Question

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.

Answer 1

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.