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Clark Method for Hydrograph Calculations

Understand the Clark method from 1945 for hydrograph analysis. Calculate accumulated areas and discharge coordinates. Utilize the method in practical examples for watershed management.

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Clark Method for Hydrograph Calculations

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  1. Método de Clark • 1945 • A=?

  2. Método de Clark: áreas acumuladas Ac= 1,414 . A.(t/ tc) 1,5 para 0≤t/tc < 0,5 Ac= A[ 1- 1,414 ( 1- t/tc) 1,5] para 0,5≤ t/tc ≤1,0 Sendo: Ac= área acumulada (m2) t= tempo (min) tc= tempo de concentração (min) A= área total (m2)

  3. Método de Clark Ii = (Ac x 0,01m) / (∆t(h) x 60 x 60) Sendo: Ii= ordenada da descarga (m3/s) Ac= area acumulada (m2) ∆t(horas)= intervalo de tempo em horas adotada. Nota: temos que transformar em segundos multiplicando por 60min e depois por 60 segundos. 0,01m= é o 1cm que usaremos no diagrama unitário. Combinando as equações acima teremos: Qt= C1. It + C2. Q t-1 C1= ∆t/ (K + 0,5.∆t) K=0,6.tc C2= 1- C1 No tempo ∆t tomamos a média: Qt= (Q t-1 + Qt) / 2

  4. Método de Clark • Exemplo: baseado em Gupta • A= 595 km2Tc= 10h ∆t= 1h K=0,6.tc=6h C1= ∆t/ (K + 0,5.∆t) C1= 1/ (6 + 0,5x 1) =0,154 C2= 1-0,154= 0,846 Ac= 1,414 x595 (t/ 10) 1,5 para 0≤t/tc < 0,5 Ac= 595[ 1- 1,414 ( 1- t/10) 1,5] para 0,5≤ t/tc ≤1,0

  5. Método de Clark • Truque: • Ac x 0,01/ (∆tx60x60)= ...m3/s/cm • Exemplo: • 26605191 x 0,01/ (1,0x60x60)= 73,9 m3/s/cm

  6. Método de Clark

  7. Método de Clark

  8. Método de Clark • Feito o hidrograma unitário • 1) Chuva excedente conforme CN e com intervalo de tempo de 1h. • 2) Fazer a CONVOLUÇÃO • 3) Achamos o hidrograma final e o pico de vazão

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