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.

0.319 m³/s

Step-by-step explanation:

Data provided:

Initial volume in the river = 255 m³

Final volume required in the storage = 325 m³

Average outflow = 0.30 m³/s

Duration for raising the level = 1 hour = 3600 seconds

Now,

The actual volume required to raise the volume to 325 m³

= Final volume - Initial volume

= 325 m³ - 255 m³

= 70 m³

also,

the amount of outflow in 1 hour = Average rate of outflow × Time

= 0.30 m³/s × 3600 s

= 1080 m³

Therefore,

the total volume required = 1080 m³ + 70 m³ = 1150 m³

Now,

the average rate of inflow required =

=

= 0.319 m³/s

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1