Chemical Engineering - Fluid Mechanics - Discussion

The answer is A.

Rh = cross sectional are divided by wetted perimeter.

Thus, Rh = (X * L)/(2X + L).

Since L>>>>X, Rh = X * L/(L) = X.

Rh=flow area / wetted perimeter = (W * X) / (2 * (X+W)).
Since W (width)>>>X , Rh= (W * X)/(2 * W)= X/2.
So, I think the answer is B.

Rh=flow area / wetted perimeter = (W * X) / (2 * (X+W)).
Since W (width)>>>X , Rh= (W * X)/(2 * W)= X/2.

So, I think the answer is B.
According to this, B should be the right answer.

The hydraulic radius, R, is defined as the ratio of the cross-sectional area of the flow, A, to the wetted perimeter of the channel, P.
Note that 4x the hydraulic radius gives the hydraulic diameter.

R = A/P (where Dh = 4A/P).

In the case of a rectangular channel, the hydraulic radius is simply equal to the rectangle's area divided by the wetted perimeter:
R = A/P = bx / (x + b + x) = bx/(b + 2x).

In the case of an open channel with a very large width b:
lim b→∞ R = lim b→∞ x * [ b/(b + 2x) ] = x * 1 = x.
Therefore, R = x.