A new residential subdivision is estimated to have a peak sewage
flow rate of . 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.

Ask by Pritchard Mcfarlane. in Canada

Dec 09,2024

Question

A new residential subdivision is estimated to have a peak sewage
flow rate of . 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.

Ask by Pritchard Mcfarlane. in Canada

Dec 09,2024

UpStudy AI · Accepted Answer

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 $$ R $$

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

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

Where $$ P $$ is the wetted perimeter, which for a full circular pipe is equal to the circumference $$ \pi D $$.

### 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: Rearranging for $$ D $$

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

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

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

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

A typical value for the Manning's roughness coefficient $$ n $$ for concrete pipes is approximately $$ 0.013 $$.

### Step 6: Substitute values into the equation

Substituting the known values:

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

Calculating $$ S^{1/2} $$:

$$
S^{1/2} = \sqrt{0.00218} \approx 0.0467
$$

Now substituting into the equation for $$ D $$:

$$
D = \left(\frac{16 \times 0.013 \times 4^{2/3} \times 0.2474}{\pi \times 0.0467}\right)^{3/8}
$$

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

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

Now substituting this value:

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

Calculating the numerator:

$$
16 \times 0.013 \times 2.5198 \times 0.2474 \approx 0.0202
$$

Calculating the denominator:

$$
\pi \times 0.0467 \approx 0.1460
$$

Now substituting these values:

$$
D = \left(\frac{0.0202}{0.1460}\right)^{3/8} \approx \left(0.1385\right)^{3/8}
$$

Calculating $$ D $$:

$$
D \approx 0.0631 \, \text{m} \approx 63.1 \, \text{mm}
$$

### Final Step: Rounding to the nearest mm

The diameter of the sewer pipe required to carry the wastewater is approximately:

$$
\boxed{63 \, \text{mm}}
$$
The diameter of the sewer pipe required is approximately 63 mm.

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