as in the loop pressure-drop calculation, and the head demanded is Δp ÷ (ρ·g).
4. The pump curve falls with Q and the loop curve rises with it, so they cross exactly once between zero and Q₀. Find the crossing by bisection.
5. Report the flow, the velocity it gives in the bore, and the head at the crossing, and compare the velocity with the 50 in/s the holding specifies.

Assumptions

• The curve shape is assumed from one rated point. A real centrifugal pump's shutoff head is typically 1.5 to 2.5 times its best-efficiency head and its free delivery 1.3 to 1.6 times its best-efficiency flow; the parabola through the rated point is the middle of that range. A curve read from the pump's datasheet is better, and replaces this page when there is one.
• The rated point is the pump's point on water. A denser medium raises the pressure the pump produces in proportion, which is why the curve is kept in head rather than pressure; a much more viscous one — above about 50 mPa·s — pulls the whole curve down, which this does not model.
• The loop's loss is as the pressure-drop calculation gives it: a smooth tube, a Newtonian medium, handbook minor losses.
• Nothing here is Meyer's except the pump rating and the 50 in/s the EPG #1 holding gives. The curve shape and the loop model are ordinary pump engineering.