1. Manometer: A manometer is a tube open at both the ends and bent into the shape of a “U” and is partially filled with mercury. When one end of the tube is subjected to an unknown pressure p, the mercury level drops on that side of the tube and rises on the other so that the difference in mercury level is h.U shaped manometer tube connected to a vessel.
According to Pascal’s Law, when we move down in a fluid pressure increases with depth and when we move up the pressure decreases with height. When we move horizontally in a fluid pressure remains constant. Therefore, p + ρ₀gh₀ – ρmgh = p₀
Where ρ₀ the atmospheric pressure, and is the density of the fluid inside the vessel
2. The Mercury Barometer: It is a straight glass tube (closed at one end) completely filled with mercury and inserted into a dish which is also filled with mercury as shown in the figure. Atmospheric pressure supports the column of mercury in the tube to a height h. The pressure between the closed end of the tube and the column of mercury is zero.
Pressure at points A and B must be equal.P₀ = 0 + ρmgh
At the sea level, p0 can support a column of mercury about 76 cm in height
p₀ = (13.6 x 10³) (9.81) (0.76) = 1.01 x 10⁵ Nm⁻² or Pa
Example: Find the pressure in the air column at which the piston remains in equilibrium. Assume the piston to be massless and frictionless.
Solution: Let pa be the air pressure above the piston.Applying Pascal’s law
At point A and B.
pₐtm + ρw g (5) = pₐ + ρk g (1.73)√3/2
Example: A weighted piston confines a fluid of density ρ in a closed container, as shown in the figure. The combined weight of piston and weight is W = 200 N, and the cross-sectional area of the piston is A= 8 cm². Find the total pressure at point B if the fluid is mercury and h = 25 cm (ρm = 13600 kg/m³). What would an ordinary pressure gauge read at B?Solution: Pascal’s principle tells us about the pressure applied to the fluid by the piston and atmosphere. This added pressure is applied at all points within the fluid. Therefore the total pressure at B is composed of three parts:
Pressure of atmosphere = 1.0 x 10⁵ Pa
W/A = 200 N/ 8 x 10⁻⁴ m² = 2.5 x 10⁵ Pa
hρg= 0.33 x 10⁵ Pa
In this case, the pressure of the fluid itself is relatively small. We have
Total pressure at B = 3.83 x 10⁵ Pa = 383 kPa
The gauge pressure does not include atmospheric pressure. Therefore,
Gauge pressure at B = 283 kPa
Example: For the system shown in figure, the cylinder on the left, at L, has a mass of 600 kg and a cross-sectional area of 800 cm².The piston on the right at S, has cross-sectional area 25 cm² and negligible weight. If the apparatus is filled with oil (ρ = 0.78 g/ cm³), find the force F required to hold the system in equilibrium.Solution: The pressures at point H₁ and H₂ are equal because they are at the same level in the single connected fluid. Therefore,
Pressure at H₁ = Pressure at H₂
(Pressure due to left piston) = Pressure due to F + Pressure due to liquid column
(600) (9.8) N/ (0.08 m²) = F/ 25 x 10⁻⁴m² + (8m) (780 kg/m³) (9.8)
After solving, we get, F = 31 N.