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CPD Test Prep

Flow in Drainage Systems

(Summary)

 

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  1. Drainage System Design for Self-Cleansing:

    • A well-designed drainage system maintains a flow velocity of about 2 feet per second. This speed is crucial to keep solids in suspension, avoiding sedimentation and blockages.
    • Designers must consider the diameter and slope of pipes, which should be proportional to the anticipated flow rate.

  2. Minimum Slope Requirements:

    • Different pipe diameters require different minimum slopes to ensure adequate flow. For instance, a 4-inch pipe might need a slope of 1/8 inch per foot, whereas a 3-inch pipe might need 1/4 inch per foot.
    • Insufficient slope can lead to slow water movement and potential blockages.

  3. Pipe Material and Size Selection:

    • The choice of material for pipes depends on factors like the type of waste (chemical composition, temperature), flow rate, and environmental considerations.
    • Sizing of pipes is guided by estimating the peak discharge and ensuring the diameter can handle this flow without excessive velocity or pressure build-up.

  4. Venting Requirements:

    • Vents prevent the loss of trap seals due to pressure variations within the drainage system. These seals are crucial for blocking sewer gases from entering buildings.
    • Proper venting involves strategic placement and sizing of vent pipes to balance the air pressure within the drainage system.

  5. Impact of Fittings on Flow:

    • Fittings like elbows and tees can create turbulence, reducing flow efficiency. Reducers can change flow velocity.
    • The design should minimize sharp bends and use fittings that maintain a smooth flow path.

  6. Handling of Special Wastes:

    • Special wastes like chemicals or high-temperature effluents might require pipes made from specific materials (like CPVC for chemical resistance or temperature tolerance).
    • The system might also need additional components like neutralization tanks or dilution pits.

  7. Maintenance and Accessibility:

    • Drainage systems should be designed for easy access for cleaning and maintenance. This involves strategic placement of cleanouts and access points.
    • Regular maintenance is key to prevent blockages and ensure system longevity.

  8. Compliance with Codes and Standards:

    • Adherence to local building codes and plumbing standards is mandatory. These regulations specify requirements for materials, design, installation, and maintenance.

  9. Use of Cleanouts:

    • Cleanouts allow for easy access to the drainage system for removal of blockages and routine cleaning.
    • They are typically installed at intervals specified by code, at changes in direction, and at the base of stacks.

Key Equations:

  • Manning Formula for velocity of flow (V): V = 1.486 * R^(2/3) * S^(1/2) / n

    • V: velocity of flow (ft/s)
    • R: hydraulic radius (ft)
    • S: hydraulic slope (ft/ft)
    • n: Manning's roughness coefficient

  • Quantity rate of flow (Q): Q = A * V

    • Q: quantity rate of flow (cfs or gpm)
    • A: cross-sectional area of flow (ft²)

  • Hydraulic radius (R): R = A / P

    • P: wetted perimeter (ft)

Steps for Calculations:

  1. Identify the given values: pipe size, slope, roughness coefficient, and flow condition (full or half-full).
  2. Calculate hydraulic radius (R):
    • For full flow: R = D/4 (D = pipe diameter)
    • For half-full flow: R = πD²/8
  3. Find R^(2/3) from Tables 1-1 or 1-2. (shown below)
  4. Calculate velocity of flow (V) using Manning's formula.
  5. Find cross-sectional area of flow (A) from Tables 1-1 or 1-2.
  6. Calculate quantity rate of flow (Q) using Q = A * V.

Practical Examples:

  1. Calculate the velocity of flow in a 6-inch diameter pipe with a slope of 0.01 and a roughness coefficient of 0.013, flowing full.

    • R = D/4 = 0.125 ft
    • R^(2/3) = 0.25 ft (from Table 1-1)
    • V = 1.486 * 0.25 * 0.01^(1/2) / 0.013 ≈ 4.54 ft/s

  2. Determine the flow rate in gallons per minute for the same pipe in example 1.

    • A = 0.1964 ft² (from Table 1-1)
    • Q = 0.1964 * 4.54 = 0.89 cfs
    • Q = 0.89 * 450 ≈ 400 gpm

  3. Calculate the velocity and flow rate for the same pipe flowing half-full.

    • R = πD²/8 = 0.0982 ft
    • R^(2/3) = 0.25 ft (still the same)
    • V ≈ 4.54 ft/s (velocity remains the same for half-full flow)
    • A = 0.0982 ft² (from Table 1-2)
    • Q ≈ 0.44 cfs ≈ 200 gpm (flow rate is reduced for half-full flow)

Remember:

  • Use consistent units throughout the calculations.
  • Tables 1-1 and 1-2 provide values for R, R^(2/3), and A, saving calculation steps. These tables will be provided in the mep guy study/crib sheet as we get closer to the test.
  • Consider the flow condition (full or half-full) when determining hydraulic radius and cross-sectional area.
 

Table 1-1: Values of R, R^(2/3), and A for Full Flow

 
 
Pipe Size (in) R = D/4 (ft) R^(2/3) (ft) A (ft²)
1½ 0.0335 0.1040 0.01412
2 0.0417 0.1200 0.02180
2½ 0.0521 0.1396 0.03408
3 0.0625 0.1570 0.04910
4 0.0833 0.1910 0.08730
5 0.1040 0.2210 0.13640
6 0.1250 0.2500 0.19640
8 0.1670 0.3030 0.34920
10 0.2080 0.3510 0.54540
12 0.2500 0.3970 0.78540
15 0.3125 0.4610 1.22700
 
 

Table 1-2: Values of R, R^(2/3), and A for Half-Full Flow

 
 
Pipe Size (in) R = D/4 (ft) R^(2/3) (ft) A (ft²)
1½ 0.0335 0.1040 0.00706
2 0.0417 0.1200 0.01090
2½ 0.0521 0.1396 0.01704
3 0.0625 0.1570 0.02455
4 0.0833 0.1910 0.04365
5 0.1040 0.2210 0.06820
6 0.1250 0.2500 0.09820
8 0.1670 0.3030 0.17460
10 0.2080 0.3510 0.27270
12 0.2500 0.3970 0.39270
15 0.3125 0.4610 0.61350
 
 
 
 

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