Hydraulic Machines

Hydraulic machines are very powerful tools that work on the principles of fluid mechanics to amplify force and execute a variety of tasks. These machines can move heavy loads, lift objects, or operate complex systems by applying pressure to a fluid, such as oil or water.

1.0What is a Hydraulic Machine?

A hydraulic machine is a device that uses fluid to generate, control, and transmit power. Hydraulic machines can be found in many industries nowadays, used to lift, move, or shape heavy loads based on the principles of fluid mechanics.

2.0Working Principle of Hydraulic Machines

There are two laws on which the Hydraulic machines work:

Pascal’s law:

Pascal's Law states that any confined fluid that has pressure applied to it has the same pressure in all directions. This law primarily helps in understanding the behavior of fluids at rest.

When a fluid is at rest in an enclosed container such as a closed pipe, hydraulic press, or piston), any pressure change at a point in the fluid is propagated to all points in the fluid without loss. This implies that the pressure exerted is equally distributed in every direction throughout the fluid. This results from the incompressibility of most fluids and the efficiency with which fluids are able to propagate pressure.

Mathematically Pascal’s law can be expressed as:

$P_{1} = P_{2} = P_{3} = ....$

Here P₁, P₂, P₃, … are pressure of the fluid at rest at different points where, $P ress u re = \frac{F orce}{A re a}$ or $P = \frac{F}{A}$

Law of Continuity:

This law states that the flow of mass and the mass flow rate of an incompressible fluid remains constant throughout a pipe or duct, regardless of changes in cross-sectional area. In simple words, the law asserts that the mass (water) entering a pipe is equal to the mass exiting the system. This law is obtained from the law of conservation of mass.

Mathematically, it can be expressed as:

$A_{1} V_{1} = A_{2} V_{2}$

where

A₁ and A₂ are areas at two different points,
V₁ and V₂ are the velocities of the fluid at those points.

These principles allow the force to be transferred through hydraulic fluid and force amplification at the output end.

3.0Types of Hydraulic Machines

There are different types of hydraulic machines, and each serves a different purpose. Some of the most common types include:

Types

Diagram

Hydraulic Press:

A hydraulic press is a machine that applies hydraulic pressure to compress, mould, or shape materials. It works by applying a small force to a small piston, which is then transmitted to a larger piston, resulting in a much greater force at the output.

Hydraulic Lift:

Hydraulic lifts work by lifting heavy objects through hydraulic pressure. For instance, in garages, car lifts use the same mechanism to raise the vehicle for servicing or repair. A small piston at the input side produces pressure that is transferred to a larger piston to lift the load.

Hydraulic Crane:

Hydraulic cranes operate by the hydraulic power that moves heavy loads. Hydraulic cranes consist of a hydraulic pump and hydraulic cylinders controlling the movement of the boom and lifting materials.

Hydraulic Jack:

A hydraulic jack is a tool used in lifting heavy loads, such as cars, through hydraulic pressure force. It applies force on a small piston that is transmitted to a large piston to raise the load.

Hydraulic Motor:

Hydraulic motors convert hydraulic energy into mechanical energy. The motors are used for driving machinery in industrial applications. Hydraulic motors work by converting the energy of the pressurized fluid into rotational motion.

4.0Advantages of Hydraulic Machines

Hydraulic machines provide the following benefits that make them truly valuable in most industries today:

Force Multiplication: A hydraulic system is able to multiply forces so that small inputs create a big force and will be suitable for lifts to be able to bring massive objects up into place through the least effort.
Precision: Hydraulic systems ensure smooth and precise control of force, motion, and speed, which is suitable for tasks requiring high precision.
Compact and Efficient: Hydraulic machines are compact, lightweight, and can transmit large amounts of power in relatively small systems, making them efficient for a variety of applications.
Durable and reliable: There is durability and reliability in hydraulic systems, particularly in their performance under grueling working conditions, with consistent performance over time.

5.0Solved Examples

Problem 1: A hydraulic lift is used to raise a car that weighs 12,000 N. The area of the large piston is 0.4 m², and the area of the small piston is 0.01 m². If a force of 500 N is applied to the small piston, calculate the force exerted by the large piston.

Solution: According to Pascal’s Law, the Force exerted by a large piston is:

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$

F₁= 500N, A₁ = 0.01m², A₂ = 0.4m²

$\frac{500}{0.01} = \frac{F _{2}}{0.4}$

$F_{2} = 50000 \times 0.4 = 20000 N$

Problem 2: A hydraulic press is used to lift a load of 1,000 kg using two pistons. The area of the small piston is 0.02 m², and the area of the large piston is 0.5 m². Find:

The force required to apply on the small piston to lift the load.
The distance by which the large piston moves when the small piston is pressed down by 0.1 m.

Solution:

The force required to apply on the small piston to lift the load.

Weight of the load = $F_{2} = m g = 1000 \times 9.8 = 9800 N$

According to Pascal’s Law, the Force exerted by a large piston is:

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$

Here, we need to find the F₁ that is the force on the small piston:

$\frac{F _{1}}{0.02} = \frac{9800}{0.5}$

$F_{1} = \frac{9800 \times 0.02}{0.5} = \frac{196}{0.5} = 392 N$

Distance moved by the large piston

Principle of conservation of volume, we know:

$A_{1} \times h_{1} = A_{2} h_{2}$

$0.02 \times 0.1 = 0.5 \times h_{2}$

$0.002 = 0.5 \times h_{2}$

$h_{2} = \frac{0.002}{0.5} = 0.004 m = 4 mm$

Problem 3: An industrial machine rotates a wheel by employing a hydraulic motor. The hydraulic motor operates at 10 MPa with hydraulic fluid, and the piston area is 0.04 m². At the applied pressure, the rotational speed is 150 RPM (revolutions per minute) from the motor.

Calculate the force exerted by the hydraulic fluid on the piston of the hydraulic motor.
If the motor shaft requires a torque of 500 N·m to turn at 150 RPM, calculate the power output of the hydraulic motor in kW.

Solution:

The force exerted by the hydraulic fluid on the piston

$F = P \times A$

$F = 10 \times 1 0^{6} \times 0.04 = 400000 N$

Power output of the hydraulic motor

We have the formula for power output as:

$P = τ \times ω$

So, firstly we need to calculate

$ω = RPM \times \frac{2 π}{60}$

$ω = 150 \times 2 π 60 = 5 π \approx 15.7 r a d / s$

$P o w er (P) = 500 \times 15.7 = 7850 W$

$P o w er = \frac{7850}{1000} = 7.85 kW$