Question 14.1: A roof of area 80 m² drains to a rectangular soakaway like t......
A roof of area 80 m² drains to a rectangular soakaway like that in Fig. 14.6. It is to be designed for rainfall with a return period of 30 years. Assume W = 1.2 m, d_{S} = 1.5 m, the void space (porosity) of the granular fill is 25% (i.e. 0.25), and the soil percolation rate (f ) is 2.78 × 10^- 5 m/s. Calculations need to be undertaken for range of rainfall depths (R) to find the maximum L.
In this example R can be obtained indirectly from Table 12.5. To illustrate the procedure, say the storm duration t = 15 min, so with a return period of 30 years the average point intensity i = 1780/(t + 8) mm/h. Thus i = 1780/23 = 77.4 mm/h. This is the depth in 1 h, so in 15 min the depth will be 77.4 × (15/60) = 19.4 mm or R = 0.0194 m. Note the following.
• The 19.4 mm depth is the equivalent of the M30–15 (i.e. MT-D) value in the Wallingford Procedure – here a generalised, non-location specific value is adopted for simplicity.
• Remember, with t > 20 min, i = 2240/(t + 15) mm/h in Table 12.5 and below.
Equation (14.1) can be rearranged to solve directly for L.
A_{ IMP } \times R-\left[2 d_{ S 50}(W+L) \times f \times t\right]=d_{ S } \times W \times L \times P (14.1)
L=\frac{\left(A_{1 MP } \times R\right) – \left(2 d_{ S50 } \times W \times f \times t\right)}{\left(d_{ S } \times W \times P\right) + \left(2 d_{ S50 } \times f \times t\right)}=\frac{(80 \times 0.0194) – \left(2 \times 0.75 \times 1.2 \times 2.78 \times 10^{- 5} \times 900\right)}{(1.5 \times 1.2 \times 0.25) + \left(2 \times 0.75 \times 2.78 \times 10^{- 5} \times 900\right)}L = 3.1 m
Repeating the calculations with different rainfall durations and depths gives the values below. The maximum L is when t = 60 min, so the soakaway should be 1.2 m × 3.7 m in plan.
Check the time to empty from full to half volume when L = 3.7 m.
t_{ S 50}=\frac{0.5 d_{ S } \times W \times L \times P}{2 d_{ S 50}(W+L) \times f}=\frac{0.5 \times 1.5 \times 1.2 \times 3.7 \times 0.25}{2 \times 0.75(1.2+3.7) \times 2.78 \times 10^{- 5}}=4074 s =1.1 hThis is much less than 24 h and very satisfactory.
| t (min) | R (m) | L (m) |
| 10 | 0.0165 | 2.7 |
| 15 | 0.0194 | 3.1 |
| 30 | 0.0249 | 3.6 |
| 60 | 0.0299 | 3.7 |
| 120 | 0.0332 | 3.1 |
| Table 12.5 Typical point intensity–duration equations for central England (i = average rainfall intensity during a storm of duration t mins). With large catchments the average areal intensity can be obtained using the areal reduction factors in Fig. 12.9. For example, with a 1 in 5 year event, t = 30 min and catchment area = 5 km², Table 12.5 gives i = 34.0 mm/h; the ARF = 0.91, so the average intensity over the catchment is 34.0 × 0.91 = 30.9 mm/h. | ||
| Return period | t = 4 to 20 min | t = 20 to 120 min |
| 1 in 1 year | i=\frac{690}{t+7} mm / h | i=\frac{1000}{t+19} mm / h |
| 1 in 2 years | i=\frac{950}{t+8} mm / h | i=\frac{1210}{t+16} mm / h |
| 1 in 5 years | i=\frac{1230}{t+8} mm / h | i=\frac{1530}{t+15} mm / h |
| 1 in 30 years | i=\frac{1780}{t+8} mm / h | i=\frac{2240}{t+15} mm / h |
| 1 in 100 years | i=\frac{2420}{t+9} mm / h | i=\frac{2990}{t+16} mm / h |
