V.4. Pascal's law.

👀 Experiment: Pascal's law.

Required materials:
Communicating vessels, water, oil.

Experiment description:

• Put water in one of the communicating vessels.
• Add oil to the middle vessel.
• What do you notice?

  The water in the other vessels changes its level and equals the one in the middle with water and oil.

Experiment conclusion:
The pressure of the oil column causes an external pressure (the oil is immiscible with water) on the water in the middle vessel, which is transmitted throughout the mass (quantity) of water and in all directions (Pascal's Law).

👀 Experiment: What are the fluids?

Pascal's law statement:

External pressure on a fluid is transmitted in all directions and across the fluid mass.

Pascal's Law Applications: Hydraulic Press and Pumps

I. Hydraulic press

The hydraulic press consists of two cylinders with liquid (oil), of different sections, provided with a piston each and which communicates at the bottom.

The man presses on the small piston by means of a lever.

The pressure exerted by the small piston (p₁) is transmitted entirely by the piston liquid to the large piston (p₂), i.e.

Principle of operation of the press:

The number of times the cross-sectional area of the large cylinder (S₂) is larger than the area of the small cylinder (S₁), the same number of times the force transmitted by the liquid to the large piston (F₂) is greater than the force with which the man acts on the small piston (F₁). So if we want the press to amplify our force F₁ 10 times, that is F₂ = 10F₁, then we choose a press so that the area of the large cylinder is 10 times larger than the area of the small cylinder.

👀 Experiment: Hydraulic press.

Required materials:
Two syringes of different sections, connecting tube.

Experimemt description:

• Fill the small syringe with liquid and connect the small syringe with another larger syringe through a tube.
• Press the small piston.
• What do you notice?

  The large piston rises when we press the small piston.

Experiment conclusion:
The external pressure exerted by us on the small piston is transmitted by the liquid to the large piston.

Uses of hydraulic press

• Hard rock crushing

• Obtaining oil by pressing seeds

• Stamping of metal objects

• Hydraulic elevator (jack)

• Car foot brake

• Pressing straws and recyclable materials into bales

• Cutting the metal sheets

• Dental or barbershop chairs

• The excavator

• Power steering and power brake systems

&#128221 Pascal's Law Applications: Hydraulic Press and Pumps

II. Pumps are used to compress gases and ensure the circulation of liquids. Vacuum pumps evacuate air from an enclosure.

Water pumps are used for water supply in household systems for transferring liquids and emptying tanks, in gardening or can be connected to water pumps. Hydraulic pumps move a fluid from the lower downstream pressure (eg a lower hydraulic level) to the upper upstream pressure (eg a higher hydraulic level). The pressure difference that the pump overcomes, usually expressed in meters of water column, is the lift height of the pump, which is greater than the difference between the upstream and downstream pressures due to losses in the pump and its pipes.

The human body has two pumps:

• heart (pumping blood) and

• lungs (which pumps air).

Pump uses

• Injection pump (for internal combustion engines)

• Refrigerator or air conditioning compressor

• Inflating tires and mattresses

• Tree spraying and irrigation

• Painting

• Mechanical milkers

• Spray sprayers (compressed air presses the liquid out)

• Gas pump

🔓 Solved problems

1. A 1000 kg machine is lifted using a hydraulic lift. Knowing the radii of the pistons of 10 cm and 50 cm, respectively, calculate:

a) the force necessary to lift the machine;

b) the stroke of the large piston, knowing the stroke of the small piston of 30 cm.

Solution:

We write down the data of the problem and turn it into SI:
m = 1000 kg
R₁ = 10 cm = 0,1 m
R₂ = 50 cm = 0,5 m
h₁ = 30 cm = 0,3 m
F₁ = ?
h₂ = ?

We calculate the weight of the machine, which is the force exerted by the liquid on the large piston, ie F₂:
F₂ = G = m ∙ g = 1.000 kg ∙ 10 N/kg = 10.000 N

We write the equation of the operating principle of the hydraulic press:

We calculate the cross-sectional areas of the two cylinders, which are circles:

S₁ = π ∙ R₁² = π ∙ (0,1 m)² = π ∙ 0,01 m²
S₂ = π ∙ R₂² = π ∙ (0,5 m)² = π ∙ 0,25 m²

We extract the unknown, F₁ from the equation of the operating principle of the hydraulic press and replace the data:

b) According to Pascal's law, the volume of liquid in the small cylinder is equal to the volume of liquid in the large cylinder:

2. Into a U tube is poured water with a density of 1000 kg/m₃ and then a column of oil with a height h₁ = 10 cm is poured into the left branch. As the oil is immiscible with water, there is an unevenness in the left branch of 2 cm. Determines the density of the oil used. Give the air pressure, p₀ = 10.000 Pa

Solution:

Write down the problem data and turn it into SI:
ρ_water = 1000 kg/m³
h₁ = 10 cm = 0,1 m (height of the oil column)
Δh = 2 cm = 0,02 m (unevenness of liquids in the two branches) p₀ = 10.000 Pa
ρ_oil = ?

The external pressure exerted by the oil, p_A, is transmitted entirely to the water in the right branch, p_B, according to Pascal's Law:
p_A = p_B

We calculate the two pressures, taking into account the atmospheric pressure exerted on the two branches of the U tube:
Δh = h₁ – h₂ = 0,02 m
h₂ = h₁ – Δh = 0,1 - 0,02 = 0,08 m (height of the water column above level B, which is the same as level A)
p_A = p₀ + ρ_oil ∙ g ∙ h₁ = 100.000 + ρ_oil ∙ 10 ∙ 0,1 = 100.000 + ρ_oil
p_B = p₀ + ρ_water ∙ g ∙ h₂ = 100.000 + 1000 ∙ 10 ∙ 0,08 = 100.000 + 800 = 100.800
p_A = p_B
100.000 + ρ_oil = 100.800
ρ_oil = 100.800 – 100.000 = 800 kg/m³