A drain has a rectangular profile. W is the width of the drain in meters, and
D is the depth of the water in meters.
In the case that the depth is much less than the width the amount of water that flows per second through a rectangular drain is given by the following formula:
Q = (A/n)*(D^2/3)*(S^1/2) ------------- (1)
In which:
- Q is the volume of water that
flows per second through the
drain (in m3/s);
- A = W×D, the cross-sectional
area of the drain up to the level
of the water (in m2)
- n is a parameter describing the
resistance to the flow of water;
- S is the gradient of the river
(in m/m).
Use formula 1 to derive the units of n.
1) Solve (1) with respect of n:
"n={A\\cdot D^\\frac{2}{3}\\cdot S^\\frac{1}{2}\\over Q}" (2);
2) Insert in(2) units:
[n]="{[A]\\cdot [D]^\\frac{2}{3}\\cdot [S]^\\frac{1}{2}\\over [Q]}" ="{[m^2]\\cdot [m]^\\frac{2}{3}\\cdot [\\frac{m}{m}]^\\frac{1}{2}\\over [\\frac{m^3}{s}]}" ="[\\frac{s}{\\sqrt[3]m}]"
So units of n is sec divided by 3- root of meter.
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