Pressure drop round the loop

What pressure does it take to push the medium round the loop at this speed, and what hydraulic power is that?

Δp = \left( f \cdot \frac{L}{D} + K \right) \cdot \frac{ρ \cdot \left( v \right)^2}{2}

Method

1. Find the Reynolds number, ρ·v·D ÷ η.
2. Choose the friction factor by regime: 64 ÷ Re below 2300 (Poiseuille — the loss is proportional to velocity); Blasius's 0.316·Re^−¼ above 4000 (the loss goes nearly as the velocity squared), or Haaland's smooth-pipe form above Re = 10⁵; a straight line between the two across the transition band.
3. The dynamic pressure is ρ·v² ÷ 2. The straight run costs f·L ÷ D of it and the bends, fittings and pump housing cost K of it; add the two and multiply.
4. The flow rate is the velocity times the bore's area, π·D² ÷ 4. Pressure times flow rate is power: the hydraulic work the pump does on the fluid each second, before the pump's own efficiency — a small centrifugal pump turns perhaps a fifth of its shaft power into this.

Assumptions

• A smooth tube: glass, acrylic or drawn metal. Roughness raises the turbulent friction factor and Haaland's full form takes it; a rubber hose or a cast fitting can be half again as lossy.
• A Newtonian medium. A slurry above about ten per cent solids, or any magnetised slurry inside the coil, has a yield stress and a shear-thinning viscosity that this cannot follow; the pressure drop is then a floor.
• The minor-loss coefficients are handbook values for water in ordinary fittings. The loss in a coil-wound section, where the bore may step, is the reader's to estimate.
• Nothing here is Meyer's except the velocity the EPG #1 holding specifies and the pump rating it gives. Darcy–Weisbach is nineteenth-century hydraulics, and the friction factors are Poiseuille's and Blasius's.