---
title: "Bernoulli for real flow with HGL and EGL"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:bernoulli-real-flow-hgl-egl"
source: https://www.bctengineer.com/q/semester_6/hydro_power/bernoulli-real-flow-hgl-egl
site: BCT Engineering
---

# Bernoulli for real flow with HGL and EGL

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2075 Bhadra, 2+2 marks
- Video: [Hydraulic and Energy Grade Line ? with animation [ HGL and EGL ]](https://www.youtube.com/watch?v=moI4DQNirAw) — NiLTime
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/bernoulli-real-flow-hgl-egl
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Derive Bernoulli's equation with the help of Euler's equation of fluid motion for a real flow. Sketch HGL and EGL for the horizontal circular pipe of gradually expanding cross-section towards the direction of flow.

## Short answer

Real flow Bernoulli includes head loss hL and kinetic energy factor α: p1/ρg + α1V1²/2g + z1 = p2/ρg + α2V2²/2g + z2 + hL. For horizontal expanding pipe, EGL falls due to losses, HGL rises due to pressure recovery, gap between them narrows downstream.

## Answer

#### Derivation from Euler's equation
Consider a fluid element of area $dA$ and length $ds$ along a streamline, rising $dz$ over the length $ds$. Forces along the streamline:
- net pressure force $= p\,dA - (p + dp)\,dA = -dp\,dA$;
- weight component $= -\rho g\,dA\,ds\,\dfrac{dz}{ds} = -\rho g\,dA\,dz$.

For steady flow the acceleration is $V\,dV/ds$, so Newton's second law gives
$$
-dp\,dA - \rho g\,dA\,dz = \rho\,dA\,ds\,V\frac{dV}{ds}
$$
Dividing by $\rho\,dA$ gives **Euler's equation**:
$$
\frac{dp}{\rho} + g\,dz + V\,dV = 0
$$
Integrating for incompressible flow and dividing by $g$:
$$
\frac{p}{\rho g} + \frac{V^2}{2g} + z = \text{constant}
$$
**Real flow:** a real fluid is viscous, so part of the energy is lost to friction and turbulence between two sections. Also, the velocity is not uniform across a section, so a kinetic energy correction factor $\alpha$ is used. Between sections 1 and 2:
$$
\frac{p_1}{\rho g} + \alpha_1\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \alpha_2\frac{V_2^2}{2g} + z_2 + h_L
$$
where $h_L$ is the head loss between 1 and 2.

#### HGL and EGL for a horizontal, gradually expanding pipe
- The **EGL** (total head) always falls in the direction of flow because of the loss $h_L$.
- As the pipe expands, the velocity falls, so the velocity head $V^2/2g$ decreases. Part of it is converted to pressure head, so the **HGL rises** in the direction of flow (pressure recovery), staying below the EGL by $V^2/2g$.
- The gap between the EGL and HGL narrows downstream.

```text
 head
  ^
  |  EGL (falls slowly due to losses)
  |  ___
  |   |  ''-.___
  |   | V1^2/2g ''-.___          EGL
  |   |               ''-.___
  |   |          _____..----|----- HGL
  |   |   __.--''   V2^2/2g |
  |  HGL--   (HGL rises as V falls)
  |
  |  pipe:  ==<  small  ......  large  >==
  +-----------------------------------> flow
```

## Check yourself

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

1. What is added to Bernoulli's equation for a real fluid?
   A. A head loss term h_L on the downstream side
   B. A density factor
   C. Nothing
   D. A pressure term on the upstream side
   Answer: A. Viscosity causes energy loss, so total head at 1 equals total head at 2 plus h_L. (Key point: Real flow modification)

2. Why is the factor alpha introduced for velocity head?
   A. Density varies
   B. Pressure varies with time
   C. The pipe is rough
   D. Velocity is not uniform across the section
   Answer: D. The true kinetic energy per unit weight with a non-uniform profile differs from V^2/2g, so alpha corrects it. (Key point: Kinetic energy correction)

3. What does the EGL do along a horizontal expanding pipe?
   A. Falls in the flow direction due to losses
   B. Rises
   C. Equals the HGL
   D. Stays horizontal exactly
   Answer: A. Total head can only fall because of the loss h_L (no energy added). (Key point: EGL in an expanding pipe)

4. Why does the HGL rise as the pipe expands?
   A. Velocity rises
   B. Velocity falls so velocity head converts to pressure head
   C. Pressure is always constant
   D. Friction adds energy
   Answer: B. By continuity V falls, so V^2/2g decreases; the HGL (p/rho g + z) rises as pressure recovers, and the gap to the EGL narrows. (Key point: HGL in an expanding pipe)

## Related in Hydro Power

- [Deriving Bernoulli's equation from Euler's equation](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeuler-bernoulli-derivation) — Hydro Power, chapter 1
- [Bernoulli's equation in real practice](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Abernoulli-real-practice) — Hydro Power, chapter 1
- [Head loss and flow direction in tapering oil pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atapering-pipe-oil-flow) — Hydro Power, chapter 1
- [Bernoulli's equation limitations and application](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Abernoulli-limitations-application) — Hydro Power, chapter 1
- [Continuity equation basis and Bernoulli equation](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acontinuity-and-bernoulli) — Hydro Power, chapter 1
- [Discharge and head loss in horizontal pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apipe-discharge-head-loss) — Hydro Power, chapter 1

## Related in other subjects

- [Euler equation, Bernoulli and Pitot tube](https://www.bctengineer.com/question.md?key=semester_7%2Fapplied_computational_fluid_dynamics%3Aeuler-equation-pitot-numerical) — Applied Computational Fluid Dynamics, chapter 2
- [Derive continuity equation for unsteady pipe flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acontinuity-equation-unsteady-pipe-flow) — Hydraulics, chapter 4
- [HGL and EGL for pipes in series](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahgl-egl-three-pipes-series) — Hydraulics, chapter 1
- [Bernoulli's equation and four applications](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Abernoulli-from-euler-applications) — Fluid Mechanics, chapter 5