calculate the ideal spacing interval for a dn150 inlet opening located at the edge of  the road and combined with the kerbs. The flow rate is 0.06m3/s and the longitudinal slope of the road is 3.5%. The width of the road is 9m

To calculate the ideal spacing interval for a DN150 inlet opening, we need to consider several factors including the flow rate, the road's longitudinal slope, and the width of the road. Here's a step-by-step solution:

### Step 1: Calculate the flow per meter width of the road

Given:
- Flow rate (\(Q\)) = 0.06 m\(^3\)/s
- Road width (\(w\)) = 9 m

Flow per meter width of the road:
\[ q = \frac{Q}{w} = \frac{0.06}{9} = 0.00667 \text{ m}^3/\text{s per meter} \]

### Step 2: Calculate the capacity of the DN150 inlet

The capacity of a DN150 (150 mm diameter) pipe depends on several factors, including the slope and roughness coefficient. For simplicity, we'll assume the pipe is running at full capacity, and we use the Manning equation for a circular pipe:

\[ Q_{\text{full}} = \frac{1}{n} \cdot A \cdot R^{2/3} \cdot S^{1/2} \]

Where:
- \( A \) is the cross-sectional area of the pipe
- \( R \) is the hydraulic radius
- \( S \) is the slope (3.5% = 0.035)
- \( n \) is the Manning roughness coefficient (typically 0.013 for concrete pipes)

For a DN150 pipe:
- Diameter \(d = 0.15\) m
- Area \(A = \frac{\pi d^2}{4} = \frac{\pi (0.15)^2}{4} \approx 0.01767 \text{ m}^2 \)
- Hydraulic radius \(R = \frac{A}{P} = \frac{\text{Area}}{\text{Wetted perimeter}} = \frac{0.01767}{\pi \times 0.15} \approx 0.0375 \text{ m} \)

Now, calculate \( Q_{\text{full}} \):
\[ Q_{\text{full}} = \frac{1}{0.013} \cdot 0.01767 \cdot (0.0375)^{2/3} \cdot (0.035)^{1/2} \]
\[ Q_{\text{full}} \approx 0.0248 \text{ m}^3/\text{s} \]

### Step 3: Calculate the number of inlets required

If each inlet can handle \( Q_{\text{full}} \):
\[ N = \frac{Q}{Q_{\text{full}}} = \frac{0.06}{0.0248} \approx 2.42 \]

Since we can't have a fraction of an inlet, we round up to 3 inlets.

### Step 4: Determine the spacing interval

Assuming evenly spaced inlets along the road length (\(L\)):
\[ \text{Spacing interval} = \frac{L}{N} \]

However, since \(L\) isn't provided, we can only determine that 3 inlets are required for the given flow rate.

### Conclusion

To handle a flow rate of 0.06 m\(^3\)/s with a longitudinal slope of 3.5% on a 9-meter wide road, you need approximately 3 DN150 inlets. The exact spacing interval depends on the total length of the road section being considered. If you provide the road section length, I can calculate the specific spacing interval.

Would you like further details on any part of this calculation, or do you have any specific questions? Here are some additional questions you might be interested in:

1. How do you calculate the Manning roughness coefficient?
2. What is the effect of varying the slope on the number of inlets required?
3. How to calculate the hydraulic radius for different pipe shapes?
4. What are the standard dimensions and capacities of other common inlet sizes?
5. How to account for additional factors like debris and partial blockages in inlet design?

### Tip
When designing drainage systems, always consider potential debris and blockages, and ensure regular maintenance to maintain efficiency.

Learn how to calculate the ideal spacing interval for a DN150 inlet opening on a 9-meter wide road with a flow rate of 0.06 m³/s and a longitudinal slope of 3.5%. Understand the application of the Manning equation and pipe flow calculations in drainage design.

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