Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diameter orifice at a constant average velocity of 6.3 m/s. if the water level in the pool rises at a rate of 1.5 cm/min, determine the rate at which water is supplied to the pool, in m3/s.

Given:

Area of pool = 3m*4m

Diameter of orifice = 0.076m

Outlet Velocity = 6.3m/s

Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate

Accumulation velocity * Area of pool = Inlet flowrate - Outlet velocity * Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min * (1min/60s) * (1m/100cm)

Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 * 4 m²

Area of pool = 12m²

Area of orifice = πd²/4 = π (0.076m) ²/4

Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity * Area of pool = Inlet flowrate - Outlet velocity * Area of orifice

0.00025 m/s * 12m² = Inlet flowrate - 6.3m/s * 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s