Critical flow at maximum discharge for given energy

Specific energy

Specific energy is the energy per unit weight of water at a section, measured above the channel bed:

$$E=y+\frac{V^2}{2g}\qquad(\alpha\approx1)$$

Proof: critical flow gives maximum discharge for given $E$

Consider a rectangular channel (width $b$, $q=Q/b$). With $V=q/y$:

$$E=y+\frac{q^2}{2gy^2}\;\Rightarrow\;q=y\sqrt{2g(E-y)}$$

For constant $E$, $q$ is a function of $y$ only. Differentiate $q^2=2gy^2(E-y)$:

$$\frac{d(q^2)}{dy}=2g\left(2yE-3y^2\right)=0\;\Rightarrow\;y=\frac{2E}{3}$$

At this depth the second derivative is negative, so $q$ is a maximum. Then

$$q_{max}=\frac{2E}{3}\sqrt{2g\left(E-\frac{2E}{3}\right)}=\sqrt g\left(\frac{2E}{3}\right)^{3/2}$$

The critical depth in a rectangular channel is $y_c=\sqrt[3]{q^2/g}$, i.e. $q^2=gy_c^3$. Comparing: $y=\dfrac{2E}{3}=y_c$ and $E=\dfrac32y_c=E_{min}$. Also

$$V=\frac{q}{y}=\sqrt{\frac{g\cdot y^3}{y^2}}=\sqrt{gy}\;\Rightarrow\;Fr=1$$

Hence, for a given specific energy, the discharge is maximum when the flow is critical. (For any section, $dQ/dy=0$ gives $Q^2T/(gA^3)=1$, the same critical-flow condition.)

  y ^
  E |*            q = 0 at y = E
    |   .
    |      .        upper part: subcritical
 yc |--------*  q_max (critical flow)
    |      .        lower part: supercritical
    |   .
    |.                q = 0 at y = 0
    +--------+--------> q   (E constant)
           q_max