Example 62. Compute the normal depth and velocity of the trapezoidal channel shown above, carrying 400 cfs, with b = 20 ft, z = 2, S_{o} = 0.0016, and n = 0.025. Solution. The following proportion holds: z/1 = x/y From which: x = zy T = b + 2x = b + 2zy A= (1/2) (b + T) y = (1/2) (b + b + 2zy) y = (b + zy) y P = b + 2 (y^{2} + z^{2}y^{2}) ^{1/2} = b + 2y (1 + z^{2})^{1/2} R = { [(b/2) + (zy)/2] y} / [ (b/2) + y (1 + z^{2})^{1/2} ] Q = (1.486/n) A R^{2/3} S^{1/2} (Qn) / (1.486 S^{1/2}) = A R^{2/3}
(Qn) / (1.486 S^{1/2}) = {2[(b/2) + (zy)/2] y} {[(b/2) + (zy)/2]y } ^{2/3} / [ (b/2) + y (1 + z^{2})^{1/2} ]^{2/3}
[(Qn) / (1.486 S^{1/2})] [ (b/2) + y (1 + z^{2})^{1/2} ]^{2/3} = 2 {[(b/2) + (zy)/2] y } ^{5/3}
[(Qn) / (2 × 1.486 S^{1/2})] [ (b/2) + y (1 + z^{2})^{1/2} ]^{2/3} = {[(b/2) + (zy)/2] y } ^{5/3}
[(Qn) / (2 × 1.486 S^{1/2})]^{3/2} [ (b/2) + y (1 + z^{2})^{1/2} ] = {[(b/2) + (zy)/2] y} ^{5/2}
{ [(b/2) + (zy)/2] y} ^{5/2}  { [(Qn) / (2 × 1.486 S^{1/2})]^{3/2} [(1 + z^{2})^{1/2}]y}  [(Qn) / (2 × 1.486 S^{1/2})]^{3/2} (b/2) = 0 [(Qn) / (2 × 1.486 S^{1/2})]^{3/2} = 771.5 (1 + z^{2})^{1/2} = 2.236 { [(b/2) + (zy)/2] y} ^{5/2}  [(771.5 × 2.236)y]  771.5 (b/2) = 0 [(10 + y)y]^{5/2}  1725y  7715 = 0 By trial and error: y_{n} = 3.36 ft. A_{n} = [(b + zy_{n}) y_{n}] = 89.78 ft. V_{n} = Q/A_{n} = 400/89.78 = 4.46 fps.
The normal flow depth equation is:
f(y) = { [(b/2) + (zy)/2] y } ^{5/2}  { [(Qn) / (2 × 1.486 S^{1/2})]^{3/2} [(1 + z^{2})^{1/2}]y}  [(Qn) / (2 × 1.486 S^{1/2})]^{3/2} (b/2) = 0
f(y) = { [(b/2) + (zy)/2] y } ^{5/2}  { [(Qn) / (2.972 S^{1/2})]^{3/2} [(1 + z^{2})^{1/2}]y}  [(Qn) / (2.972 S^{1/2})]^{3/2} (b/2) = 0 Changing variable to x for simplicity: f(x) = { [(b/2) + (zx)/2] x } ^{5/2}  { [(Qn) / (2.972 S^{1/2})]^{3/2} [(1 + z^{2})^{1/2}]x}  [(Qn) / (2.972 S^{1/2})]^{3/2} (b/2) = 0
To solve this equation by trial and error (brute force method), the following simple algorithm is suggested:
Solution by using Newton's approximation:
f(x) = { [(b/2) + (zx)/2] x} ^{5/2}  { [(Qn) / (2.972 S^{1/2})]^{3/2} [(1 + z^{2})^{1/2}]x}  [(Qn) / (2.972 S^{1/2})]^{3/2} (b/2)
f'(x) = x^{5/2} (5/2) [(b/2) + (zx)/2] ^{3/2} (z/2) + [(b/2) + (zx)/2] ^{5/2} (5/2) x^{3/2}  [(Qn) / (2.972 S^{1/2})]^{3/2} (1 + z^{2})^{1/2} f'(x) = (5/4) zx^{5/2} [(b/2) + (zx)/2]^{3/2} + (5/2) x^{3/2} [(b/2) + (zx)/2]^{5/2}  [(Qn) / (2.972 S^{1/2})]^{3/2} (1 + z^{2})^{1/2}
[When the root x_{r} is approached from the left (x_{o} < x_{r}), the slope f'(x_{o}) is positive, f(x_{o}) is negative, and the denominator (x_{o}  x_{r}) is negative].
[When the root x_{r} is approached from the right (x_{o} > x_{r}) (the usual case), the slope f'(x_{o}) is positive, f(x_{o}) is positive, and the denominator (x_{o}  x_{r}) is positive].
f'(x_{o}) = f(x_{o}) / (x_{o}  x_{r})
The root is: x_{r} = x_{o}  f(x_{o}) / f'(x_{o})
To solve the normal flow depth equation using Newton's approximation, the following algorithm is suggested:
