An 8 -ft-diameter concrete drainage pipe that flows half full is to be replaced by a concrete-lined V-shaped open channel having an interior angle of \(90^{\circ} .\) Determine the depth of fluid that will exist in the \(V\) -shaped channel if it is laid on the same slope and carries the same discharge as the drainage pipe.

The cross-sectional area \(A_1\) of the flow inside the drainage pipe can be computed using the formula: \(A_1 = \frac{1}{2} * \pi * r^2\), where \(r = \frac{8 ft}{2} = 4ft\) is the radius. Let's designated the hydraulic radius \(R_1\) to be the ratio of the cross-sectional area to the wetted perimeter of the circular pipe (half of the pipe's circumference), which is: \(R_1 = \frac{A_1}{\pi*r}.\) The flow rate or the discharge should be kept constant and is given by \(Q = A_1 * V_1,\) where \(V_1\) is the velocity of flow in the pipe. The velocity can be evaluated using the Manning's formula: \(V_1 = \frac{1}{n} * R_1^{\frac{2}{3}} * S^{\frac{1}{2}},\) where \(n\) is the Manning roughness coefficient and \(S\) is the slope.

Now, let's consider the V-shaped open channel with interior angle of \(90^{\circ}\). For each cross section, the area is \(A_2 = d^2\), where \(d\) is the depth of the channel. The wetted perimeter for the channel, \(P_2\) is \(P_2 = 2 * d * \sqrt{2},\) therefore the hydraulic radius of the flow \(R_2 = \frac{A_2}{P_2}.\) The velocity of flow in the channel, \(V_2\) can also be evaluated using the Manning's formula: \(V_2 = \frac{1}{n} * R_2^{\frac{2}{3}} * S^{\frac{1}{2}},\) and hence the discharge in the V-shaped channel \(Q = A_2 * V_2.\)

Since the same discharge must be maintained, the flow rates in the pipe and the V-shaped channel must be the same, so we can set them equal and solve for the unknown which is \(d,\) i.e., \(A_1 * V_1 = A_2 * V_2.\)