I'm trying to calculate the hydraulic radius but I think the solution in my textbook may be wrong. But I wanted to see if anyone here might be able to point out my mistake, if any.
A trapezoidal channel has a bottom width of 3 ft and side slopes of 2 ft vertical and 1 ft horizontal. Assume a depth of 2 ft.
SInce the depth is 2 ft, the "base" of the triangles should be 1 ft as per the side slopes given. To find the length of the side, I did 1^2 + 2^2 = c^2.
c was calculated to be 2.236
This will give a wetted perimeter of 2.236(2) + 3 = 7.472
Area of triangle: 1/2 x 1 ft x 2ft x 2 triangles = 2 ft
Area of rectangle: 3 ft x 2 ft = 6 ft
So the hydraulic area = 6 ft + 2 ft = 8
Hydraulic radius = A/P = 8/7.472 = 1.07 ft.
My textbook says Rh should be 1.72 ft. Am I doing something wrong?
A trapezoidal channel has a bottom width of 3 ft and side slopes of 2 ft vertical and 1 ft horizontal. Assume a depth of 2 ft.
SInce the depth is 2 ft, the "base" of the triangles should be 1 ft as per the side slopes given. To find the length of the side, I did 1^2 + 2^2 = c^2.
c was calculated to be 2.236
This will give a wetted perimeter of 2.236(2) + 3 = 7.472
Area of triangle: 1/2 x 1 ft x 2ft x 2 triangles = 2 ft
Area of rectangle: 3 ft x 2 ft = 6 ft
So the hydraulic area = 6 ft + 2 ft = 8
Hydraulic radius = A/P = 8/7.472 = 1.07 ft.
My textbook says Rh should be 1.72 ft. Am I doing something wrong?
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