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Evaluating losses in intakes

Short answer

Total intake loss is the sum of trash rack, entrance, gate slot, transition, bend and friction losses. Each is a coefficient times velocity head, except friction. Typical RoR intake total loss is about 0.1-0.3 m, subtracted from gross head.

How do you evaluate losses in intakes?

The head loss in an intake is the sum of the following parts. Each is computed as a coefficient times a velocity head, except friction.

  1. Trash rack loss (Kirschmer):
ht=β(sb)4/3Va22gsin⁡θh_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V_a^2}{2g}\sin\theta

with β\beta = bar shape factor (2.42 rectangular, 1.79 circular), ss = bar thickness, bb = clear spacing, VaV_a = approach velocity and θ\theta = rack inclination to the horizontal. For a partly clogged rack, increase the loss by using the blocked area (net velocity).

  1. Entrance loss: he=keV22gh_e=k_e\dfrac{V^2}{2g}, where ke=0.5k_e=0.5 for a sharp-edged entrance, 0.2 for a rounded one, and 0.05-0.1 for a well-shaped bell-mouth.

  2. Gate slot loss: hg=kgV22gh_g=k_g\dfrac{V^2}{2g}, with kg≈0.1k_g\approx0.1 (nil for a fully open gate with flush slots).

  3. Transition loss (contraction or expansion): htr=k∣V22−V12∣2gh_{tr}=k\dfrac{\left|V_2^2-V_1^2\right|}{2g}, with k≈0.1k\approx0.1 for a gradual contraction and 0.2-0.3 for a gradual expansion.

  4. Bend loss: hb=kbV22gh_b=k_b\dfrac{V^2}{2g}, where kb≈0.1−0.25k_b\approx0.1-0.25 depends on the bend radius and angle.

  5. Friction loss in the intake conduit (Manning or Darcy):

hf=n2V2LR4/3orhf=fLDV22gh_f=\frac{n^2V^2L}{R^{4/3}}\quad\text{or}\quad h_f=f\frac{L}{D}\frac{V^2}{2g}

Total loss

hL=ht+he+hg+htr+hb+hfh_L=h_t+h_e+h_g+h_{tr}+h_b+h_f

The loss is kept small by a low velocity at the rack (about 0.8-1.0 m/s), a bell-mouthed entrance, smooth gradual transitions and few gate slots. For a typical RoR intake the total is of the order of 0.1-0.3 m; it is subtracted from the gross head when the net head of the plant is worked out.

✓ Answer checked by Claude Opus 5.5 (high effort)

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