# Hydraulic Machines

Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points [Fig. 1]. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all.If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.

Fig 1 Whenever external pressure is applied on any part of a fluid in a vessel, it is equally transmitted in all directions.

This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. We can say whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is another form of the Pascal’s law and it has many applications in daily life.

Fig 2 Schematic diagram illustrating the principle behind the hydraulic lift, a device used to lift heavy loads.

A number of devices, such as hydraulic lift and hydraulic brakes, are based on the Pascal’s law. In these devices, fluids are used for transmitting pressure. In a hydraulic lift, as shown in Fig. 2, two pistons are separated by the space filled with a liquid. A piston of small cross-section A$_1$  is used to exert a force F$_1$  directly on the liquid. The pressure P = $\frac{F_1}{A_1}$  is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A$_2$ , which results in an upward force of P × A$_2$ . Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform) F$_2$ = P$A_2$ = $\frac{F_1}{A_1}\times A_2$ . By changing the force at A$_1$ , the platform can be moved up or down. Thus, the applied force has been increased by a factor of $\frac{A_2}{A_1}$  and this factor is the mechanical advantage of the device. The example below clarifies it.

Example 1

Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively.

(a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston.

(b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?

(a) Since pressure is transmitted undiminished throughout the fluid,

$F_2=\frac{A_2}{A_1}F_2$

$F_2=\frac{\pi (3/2 \times 10^{-2} \ m)^2}{\pi(1/2 \times 10^{-2} \ m)^2} \times 10 \ N$

= 90 N

(b) Water is considered to be per fectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.

$L_1A_1=L_2A_2$

$L_2=\frac{L_2A_2}{A_1}$

$L_2=\frac{\pi (1/2 \times 10^{-2} \ m)^2}{\pi(3/2 \times 10^{-2} \ m)^2} \times 6 \times 10^{-2} \ m^2$

= 0.67 × 10$^{-2}$ m

= 0.67 cm

Note, atmospheric pressure is common to both pistons and has been ignored.

Example 2

In a car lift compressed air exerts a force F$_1$  on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 10.7). If the mass of the car to be lifted is 1350 kg, calculate F$_1$ . What is the pressure necessary to accomplish this task? (g = 9.8 m.s$^{-2}$).

Since pressure is transmitted undiminished throughout the fluid,

$F_1=\frac{A_1}{A_2}F_2$

$F_2=\frac{\pi (5 \times 10^{-2} \ m)^2}{\pi(15 \times 10^{-2} \ m)^2} \times (1350 \ kg \times 9.8 \ m.s^{-2})$

= 1470 N

The air pressure that will produce this force is

$P=\frac{F_1}{A_1}$

$=\frac{1.5 \times 10^3 \ N}{\pi (5 \times 10^{-2} \ m)^2}=1.9 \times 10^5$ Pa

This is almost double the atmospheric pressure.

Hydraulic brakes in automobiles also work on the same principle. When we apply a little force on the pedal with our foot the master piston moves inside the master cylinder, and the pressure caused is transmitted through the brake oil to act on a piston of larger area. A large force acts on the piston and is pushed down expanding the brake shoes against brake lining. In this way, a small force on the pedal produces a large retarding force on the wheel. An important advantage of the system is that the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on all wheels.