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When the hydraulic conductivity Ks = 10 mm/hr; effective matrix potential Ns = 20 mm, and rainfall intensity I = 30 mm/hr , determine the amount of runoff generated when the runoff rate reaches 15 mm/hr?( 2.7 mm or 0.21 mm or 18 mm or 0.67 mm)

Respuesta :

Answer:

[tex]t = 0.75 \ hr = 0.75 *60 = 45 \ minutes[/tex]

Explanation:

From the question we are told that

   The  hydraulic conductivity is  [tex]Ks = 10\ mm/hr[/tex]

    The  effective matrix potential [tex]Ns = 20 \ mm[/tex]

    The  intensity of  rainfall is  [tex]I = 30 \ mm/hr[/tex]

    The runoff rate [tex]R = 15 \ mm/hr[/tex]

Generally the run off rate is mathematically represented as  

      [tex]R = (I - Ks) t[/tex]

Here t is the amount of runoff generated

=>   [tex]15 = (30 - 10) t[/tex]

=>  [tex]t = 0.75 \ hr[/tex]

converting to minutes

    [tex]t = 0.75 \ hr = 0.75 *60 = 45 \ minutes[/tex]

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