Water flows in a pipe of 23-cm diameter with an average velocity of 0.73 m/s. The diameter of the pipe is reduced and the average velocity of the water increases to 1.66 m/s. What is the diameter of the smaller pipe? What is the flow rate? Provide me a step by step solution 

To solve this problem, we need to use the principle of conservation of mass for an incompressible fluid, which is represented by the continuity equation. The continuity equation states:

\[
A_1 v_1 = A_2 v_2
\]

Where:
- \( A_1 \) and \( A_2 \) are the cross-sectional areas of the larger and smaller pipes, respectively,
- \( v_1 \) and \( v_2 \) are the average velocities in the larger and smaller pipes, respectively.

### Given:
- Diameter of the larger pipe, \( d_1 = 23 \, \text{cm} = 0.23 \, \text{m} \),
- Velocity in the larger pipe, \( v_1 = 0.73 \, \text{m/s} \),
- Velocity in the smaller pipe, \( v_2 = 1.66 \, \text{m/s} \).

We are required to find:
1. The diameter of the smaller pipe (\( d_2 \)),
2. The flow rate (\( Q \)).

### Step 1: Find the cross-sectional area of the larger pipe (\( A_1 \))

The cross-sectional area of a pipe is given by the formula:

\[
A = \pi \left(\frac{d}{2}\right)^2
\]

For the larger pipe:

\[
A_1 = \pi \left(\frac{0.23}{2}\right)^2 = \pi \times (0.115)^2 = \pi \times 0.013225 \, \text{m}^2
\]

\[
A_1 = 0.04154 \, \text{m}^2
\]

### Step 2: Use the continuity equation to find \( A_2 \)

From the continuity equation:

\[
A_1 v_1 = A_2 v_2
\]

Solving for \( A_2 \):

\[
A_2 = \frac{A_1 v_1}{v_2} = \frac{0.04154 \times 0.73}{1.66} = \frac{0.0303242}{1.66} = 0.01827 \, \text{m}^2
\]

### Step 3: Find the diameter of the smaller pipe (\( d_2 \))

Now, use the area formula to find \( d_2 \):

\[
A_2 = \pi \left(\frac{d_2}{2}\right)^2
\]

Solving for \( d_2 \):

\[
\left(\frac{d_2}{2}\right)^2 = \frac{A_2}{\pi} = \frac{0.01827}{\pi} = 0.005815 \, \text{m}^2
\]

\[
\frac{d_2}{2} = \sqrt{0.005815} = 0.07625 \, \text{m}
\]

\[
d_2 = 2 \times 0.07625 = 0.1525 \, \text{m} = 15.25 \, \text{cm}
\]

So, the diameter of the smaller pipe is approximately **15.25 cm**.

### Step 4: Calculate the flow rate \( Q \)

The flow rate \( Q \) is the product of the cross-sectional area and the velocity:

\[
Q = A_1 v_1 = 0.04154 \times 0.73 = 0.0303 \, \text{m}^3/\text{s}
\]

Thus, the flow rate is approximately **0.0303 cubic meters per second**.

---

### Final Answers:
- The diameter of the smaller pipe is **15.25 cm**.
- The flow rate is **0.0303 m³/s**.

Would you like further details or have any questions?

Here are 5 related questions for further practice:
1. How does the velocity change if the diameter of the smaller pipe is reduced further?
2. What would the flow rate be if the larger pipe had a diameter of 30 cm instead of 23 cm?
3. How would the diameter of the smaller pipe change if the velocity increased to 2 m/s?
4. What happens to the flow rate if the velocity in both pipes doubles?
5. How does the flow rate depend on the cross-sectional area in a steady flow?

**Tip:** The continuity equation is crucial for analyzing fluid dynamics, as it reflects the conservation of mass in fluid flow systems.

Water flows in a pipe of 23-cm diameter with an average velocity of 0.73 m/s. The diameter of the pipe is reduced and the average velocity of the water increases to 1.66 m/s. What is the diameter of the smaller pipe? What is the flow rate? Provide me a step by step solution.

Solve a fluid dynamics problem where water flows in a 23-cm diameter pipe at 0.73 m/s, and the diameter is reduced with the velocity increasing to 1.66 m/s. Learn step-by-step how to calculate the smaller pipe's diameter and the flow rate, using the continuity equation. Perfect for Grades 10-12.

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