Question 4: The storage in a reach of a river at a point in time is 255 m3 (meters cubed) . Determine the average rate of inflow (m3 s -1 ) to the river reach required to raise the storage to 325 m 3 if the average outflow was 0.30 m3s -1 (meters cubed/ second) over a 1-hour period.

Answer:0.319 m³/sStep-by-step explanation:Data provided:Initial volume in the river = 255 m³Final volume required in the storage = 325 m³Average outflow = 0.30 m³/sDuration for raising the level = 1 hour = 3600 secondsNow,The actual volume required to raise the volume to 325 m³= Final volume - Initial volume = 325 m³ - 255 m³= 70 m³also,the amount of outflow in 1 hour = Average rate of outflow × Time= 0.30 m³/s × 3600 s= 1080 m³Therefore,the total volume required = 1080 m³ + 70 m³ = 1150 m³Now,the average rate of inflow required = [tex]\frac{\textup{Total volume required}}{\textup{Time}}[/tex] = [tex]\frac{\textup{1150}}{\textup{3600}}[/tex] = 0.319 m³/s