Simple Fluid Mechanics Experiments and Calculations

This article tells you the concept of a simple fluid mechanics experiment and its calculation as its application in the marine field and other fields involving hydraulics. This simple fluid mechanics experiment is based on the principle of Pascal’s law.

Fluid Mechanics Experiment:

Let there be a fluid system consisting of two cylinders of different diameters. One cylinder is of larger diameter and contains a ram in it that can be moved up and down. Another cylinder is of smaller and contains a plunger that can move as well. The large cylinder and small cylinder are connected by a pipe of suitable thickness in order to withstand the pressure created inside. The cylinders and pipe contain liquid or hydraulic oil via which pressure can be transmitted.

On application of a small force on the plunger in a downward direction, a pressure is produced on the liquid in contact with the plunger. This pressure is transmitted equally in all the directions and acts on the large cylinder ram to move it.

It is observed that the movement of the plunger in the small cylinder due to the small force applied displaces the hydraulic fluid in it. Since the hydraulic fluid or oil cannot be compressed, pressure builds up inside the cylinder which depends upon the force applied and the size of the plunger. The pressure created inside the cylinder will act equally in all directions as stated by pascal’s law. The same pressure acts on the ram that is in contact with the hydraulic fluid. This will exert a force on the ram and moves the ram in the desired direction. The force which that moves the ram depend on the diameter of the ram and the pressure developed in the hydraulic fluid due to the force applied manually on the plunger. Finally, the force applied manually on the plunger is converted into the desired motion on the ram which is the system output.

Let’s consider this experiment using Pascal’s Law

Calculations Related to above Experiment

  • Let f1 be the force applied on the plunger
  • f2 be the force acting on the ram
  • Let A1 be the area of the plunger
  • A2 be the area of the ram
  • p will be the intensity of pressure produced by the force f1

Therefore

Pressure intensity produced by the force f1 = (force applied on the plunger f1) / (area of the plunger A1)

As per Pascal’s law, the pressure intensity in a static fluid is transmitted equally in all directions, so the above intensity of pressure produced by the plunger will be equally transmitted in all directions. Therefore the pressure intensity at the ram will be equal to the force applied on the plunger f1 divided by area of the plunger A1.

Also pressure intensity on the ram = (force acting on ram f2) / (area of the ram A2).

Equating the pressure intensity on the ram, we get

(f1 / A1) = (f2 / A2).

The total force acting on the ram f2 = (force applied on the plunger f1 × area of the ram A2) / area of the plunger A1

Example for the above Derivation

Let us consider a ram with 300 mm diameter to be moved, the diameter of plunger is 20mm, and the force applied on the plunger is 100N.

Next the Force available at the ram = (Force applied on the plunger × Area of the ram) / Area of the plunger

Area of the ram = ( 3.14×0.32 ) / 4

Area of the ram=0.07068 mm2

Area of the plunger or piston = ( 3.14 ×.022 ) / 4

Area of the plunger or piston =0.00031 mm2

Force available at the ram = (100 × 0.07068 ) / 0.00031

Force available at the ram =22509.55 N

From the above example, it is proved that the force applied manually on the system is only 100N and which gets converted to 22510N and can be able to operate the heavy system that cannot be operated manually.

Application

The above explained simple experiment is used in hydraulic steering gear. The small movement of the steering wheel in the navigation bridge might actuate the rudder to turn towards starboard or port side due to oil pressure acting on the ram.

Hydraulic Steering Gear

Working

When wheel 1 is turned anticlockwise, the pinion 2 moves the toothed rack 3 downward and moves the toothed rack 4 upward. As it is fixed to the two pistons 5 and 6, the piston also moves correspondingly. As these two cylinders 7 & 8 are filled with oil, the movement of the pistons yields in oil pressure being applied to the bottom of the piston 10 and moves it upward and these forces the oil in upper part of cylinder 9 up in to the cylinder 8.

Piston 10 has a piston rod connected to a slide valve 11. In its middle position, the slide valve just closes the ports 12, 13, 14 in the slide valve housing 15. As the piston 10 moves upward, the slide valve 11 moves along with it and opens port 12 and 14. These results oil from the pressure vessel to come under side of the piston 20 and the oil above piston 20 is forced in to the slide valve housing 15 and out through the port 12 to the discharge tank 16. As a result the piston 21 moves upward along with the piston 20 since both these piston are connected together by piston rod. These upward movements of the two pistons impart movement to the tiller arm which is mounted on the rudder stock and hence moves the rudder.

So now we conclude the discussion. Keep looking for this space for the second part.

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