To determine the speed of water as it exits the sprinkler holes, we can use the principle of conservation of mass, specifically the equation of continuity. This principle states that the mass flow rate must remain constant in a closed system, which means that the product of the cross-sectional area and the velocity of the fluid must be the same at different points in the system.
Understanding the Problem
We have a hosepipe with a larger cross-sectional area and a sprinkler with multiple smaller holes. The water flows from the hosepipe into the sprinkler, and we need to find out how fast the water exits through the holes.
Given Data
- Cross-sectional area of the hosepipe, A_h = 2.4 cm²
- Speed of water in the hosepipe, v_h = 1.5 m/s
- Number of holes in the sprinkler = 20
- Cross-sectional area of each hole, A_s = 2 × 10⁻² cm²
Calculating Total Area of Sprinkler Holes
First, we need to find the total cross-sectional area of all the holes in the sprinkler:
Total area of holes, A_total = Number of holes × Area of each hole
A_total = 20 × (2 × 10⁻² cm²) = 20 × 0.02 cm² = 0.4 cm²
Converting Units
Since we are working with different units, it’s essential to convert the areas from cm² to m² for consistency:
A_total = 0.4 cm² × (1 m² / 10,000 cm²) = 0.00004 m²
A_h = 2.4 cm² × (1 m² / 10,000 cm²) = 0.00024 m²
Applying the Continuity Equation
The continuity equation can be expressed as:
A_h × v_h = A_total × v_s
Where v_s is the speed of water exiting the sprinkler holes. Rearranging this gives us:
v_s = (A_h × v_h) / A_total
Substituting the Values
Now, we can substitute the known values into the equation:
v_s = (0.00024 m² × 1.5 m/s) / 0.00004 m²
v_s = (0.00036 m³/s) / 0.00004 m²
v_s = 9 m/s
Final Result
The speed of water as it exits from the holes of the sprinkler is approximately 9 meters per second. This illustrates how the water accelerates as it moves from a larger area (the hosepipe) to smaller areas (the holes), which is a common phenomenon in fluid dynamics.
