---
title: "Log-Pearson III 50-year flood, 1990-2006"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:log-pearson-50yr-1990-2006"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/log-pearson-50yr-1990-2006
site: BCT Engineering
---

# Log-Pearson III 50-year flood, 1990-2006

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2070 Asar, 10 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/log-pearson-50yr-1990-2006
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

The following are the annual peak flow data (m$^3$/s) of a river from 1990 to 2006:

| Year | 1990 | 1991 | 1992 | 1993 | 1994 | 1995 | 1996 | 1997 | 1998 |
|---|---|---|---|---|---|---|---|---|---|
| Peak discharge (m$^3$/s) | 1400 | 4160 | 2580 | 2910 | 2250 | 1360 | 2280 | 2540 | 3900 |

| Year | 1999 | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 |
|---|---|---|---|---|---|---|---|---|
| Peak discharge (m$^3$/s) | 3420 | 6170 | 2160 | 1360 | 5440 | 1340 | 3360 | 2800 |

Compute flood magnitude with 50 year return period (T) using Log-Pearson type III distribution. For T = 50 year, obtain frequency factor ($K_T$) for the computed coefficient of skewness ($C_s$) using the following table.

| $C_s$ | 0 | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| $K_T$ | 2.054 | 2.107 | 2.159 | 2.211 | 2.261 | 2.311 | 2.359 | 2.407 | 2.453 | 2.498 | 2.542 |

| $C_s$ | 1.2 | 1.4 | 1.6 | 1.8 | 2 | 2.2 | 2.5 | 3 |
|---|---|---|---|---|---|---|---|---|
| $K_T$ | 2.626 | 2.706 | 2.78 | 2.848 | 2.912 | 2.970 | 3.048 | 3.152 |

## Short answer

Log-Pearson type III distribution applied to log10 of data yields a coefficient of skewness 0.063 and frequency factor 2.088 for T=50 years. The 50-year flood is estimated as 7022 m³/s.

## Answer

**Method.** Take $z=\log_{10}Q$, find its mean, standard deviation and skewness, then $z_T=\bar z+K_TS_z$ and $Q_T=10^{z_T}$.

| Year | Q (m³/s) | z = log Q | z − z̄ | (z − z̄)² | (z − z̄)³ |
|---|---|---|---|---|---|
| 1990 | 1400 | 3.1461 | -0.2720 | 0.0740 | -0.02013 |
| 1991 | 4160 | 3.6191 | 0.2009 | 0.0404 | 0.00811 |
| 1992 | 2580 | 3.4116 | -0.0065 | 0.0000 | 0.00000 |
| 1993 | 2910 | 3.4639 | 0.0457 | 0.0021 | 0.00010 |
| 1994 | 2250 | 3.3522 | -0.0660 | 0.0044 | -0.00029 |
| 1995 | 1360 | 3.1335 | -0.2846 | 0.0810 | -0.02306 |
| 1996 | 2280 | 3.3579 | -0.0602 | 0.0036 | -0.00022 |
| 1997 | 2540 | 3.4048 | -0.0133 | 0.0002 | 0.00000 |
| 1998 | 3900 | 3.5911 | 0.1729 | 0.0299 | 0.00517 |
| 1999 | 3420 | 3.5340 | 0.1159 | 0.0134 | 0.00156 |
| 2000 | 6170 | 3.7903 | 0.3721 | 0.1385 | 0.05153 |
| 2001 | 2160 | 3.3345 | -0.0837 | 0.0070 | -0.00059 |
| 2002 | 1360 | 3.1335 | -0.2846 | 0.0810 | -0.02306 |
| 2003 | 5440 | 3.7356 | 0.3174 | 0.1008 | 0.03199 |
| 2004 | 1340 | 3.1271 | -0.2911 | 0.0847 | -0.02466 |
| 2005 | 3360 | 3.5263 | 0.1082 | 0.0117 | 0.00127 |
| 2006 | 2800 | 3.4472 | 0.0290 | 0.0008 | 0.00002 |
| Sum |  | 58.1088 | 0 | 0.6735 | 0.00774 |

n = 17.

$$
\bar z=\frac{\sum z}{n}=\frac{58.1088}{17}=3.4182
$$

$$
S_z=\sqrt{\frac{\sum(z-\bar z)^2}{n-1}}=\sqrt{\frac{0.6735}{16}}=0.2052
$$

$$
C_s=\frac{n\sum(z-\bar z)^3}{(n-1)(n-2)S_z^3}=\frac{17\times(0.00774)}{16\times15\times0.00864}=0.063
$$

**Frequency factor for T = 50 years.** From the table, $C_s=0$ gives 2.054 and $C_s=0.1$ gives 2.107. Interpolating for $C_s=0.063$:

$$
K_T=2.054+\frac{0.063-0}{0.1-0}(2.107-2.054)=2.088
$$

**Flood magnitude.**

$$
z_{50}=\bar z+K_TS_z=3.4182+2.088\times0.2052=3.8465
$$

$$
Q_{50}=10^{3.8465}=7022 \text{ m}^3/\text{s}
$$

**Answer: the 50-year flood ≈ 7022 m³/s.**

## Check yourself

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

1. In the log-Pearson III method, which quantity is analysed statistically?
   A. Q^2
   B. 1/Q
   C. Q itself
   D. z = log10 Q
   Answer: D. The mean, standard deviation and skewness are computed for z = log10 Q, and the result is converted back at the end. (Key point: Variable analysed)

2. Which divisor is used for the standard deviation S_z of the logs (n = 17)?
   A. n - 1 = 16
   B. n - 2 = 15
   C. n = 17
   D. n + 1 = 18
   Answer: A. Sample standard deviation uses n - 1: sqrt(0.6735/16) = 0.2052. (Key point: Standard deviation)

3. Cs is computed as 0.063. How is K_T for T = 50 obtained?
   A. Interpolate between 2.054 (Cs = 0) and 2.107 (Cs = 0.1) to get about 2.088
   B. Use K_T = 2.326 for the normal distribution
   C. Take 2.107 because Cs rounds up
   D. Directly K_T = Cs
   Answer: A. Cs lies between table entries 0 and 0.1, so linear interpolation gives K_T = 2.088. (Key point: Skewness and K_T)

4. With z-bar = 3.4182, S_z = 0.2052, K_T = 2.088, how is Q50 found?
   A. Q = 3.4182 + 2.088 x 0.2052 = 3.85 m^3/s
   B. z50 = 3.4182 + 2.088 x 0.2052 = 3.8465, Q50 = 10^3.8465 = 7022 m^3/s
   C. Q = 2.088 x 3.4182 x 0.2052
   D. Q = 10^3.4182 x 2.088
   Answer: B. z_T = z-bar + K_T S_z is a logarithm, so it must be converted back using 10^z_T. (Key point: Design flood)

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