---
title: "Tile size (2 cm/day) and spacing (3 cm/day)"
subject: "Irrigation and Drainage Engineering"
chapter: 9
key: "semester_6/irrigation_and_drainage_engineering:lateral-tile-size-and-spacing-3cmday"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lateral-tile-size-and-spacing-3cmday
site: BCT Engineering
---

# Tile size (2 cm/day) and spacing (3 cm/day)

Irrigation and Drainage Engineering · Chapter 9: Water Logging and Drainage

- Asked in: 2078 Chaitra, 6 marks
- Video: [Sub Surface Tile Drainage Size GATE 2016 Q-37](https://www.youtube.com/watch?v=0UlBzmVMrE0) — Agricultural Engineering Online
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lateral-tile-size-and-spacing-3cmday
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-9

## Question

In a sub-surface drainage system, 200 m long laterals were laid out 50 m apart. The laterals have a grade of 0.3%. (i) If the drainage coefficient of the area is 2 cm/day, what size of tiles would you recommend? (ii) If the drainage coefficient is increased to 3 cm/day, what will be the spacing of the laterals? Assume the rugosity coefficient of the tile drain materials as 0.011.

## Short answer

(i) Using Manning's formula, required diameter is 87 mm; adopt 100 mm standard tile (capacity 3.34 L/s). (ii) For 3 cm/day, 100 mm tile capacity allows 48 m spacing (area 9630 m² / 200 m length).

## Answer

#### (i) Size of tiles for 2 cm/day
Each lateral drains 200 m × 50 m = 1 ha. The discharge is

$$
Q = 0.02\times\frac{10000}{86400} = 0.00231\ \text{m}^3/\text{s} = 2.31\ \text{l/s}
$$

For a pipe flowing full ($A=\pi D^2/4$, $R=D/4$, $n$ = 0.011, $S$ = 0.003):

$$
D=\left(\frac{4^{5/3}\,nQ}{\pi\sqrt{S}}\right)^{3/8}=\left(\frac{4^{5/3}\times0.011\times0.00231}{\pi\times\sqrt{0.003}}\right)^{3/8} = 87\ \text{mm}
$$

Adopt the next standard size, **100 mm**. Full-flow capacity of the 100 mm tile:

$$
Q_{full}=\frac{1}{0.011}\times\frac{\pi(0.1)^2}{4}\times\left(\frac{0.1}{4}\right)^{2/3}\times\sqrt{0.003} = 0.00334\ \text{m}^3/\text{s} = 3.34\ \text{l/s}
$$

#### (ii) Spacing at 3 cm/day
The same 100 mm tile at the same grade can carry only 3.34 l/s. The area it can serve at a drainage coefficient of 0.03 m/day is

$$
A = \frac{Q_{full}}{q} = \frac{0.00334}{0.03/86400} = 9630\ \text{m}^2
$$

For a lateral of length 200 m:

$$
\text{Spacing}=\frac{9630}{200} = 48.1\ \text{m}
$$

So the laterals must be placed about **48 m** apart (slightly closer than 50 m). If the 50 m spacing is to be kept, the tile size must be increased: at 3 cm/day the discharge is 3.47 l/s, which needs the next size, 125 mm.

**Answer: (i) 100 mm tiles; (ii) spacing ≈ 48 m for 100 mm tiles.**

## Check yourself

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

1. What discharge must the lateral carry at 2 cm/day?
   A. 2.31 l/s from 1 ha
   B. 23.1 l/s
   C. 0.231 l/s
   D. 3.47 l/s
   Answer: A. Q = 0.02 x 10000/86400 = 2.31 l/s. (Key point: Discharge at 2 cm/day)

2. What tile is adopted for part (i) and why?
   A. 87 mm, the exact computed size
   B. 100 mm, the next standard size above 87 mm; full capacity 3.34 l/s
   C. 50 mm, to keep velocity low
   D. 150 mm because n = 0.011
   Answer: B. Standard sizes are used; 100 mm carries 3.34 l/s which exceeds 2.31 l/s. (Key point: Tile size)

3. How is the new spacing found?
   A. Spacing is unchanged at 50 m
   B. Spacing = 50 x 3/2 = 75 m
   C. Area served = Qfull/q = 0.00334/(0.03/86400) = 9630 m2, divided by 200 m length = 48 m
   D. Spacing = 9630 x 200 = 1.9 million m
   Answer: C. The same tile can serve a limited area; dividing by the lateral length gives spacing. (Key point: Spacing at 3 cm/day)

4. How could 50 m spacing be kept at 3 cm/day?
   A. Use a smaller tile
   B. Reduce the grade to zero
   C. Nothing is needed; 100 mm is enough
   D. Use a larger tile, 125 mm, since the discharge becomes 3.47 l/s
   Answer: D. 3.47 l/s exceeds the 3.34 l/s capacity of 100 mm, so the next size 125 mm is needed. (Key point: Alternative)

## Close variants set in other exams

- [Tile size for laterals, 2 cm/day](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alateral-tile-size-2cmday-grade03) — Irrigation and Drainage Engineering, chapter 9, asked 2×

## Related in Irrigation and Drainage Engineering

- [Tile drain spacing with drainage coefficient 0.012](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Atile-drain-spacing-drainage-coeff-0012) — Irrigation and Drainage Engineering, chapter 9
- [Tile drain spacing: rainfall 80 cm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Atile-drain-rainfall-80cm-impervious9) — Irrigation and Drainage Engineering, chapter 9, asked 2×