Question 10.6: The 50° triangular channel in Fig. E10.6 has a flow rate Q =...

Part (a)

This is an easy cross section because all geometric quantities can be written directly in terms of depth y:

P = 2y csc 50°                      A = y^2 cot 50°

R_h = \frac{1}{2}y \cos 50^{\circ}                          b_0 = 2y \cot 50^{\circ}                                       (1)

The critical flow condition satisfies Eq. (10.37a):

A_c = \left(\frac{b_0Q^2}{g}\right)^{1/3}                                              (10.37a)

or                      g(y^2_c \cot 50^{\circ})^3 = (2y_c \cot 50^{\circ})Q^2

Part (b)

With y_c known, from Eqs. (1) we compute P_c = 6.18  m,  R_{hc} = 0.760  m,  A_c = 4.70  m^2,  and  b_{0c} = 3.97  m. The critical velocity from Eq. (10.37b) is

V_c = \frac{Q}{A_c} = \left(\frac{gA_c}{b_0}\right)^{1/2}                                                  (10.37b)

Part (c)

With n = 0.018, we compute from Eq. (10.38) a critical slope:

S_c = \frac{n^2gA_c}{\alpha^2 b_0 R^{4/3}_{hc}} = \frac{n^2 V^2_c}{\alpha^2 R_{hc}^{4/3}} = \frac{n^2g}{\alpha^2 R^{1/3}_{hc}} \frac{P}{b_0} = \frac{f}{8} \frac{P}{b_0}                                              (10.38)