Pascal's Law
Pascal's Law
Statement:
It states that the pressure or intensity of pressure at a point in a static
fluid is equal in all directions. This is proved as:
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Fig. Force on a Fluid Element |
Then the
forces acting on the element are:
1. Pressure
forces normal to the surfaces, and
2. Weight
of element in the vertical direction.
The
forces on the faces are:
Force
on the face AB = Px
× Area of face AB
= Px × dy × 1
Similarly
force on the face AC = Py
× dx × 1
Force
on the face BC = Pz × ds × 1
Weight
of element =
(Mass of element) × g
= (Volume of ρ) × g
= ((AB × AC × 1)/2) × ρ × g,
Where
ρ = density of fluid.
Resolving
the forces in x-direction, we have
Px × dy × 1 - P (ds
× 1) sin (90० - Ɵ) = 0
Px × dy × 1 - Pz
ds × 1 cosƟ = 0.
But from Fig. ds cosƟ = AB = dy
∴ Px × dx × 1-Pz
× ds × 1 - Pz × ds × 1 = 0
Px
= Pz
Similarly, resolving the forces in
y-direction, we get
Py
× dx × 1 – Pz × ds × 1 cos(90० - Ɵ)- (dx × dy)/2 × 1 × ρ
× g = 0
Py
× dx – Pz ds sinƟ – (dxdy)/2
× ρ × g = 0
But
ds sinƟ = dx and also the element is very small and hence
weight id negligible.
∴ Pydx – Pz
× dx = 0
Py
= Pz
The
above equation shows that the pressure at any point in x, y, z directions is
equal.
Since
the choice of fluid element was completely arbitrary, which means the pressure
at any point is the same in all directions.
Problem
1. A hydraulic press has a ram of 20 cm diameter and a plunger of 3 cm diameter.
It is used for lifting a weight of 30 KN. Find the force required at the
plunger.
Solution.
Given:
Dia.
Of ram, D = 20 cm =0.2 m
∴ Area of ram, A = π/4 D2 = 0.0314 m2
Dia.
Of plunger d = 3 cm = 0.03 m
∴ Area of plunger, a = π/4 (0.03)2 = 7.068 × 10-4
m2
Weight
lifted, W = 30 KN = 30000 N.
Pressure
intensity developed due to plunger = Force/Area = F/A
By
Pascal’s Law, this pressure is transmitted equally in all directions
Hence
pressure transmitted at the ram = F/A
∴ Force acting on ram = pressure intensity ×
Area of ram
= F/a × A = (F × 0.0314)/ (7.068 × 10-4
) N
But
force acting on ram = Weight lifted = 30000N
30000 = (F ×
0.0314)/ (7.068 × 10-4 )
F = (30000 ×
7.068 × 10-4)/ (0.0314) = 675.2 N.
Problem
2. An oil sp. gr. 0.9 is contained in a vessel. At a point the height of oil is
40 m. Find the corresponding height of water at the point.
Solution.
Given:
Sp.
gr. Of oil, S0
= 0.9
Height
of oil, Z0
= 40 m
Density
of oil, ρ0
= Sp. gr. Pf oil × Density of water = 0.9 ×1000 = 900 kg/m3
Intensity
of pressure, P = ρ0
× g × Z0 = 900 × 9.81 × 40 N/m2
∴ Corresponding height of water = P/Density of water × g
= (900 × 9.81 × 40)/ (1000 × 9.81) = 36 m of water.
Absolute, Gauge, Vacuum, Atmospheric Pressure
Atmospheric Pressure is
defined as the pressure exerted by atmosphere on the surface of earth.
Patm = 1.013 bar = 1.013 × 105
N/m2 = 10.33 m of water = 760 mm of Hg
Gauge Pressure is
the pressure measured with the help pg pressure measuring instrument in which
the atmospheric pressure is taken as datum.
Vacuum Pressure
is defined as the pressure below the atmospheric pressure. The vacuum pressure
is also known as negative gauge pressure.
Absolute Pressure
is measured above the absolute zero.
Absolute
Pressure = Atmospheric Pressure + Gauge Pressure
Vacuum
Gauge Pressure = Atmospheric Pressure – Absolute Pressure.
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