---
title: "Number of additional rain gauges required"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:optimum-raingauges-7stn"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/optimum-raingauges-7stn
site: BCT Engineering
---

# Number of additional rain gauges required

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2078 Bhadra, 5 marks
- Video: [WRE Module1 [PART05]- Optimum Number of Rain Gauges - Rain Gauge Network - Equations & Problem](https://www.youtube.com/watch?v=LQJD3hmmbU0) — AM CIVIL ACADEMY
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/optimum-raingauges-7stn
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

Mean and standard deviation of annual rainfall estimated based on seven stations in a catchment are 143 mm and 31.6 mm, respectively. For a 7% error in the estimation of mean rainfall, do we need additional rain gauges in the catchment? If yes, what is the minimum number of additional rain gauges required to be established in the watershed?

## Short answer

Coefficient of variation Cv = (100 × 31.6)/143 = 22.10%; required gauges N = (22.10/7)² = 9.97 ≈ 10. Since 7 exist, 3 additional gauges are needed for a total of at least 10.

## Answer

The number of gauges needed for a permissible error $\varepsilon$ in the mean is

$$
N = \left(\frac{C_v}{\varepsilon}\right)^2, \quad C_v = \frac{100\,\sigma}{\bar P}
$$

**Given:** $\bar P = 143$ mm, $\sigma = 31.6$ mm, existing $n = 7$, $\varepsilon = 7\%$.

$$
C_v = \frac{100 \times 31.6}{143} = 22.10\%
$$

$$
N = \left(\frac{22.10}{7}\right)^2 = 9.97 \approx 10
$$

Since $N = 10 > 7$, **additional gauges are needed**.

Additional gauges $= 10 - 7 = 3$.

**Answer: Yes. A minimum of 3 additional rain gauges (10 in total) must be installed.**

## Check yourself

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

1. Which formula gives the number of gauges for permissible error e?
   A. N = Cv/e
   B. N = (Cv/e)^2
   C. N = (e/Cv)^2
   D. N = sigma x e
   Answer: B. Statistically N = (Cv/e)^2 with Cv = 100 sigma / mean. (Key point: Formula for number of gauges)

2. What is Cv for mean 143 mm and sigma 31.6 mm?
   A. 4.5 %
   B. 31.6 %
   C. 22.1 %
   D. 143 %
   Answer: C. Cv = 100 x 31.6/143 = 22.1 %. (Key point: Coefficient of variation)

3. With N = (22.10/7)^2 = about 10 and 7 existing gauges, how many more are needed?
   A. 0
   B. 7
   C. 10
   D. 3
   Answer: D. Additional = N - n = 10 - 7 = 3. (Key point: Additional gauges)

## Related in Engineering Hydrology

- [Additional rain gauges for 5% error, seven stations](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aoptimum-raingauges-7stn-130) — Engineering Hydrology, chapter 2, asked 2×
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