A new residential subdivision is estimated to have a peak sewage flow rate (Q) of $$ 0.2474 \mathrm{~m}^{3} / \mathrm{s} $$. The proposed concrete wastewater sewer carrying sewage from the subdivision drops 0.622 m in elevation over a distance of 285 m to connect to a larger sanitary trunk sewer. Determine the EXACT diameter in mm (accurate to the nearest mm ) of sewer pipe required to carry this wastewater.

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To determine the diameter of the sewer pipe required to carry the wastewater, we can use the Manning's equation for open channel flow, which is commonly used for sewer design. The equation is given by:

$$
Q = \frac{1}{n} A R^{2/3} S^{1/2}
$$

Where:
- $$ Q $$ = flow rate (m³/s)
- $$ n $$ = Manning's roughness coefficient (dimensionless)
- $$ A $$ = cross-sectional area of flow (m²)
- $$ R $$ = hydraulic radius (m)
- $$ S $$ = slope of the energy grade line (dimensionless)

### Step 1: Calculate the slope $$ S $$

The slope $$ S $$ can be calculated as:

$$
S = \frac{\text{drop in elevation}}{\text{length of the sewer}}
$$

Given:
- Drop in elevation = 0.622 m
- Length of the sewer = 285 m

$$
S = \frac{0.622}{285} \approx 0.00218
$$

### Step 2: Assume a pipe diameter and calculate $$ A $$ and $$ R $$

Assuming the pipe is circular, the cross-sectional area $$ A $$ and hydraulic radius $$ R $$ can be expressed in terms of the diameter $$ D $$:

$$
A = \frac{\pi D^2}{4}
$$
$$
R = \frac{A}{P} = \frac{A}{\pi D} = \frac{\frac{\pi D^2}{4}}{\pi D} = \frac{D}{4}
$$

### Step 3: Substitute $$ A $$ and $$ R $$ into Manning's equation

Substituting $$ A $$ and $$ R $$ into Manning's equation gives:

$$
Q = \frac{1}{n} \left(\frac{\pi D^2}{4}\right) \left(\frac{D}{4}\right)^{2/3} S^{1/2}
$$

This simplifies to:

$$
Q = \frac{1}{n} \left(\frac{\pi D^2}{4}\right) \left(\frac{D^{2/3}}{4^{2/3}}\right) S^{1/2}
$$

$$
Q = \frac{\pi D^{8/3}}{16 n 4^{2/3}} S^{1/2}
$$

### Step 4: Solve for $$ D $$

Rearranging the equation to solve for $$ D $$:

$$
D^{8/3} = \frac{16 n Q 4^{2/3}}{\pi S^{1/2}}
$$

Taking the cube root:

$$
D = \left(\frac{16 n Q 4^{2/3}}{\pi S^{1/2}}\right)^{3/8}
$$

### Step 5: Choose a value for $$ n $$

A common value for $$ n $$ for concrete pipes is approximately $$ 0.013 $$.

### Step 6: Substitute values into the equation

Substituting $$ Q = 0.2474 \, \text{m}^3/\text{s} $$, $$ n = 0.013 $$, and $$ S = 0.00218 $$:

$$
D = \left(\frac{16 \times 0.013 \times 0.2474 \times 4^{2/3}}{\pi \sqrt{0.00218}}\right)^{3/8}
$$

Calculating $$ 4^{2/3} $$:

$$
4^{2/3} \approx 2.5198
$$

Now substituting:

$$
D = \left(\frac{16 \times 0.013 \times 0.2474 \times 2.5198}{\pi \sqrt{0.00218}}\right)^{3/8}
$$

Calculating $$ \sqrt{0.00218} $$:

$$
\sqrt{0.00218} \approx 0.0467
$$

Now substituting everything into the equation:

$$
D = \left(\frac{16 \times 0.013 \times 0.2474 \times 2.5198}{\pi \times 0.0467}\right)^{3/8}
$$

Calculating the numerator:

$$
16 \times 0.013 \times 0.2474 \times 2.5198 \approx 0.0204
$$

Calculating the denominator:

$$
\pi \times 0.0467 \approx 0.1462
$$

Now substituting:

$$
D = \left(\frac{0.0204}{0.1462}\right)^{3/8}
$$

Calculating the fraction:

$$
\frac{0.0204}{0.1462} \approx 0.139
$$

Now taking the $$ \frac{3}{8} $$ power:

$$
D \approx (0.139)^{0.375} \approx 0.574
$$

### Step 7: Convert to mm

To convert to mm:

$$
D \approx 0.574 \, \text{m} \times 1000 \approx 574 \, \text{mm}
$$

### Final Answer

The required diameter of the sewer pipe is approximately **574 mm**.
The required diameter of the sewer pipe is approximately 574 mm.

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