---
title: "Hydraulic and energy grade lines"
subject: "Fluid Mechanics"
chapter: 8
key: "semester_4/fluid_mechanics_mechanical:hgl-egl-reservoir-pump-turbine"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/hgl-egl-reservoir-pump-turbine
site: BCT Engineering
---

# Hydraulic and energy grade lines

Fluid Mechanics · Chapter 8: Flow Measurement

- Asked in: year not given, 6 marks
- Video: [Fluid Mechanics: Topic 7.3.1 - Energy grade line (EGL) & Hydraulic grade line (HGL)](https://www.youtube.com/watch?v=-oecDDrYfyY) — CPPMechEngTutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/hgl-egl-reservoir-pump-turbine
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-8

## Question

Define hydraulic grade line (HGL) and energy grade line (EGL). Sketch both for a pipe system that carries water from a lower reservoir to an upper reservoir through a pump, and explain their features at the reservoirs, along the pipe, at the pump and at a turbine.

## Short answer

EGL shows total head, HGL shows piezometric head (EGL minus velocity head). At reservoirs both lines coincide with free surface. Along pipes they drop due to friction; a pump adds head, a turbine extracts head.

## Answer

#### Definitions
- **Energy grade line (EGL):** line showing the total head at each section: $H = \dfrac{p}{\rho g} + \dfrac{V^2}{2g} + z$.
- **Hydraulic grade line (HGL):** line showing the piezometric head: $\dfrac{p}{\rho g} + z$. It is the level to which water would rise in a piezometer tube.

$$
\text{EGL} = \text{HGL} + \frac{V^2}{2g}
$$

For a pipe of uniform diameter, the EGL and HGL are parallel and the vertical gap equals the velocity head.

#### Sketch (lower reservoir, pump, upper reservoir)
```text
 head
  ^                                    upper reservoir
  |                                   ___ EGL = HGL
  |                           ,------'
  |                         ,'  <- Hp added by pump
  | ____ EGL = HGL         /
  |      `--._______      |   (EGL falls with friction,
  |  lower res.     `-.___|    HGL one V^2/2g below EGL)
  +-------------------[P]---------------------> x
```

#### Features
1. **At a reservoir:** velocity is nearly zero and pressure is atmospheric (gauge zero), so EGL and HGL both coincide with the free surface level.
2. **Along the pipe:** the EGL falls continuously because of friction losses, with slope $h_f/L = fV^2/(2gD)$. The HGL falls parallel to it for constant diameter.
3. **At an entrance or valve (minor loss):** the EGL drops suddenly by $K\,V^2/2g$.
4. **At a pump:** the pump adds head $H_p$, so both EGL and HGL rise suddenly by $H_p$ across it.
5. **At a turbine:** the turbine extracts head $H_t$, so both lines fall suddenly by $H_t$.
6. **At a change of diameter:** the gap between EGL and HGL changes, as $V^2/2g$ changes.
7. **At a free discharge:** the HGL passes through the centre of the outlet (pressure atmospheric) and the EGL lies a velocity head above it.
8. **Pressure:** the pressure head at any point is the vertical distance between the pipe centre line and the HGL. If the pipe is above the HGL, pressure is below atmospheric (negative); if it falls to vapour pressure, cavitation or siphon breaking can occur.

## Check yourself

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

1. How are the EGL and HGL related?
   A. HGL = EGL + V^2/2g
   B. EGL = HGL + V^2/2g, where HGL = p/(rho g) + z
   C. EGL = HGL always
   D. HGL includes friction loss, EGL does not
   Answer: B. The EGL shows total head and the HGL shows piezometric head; they differ by the velocity head. (Key point: Definitions)

2. What happens to the lines at a pump and a turbine?
   A. Both fall at a pump and rise at a turbine
   B. Both rise suddenly by H_p at a pump and fall suddenly by H_t at a turbine
   C. Only EGL changes
   D. No change in either
   Answer: B. A pump adds head so both lines rise; a turbine extracts head so both fall. (Key point: Pump and turbine)

3. What does it mean if part of the pipe lies above the HGL?
   A. Pressure there is above atmospheric
   B. Velocity is zero
   C. The EGL is below the HGL
   D. Pressure there is below atmospheric (negative), with a risk of cavitation
   Answer: D. Pressure head is the vertical distance from pipe centre line to HGL, so a pipe above the HGL has negative gauge pressure. (Key point: Pipe above HGL)

## Related in Fluid Mechanics

- [Pump head and power between two reservoirs](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Apumped-system-power-between-reservoirs) — Fluid Mechanics, chapter 8

## Related in other subjects

- [HGL and TEL for two-pipe system](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahgl-tel-two-pipe-reservoir) — Hydraulics, chapter 1
- [Pump horsepower, TEL and HGL](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apump-horsepower-tel-hgl) — Fluid Mechanics, chapter 5
- [Three reservoirs with pump in pipe 2](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apump-three-reservoir-flow-and-head-j) — Hydraulics, chapter 3
- [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