In the previous chapter, we have studied the pressure exerted by fluid on the wall of a container is calculated by using mathematical formula. In this chapter, we learn how to measure pressure using manometers and other mechanical gauges.

**Manometers**

Several instruments are used to measure pressure. A manometer is a device is used to measure pressure of a liquid by using the column of liquid. This column of liquid can be measured in vertical or inclined tubes. These devices are used for measuring different types of pressures that include absolute pressure, gauge pressure, atmospheric pressure and differential pressure.

There are different types of manometers available to measure these types of pressure. But manometers are mainly classified into two types:

- Simple Manometer
- Differential Manometer

**Simple Manometer**

Simple manometers are just tubes with open top. These are attached to the vessel at its top and used to measure pressure as compare to atmospheric pressure. Simple manometers are of different types. It can be classified into:

**Piezometer:**

Piezometer is one of the simplest manometers with open top which is attached to a vessel containing liquid. The pressure is exerted by the liquid increases the level of liquid in the tube as shown in Fig. 3.1

Let us consider the height of point A from the upper surface of liquid in the piezometer be h_{1}

Height of point B from the upper surface of liquid in the piezometer be h_{2}

P_{a} be the atmospheric pressure

The fluid pressure at point A can be given by,

P_{A} = h_{1}w + P_{a}

The fluid pressure at point B can be given by,

P_{B} = h_{2}w + P_{a}

When both these pressures at point A and B need to be expressed in terms of gauge pressure, the term P_{a} will be zero. Hence, we get-

P_{A} = h_{1}w

P_{B} = h_{2}w

**U-tube Manometer:**

The name suggests that it is of U-shape and the raise of liquid in both the tube is considered to measure the fluid pressure as shown in Fig. 3.2 (a) Double Column Manometer (P_{x}> P_{a})

Let us consider the gauge pressure in the container be P_{x}

P_{a} be the atmospheric pressure exerts on the liquid in the U-tube manometer

w_{1}be the specific gravity of liquid in the container

w_{2}be the specific gravity of liquid in the U-tube manometer

h_{1} be the height of liquid in the container up to the datum line AB

h_{2} be the height of liquid in the U-tube manometer up to the datum line AB

So,

The pressure exerted by liquid at A is given by,

P_{A} = P_{x} + w_{1}h_{1}

Similarly, pressure exerted by liquid at B is given by

P_{B} = P_{a}+ w_{2}h_{2}

Or,

P_{B} = w_{2}h_{2} (P_{a}= 0)

We know that,

P_{A} = P_{B}

P_{x} + w_{1}h_{1} = w_{2}h_{2}

Or,

P_{x} = (w_{2}h_{2} – w_{1}h_{1})

If the head of water column is considered, then the specific gravity can be given by-

P_{x}/ w = (w_{2}h_{2} – w_{1}h_{1})/ w (w = specific weight of water)

Or,

P_{x}/ w = (S_{2}h_{2} – S_{1}h_{1})

Here,

S_{1} and S_{2} = Specific Gravity of both liquids

Now,

Considering Fig. 3.2 (b) Double Column Manometer (P_{x}< P_{a})

The pressure exerted by liquid at A is given by,

P_{A} = P_{x} + w_{1}h_{1} + w_{2}h_{2}

Similarly, pressure exerted by liquid at B is given by

P_{B} = 0

We know that,

P_{A} = P_{B}

P_{x} + w_{1}h_{1} + w_{2}h_{2} = 0

Or,

P_{x} = – (w_{1}h_{1}+w_{2}h_{2})

If the head of water column is considered, then the specific gravity can be given by-

P_{x}/ w = – (w_{1}h_{1}+w_{2}h_{2})/ w (w = specific weight of water)

Or,

P_{x}/ w = – (S_{1}h_{1}+S_{2}h_{2})

Here,

S_{1} and S_{2} = Specific Gravity of both liquids

**Single Column Manometer:**

Single Column Manometer is just like u-tube manometer whose one end is replaced with a reservoir of large cross-sectional area as compare to other limb as shown in Fig. 3.3 (a)

Let us consider both the limbs of manometer are exposed to atmospheric pressure, P_{a}

When the limb (left one) with reservoir is connected to the container, the pressure of liquid exerts on it to push it down from A_{1}B_{1} to A_{2}B_{2}.

The change in the level of liquid be h

h_{1} be the height of liquid in the container up to the line A_{1}B_{1}

h_{2} be the height of liquid in the manometer up to the line A_{1}B_{1}

w_{1}be the specific gravity of liquid in the container

w_{2}be the specific gravity of liquid in the manometer

So,

Mass of liquid displaced by liquid is given by,

m = h x A x P_{x}

Similarly, Mass of liquid rise in the right limb is given by-

m = h_{2} x a x P_{x}

According to law of conservation of masses, it can be written as-

h x A x P_{x} = h_{2} x a x P_{x}

Or,

h = h_{2} x a/ A

Now,

Gauge pressure at A_{2} is given by,

PA_{2} = P_{x} + (h_{1} + h) w_{1}

Similarly,

Gauge pressure at B_{2} is given by,

Pa_{2} = P_{a} + (h_{2} + h) w_{2}

We know that,

PA_{2} = Pa_{2}

P_{x} + (h_{1} + h) w_{1} = P_{a} + (h_{2} + h) w_{2} (P_{a} = 0)

P_{x} = (w_{2}h_{2} – w_{1}h_{1}) + h(w_{2} – w_{1})

Or,

P_{x} = w_{2}h_{2} (1 + a/ A) – w_{1} (h_{1} + h_{2}.a/ A)

a/A is negligible as its value is nearly zero

Hence,

P_{x} = (w_{2}h_{2} – w_{1}h_{1})

Considering the Fig. 3.3 (b)

It shows an inclined manometer with an angle with horizontal then,

We know that,

P_{x} = (w_{2}h_{2} – w_{1}h_{1})

= w_{2} sin – w_{1}h_{1}

** **

**Note:**

Inclined manometer is considered as 25 times more sensitive than double column U-tube manometer.

**Differential Manometers**

Differential manometers measure the pressure difference between two different points in two different containers. It can be classified into three differential manometers:

**Differential U-tube manometer:**

Differential U-tube manometer is similar to U-tube manometer and is used to measure pressure difference directly as shown in Fig. 3.4 (a)

Let us consider heavy liquid say mercury is present in the U-tube and it is attached to the ends of two different reservoirs.(Container A and B are placed at different heights)

w_{1}be the specific gravity of liquid in container A

w_{2}be the specific gravity of liquid in container B

w_{g}be the specific gravity of mercury in U-tube differential manometer

h_{1} be the height of mercury from the CD in U-tube differential manometer

h_{2} be the height of liquid in the container B above the mercury level

h_{3} be the height of liquid in the container Aup to the level of h_{2}

We know that,

Pressure at A, P_{A} = Pressure at B, P_{B}

Similarly,

Pressure at C, P_{C} = Pressure at D, P_{D} (For Equipressure line)

P_{A} + w_{1}(h_{1} + h_{2}+ h_{3}) = P_{B}+ w_{g}h_{1}+w_{2}h_{2}

P_{A} –P_{B} = w_{g}h_{1}+w_{2}h_{2} – w_{1}(h_{1} + h_{2} + h_{3})

= h_{1}(w_{g}– w_{1}) + h_{2}(w_{2}– w_{1}) – w_{1}h_{3}

**This is the pressure difference between A and B**

Suppose container A and B are in the same level as shown in Fig. 3.4 (b)

Here,

h_{1} be the height of mercury from the CD in U-tube differential manometer

h_{2} be the height of liquid in the above the mercury level

Now,

Pressure at C, P_{C} = Pressure at D, P_{D} (For Equipressure line)

P_{A} + w_{1}(h_{1} + h_{2}) = P_{B} + w_{g}h_{1}+w_{2}h_{2}

P_{A} –P_{B}= w_{g}h_{1}+w_{2}h_{2} – w_{1}(h_{1} + h_{2})

= h_{1}(w_{g }– w_{1}) + h_{2 }(w_{2 }– w_{1})

**Note:**

If we put the value of h_{3} = 0, then same equation can be achieved from the above equation for Fig. 3.4 (a).

**Inverted U-tube differential manometer**

Inverted U-tube differential manometer is used to measure low pressure difference as shown in Fig. 3.5

Let us consider heavy liquid say mercury is present in the U-tube and it is attached to the ends of two different reservoirs similar to U-tube differential manometer. (Container A and B are placed at different heights)

w_{1}be the specific gravity of liquid in container A

w_{2}be the specific gravity of liquid in container B

w_{g}be the specific gravity of mercury in inverted U-tube differential manometer

h_{1} be the height of liquid in the container A up to the level of h_{2}

h_{2} be the height of liquid in the container B below the mercury level

h_{3} be the height of mercury from the CD in inverted U-tube differential manometer

We know that,

Pressure at C, P_{C} = Pressure at D, P_{D} (For Equipressure line)

P_{A} – w_{1}(h_{1} + h_{2} + h_{3}) = P_{B}– w_{g}h_{1}–w_{2}h_{2}

P_{A} –P_{B}= w_{1}(h_{1} + h_{2} + h_{3})– w_{g}h_{1}–w_{2}h_{2}

= h_{3}(w_{1}– w_{g}) + h_{2 }(w_{1}– w_{2}) – w_{1}h_{1}

**This is the pressure difference between A and B**

**Micro manometer:**

This type of manometers consists of two reservoirs having large cross-sections and contains two different liquids of different specific gravity as shown in Fig. 3.6

Let us consider standard liquid is present in the tube and it is attached to the ends of two different reservoirs. (Reservoir A and B are placed at same heights)

w_{1 }be the specific gravity of liquid in reservoir A

w_{2 }be the specific gravity of liquid in reservoir B

w be the specific gravity of standard liquid in the tube

h be the height of liquid (fall) in the reservoir A

h_{1} be the height of standard liquid (rise) in the reservoir B

h_{2} be the height of liquid in the reservoir B up to the initial level mark

h_{3} be the height of line AB up to the initial level mark

Now,

Fall of standard liquid in reservoir A = Rise of standard liquid in reservoir B = h. a/ A

We know that,

Pressure at C, P_{C} = Pressure at D, P_{D} (For Equipressure line)

P_{A}+ w_{1}(h_{1} + h_{2}– h. a/ A) + w (h_{3} +h. a/ A) = P_{B}+ w_{1} (h_{2} +h. a/ A) + w_{2}h_{1}+w (h_{3} –h. a/ A)

Consider h. a/ A is negligible as its value is nearly 0, then-

P_{A} + w_{1}(h_{1} + h_{2}) + w (h_{3}) = P_{B} + w_{1} (h_{2}) + w_{2}h_{1}+w (h_{3})

P_{A} –P_{B} = w_{1}h_{2} + w_{2}h_{1}+wh_{3} – w_{1}(h_{1} + h_{2})– wh_{3}

= h_{1}(w_{2}– w_{1})

Or,

P_{A} –P_{B} = 2h(w_{2 }– w_{1}) (h_{1} = 2h)

**This is the pressure difference between A and B
**

**Some Basic Problems on Manometers**

**Example 3.1:**

Consider the Fig. 3.7 to calculate height of liquid column in three different piezometer tubes.

Take specific gravity of three liquids as 0.8 for A, 0.85 for B and 0.95 for C.

**Solution:**

We know that,

Pressure, P = w x h (w = S x 1000 x 9.81 N/m^{3})

Where,

S = Specific Gravity of liquid

**Pressure at point 1:**

P_{1} = 0.8 x 1000 x 9.81 x 2

= 15696 N/m^{2}

Thus,

Height of liquid column will be given by,

h_{1 }= 15696/ (0.8 x 1000 x 9.81)

= 2 m

** **

**Pressure at point 2:**

P_{2} = (0.85 x 1000 x 9.81 x 2.5) + (0.8 x 1000 x 9.81 x 2)

= 36542 N/m^{2}

Thus,

Height of liquid column will be given by,

h_{2}= 36542/ (0.85 x 1000 x 9.81)

= 4.4 m

Pressure at point 3:

P_{3} = (0.95 x 1000 x 9.81 x 3) + (0.85 x 1000 x 9.81 x 2.5) + (0.8 x 1000 x 9.81 x 2)

= 64501 N/m^{2}

Thus,

Height of liquid column,h_{3}= 64501/ (0.95 x 1000 x 9.81)

= 6.92 m

**Example 3.2:**

A U-tube manometer is shown in Fig. 3.8. Find the vacuum pressure at the point A.

**Solution:**

From the given data, the equation for finding pressure at point A can be written as-

P_{A} + 0.2 x 0.9 x 9810 + 0.5 x 13.6 x 9810 = 0

Hence,

P_{A} = – 98474 N/m^{2}