---
title: "Specific force and venturi flume relations"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:specific-force-discharge-max-venturi-flume"
source: https://www.bctengineer.com/q/semester_4/hydraulics/specific-force-discharge-max-venturi-flume
site: BCT Engineering
---

# Specific force and venturi flume relations

Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow

- Asked in: 2069 Bhadra, 2+4+3+3 marks
- Video: [Experiment 8: Generalized Specific Energy and Specific Force Curve](https://www.youtube.com/watch?v=ojVnXtuF0pU) — Alim Bro
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/specific-force-discharge-max-venturi-flume
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

What is specific force? Prove that for a given specific force the discharge in a given channel section is maximum when the flow is in the critical state. A venturiflume in a rectangular channel of width B has the throat width of b. The depth of liquid at entry is H and at the throat is h. Prove that following relation exists for the discharge and width ratio: $Q = 3.13\,bH^{3/2}\left(\frac{h}{H}\right)^{3/2}$ and $\frac{b}{B} = \sqrt{3}\left(\frac{h}{H}\right) - \sqrt{3}\left(\frac{h}{H}\right)^{3/2}$.

## Short answer

Specific force is F = Q^2/(gA) + Aȳ (momentum function). For given F, Q is max at critical state because dQ/dy=0 yields Fr=1. Applying this to a venturi flume with critical flow at the throat gives Q = 3.13 b H^(3/2)(h/H)^(3/2) and b/B = (H/h)√(3 − 2 H/h).

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Specific force at a section is:
   A. y + V^2/2g
   B. Q^2/(gA) + A*ybar (momentum flux plus hydrostatic force per unit weight)
   C. Q^2/(gA^3)
   D. rho Q V
   Answer: B. F = Q^2/(gA) + A ybar, the sum of momentum flux and pressure force per unit weight. (Key point: Definition of specific force)

2. Setting dQ^2/dy = 0 at fixed F leads to which condition?
   A. Q^2 T/(g A^3) = 1, i.e. Fr = 1
   B. A = T y
   C. S0 = Sf
   D. Q^2/(gA) = A ybar only for rectangles
   Answer: A. T(F - A ybar) = A^2 with F - A ybar = Q^2/(gA) gives Q^2 T/(gA^3) = 1, the critical flow condition. (Key point: Maximum discharge condition)

3. Why is Q = 3.13 b h^(3/2) at the throat?
   A. Flow is assumed critical at the throat, so q = sqrt(g) h^(3/2)
   B. Flow is uniform in the throat
   C. The throat depth equals the entry depth H
   D. Head loss is assumed equal to velocity head
   Answer: A. With h = yc, Q = b sqrt(g h^3) and sqrt(9.81) = 3.13. (Key point: Venturi throat discharge)

4. Which energy balance is used to relate b/B to H and h?
   A. H + V^2/2g = h
   B. H + Q^2/(2gB^2H^2) = 1.5 h, with Q^2 = g b^2 h^3
   C. H = 1.5h
   D. Specific force equal at both sections
   Answer: B. Entry energy equals the critical energy 1.5h at the throat; substituting Q^2 = g b^2 h^3 gives b/B in terms of H/h. (Key point: Energy equation for width ratio)

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