### Table of Contents

### Definition

**Hydraulic energy** is the energy produced by the water which is stored in reservoirs and lakes at a high altitude (in order to have gravitational potential energy). If at a given moment the water falls to a lower level, this hydraulic energy is transformed into kinetic energy and afterwards into electrical energy by electric generators in the hydroelectric plant.

Compared with other sources of energy, hydraulic energy is a clean energy source due to the fact that does not produce any waste products. Another advantage is that it’s easy to store (in reservoirs or lakes). Also, the water stored in reservoirs situated at altitude permits the regulation of the flow of the river.

### Formula

**Hydraulic energy** is the product between hydraulic power (also called hydropower) and time.

_{h}= P

_{h}· t

where:

- E
_{h}[J] – hydraulic energy - P
_{h}[W] – hydraulic power - t [s] – time

**Hydraulic power** or hydropower is calculated as:

_{h}= m

_{f}· g · Δh

where:

- m
_{f}[kg/s] – water mass flow rate - g [m/s
^{2}] – gravitational acceleration - Δh [m] – hydraulic head

If instead of the water mass flow rate we know the volumetric flow rate, the hydraulic power can be calculated as:

_{h}= ρ · V

_{f}· g · Δh

where:

- ρ [kg/m
^{3}] – water density, which is 1000 kg/m^{3} - V
_{f}[m^{3}/s] – volumetric flow rate of water

Hydraulic head is the height difference between the level where the water enters into the hydro system and the level where it leaves, measured in metres.

Replacing equations (2) and (3) in (1) gives the formula (equation) for **hydraulic energy**:

_{h}= m

_{f}· g · Δh · t = ρ · V

_{f}· g · Δh · t

The unit of measurement of **hydraulic energy** is **joule** [J].

### Example

Calculate what is the theoretical capacity of hydraulic energy for a hydroplant with a volumetric flow rate of 2800 cubic feet per second and a hydraulic head of 150 meters, during 1 continuous hour of work. The density of water is assumed to be 1000 kilograms per cubic meter.

**Step 1**. Convert the volumetric flow rate from [ft^{3}/s] to [m^{3}/s], by dividing the [ft^{3}/s] value to 35.315:

_{f}= 2800 / 35.315 = 79.2864 m

^{3}/s

**Step 2**. Calculate the hydraulic energy E_{h} [J] of the hydroplant using equation (4):

_{h}= ρ · V

_{f}· g · Δh · t = 1000 · 79.2864 · 9.81 · 150 · 3600 = 420011775360 J = 420 GJ

### Calculator

The hydraulic energy calculator allows you to calculate the hydraulic energy of a hydroplant. You need to enter the volumetric flow rate, hydraulic head, working time and choose the desired unit of measurement. The gravitational acceleration is assumed to be 9.81 [m/s^{2}] and water density 1000 [kg/m^{3}].

The default unit of measurement for energy is **Joule**. If you want the result displayed in another unit, use the dropt down list to choose and click the CALCULATE button again.

### References

[1] David Halliday, Robert Resnick, Jearl Walker, Fundamentals of Physics, 7th edition, John Wiley & Sons, 2004.

[2] Benjamin Crowell, Light and Matter – Physics, 2007.

[3] Raymond A. Serway and John W. Jr. Jewett, Physics for Scientists and Engineers, 6th edition, Brooks/Cole Publishing Co.,2004

[4] Jiansong Li, Jiyun Zhao, and Xiaochun Zhang, A Novel Energy Recovery System Integrating Flywheel and Flow Regeneration for a Hydraulic Excavator Boom System, Energies 2020.

[5] Leo H. Holthuijsen, Waves in oceanic and coastal waters, Cambridge University Press, 2007.

[6] Kira Grogg, Harvesting the Wind: The Physics of Wind Turbines, Carleton College, 2005.