Introducing Transmission Of Fluid Power

Transmission of Fluid Power

Fig. 1-7. Transmission of Force

Using a hydraulic fluid to accomplish work requires the application of all of the points covered so far. as shown in Fig. 1-7(A), if  a force of 10 lbs is applied to piston 1, it will be transmitted through the liquid in the cylinder to piston 2. Pascal's Law states that pressure developed in  a confined fluid is equal at every point. Therefore, the internal fluid pressure developed by piston 1 acts on piston 2. If the area of each piston is the same, the force developed on piston 2 is the same as the force applied by piston 1, discounting friction losses. This is the principle  upon which hydraulic power transmission system are based.

The single cylinder in Fig. 1-7(A) has been replaced by two individual cylinders in Fig. 1-7(B). Both are of the same diameter and are connected with a suitable hydraulic line. The conditions present in Fig. 1-7(B) are not changed, because the hydraulic system has not been changed. The force applied to piston 1 is transmitted through the fluid to piston 2. Some frictional losses are present in any operating system, but we will forget about it for now.

Fig. 1-8. Unequal piston areas

A similar arrangement of two pistons connected by a tube is shown in Fig. 1-8. However, the piston are placed in a vertical position and are different sizes. If a force of 100 pounds is applied to the 10square inch area of piston 1, a hydraulic pressure of 10 psi (100 pounds:10sq in) is built up under piston 1, in the connecting tubing, and under the 50 square-inch area of piston 2. The 10 psi therefore exerts a total force of 500 pounds on piston2 (10psi x 50 sq in). This increase in power is hydraulic leverage, and occurs in all similar applications.

However, if the applied force is reversed and the 500 pounds in Fig. 1-8. is apllied Against piston 2, the output force on piston 1 is reduced to 100 pounds. The calculation remain the same:

500 lbs : 50 sq in = 10 psi.

10 psi x 10 sq in =100 pounds.

Both of these example demonstrate how force can be increased or decreased in a hydraulic system by leverage. There is another principle of leverage that must be remembered. That is, for every force increase in a two piston system, there is a corresponding movement decrease. If piston 1 in Fig. 1-8. move 5 inches, it diplaces50 cubic inches of fluid (5 in x 10 sq in) = 50 cu in). The 50 cubic inches of hydraulic fluid is transmitted through the system to piston 2. The 50 cubic inches of fluid acts on the square inch area of piston 2, causing it to move 1 inch (50 cu in : 50 sq in = 1 in).

The arrangement of pistons shown in Fig. 1-8. provides a ratio of  5 to 1 for any force applied on piston 1. At the same time the amount of movement of piston 2 is 1/5 the movement of piston 1. The velocity or speed of piston 2 is also 1/5 the velocity of piston 1. No matter what the ratio, if you want to multiply the hydralic force of the system, you will reduce the amount and speed of movement. On the other hand, if the force is apllied to the larger piston, you increase the amount and speed of movement, but you reduce the force exerted by the system.