Question

2 pipes with different diameters are connected. Inside the pipes is flowing water. The velocity of the water in the larger pipe is 5 m/s while the velocity of the water in the smaller pipe is thrice as fast as the larger pipe. The cross-sectional area of the smaller pipe is 0.0625 cm^2. Find the radius of the larger pipe.

Answer 1

Answer on Question 69323, Physics, Other

Question:

2 pipes with different diameters are connected. Inside the pipes is flowing water. The velocity of the water in the larger pipe is $5\,m/s$ while the velocity of the water in the smaller pipe is thrice as fast as the larger pipe. The cross-sectional area of the smaller pipe is $0.0625\,cm^2$. Find the radius of the larger pipe.

Solution:

By the definition of the Law of Continuity we have:

$$A_1 v_1 = A_2 v_2,$$

here, $A_1$, $A_2$ are the cross-sectional areas of the larger and smaller pipes, respectively, $v_1, v_2$ are the velocities of the water flowing through the larger and smaller parts of the pipes, respectively.

From this formula we can find the cross-sectional area of the larger pipe:

$$A_1 = \frac{A_2 v_2}{v_1}.$$

From the other hand:

$$A_1 = \pi r_1^2.$$

Therefore, equating these two formulas, we can find the radius of the larger pipe, $r_1$:

$$\pi r_1^2 = \frac{A_2 v_2}{v_1},$$ $$r_1 = \sqrt{\frac{A_2 v_2}{\pi v_1}} = \sqrt{\frac{0.0625\,cm^2 \cdot \frac{1\,m^2}{10000\,cm^2} \cdot 15\, \frac{m}{s}}{\pi \cdot 5\, \frac{m}{s}}} = 2.4 \cdot 10^{-3}\,m.$$

Answer:

$$r_1 = 2.4 \cdot 10^{-3}\,m.$$

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