---
title: "Specific energy, minimum value and critical flow"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:specific-energy-minimum-critical-curve"
source: https://www.bctengineer.com/q/semester_4/hydraulics/specific-energy-minimum-critical-curve
site: BCT Engineering
---

# Specific energy, minimum value and critical flow

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

- Asked in: 2079 Chaitra, 1+3+2 marks; 2071 Magh, 6 marks (asked 2 times)
- Video: [Specific Energy and Specific Energy Curve - Flow through Open Channels - Applied Hydraulics](https://www.youtube.com/watch?v=oiahLm1_se4) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/specific-energy-minimum-critical-curve
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

Define specific energy. Prove that for a given discharge, the specific energy will be minimum when the flow in the channel is critical. Draw specific energy curve and show the alternate depths, critical specific energy, subcritical and supercritical zones.

## Short answer

Specific energy E = y + Q²/(2gA²). dE/dy = 1 − Q²T/(gA³) = 0 at critical flow (Fr=1), giving minimum E. The specific energy curve has a subcritical upper limb and supercritical lower limb; for E > E_min two alternate depths exist. For a rectangular channel, E_min = 1.5y_c.

## Answer

#### Specific energy
**Specific energy** $E$ is the energy per unit weight of water measured with the channel bed as datum:
$$
E=y+\frac{\alpha V^2}{2g}\approx y+\frac{V^2}{2g}=y+\frac{Q^2}{2gA^2}
$$
It has the unit of length and, unlike the total head, it can increase or decrease along the flow.

#### Proof of minimum energy at critical flow
For constant $Q$, $E$ is a function of $y$ only. Differentiate with respect to $y$ ($dA=T\,dy$):
$$
\frac{dE}{dy}=1-\frac{Q^2}{gA^3}\frac{dA}{dy}=1-\frac{Q^2T}{gA^3}
$$
For minimum $E$, $dE/dy=0$:
$$
\frac{Q^2T}{gA^3}=1\;\Rightarrow\;\frac{V^2}{g(A/T)}=1\;\Rightarrow\;Fr=\frac{V}{\sqrt{gD_h}}=1
$$
Since $\dfrac{d^2E}{dy^2}=\dfrac{3Q^2T^2}{gA^4}-\dfrac{Q^2}{gA^3}\dfrac{dT}{dy}>0$ for usual sections, this is a minimum. The condition $Fr=1$ is critical flow. Hence **the specific energy is minimum at critical flow**.

For a rectangular channel: $y_c=\sqrt[3]{q^2/g}$ and $E_{min}=\dfrac32y_c$.

#### Specific energy curve
```text
  y ^
    |                    .   upper limb: subcritical (Fr < 1),
    |                  .     approaches the line E = y
 y1 |----------------*
    |             .
    |         .
 yc |-------*  <- critical point: y = yc, E = E_min
    |         .
    |             .
 y2 |----------------*       lower limb: supercritical (Fr > 1),
    |                    .   approaches the E-axis
    +-------+--------+-------> E
          E_min      E
```
- The curve has two branches for a given $E>E_{min}$ and a single point at $E=E_{min}$.
- **Alternate depths:** the two depths $y_1$ (subcritical, upper limb) and $y_2$ (supercritical, lower limb) have the same specific energy $E$.
- **Critical depth $y_c$ and $E_{min}$** are at the nose of the curve.
- **Subcritical zone:** $y>y_c$, $V<\sqrt{gD_h}$, upper limb (asymptotic to the 45° line $E=y$).
- **Supercritical zone:** $y<y_c$, $V>\sqrt{gD_h}$, lower limb (asymptotic to the $E$-axis).
- For a rectangular section the curve is a cubic: $y^3-Ey^2+\dfrac{q^2}{2g}=0$.

## Check yourself

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

1. Specific energy is:
   A. Energy per unit weight measured with the channel bed as datum, E = y + V^2/2g
   B. Energy per unit weight measured from a fixed datum
   C. Momentum per unit weight
   D. The pressure head alone
   Answer: A. E = y + V^2/2g relative to the bed; unlike total head it can rise or fall along the flow. (Key point: Definition)

2. Setting dE/dy = 1 - Q^2 T/(g A^3) = 0 means:
   A. Fr = 0
   B. Fr = 1, the flow is critical
   C. Fr > 1
   D. V = 0
   Answer: B. Q^2 T/(g A^3) = Fr^2 = 1, so E is minimum at critical flow. (Key point: Proof of minimum)

3. On the specific energy curve, alternate depths are:
   A. Two depths with the same specific force
   B. Two depths with the same velocity
   C. The critical depth and normal depth
   D. Two depths with the same E: a subcritical (upper limb) and a supercritical (lower limb)
   Answer: D. For E > Emin, the curve gives one depth on each limb. (Key point: Alternate depths)

4. For a rectangular channel, y_c and E_min are:
   A. y_c = (q^2/g)^(1/3), E_min = 1.5 y_c
   B. y_c = q^2/g, E_min = 2 y_c
   C. y_c = (q^2/g)^(1/3), E_min = y_c
   D. y_c = (q^2/g)^(1/2), E_min = 3 y_c
   Answer: A. Critical depth is cube root of q^2/g and E_min = y_c + y_c/2 = 3 y_c/2. (Key point: Rectangular channel results)

## Related in Hydraulics

- [Specific energy from alternate depths](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aspecific-energy-in-terms-of-alternate-depths) — Hydraulics, chapter 7
- [Specific force and its minimum at critical flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aspecific-force-expression-minimum-critical) — Hydraulics, chapter 7, asked 3×
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- [Minimum specific energy for triangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aminimum-specific-energy-triangular-5-4-yc) — Hydraulics, chapter 7
- [Program for alternate depths in rectangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aalgorithm-alternate-depths-rectangular) — Hydraulics, chapter 7

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