To solve this problem we will use the concepts related to the volumetric flow rate, which describes the amount of volume per unit of time, through the Area of the section and the speed of the fluid over it. Mathematically this is:
[tex]Q = A*V \rightarrow V = \frac{Q}{A}[/tex]
Where,
Q = Discharge
A= Cross sectional Area
V = Velocity
Our values are given as,
[tex]A = 18ft^2[/tex]
[tex]Q = 0.7ft^3/min[/tex]
Replacing,
[tex]V = \frac{0.7}{18}[/tex]
[tex]V = 0.038ft/min[/tex]
Therefore the water level rises to a velocity of 0.038ft/min
