a tap discharge water evenly in a jet at a velocity of 2.6 m/s at the tap outlet, the diameter of the jet at this point being 15 mm. the jet flows down vertically in a smooth stream. determine the velocity and the diameter of the jet at 0.6 m below the tap outlet.
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To determine the velocity and diameter of the water jet at a specific point below the tap outlet, we can utilize the principle of conservation of mass and the principle of conservation of energy.
1. Conservation of Mass:
According to the principle of conservation of mass, the mass flow rate of water remains constant at any given point along the stream. This means that the product of velocity and cross-sectional area at one point should be equal to the product of velocity and cross-sectional area at another point.
2. Conservation of Energy:
According to the principle of conservation of energy, the total mechanical energy of the water remains constant in the absence of any external forces like friction or air resistance. Since the water is flowing down vertically, we can assume there is no change in elevation, and we only need to consider velocity changes.
Now, let's calculate the velocity and diameter of the water jet at 0.6 m below the tap outlet:
Step 1: Calculate the cross-sectional area at the tap outlet:
The diameter of the jet at the tap outlet is given as 15 mm, which is equal to 0.015 m.
The cross-sectional area (A1) at the tap outlet can be calculated using the formula:
A1 = π*(d1/2)^2
= π*(0.015/2)^2
= 0.00017671 m^2
Step 2: Calculate the velocity at the tap outlet:
The velocity at the tap outlet is given as 2.6 m/s.
Step 3: Apply conservation of mass:
According to the conservation of mass principle, the mass flow rate (m1) at the tap outlet is equal to the mass flow rate (m2) at any other point below.
m1 = m2
ρ1 * A1 * V1 = ρ2 * A2 * V2
Where:
ρ1 is the density of water (assumed to be constant)
V1 is the velocity at the tap outlet
A2 is the cross-sectional area below the tap outlet
V2 is the velocity below the tap outlet (to be determined)
Step 4: Solve for A2:
Since the water jet flows down vertically, the area remains constant. Therefore, A2 is equal to A1. So, A2 = A1 = 0.00017671 m^2.
Step 5: Solve for V2:
ρ1 * A1 * V1 = ρ2 * A2 * V2
V2 = (ρ1 * A1 * V1) / (ρ2 * A2)
To continue, we need the density (ρ2) of water at the specific point below the tap outlet.