Minimum specific energy for triangular channel

Condition for critical flow

Minimum specific energy occurs at critical flow:

$$\frac{Q^2T}{gA^3}=1\quad\text{or}\quad \frac{Q^2}{gA^2}=\frac{A}{T}\qquad(1)$$

Triangular section

Side slope $z$ (H:V), depth $y$:

$$A=zy^2,\qquad T=2zy,\qquad \frac{A}{T}=\frac{zy^2}{2zy}=\frac y2$$

 T = 2zy
 \~~~~~~~/
  \     /  y
   \   /
    \ /

At critical depth $y=y_c$

From (1):

$$\frac{Q^2}{gA^2}=\frac{A}{T}=\frac{y_c}{2}\;\Rightarrow\;\frac{V_c^2}{2g}=\frac{Q^2}{2gA^2}=\frac{y_c}{4}$$

(the velocity head at critical flow is half the hydraulic depth, $\dfrac{A}{2T}=\dfrac{y_c}{4}$).

Minimum specific energy

$$E_c=y_c+\frac{V_c^2}{2g}=y_c+\frac{y_c}{4}=\frac54\,y_c$$ $$\boxed{E_c=\frac54\,y_c}$$

Hence proved. (Compare: rectangular channel $E_c=\tfrac32y_c$; parabolic channel $E_c=\tfrac43y_c$. In general $E_c=y_c+\dfrac{A_c}{2T_c}$.)