The mean velocity V of flow in straight prismatic channels is proportional to (A/p)^r, where A is the cross-sectional area of the flow, p is the wetted perimeter and r is approximately a constant (). Given the channel section for minimum flows (that is A₀ and p₀),the objective is to design a channel such that V has the same value for all larger discharges.

Assume a symmetric channel cross-section as shown below, where A₀ and p₀ are the minimum flow value of A and p. Let the shape of the channel be given by . (Note that in this application y, the height of the surface above the datum line, is the independent variable.) Then we want to find the function f(y)such that the mean flow velocity is independent of y. this implies that .

 Solution:

The area A is given by the integral of f(y). Thus , where x=f(y) and h>0.

The wetted perimeter p is given by (h>0)

Since A/A₀=p/p₀, we deduce that

Rearranging the integrals under a common integral sign gives

(h>0)since this is true for all h>0, it implies that the integrand must be identically zero. Thus x=f(y) satisfies the differential equation which implies giving

Integrating with respect to y then gives

Using the substitution u = cosh (p₀x/A₀) on the left-hand side gives

if the channel had width 2b where y=0, we can obtain the value of the constant of integration c as and deduce the formula for a suitable channel shape as . This solution is not unique and we note that the differential equation is satisfied by .

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