Question

375 yd3 gasoline-contaminated soil will be treated using bioremediation. The soil has low organic content. It is desired to bring the water moisture level to 60% to maintain high reaction rates. Calculate how much water (gals) should be added for the following conditions: Soil porosity, n = 30% Initial saturation, S = 20% Desired water content = 60

Answer 1

Answer:

The required water is 9090 gallon

Explanation:

The objective of this solution is obtain how much water for the condition stated in the question to be attained

       From the question the amount of gasoline is $G=$ 375 $yd^3$

       The porosity is $n = 30$% = $0.3$

       The initial saturation $s = 20$% = $0.2$

       The desired water content $d = 60$% = $0.6$

The amount of water needed can be mathematically represented as

          Amount of water needed(W)   $=G *n*(d-s)$

                                                               $= 375 *0.3 *(0.6-0.2)$

                                                               $=45 yd^3$

   Now one cubic yard ($yd^3$) $\approx$  202 US liquid gallon

So to convert 45$yd^3$ to gallon would require multiplying it by  202

                                                              $=9090 gallon$