---
title: "Evaluating losses in intakes"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:intake-losses-evaluation"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/intake-losses-evaluation
site: BCT Engineering
---

# Evaluating losses in intakes

Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants

- Asked in: 2080 Baishakh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/intake-losses-evaluation
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

How do you evaluate 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.

## Answer

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):
$$
h_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V_a^2}{2g}\sin\theta
$$
with $\beta$ = bar shape factor (2.42 rectangular, 1.79 circular), $s$ = bar thickness, $b$ = clear spacing, $V_a$ = approach velocity and $\theta$ = rack inclination to the horizontal. For a partly clogged rack, increase the loss by using the blocked area (net velocity).

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

3. **Gate slot loss**: $h_g=k_g\dfrac{V^2}{2g}$, with $k_g\approx0.1$ (nil for a fully open gate with flush slots).

4. **Transition loss** (contraction or expansion): $h_{tr}=k\dfrac{\left|V_2^2-V_1^2\right|}{2g}$, with $k\approx0.1$ for a gradual contraction and 0.2-0.3 for a gradual expansion.

5. **Bend loss**: $h_b=k_b\dfrac{V^2}{2g}$, where $k_b\approx0.1-0.25$ depends on the bend radius and angle.

6. **Friction loss** in the intake conduit (Manning or Darcy):
$$
h_f=\frac{n^2V^2L}{R^{4/3}}\quad\text{or}\quad h_f=f\frac{L}{D}\frac{V^2}{2g}
$$

**Total loss**
$$
h_L=h_t+h_e+h_g+h_{tr}+h_b+h_f
$$

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which list correctly gives the intake loss components?
   A. Rack and friction only
   B. Entrance and exit only
   C. Bend and surge
   D. Rack, entrance, gate slot, transition, bend and friction
   Answer: D. Total loss is the sum of these parts. (Key point: Loss components)

2. Which entrance gives the lowest ke?
   A. Well-shaped bell-mouth (0.05-0.1)
   B. Sharp-edged (0.5)
   C. Rounded (0.2)
   D. All the same
   Answer: A. A bell-mouth minimises entry loss. (Key point: Entrance coefficient)

3. How are most losses computed?
   A. As coefficient times V
   B. As a coefficient times a velocity head V^2/2g
   C. Using the Froude number
   D. As a fixed 1 m
   Answer: B. Except friction, each loss is k V^2/2g. (Key point: Velocity head)

4. What reduces the rack loss most directly?
   A. High velocity
   B. More gate slots
   C. Low rack velocity (0.8-1.0 m/s) since loss varies with V^2
   D. Sharp entrance
   Answer: C. Loss varies with the square of velocity. (Key point: Reduce losses)

## Related in Hydropower Engineering

- [Common RoR intakes in Nepal; site selection](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-common-intakes-nepal-location) — Hydropower Engineering, chapter 5, asked 2×

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