Question 13.8: A plan of a sewer network for a housing estate is shown in F......

From Table 13.11, take the time of entry as 7 min. Since impermeable areas are given, assume C = 1.0. The HR chart in Fig. 6.15 can be used to obtain Q_{FULL} and V. The calculations are shown in Table 13.12 and a commentary is given below.

Column 1. Draw a plan of the sewer network (Fig. 13.18). Working back from the outfall, the longest run of pipes is the primary sewer, the most upstream length of which is designated as 1.0, the next downstream as 1.1, then 1.2, and so on. The sewers in the most upstream branch are numbered 2.0, 2.1, 2.2, etc., and the next branch 3.0, 3.1, 3.2, etc. Thus the first number indicates relative position in the network, and the second number after the decimal point the location within the particular branch.

The design process starts with pipe 1.0, then progresses downstream, picking up branches like 2.0 and 3.0, 3.1 as they occur. Each numbered length is designed individually, and its design point is always at the downstream end, except for the special case of trunk or carrier sewers (see Pipe 1.2, after Column 12 below) when it is at the upstream end. At each design point the value of t_{C} is calculated and used to obtain i.

In this example, pipe 1.0 is designed first, then 2.0 (the branch), then 1.1 which carries runoff from all 3 drainage areas.

Column 2. Determine the length (L) of the sewers. This is normally the plan length measured off a drawing, not the exact slope length (the difference is usually unimportant). The lengths are given in this example.

Column 3. When possible, make the gradient of the sewer equal to the slope of the ground surface so that the depth of excavation is constant with at least the required 0.9 – 1.2 m of cover. Generally pipes should be as shallow as possible since 60% of the cost of construction may arise from excavating and backfilling the trench, and this increases with depth. The gradient is given in Example 13.8, but it must be expressed in m/100 m to be consistent with the right-hand scale of the HR Wallingford pipe design chart in Fig. 6.15. Thus for pipe 1.0 the gradient is 1 in 250 or (1/250) × 100 = 0.4 m/100 m.

Column 4. Guess the diameter of pipe required from the range specified in the question. The smallest is 0.150 m so, for sewer 1.0, start with that. (In professional practice, check what materials and sizes are available – as a result of European harmonisation, after 0.375 m some manufacturers’ standard diameters may be 0.400 m, 0.450 m, 0.500 m, 0.600 m.)

Within a particular branch, as the flow increases, pipe diameters increase in the downstream direction. Within a branch, never make the downstream pipe smaller than the one upstream, even if the calculations suggest it as a possibility (e.g. pipe 1.1 is 0.375 m diameter, the same as pipe 1.0, despite 1.1 having a steeper gradient that might allow a smaller pipe).

If the wrong diameter is guessed and the pipe fails, just repeat the calculation on the line below. It should never be necessary to repeat the calculation more than once. For instance, with sewer 1.0 a 0.150 m diameter pipe fails, but we know from column 12 that Q_{P} = 0.095 m³/s, so from the HR chart in Fig. 6.15 it is possible to determine directly which of the standard pipe sizes could accommodate this. Here it is a 0.375 m pipe.

Column 5. Knowing the pipe gradient and its guessed diameter, the value of Q_{FULL} is obtained from the HR chart (i.e. Q in Fig. 6.15; note 1 m³ = 1000 litres). In Table 13.12, with pipe 1 at 0.150 m diameter, Q_{FULL} = 11 litres/s = 0.011 m³/s. This is the discharge capacity of the pipe when flowing full, so the value of Q_{P} estimated from the rational equation in column 12 has to be less than this. If it isn’t, then a larger pipe must be adopted and the calculation repeated.

Column 6. Knowing the pipe gradient and its guessed diameter, the mean flow velocity (V) in the full pipe is obtained from Fig. 6.15. Remember V must be ≥ 1 m/s. Pipe 1.0 has V = 0.63 m/s < 1 m/s when the diameter is 0.150 m, but this does not matter since the design fails. A 0.375 m diameter gives V = 1.15 m/s and is OK.

Column 7. The time of flow in a particular sewer length t_{F} = (L/60V) min. Thus when pipe 1.0 is 0.150 m diameter, t_{F} = 80/(60 × 0.63) = 2.12 min.

Column 8. For the first sewer of a branch (i.e. 1.0, 2.0, 3.0, etc.) the time of concentration is t_{ C }=t_{ E }+t_{ F } where in this example t_{E} = 7 min (see Table 13.11). For the 0.150 m diameter pipe 1.0, t_{C} = 7.00 + 2.12 = 9.12 min.

For the second sewer (i.e. 1.1, 2.1, 3.1, etc.), and all subsequent sewers in that branch, just keep adding the new value of t_{F} to the previous value of t_{C}. The exception to this is a carrier or trunk sewer (see below). Where two sewers join, to reduce the tendency to overdesign, use the longest upstream value of t_{C}. Thus in Table 13.12, for sewer 1.1, use 8.16 min from pipe 1.0 (not 7.79 min from 2.0) so that t_{C} = 8.16 + 0.56 = 8.72 min.

Column 9. In professional practice the location specific rainfall intensity for a given return period would be obtained from maps, tables, the Wallingford Procedure or software, but for simplicity the formula in Table 12.5 is used here. The rational method assumes t = t_{C}. For the 0.150 m diameter pipe 1.0, i = 690/(9.12 + 7) = 42.8 mm/h.

Column 10. This is the impermeable area (A_{IMP}) drained by each individual sewer length, and is given in the question. Otherwise, it may be measured from plans, or estimated from Table 13.10 or similar.

Column 11. This is the total impermeable area (ΣA_{IMP}) drained by all of the sewers upstream of the design point. e.g. for pipe 1.1 we have ΣA_{IMP} = 0.008 + 0.001 + 0.003 = 0.012 km².

Column 12. With C = 1.0 the peak flow at any design point is Q_{P} = 0.278  i  \Sigma A_{ IMP } m³/s. If Q_{ P } < Q_{\text {fULL }} then the assumed diameter is OK; if Q_p > Q_{\text {FULL }} then the assumed diameter is too small and should be increased. In Table 13.12, with pipe 1.0 at 0.150 m diameter, Q_{\text {FULL }} = 0.011  m ^3 / s <Q_p = 0.095  m ^3 / s, so it fails. For this Q_{P}, Fig. 6.15 shows that a 0.375 m pipe is needed. Re-calculating, now Q_{\text {FULL. }} = 0.128  m ^3 / s />Q_p=0.101  m ^3 / s, so it is OK.

Note that increasing the diameter by one increment can increase Q_{\text {FULL. }}by a relatively large amount, so a marginal failure may sometimes be acceptable. However, this does not allow for any future increase in runoff, and could result in the sewer’s design capacity being exceeded slightly more often than originally intended when selecting the return period, perhaps with some local surface flooding. On a housing estate this may not matter; on a motorway it may be dangerous. An undersize sewer may also cause the water level in the upstream manhole to rise above the crown of the pipe, turning the normal gravity flow into pressurised flow. This would increase the hydraulic gradient and hence Q_{\text {FULL. }}, but could also result in water spilling out of the manhole (Fig. 13.17).

Pipe 1.2. This is a carrier or trunk sewer that merely transfers the flow between two points and does not have any area draining directly to it (e.g. a pipe passing under a road embankment). In this special case the design point is assumed to be at the upstream end, so all that is really required is to ensure that the gradient and diameter of the pipe are sufficient to carry the flow from upstream (0.146 m³/s in this example).

The time of flow in the carrier sewer is not needed to design it, but t_{F} is calculated and used to obtain t_{C} for the pipes downstream. In Table 13.12, sewer 1.2 at 0.450 m diameter has a time of flow of 1.30 min, giving t_{C} = 8.72 + 1.30 = 10.02 min. It is shown in brackets because it is not needed on this line, but when Example 13.8 is continued in Self Test Question 13.5, pipe 1.3 immediately downstream of the carrier sewer has t_{C} = 10.02 + 0.82 = 10.84 min.

Table 13.12 shows the complete solution to Example 13.18. In practice, for new developments consideration would now be given to the underground storage required in the sewer system to prevent surface flooding during the 1 in 30 years storm, and to limit the peak discharge to the rate of runoff from a greenfield site of equal size.