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Benefits
Design
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Hydrologic Design of Pervious Concrete
Appendix E: Slope Effects
►Effects of Slope on the Storage Capacity of the Slab◄
Figure E1 shows the volume that can be filled (Volf, slab) of a pervious concrete slab on a slope greater than d/L. This volume does not include the effects of porosity. The volume is:
where w is width (not shown), d is depth and r is that portion of the length of the slab (L) which holds runoff (w, d, and L all in consistent units).
The slope (s) is defined as h/L. From similar triangles, h/L is approximately equal to d/r (the approximation is good for slopes less than 12%), so, r = d/s (approximately), and Equation E1 can be expressed as:
The filled volume can be expressed as a percentage of the nominal volume (w d L) by dividing Volf, slab by (w d L), that is, %Vol = ? w d² / (s w d L), or, with simplification:
where d and L are depth and length of the slab, respectively, in consistent units and s is slope.
Example E1*
Consider a 6-in. (150mm) deep, 100 ft (30.5m) long pervious concrete slab with a 1% slope. The slope (0.01) is greater than d/L ((6 in./12 in./ft) / 100 ft = 0.005), so Equation E3 is applicable:
%Vol = (6 in. / 12 in./ft) / [(2) (0.01) (100 ft)] = 25%.
%Vol = (150mm / 1000mm/m) / [2 (0.01) (30.5m) = 25%.
These are slab volumes and must be multiplied by the effective porosity of the pervious concrete to determine the storage capacity.
When the slope is less than d/L, a different equation should be used. Figure E2 shows the additional storage found with flatter slopes. In this case,
Volf, slab = 1/2 u w L + v w L (approx.)
The slope (s) is given (approximately) by u/L, and v + u = d. Substituting for u and v leads to:
which can be simplified to
As above, the filled volume can be expressed as a percentage of the nominal volume (w d L) by dividing Volf, slab by (w d L), or, with simplification,
where d and L are depth and length of the slab, respectively, in consistent units and s is slope.
Example E2*
Consider a 6-in. (150mm) deep, 100 ft (30.5m) long pervious concrete slab with a 0.25% slope. The slope (0.0025) is less than d/L ((6 in. /12 in./ft) / 100 ft = 0.005), so Equation E6 is applicable. In the other direction, the slab is level and extends 300 ft (91m).
% Vol = 1 - (0.0025) (100 ft) / [(2) (6 in. / 12 in./ft)] = 75%.
% Vol = 1 - (0.0025) (30.5m) / [(2) (150mm / 1000mm/m)] = 75%.
A pervious concrete slab of these dimensions with an effective porosity of 15% would be able to store 11,250 ft3 (46.8m3) of stormwater. The effective storage capacity is:
(slope effect) (porosity) (nominal volume = w d L)
(0.75) (0.15) (100 ft) (300 ft) (6 in / 12 in/ft) = 11,250 ft3 (0.75) (0.15) (30.5m) (91m) (0.15m) = 46.8 m3
►Effects of Slope on the Storage Capacity of the Base Course◄
The volume of the stone base should be handled separately since it has a different porosity. If the slope is equal to or less than d/L, the entire volume of the stone base will be filled at maximum capacity and no adjustment is necessary.
where bavg is the average depth of the base course, and L and w are the length and width of the pervious concrete pavement system, respectively, all in consistent units.
If the slope is greater than d/L, not all of the voids will be filled. Figure E3 shows the effect of a slope greater than d/L on storage of a pervious concrete pavement system including a stone base.
The filled volume of the base course in this case is:
where bmin is the minimum depth of the base course, m and p are the height and base of the small triangular volume of base cutoff from filling, and h, L and w are as previously defined, all in consistent units.
As noted above, the slope, s, is defined as h/L, so h = s L. In addition, m/p = h/L = s by similar triangles, so m = s p. It is possible to solve for p by noting that p + r is approximately equal to L for slopes less than about 12%, and, with r = d/s, p = L – d/s, or:
½ hL = ½ sL² and ½ m p=½ sp2=½ s(L-d/s)2
Therefore, Volf, base = bmin L w + [½ s L² - ½ s (L - d/s)²] w, or
Example E3*
Consider a 6-in. (150mm) deep, 100 ft (30.5m) long pervious concrete slab with a 1 % slope over a base course with a minimum depth of 8 in. (200mm). The slope (0.01) is greater than d/L ((6 in. / 12 in./ft) / 100 ft = 0.005), so Equation E9 is applicable. In the other direction, the slab is level and extends 300 ft (91m). The effective porosity of the slab is 15% and the porosity of the base is 40%.
Volf, base = [(bmin + d) L - d²/2s] w
Volf, base = {(8 in. / 12 in./ft) + (6 in. / 12 in./ft)} (100 ft) - (6 in. / 12 in./ft)² / {(2) (0.01)}] (300 ft) = 31,250 ft3
Volf, base = {(0.20m) + (0.15m)} (30.5m) - (0.15m)² / {(2) (0.01)}] (91m) = 869 m3
In this case, the maximum depth of base course would be 20 in. (508mm) (the minimum base course depth plus s L). The nominal capacity of the base (bavg w L) is 35,000 ft3 (990m3) so the effective capacity of the base course is 89% of the nominal capacity.
The total capacity of this pervious concrete pavement system is the sum of the capacities of the slab and the base. The percent volume of the slab was determined to be 25% (see Example E1). The effective capacity of the slab (15% porosity) is therefore
(0.25)(0.15)(100 ft)(300 ft)(6 in/12 in/ft) = 280 ft3.
(0.25)(0.15)(30.5 m)(91 m)(0.15 m) = 15.6 m3.
The effective capacity of the base course (40% porosity) is
(0.40) (31,250 ft3) = 12,500 ft3.
(0.40) (869 m3) = 348 m3.
This pervious concrete pavement system can hold approximately 12,780 ft3 (348 m3) of runoff.
The effects of slope can be accounted for in the CN method analysis by multiplying the design depth by the percentage of storage capacity determined using the equations given in this appendix. For example, if the effective porosity of the base course was determined to be 89% and the design depth was selected to be 8 in. (200 mm), the value used in the analysis would be 7.1 in. (178 mm). A separate adjustment would be needed for the pervious concrete slab.
*Note: Hard conversions of dimensions are not used in these examples so the final solutions in SI and US customary units are not exactly equal. |
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