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We can approach this siphon problem using Bernoulli's equation and the continuity equation. Let's summarize the key data from the problem:

- The siphon is made of a 50-mm diameter tube.
- The liquid is water at 15°C (the density of water, \( \rho \), at this temperature is about 999 kg/m³, and the dynamic viscosity is approximately \(1.14 \times 10^{-3} \, \text{Pa}\cdot\text{s}\)).
- The height difference between the liquid surface and the siphon outlet is 2.5 meters, while the difference between the liquid surface and the top of the siphon (point 3) is 0.6 meters.
- The minimum pressure in the tube occurs at the highest point of the siphon, point 3.

### Step 1: Use Bernoulli’s Equation

For Bernoulli’s equation, we compare two points along the streamline: 
- Point 1: at the liquid surface.
- Point 3: at the top of the siphon, where the pressure will be minimum.

We will apply Bernoulli’s equation between these two points:

\[
P_1 + \frac{1}{2} \rho v_1^2 + \rho g z_1 = P_3 + \frac{1}{2} \rho v_3^2 + \rho g z_3
\]

- At point 1, the velocity \( v_1 \) is negligible since the surface is large, so \( v_1 \approx 0 \).
- \( z_1 = 0 \) and \( z_3 = 2.5 \, \text{m} \) (height difference between points 1 and 3).
- We are asked to find the minimum pressure at point 3, which means we need to solve for \( P_3 \).

### Step 2: Apply Continuity Equation

The continuity equation ensures the flow rate remains constant throughout the siphon:

\[
A_1 v_1 = A_3 v_3
\]

Since \( A_1 \approx A_3 \), we have \( v_1 \approx v_3 \). Thus, the velocities at points 1 and 3 are approximately the same.

### Step 3: Use Hydrostatic Pressure for Point 2

At the outlet (point 2), the height is zero, so the pressure at this point can be taken as atmospheric pressure, which simplifies the calculation.

Let's calculate the flow rate and estimate the minimum pressure inside the tube using these relations.

I'll proceed with the calculations now.### Results:

1. **Minimum Pressure at Point 3**: The minimum pressure inside the tube at the top of the siphon (point 3) is approximately **76,825 Pa** (or 76.8 kPa). This is equivalent to about **-24.5 kPa** gauge pressure, considering that atmospheric pressure is around 101.3 kPa.
   
2. **Volume Flow Rate**: The volume flow rate through the siphon is approximately **0.01375 m³/s**, or **13.75 liters per second**.

Let me know if you'd like more detailed steps on the calculations or further explanations!

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Here are 5 related questions to deepen your understanding:
1. How does the diameter of the tube affect the volume flow rate?
2. What would happen to the siphon flow if the height difference between the liquid surface and the outlet is reduced?
3. How would a change in fluid density (e.g., using oil instead of water) impact the minimum pressure?
4. Why is the velocity at the liquid surface (point 1) considered negligible?
5. How does the pressure at the top of the siphon compare to atmospheric pressure, and why does it drop?

**Tip**: Always remember that the flow rate depends not just on the height differences, but also on the fluid's properties (like density) and the tube's dimensions.

The siphon shown is fabricated from 50-mm-ID drawn aluminum tubing. The liquid is water at 15°C. Compute the volume flow rate through the siphon. Estimate the minimum pressure inside the tube.

This problem involves calculating the volume flow rate through a siphon and estimating the minimum pressure inside the tube using Bernoulli's equation and the continuity equation. The siphon is fabricated from a 50-mm-ID drawn aluminum tubing, and the liquid is water at 15°C. Learn how to apply fluid mechanics principles to find the solution.

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