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Vazão mínima com Weibull

Vazão mínima com Weibull. Critérios para achar vazão mínima: Tennant : 30% da vazão média Q7,10. Exemplo: vazões mínimas de 7 dias seguidos de 1938 a 1978. Relações auxiliares de Weibull Tabela 133.2. X= média dos valores = 28,473m 3 /s S= desvio padrão= 7,590 m 3 /s

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Vazão mínima com Weibull

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  1. Vazão mínima com Weibull

  2. Critérios para achar vazão mínima: • Tennant: 30% da vazão média • Q7,10

  3. Exemplo: vazões mínimas de 7 dias seguidos de 1938 a 1978

  4. Relações auxiliares de WeibullTabela 133.2

  5. X= média dos valores =28,473m3/s • S= desvio padrão= 7,590 m3/s • Cv= coeficiente de variação= S/X=7,590/28,473= 0,2666 • XT= valor de Q7,10 • T= 10 para período de retorno de 10anos • Com o valor de Cv= S/X= 7,590/ 28,473= 0,2666 entramos na Tabela (133.2) e achamos: • A(α)= 0,9093 • β= X / A (α) • β= 28,473/ 0,9093 =31,31 • XT= β . [ -ln(1- 1/T)]1/α • Para T=10anos, temos: • X10= β . [ -ln(1- 1/10)]1/α • X10= 31,31x [ -ln(1- 1/10)]0,2364 =18,4 m3/s

  6. Para conferirOrdem crescente das vazões e divisão (41+1)/ m

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