To find the speed at which water exits the holes of the sprinkler, we can apply the principle of conservation of mass, specifically the concept of continuity in fluid dynamics. This principle states that for an incompressible fluid flowing in a closed system, the product of the cross-sectional area and the velocity must remain constant. In simpler terms, if water flows from a larger area to a smaller area, its speed must increase.
Understanding the Areas
First, we need to calculate the cross-sectional area of both the garden hose and the sprinkler holes. The formula for the area \( A \) of a circle is given by:
A = πr²
Calculating the Hose Area
The internal diameter of the hose is 0.75 inches, which gives a radius of:
r_hose = diameter / 2 = 0.75 in / 2 = 0.375 in
Now, we can find the area of the hose:
A_hose = π(0.375 in)² ≈ 0.4418 in²
Calculating the Sprinkler Hole Area
Next, we look at the sprinkler holes. Each hole has a diameter of 0.050 inches, leading to a radius of:
r_hole = diameter / 2 = 0.050 in / 2 = 0.025 in
Now we calculate the area of one hole:
A_hole = π(0.025 in)² ≈ 0.0019635 in²
Since there are 24 holes, the total area of all the holes is:
A_sprinkler = 24 × A_hole ≈ 24 × 0.0019635 in² ≈ 0.0471 in²
Applying the Continuity Equation
According to the continuity equation, the flow rate must remain constant. Therefore, we can set up the equation:
A_hose × v_hose = A_sprinkler × v_sprinkler
Substituting Known Values
We already know the following:
- A_hose ≈ 0.4418 in²
- v_hose = 3.5 ft/s
- A_sprinkler ≈ 0.0471 in²
Now we can rearrange the equation to solve for the velocity of the water exiting the sprinkler holes:
v_sprinkler = (A_hose × v_hose) / A_sprinkler
Substituting the values:
v_sprinkler = (0.4418 in² × 3.5 ft/s) / 0.0471 in²
Calculating the Exit Speed
Now we perform the calculation:
v_sprinkler ≈ (1.5473 ft²/s) / 0.0471 in²
Since our areas are in square inches and our speed in feet per second, we need to convert the sprinkler area to square feet:
0.0471 in² = 0.0471 in² × (1 ft / 12 in)² ≈ 0.000327 ft²
Thus, we have:
v_sprinkler ≈ 1.5473 ft²/s / 0.000327 ft² ≈ 4726.4 ft/s
Final Thoughts
Therefore, the speed at which water exits the sprinkler holes is approximately 4726.4 ft/s. This illustrates how a fluid speeds up when it moves from a larger cross-sectional area to a smaller one, demonstrating the conservation of mass in fluid dynamics.
