---
title: "Rainfall, lag time and time of concentration from hydrograph"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:hydrograph-figure-lag-tc"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/hydrograph-figure-lag-tc
site: BCT Engineering
---

# Rainfall, lag time and time of concentration from hydrograph

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2080 Bhadra, 6 marks
- Video: [What is lag time in Hydrology?](https://www.youtube.com/watch?v=1f8czU1HU6E) — Let's GROW!
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/hydrograph-figure-lag-tc
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

From the given hydrograph below, answer the questions:

[Figure: direct runoff hydrograph, Q (cumecs, axis 0 to 80) versus time (hr, axis 0 to 12). The curve rises from 0 at t = 0, peaks at about 70 cumecs at t = 5 hr, falls to about 10 cumecs near t = 8 hr and has a long tail reaching about 0 near t = 12 hr. An inset hyetograph (rainfall in cm/hr versus time in hr) shows three one-hour blocks: roughly 6 cm/hr in the 1st hour, 12 cm/hr in the 2nd hour and 6 cm/hr in the 3rd hour. Values are read approximately from the plot.]

(i) What is the total amount of rainfall and average rainfall intensity?
(ii) What is basin lag time and time of concentration?
(iii) What is the total time base of flow?

## Short answer

Total rainfall is 24 cm with average intensity 8 cm/h. Basn lag (centroid of rain to peak) is 3.5 h, time of concentration (end of rain to end of direct runoff) is 9 h, and time base of flow is 12 h.

## Answer

Reading the hyetograph: rainfall is 6 cm/h in hour 1, 12 cm/h in hour 2 and 6 cm/h in hour 3 (rain from t = 0 to 3 h). The hydrograph peaks at t = 5 h and ends at about t = 12 h.

#### (i) Total rainfall and average intensity
$$
P = 6\times 1 + 12\times 1 + 6\times 1 = 24 \text{ cm}
$$

$$
i_{avg} = \frac{24}{3} = 8 \text{ cm/h}
$$

**Answer: total rainfall = 24 cm, average intensity = 8 cm/h.**

#### (ii) Basin lag and time of concentration
**Basin lag** $t_L$ = time from the centre of mass of the rainfall to the peak of the hydrograph. The rain is symmetric (6, 12, 6), so its centroid is at 1.5 h.

$$
t_L = 5 - 1.5 = 3.5 \text{ h}
$$

**Time of concentration** $t_c$ = time from the end of the rainfall to the end of direct runoff (time for the farthest water to reach the outlet):

$$
t_c = 12 - 3 = 9 \text{ h}
$$

**Answer: basin lag = 3.5 h; time of concentration ≈ 9 h.** (If $t_c$ is measured only to the point of inflection on the recession limb, at about t = 8 h, then $t_c \approx 8 - 3 = 5$ h.)

#### (iii) Time base of flow
The direct runoff starts at t = 0 and ends at about t = 12 h.

**Answer: time base $t_b$ ≈ 12 h.**

## Check yourself

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

1. Hyetograph 6, 12, 6 cm/h for three 1-h blocks. Total and average intensity?
   A. 24 cm and 12 cm/h
   B. 8 cm and 8 cm/h
   C. 12 cm and 4 cm/h
   D. 24 cm and 8 cm/h
   Answer: D. P = 6+12+6 = 24 cm over 3 h. (Key point: Total rainfall)

2. How is basin lag defined and found?
   A. Centroid of rain (1.5 h) to hydrograph peak (5 h) = 3.5 h
   B. Start of rain to peak = 5 h
   C. End of rain to peak = 2 h
   D. Peak to end of runoff = 7 h
   Answer: A. Lag is measured from the centre of mass of rainfall to the peak. (Key point: Basin lag)

3. How is tc found here?
   A. From start of rain to peak = 5 h
   B. From end of rain (3 h) to end of runoff (12 h) = 9 h
   C. 12 h, the time base
   D. 3 h, the rain duration
   Answer: B. tc = tb - rain duration in this convention. (Key point: Time of concentration)

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