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4 March, 06:13

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.

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  1. 4 March, 06:32
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    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
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