---
title: "Drainage rate: 300 mm in 3 days, growth factor 1.5"
subject: "Irrigation and Drainage Engineering"
chapter: 9
key: "semester_6/irrigation_and_drainage_engineering:drainage-rate-3-day-300mm-growth"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/drainage-rate-3-day-300mm-growth
site: BCT Engineering
---

# Drainage rate: 300 mm in 3 days, growth factor 1.5

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

- Asked in: 2074 Bhadra, 6 marks
- 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/drainage-rate-3-day-300mm-growth
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-9

## Question

Determine the drainage rate required to meet the following condition. Maximum yearly precipitation for three consecutive days = 300 mm. The design rainfall is to be taken as 10 year return periods. Initial water level in field = 40 mm. Maximum water level is 300 mm, which may persist for up to one day and depth in excess of 200 mm may persist for up to 3 days. Take growth factor for 10 year return period as 1.5. Assume other suitable data if necessary.

## Short answer

WithR=450 mm,h0=40 mm, limits max300 mm1 day,>200 mm3 days. Trial givesD=60.6 mm/day (≈7.0 l/s/ha). Peak308 mm, time above200 mm=3.00 days, time above300 mm=0.23 day.

## Answer

Rice tolerates standing water only up to limits of depth and duration, so the drainage rate is chosen such that the field water level, after the design storm, stays within these limits. The 3-day rainfall of the 10-year return period is the mean annual maximum multiplied by the growth factor: $R = 1.5\times300 = 450$ mm.

#### Given data and assumptions
- Initial depth $h_0$ = 40 mm; design 3-day rainfall $R$ = 450 mm, falling uniformly, so the rain rate $r = R/3$ = 150.0 mm/day
- Maximum depth 300 mm may persist for 1 day; depth above 200 mm may persist for 3 days
- ET and deep percolation neglected (safe side); drainage $D$ (mm/day) starts with the rain and is constant

#### Water level history
During the storm the level rises at $(r-D)$ per day, and after it the level falls at $D$ per day:

$$
\begin{aligned} h_{peak} &= h_0 + R - 3D \\ t_1 &= \frac{h_e-h_0}{r-D} \quad (\text{time to reach } h_e) \\ t_2 &= \frac{h_{peak}-h_e}{D} \quad (\text{time to fall back to } h_e) \\ T_e &= (3-t_1)+t_2 \le 3 \end{aligned}
$$

The duration condition above 200 mm governs. Trial values:

| D (mm/day) | Peak level (mm) | Time above limit (days) |
|---|---|---|
| 50 | 340 | 4.20 |
| 55 | 325 | 3.59 |
| 60.6 | 308 | 3.00 |
| 65 | 295 | 2.58 |
| 70 | 280 | 2.14 |

Solving $T_e$ = 3 days gives $D$ = 60.6 mm/day. Then:
- $h_{peak}$ = 40 + 450 − 3 × 60.6 = 308 mm (slightly above 300 mm; the time above 300 mm is 0.23 day, less than 1 day, so this limit is satisfied)
- $t_1$ = 1.79 days, $t_2$ = 1.79 days, $T_e$ = (3 − 1.79) + 1.79 = 3.00 days

Conversion: 1 mm/day = 10 m³/ha/day = 0.1157 l/s/ha, so

$$
D = 60.6\times\frac{10}{86.4} = 7.01\ \text{l/s/ha}
$$


**Answer: drainage rate ≈ 60.6 mm/day ≈ 7.0 l/s/ha** (adopt 7.0 l/s/ha).

## Check yourself

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

1. What 3-day design rainfall is used and what is the rain rate?
   A. 450 mm (1.5 x 300) at 150 mm/day
   B. 300 mm at 100 mm/day
   C. 200 mm at 66.7 mm/day
   D. 450 mm at 450 mm/day
   Answer: A. The 10-year rainfall is 1.5 x 300 = 450 mm, falling uniformly over 3 days = 150 mm/day. (Key point: Design rainfall)

2. During the storm, how does the field level change?
   A. It rises at r per day throughout
   B. It rises at (r - D) per day, then falls at D per day after the rain stops
   C. It stays constant while drainage works
   D. It falls at (r - D) during the rain
   Answer: B. Drainage acts during rain so the net rise is r - D; after rain, drainage alone lowers the level at D. (Key point: Level history)

3. Which D makes the time above 200 mm exactly 3 days?
   A. 50 mm/day
   B. 70 mm/day
   C. 60.6 mm/day
   D. 45 mm/day
   Answer: C. The trial table shows 50 mm/day gives 4.20 days, 70 gives 2.14 days; 60.6 gives 3.00 days. (Key point: Result)

4. The peak at D = 60.6 is 308 mm, above 300 mm. Is the design acceptable?
   A. No, any exceedance of 300 mm is not permitted
   B. No, D must be increased until peak is under 300 mm
   C. Yes, because the 300 mm limit is ignored
   D. Yes, because the time above 300 mm is only 0.23 day, less than the 1 day allowed
   Answer: D. The 300 mm limit is a duration limit (up to 1 day), and 0.23 day satisfies it. (Key point: 300 mm limit)

## Close variants set in other exams

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