Manometers

3.1    INTRODUCTION

In the previous chapter we have calculated pressure by mathematical formula. Here we

will measure the pressure by means of manometers and mechanical gauges.

3.2     MANOMETERS

Manometer is a device which measures pressure by balancing a column of liquid against

the pressure to be measured. It can be used for measuring gauge, absolute, atmospheric

and differential pressures.

3.2.1 Types of Manometers

There are mainly two types of manometers (a) Simple manometers (b) Differential manometers.

3.2.1.1 Simple Manometers

Following are in the category of simple manometers.

(i) Piezometer, (ii) U-tube manometers and (ii) single column manometers.

(i) Piezometer : A piezometer is a vertical transparent glass tube. The lower end of it is connected to the point in the fluid container at which pressure is to be measured. The upper end is open to atmosphere (Fig. 3.1).

A and B are two gauge points. Fluid pressure at A is P A given by

PA=h1w+PaN/m2 [h1 in meter ,w => N/m3

[Pa= atmospheric pressure]

Fluid pressure at B is Ps given by

Ps = wh2 +P n , N/m2

If P A and Ps are to be expressed in gauge pressures, then atmospheric pressure line will be the datum i.e., Pn = 0

So                    P A= wh1 and P13 = wh2

(ii) U-tube manometers : Fig. 3.2 (a) shows a u-tube double column manometer shown in an arrangement where we have to measure pressure greater than atmospheric pressure.

The pressure at A is                 PA = P x + W1h1

Pressure at                               PB = Pa + w2h2

= 0 + w2h2

PA= PB

P x + w1h1 = w2h2

P x= (w2h2-w1h1)

where P x is the gauge pressure in the container. P x in terms of head of water column. (w = sp. weight of water) P x / w = w2 h2 / w- w1 h1 / w = S2 h2 – S1 h1 are the specific gravity of respective liquids.

Fig. 3.2 (b) shows a 11-tube double column manometer shown in an arrangement where we have to measure pressure less than atmospheric pressure.

p A= w1h1 + w2h2 + p x

Ps= P0 = 0

PA= PB

or           w1h1 +w2h2+Pr = 0

B

Or                                P x = – (w1h1 + w2h2) =Gauge pressure in the container.

In terms of head of water column

P x / w = – (w1 w h1 + w2 / w h2) = – (s1h1 + s2 h2)

Disadvantage of double column: In double column we have to measure h1 and h2 and that is likely to increase the errors in the measurement. (This disadvantage is overcome by using a single column manometer).

(iii) Single column manometer: One of the limbs in double column manometer is converted into a reservoir having large cross sectional area (about 100 times) with respect to the other limb.

Fig. 3.3 (a) shows a vertical single column manometer. Initially, the both limbs of the manometer are exposed to atmospheric pressure. When the left limb with reservoir is connected to container, the liquid pressure of it pushes down the level A1 B1 to A2 B2 causing !l h downward movement of liquid in the reservoir. Let h2 is the rise of manometric liquid of specific weight, w2 in the narrow limb. The mass of the liquid displaced in the reservoir

= Δ h x A x p2 where A = cross-sectional area of the reservoir.

The mass of the liquid rise in the narrow limb

= ah2 x p2 where a= cross-sectional area of the narrow limb.

From the conservation of mass we can write

P2 x Δh x A = ah2 x p2

or                                               Δh = a / A h2

The gauge pressure at A2 = P A2 = P x + (h1 +Δh) w1

The gauge pressure at B2 = Pa2 = 0 + (h2 + Δh) w2

PA2 =Pa2

or         P x + (h1+ Δh) w1 = (h2 + Δh) w2

or                                      P x = (w2 h2 -w1 h1)+ (w2 – w1)

= w2 h2 (1 + a /A ) – w1(h1+ h2 a /A)

Neglecting a /A,                 P x = w2 h2 – w1 h1

If the underline term is neglected then

P x = w2 h2, so we have to measure only h2.

Fig. 3.3 (b) shows an inclined manometer. This manometer is more sensitive because the manometric liquid will traverse more distance than that of vertical. The narrow limb in this case is inclined at angle θ with the horizontal.

P t = w2 h2– w l hl

= W2 [ sin θ – W1 h1

θ min =so

Sensitivity about 25 times that of double column U-tube.

3.2.1.2 Differential Manometers

Differential manometers are used to measure the pressure difference between two points in a pipe or in two different containers. There are three types of differential manometers: (i) U-tube differential manometers (ii) Inverted u-tube differential manometers (iii) Micro manometers or u-tube double reservoir manometers.

(i) U-tube differential manometers: It consists of a u-tube, containing a heavy liquid. The two ends are connected to two different points whose pressure difference is to be measured.

From Fig. 3.4 (a) A and Bare two points at different heights which are connected by two ends of the u-tube as per the arrangement shown in Fig. 3.4 (a).

Let

w1 = sp. weight of liquid at A

w2 = sp. weight of liquid at B

w g = sp. weight of liquid at manometric liquid say mercury

P A= Pressure at A & P8 =Pressure at B.

Taking CD as equipressure line, Pressure at C (left limb)= Pressure at D (right limb).

P A+ w1 (h1 + h2 + h3) = PB + w g h1 + w2 h2

or                                     PA –PB= h1 (wg-w1) +h2 (w2-w1)-w1 h3

Difference of pressure between A and B = h1 (w g– w1) + h2 (w2 – w1)- h3 w1.

From Fig. 3.4 (b) A and B points are in the same level.

Pressure at C (left limb) = Pressure at D (right limb)

PA + (h1 +h2)w1 = w g h1 +w2 h2 + P8

or                                     P A- PB = w g h1 + w2 h2 – w1 h1 – w1 h2

Difference in pressures between A and B = h2 (w g -w1) + h2 (w2 -w1)

[Note that if we put h3 = 0, for Fig. 3.4(a) we will reach to the same equation for Fig. 3.4(b)](H)

(ii)  Inverted u-tube differential manometers: It is used for measuring difference between

low pressures. Fig. 3.5 shows an inverted u-tube differential manometer.

Pressure at C (left limb) = Pressure at D (right limb)

PA – w1 (h1 +h2 +h3) = PB -w2 h2 – wg h3

Or                                  P A– PB = w1 (h1 + h2 + h3) – w2 h2 – wg h3

Difference in pressures between A & B.= w1 h1 + h2 (w1 -w2) + h3 (w1 – wg)

(iii) Micromanometers: In this manometer there are two reservoirs and uses two

manometric liquids of different specific gravity. Fig. 3.6 shows a micromanometer.

h = fall in liquid height of sp.wtw1 in the left limb.

The fall in reservoir level in the left limb = h a / A

= rise in the reservoir height in right limb.

Pressure at C =Pressure at D

PA + w1 ( h1 + h2 – h a /A) + w (h3 + h a/A)

= PB + w1 (h2 + h a / A) + w2 h1 + w ( h3 – h a/A)

a / A = 0 (neglected), then

P A+ w1 (h1 + h2) + w1h2 = PB+ w1 h2 + w2 h3 + wh3

PA -PB = w1 h2 + w2 h1 + wh3 – w1 h1 – w1 h2 -wh3

= w2 hl – w1 h1

= h1 (w2 – w1)

PA-PB = 2h(w2-w1)

Difference in pressure between A & B = 2 h (w2 – w1).

SOLVED EXAMPLES

Example 3.1. A, B, C are three liquids of specific gravity 0.8, 0.85 and 0.95, respectively. Calculate the height of liquid column in three piezometer tubes shown in Fig. 3.7. Take datum as bottom line.3

Solution. We use                 P = wh, formula, w = s x 1000 x 9.81 N I m3

s = sp. gravity

(i) Pressure at 3,                      P3 = (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)

P 3 = 64501 N /m2

P3 = 64501 N/m2,, thus pressure head in terms of liquid C

= 64501 / 0.95 x 1000 x 9.81 m = h3

or                                             h3 = 6.92 m.

(ii) Pressure at 2,              p 2 = (0.85 x 1000 x 9.81 x 2.5) + (0.8 x 1000 x 9.81 x 2)

P2 = 36542 N/ 1112

h2 = p2 WB = 36542 / 0.85 x 1000 x 9.81 = 4.4 m (in terms of B)

(iii) Pressure at 1,                = p1 = 0.8 x 1000 x 9.81 x 2

= 15696 N / m2

Pressure head in terms of liquid A

= 15696 / 0.8 x 1000 x 9.81 = 2m = h1

Example 3.2. Fig. 3.8 shows a u-tube manometer. Determine the pressure of the fluid in pipe.

Solution.

Pressure at C = Pressure at D

Or                    p A + 0.2 x 0.85 x 9810 = 0.3 X 13.6 X 9810

or                                                     P A= 38357.1 N/m2 . Ans.

Example 3.3. Fig. 3.9 shows a u-tube manometer in balanced position. Find out the vacuum

pressure at A.

Solution.

Pressure at C =Pressure at D

p A+ 0.2 x 0.9 x 9810 + 0.5 x 13.6 x 9810 = 0,

P A = – 68474 N/m2.

Example 3.4. Fig. 3.10 shows a u-tube differential manometer. Find out the value of x.

Solution.

Pressure at C =Pressure at D

120 x 103 + 4 x 1.6 x 9810 +x x 13.6 x 9810 = 180 x 103 + (2 +X) x0.9 x 9810

Or                                    x = 0.12, (m).

Example 3.5. fig. 3.11 shows n11 i11verted u-tube differential manometer. Calculate the pressure at A if pressure at B is 4 x 104 N/m2

Solution.

Pressure at C =Pressure at D

p A– 0.75 x 1.0 x 9810 = p B– 0.25 x 0.9 x 9810- 0.3 x 0.85 x 9810

or                                               PA-PB=2648.7

or                            p A= PB + 2648.7 = 4 x 104 + 2648.7

= 42648.7N/ m2 (gauge).

Example 3.6. Fig. 3.12 shows au-tube double reservoir manometer ,find out the value of h.

Given                          PC – PB = 1 cm of water.

Solution. For initial separation

PB= PC

w1 (h1 + h a / A) = w2 (h1 – h a / A)

For final separation

Pressure at C2 = Pressure at D2

Pc +w1 (h1 +h)= PB +w2 (h1 +h)

or                                     Pc-Pa=w2 (h+h1)-w1 (h+h1)

or                        PC – PB / w = w2 / w (h + h1) – w1 /w (h + h1)

= 1 (h + h1) – 0.9 (h + h1)

= 0.1 (h+h1)

h + h1 = 0.01 / 0.1 = 1 / 10

Again for equation (1)

0.9 + 0.9h  a / A= h1 – h a / A

0.1 h1 = 1.9h a/ A = 1.9 h x 0.25 / 10 = 1.0475 h

h1 = 0.475  h

From equation (2) we have

h + 0.475 h = 0.1

h = 0.1 / 1.475 m = 0.068 m.