- Contents :
- Differential Manometer
- U Tube Manometer
- Manometer Gauge
- Inclined Manometer
- Mercury Manometer
- Manometer Pressure
- Inverted U Tube Manometer
U-tube Manometer :-
It consists of glass tube bent in U-shape, one end of which is connected to a point at which pressure is to be measured and other end remains open to the atmosphere as shown in Fig. The tube generally contains mercury or any other liquid whose specific gravity is greater than the specific gravity of the liquid whose pressure is to be measured.
(a) For Gauge Pressure:-
Let B is the point at which pressure is to be measured, whose value is p. The datum line is A-A.
Let,
h₁ = Height of light liquid above the datum line.
h₂ = Height of heavy liquid above the datum line
S₁ = Sp. gr. of light liquid.
ρ₁ = Density of light liquid = 1000 × S₁
S₂ = Sp. gr. of heavy liquid.
ρ₂ = Density of heavy liquid = 1000 × S₂
As the pressure is the same for the horizontal surface. Hence pressure above the horizontal datum line A-A in the left column and in the right column of U-tube manometer should be same.
Pressure above A-A in the left column
= p + ρ₁× g × h₁
Pressure above A-A in the right column
= ρ₂ × g × h₂
Hence equating the two pressures.
p + ρ₁.g.h₁ = ρ₂.g.h₂
p = ( ρ₁.g.h₁ - ρ₂.g.h₂ )
(b) For Vacuum Pressure:-
For measuring vacuum pressure, the level of the heavy liquid in the manometer will be as shown in Fig. (b): Then
Pressure above A-A in the left column
= p + ρ₁.g.h₁ + ρ₂.g.h₂
Pressure head in the right column above
A-A = 0
p + ρ₁.g.h₁ + ρ₂.g.h₂ = 0
p = - ( ρ₁.g.h₁ + ρ₂.g.h₂ )
Problems on U-Tube Manometer:-
1. The right limb of a simple U-tube manometer containing mercury is open to the atmosphere while the left limb is connected to a pipe in which a fluid of sp. gr. 0.9 is flowing. The centre of the pipe is 12 cm below the level of mercury in the right limb. Find the pressure of fluid in the pipe if the difference of mercury level in the two limbs is 20 cm.
Ans. - 25977 N/m²