---
title: "Length of bottom intake rack"
subject: "Hydropower Engineering"
chapter: 5
key: "semester_7/hydropower_engineering:intake-bottom-rack-length-8cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/intake-bottom-rack-length-8cumec
site: BCT Engineering
---

# Length of bottom intake rack

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

- Asked in: 2075 Chaitra, 5 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-bottom-rack-length-8cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5

## Question

Determine the necessary length of a rack of a bottom intake with the intercepted flow of 8 m³/s and width of the rack of 10 m. Inclination of the rack is 30°. Thickness, spacing and contraction factor of the bars are 10 mm, 15 mm and 0.82 respectively.

## Short answer

Use Frank's formula with unit discharge 0.8 m2/s and critical depth 0.403 m. Rack coefficient lambda is 2.028, giving rack length 1.59 m; add 25% for clogging and adopt 2.0 m for the 10 m wide rack.

## Answer

**Given**: intercepted flow $Q=8$ m³/s; rack width $B=10$ m; rack inclination $\beta=30^\circ$; bar thickness $s=10$ mm; clear spacing $a=15$ mm; contraction factor $\mu=0.82$.

The rack length is found from **Frank's formula** for a bottom (Tyrolean) intake, in which the depth of flow falls along the rack as the water is drawn through the bars:
$$
L=\frac{2.561\,q}{\lambda\sqrt{h}},\qquad \lambda=\psi\,\mu\sqrt{2g\cos\beta},\qquad \psi=\frac{a}{a+s}
$$
where $q$ is the discharge per metre width and $h$ the depth of flow at the upstream end of the rack. The flow is assumed to arrive at the rack at critical depth (over the weir crest), so $h=h_c$.

#### Step 1: Unit discharge and approach depth
$$
q=\frac{Q}{B}=\frac{8}{10}=0.8\ \text{m}^2/\text{s},\qquad h=h_c=\left(\frac{q^2}{g}\right)^{1/3}=\left(\frac{0.8^2}{9.81}\right)^{1/3}=0.403\ \text{m}
$$

#### Step 2: Opening ratio
$$
\psi=\frac{a}{a+s}=\frac{15}{15+10}=0.60
$$

#### Step 3: Rack coefficient
$$
\lambda=\psi\mu\sqrt{2g\cos\beta}=0.60\times0.82\times\sqrt{2\times9.81\times\cos30^\circ}=0.492\times4.122=2.028\ \text{m}^{1/2}/\text{s}
$$

#### Step 4: Length of the rack
$$
L=\frac{2.561\times0.8}{2.028\times\sqrt{0.403}}=\frac{2.049}{1.287}=1.59\ \text{m}
$$

Allow about 25% extra for partial clogging by gravel and leaves: $L=1.25\times1.59=1.99$ m. Adopt about **2.0 m**.

**Answer**: required rack length $\approx1.6$ m; adopt $2.0$ m (rack 10 m wide, inclined at $30^\circ$).

## Check yourself

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

1. What assumption is used to get the specific energy at the rack (q = 0.8 m^2/s)?
   A. Flow arrives at critical depth h_c = (q^2/g)^(1/3) = 0.403 m and E = 1.5 h_c = 0.604 m, nearly constant along the rack
   B. Flow arrives at normal depth and E = h_c = 0.403 m
   C. Flow arrives at critical depth and E = 2 h_c = 0.806 m
   D. Energy drops to zero along the rack
   Answer: A. For critical flow E = 1.5 h_c, and over a short rack E is taken as constant (Noseda/Mostkow approximation). (Key point: Energy assumption)

2. With bar thickness 10 mm and clear spacing 15 mm, what is the opening ratio?
   A. epsilon = s/(b + s) = 10/25 = 0.40
   B. epsilon = b/(b + s) = 15/25 = 0.60
   C. epsilon = b/s = 1.5
   D. epsilon = (b - s)/b = 0.33
   Answer: B. The open fraction of the rack is the clear spacing divided by one bar-plus-gap pitch. (Key point: Opening ratio)

3. Which expression gives the flow per unit plan area through the inclined rack?
   A. q_b = c x epsilon x sin(alpha) x sqrt(2gE) = 0.85 m/s
   B. q_b = epsilon x sqrt(2gE) without c and cos(alpha) = 2.1 m/s
   C. q_b = c x epsilon x cos(alpha) x sqrt(2gE) = 1.467 m/s
   D. q_b = c x sqrt(2gE) / epsilon = 4.0 m/s
   Answer: C. Flow through the rack is reduced by the contraction factor, the opening ratio and cos(alpha) for inclination, times the velocity from the energy head. (Key point: Discharge through the rack per unit area)

4. How is the rack length found, and why is it increased?
   A. L = q_b/q = 1.8 m, increased 25% for flood
   B. L = q/q_b = 0.55 m, with no increase needed
   C. L = Q/q_b = 5.5 m, increased 25% for temperature
   D. L = q/q_b = 0.8/1.467 = 0.55 m, increased by 25% (to about 0.7 m) for clogging by gravel and leaves
   Answer: D. All unit flow must pass through the rack, so L = q/q_b; a clogging allowance keeps the intake working when part of the rack is blocked. (Key point: Length and clogging allowance)

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