The fluid element is of very small dimensions i.e. dx, dy, and ds.

Consider an arbitrary fluid element of wedge shape in a fluid mass at rest as shown in figure. Let, the width of the element perpendicular to the plane of the paper is unity and P_x, P_y and P_z are the pressure or intensity of pressure acting on the force AB, AC and BC respectively.

Then,

Force on side AB = P_x* Area of face AB

Or, Force on side AB = P_x*dy*1

Or, Force on side AB = P_x dy

Similarly,

Force on side AC = P_y dx

Force on side BC = P_z ds

Weight (W) = Area of triangular element*depth*Specific weight

Or, W = 1/2* AB * AC *1*

Or, W = * dxdy * ρg

Resolving the forces in x-direction

P_x dy = P_z ds sin (90- θ)

Or, P_x dy = P_z ds cos θ

Also from figure,

dy = ds cos θ

Now,

P_xdy = P_z dy

Or, P_x = P_z              ………………….             (i) 

Similarly,

Resolving the forces in y-direction

Pydx = Pzds cos(90- θ) + 1/2* dxdy * ρg

Also,

dscos θ = dx {also the element is very small and hence weight is negligible}

Now,

P_ydx = P_zdx

Or, P_y = P_z             ………………..               (ii)

From equation (i) and (ii), we have

P_x = P_y = P_z

The above equation shows that pressure at any point in x, y and z direction is equal.

Absolute pressure is a pressure that is relative to the zero pressure in the empty, air-free space of universe. This reference pressure is the ideal or absolute vacuum. Also, it is the sum of gauge pressure and atmospheric pressure.

i.e. P_ab= P_gauge+P_atm

Where,

Absolute pressure = P_ab

Gauge pressure = P_gauge

Atmospheric pressure = P_atm

Gauge pressure is defined as the difference between an absolute pressure and atmospheric pressure.

i.e. P_gauge = P_ab – P_atm

Where,

Absolute pressure = P_ab

Gauge pressure = P_gauge

Atmospheric pressure = P_atm

Atmospheric pressure is also known as barometric pressure, it is the pressure within the atmospheric of earth.

i.e. P_atm= P_ab – P_gauge

Where,

Absolute pressure = P_ab

Gauge pressure = P_gauge

Atmospheric pressure = P_atm

A pressure reading below the atmospheric pressure is known as a vacuum pressure.

The graphical representation of relationship between Absolute, gauge and atmospheric pressure at a point:

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= Specific gravity of light liquid.

s₂= Specific gravity of heavy liquid.

ρ₁ = Density of light liquid. = 1000 * s₁

ρ₂  = Density of heavy liquid. = 1000 * s₂

As the pressure is the same for horizontal surface. Hence, pressure above the horizontal datum line A-A in the left column and in 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 columns = ρ₂ g h₂

Hence, equating the two pressure

P + ρ₁ g h₁= ρ₂ g h₂

 Or,    P =    ρ₂ g h_{2 –} ρ₁ g h₁

For measuring vacuum pressure the level of the heavy liquid in the manometer will be as shown in figure.

Let,

h₁= Height of light liquid above the datum line.

h₂= Height of heavy liquid above the datum line.

s₁= Specific gravity of light liquid.

s₂= Specific gravity of heavy liquid.

ρ₁ = Density of light liquid. = 1000 * s₁

ρ₂= Density of heavy liquid. = 1000 * s₂

Then,

Pressure above A-A in the left column = P + ρ₁ g h₁+ ρ₂ g h₂

Pressure above A-A in the right columns = 0

Hence, equating the two pressure

              P + ρ₁ g h_{1 +} ρ₂ g h₂= 0

       Or,   P = – (ρ₁ g h_{1 +} ρ₂ g h₂)

h₁= Height of light liquid above the datum line.

h₂=Height of heavy liquid above the datum line.

 s₁= Specific gravity of light liquid.

s₂= Specific gravity of heavy liquid.

ρ₁= Density of light liquid.

ρ₂= Density of heavy liquid.

△h= Fall of heavy liquid.

P = Pressure at A.

A= Cross- sectional area of the reservoir.

a = Cross-sectional area of the right limb.

Fall of heavy liquid in reservoir will cause a rise of heavy liquid level in the right limb

A*△h = a * h₂

 Or, △h = (a * h₂) / A                                     (a)

Now, consider the datum line Y-Y as shown in figure.

Then,

 Pressure in the right limb above Y-Y = ρ₂g (△h + h₂)                     (b)

Pressure in the left limb above Y-Y = ρ₁g (△h + h₁) + P                      (c)

Equating equation (b) and (c) , we have

ρ₂g (△h + h₂) = ρ₁g (△h + h₁) + P                      

Or, P = ρ₂g (△h + h₂) – ρ₁g (△h + h₁)

Or, P = △h [ρ₂g – ρ₁g] + h₂ ρ₂g – h₁ ρ₁g

From equation (a)

P =  [(a * h₂) / A ] * [ρ₂g – ρ₁g] + h₂ ρ₂g – h₁ ρ₁g

As, the area A is very large compared to a, hence ratio a/A becomes very small and can be neglected.

Thus,

P= h₂ ρ₂g – h₁ ρ₁g

Above figure shows the inclined single column manometer. This manometer is more sensitive. Due to inclination the distance moved by the heavy liquid in the right limb will be more.

Let,

L= length of heavy liquid moved in right limb from X-X

θ = Inclination of right limb with horizontal

h₂ = Vertical rise of heavy liquid in right limb from X-X

    h₂  = Lsinθ

Now,

From equation of pressure at A

P= h₂ ρ₂g – h₁ ρ₁g

Substituting the value of   , we get

P=Lsinθ ρ₂g – h₁ ρ₁g

In above figure the two points A and B are at different level and also contain liquids of different specific gravity. These points are connected to the U- tube differential manometer. Assume the pressure at A and B are

P_A and P_B respectively.

Let,

h= Difference of mercury level in U-tube

y = Distance of the center of B from the mercury level in the right limb

x= Distance of the center of A from the mercury level in the right limb

ρ₁= Density of liquid at A

ρ₂= Density of liquid at B

ρ_m= Density of heavy liquid or mercury

Pressure above X-X in the left limb = ρ₁g(h+x) + P_A

Pressure above X-X in the right limb = ρ_m gh + ρ₂gy + P_B

Equating the two pressure, we have

ρ₁g(h+x) + P_A = ρ_m gh + ρ₂gy + P_B

 Or, P_A – P_B = ρ_m gh + ρ₂gy – ρ₁g(h+x

Or, P_A – P_B = hg(ρ_m – ρ₁ )+ ρ₂ gy – ρ₁gx