Cs is computed as 0.063. How is K_T for T = 50 obtained?

Interpolate between 2.054 (Cs = 0) and 2.107 (Cs = 0.1) to get about 2.088

Use K_T = 2.326 for the normal distribution

Take 2.107 because Cs rounds up

With z-bar = 3.4182, S_z = 0.2052, K_T = 2.088, how is Q50 found?

z50 = 3.4182 + 2.088 x 0.2052 = 3.8465, Q50 = 10^3.8465 = 7022 m^3/s

Q = 3.4182 + 2.088 x 0.2052 = 3.85 m^3/s

2070 Asar · 10 marks

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

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.

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

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.

✓ Answer checked by Claude Sonnet 5.5

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