Specific force and venturi flume relations

Specific force

The specific force (momentum function) is the sum of the momentum flux and the hydrostatic pressure force per unit weight of water at a section:

$$F = \frac{Q^2}{gA} + A\bar{y}$$

where $A$ is the flow area and $\bar y$ is the depth of the centroid below the free surface. For a given $Q$, $F$ is minimum at the critical depth, and the two depths with equal $F$ are the conjugate depths of a hydraulic jump.

Discharge is maximum at critical state for a given $F$

From the definition, $Q^2 = gA\,(F - A\bar y)$. For fixed $F$ and a given section, $Q$ is a function of $y$ only. For $Q$ to be a maximum, $dQ^2/dy = 0$:

$$\frac{d}{dy}\left(AF - A^2\bar y\right) = 0$$

Use $dA/dy = T$ (top width) and $d(A\bar y)/dy = A$ (the first moment of area increases by $A\,dy$ when the depth increases by $dy$):

$$\begin{aligned} TF - \left[A\cdot A + A\bar y\cdot T\right] &= 0 \\ T\,(F - A\bar y) &= A^2 \end{aligned}$$

Now $F - A\bar y = Q^2/(gA)$, so

$$\frac{Q^2 T}{gA} = A^2 \;\Rightarrow\; \frac{Q^2 T}{gA^3} = 1$$

This is exactly the condition $Fr = 1$ for critical flow. The second derivative is negative, so this is a maximum. Hence for a given specific force the discharge is maximum at the critical state.

Venturi flume: discharge

Assume the flow passes through critical depth at the throat, so $h = y_c$. For a rectangular throat of width $b$:

$$Q = b\,\sqrt{g}\,h^{3/2} = \sqrt{9.81}\,b\,h^{3/2} = 3.13\,b\,h^{3/2}$$

Writing $h^{3/2} = H^{3/2}\left(\dfrac{h}{H}\right)^{3/2}$:

$$Q = 3.13\,b\,H^{3/2}\left(\frac{h}{H}\right)^{3/2}$$

Venturi flume: width ratio

Apply the energy equation between the entry (depth $H$, width $B$) and the throat (critical, $E = \tfrac{3}{2}h$), neglecting losses and with a level bed:

$$H + \frac{Q^2}{2gB^2H^2} = \frac{3}{2}h$$

Put $Q^2 = g\,b^2h^3$ (critical flow at the throat):

$$H + \frac{b^2h^3}{2B^2H^2} = \frac{3}{2}h \;\Rightarrow\; \frac{b^2}{B^2} = \frac{2H^2}{h^3}\left(\frac{3}{2}h - H\right) = 3\left(\frac{H}{h}\right)^2 - 2\left(\frac{H}{h}\right)^3$$ $$\boxed{\frac{b}{B} = \frac{H}{h}\sqrt{3 - 2\frac{H}{h}}}$$

Check: with $H = 1\ \text{m}$, $h = 0.8\ \text{m}$ this gives $b/B = 0.884$. For $B = 3\ \text{m}$, $b = 2.652\ \text{m}$, $Q = 3.13 \times 2.652 \times 0.8^{1.5} = 5.94\ \text{m}^3/\text{s}$, and the energy at entry $= 1 + 0.2 = 1.2 = 1.5 \times 0.8$. This agrees.

Note: the second form printed in the question, $\sqrt3\,(h/H) - \sqrt3\,(h/H)^{3/2}$, does not reduce to $b/B = 1$ when there is no contraction, so the standard energy-equation result above is used. The discharge relation is exactly as given.