**FLUID STATIC **

**OBJECTIVES **

**General Objective ** : To know, understand and apply pressure measurement using
** piezometer, barometer, bourdon gauge and manometers. **
**Specific Objectives : At the end of the unit you should be able to : **

¾ explain the concept of pressure measurement using piezometer.

¾ solve problems related to piezometer.

¾ explain the concept of pressure measurement using barometer.

¾ solve problems related to barometer.

¾ sketch, label and explain the working principle of bourdon gauge

¾ explain the concept of pressure measurement using manometers.

¾ solve problems related to the use of simple manometer.

¾ solve problems related to the use of differential manometer.

¾ solve problems related to the use of inverted differential manometer.

** **

**3.0 ** **PRESSURE MEASUREMENT **

**3.1 PIEZOMETER ( Pressure Tube ) **

Figure 3.1: Piezometer inside a pipe

A Piezometer is used for measuring pressure inside a vessel or pipe in which
liquid is there. A tube may be attached to the walls of the container (or pipe) in
which the liquid resides so that liquid can rise in the tube. By determining the
*height to which liquid rises and using the relation p*1* = ρgh, gauge pressure of *

the liquid can be determined. It is important that the opening of the device is to be tangential to any fluid motion, otherwise an erroneous reading will result.

Although the Piezometer tube is a very simple and accurate pressure measuring device, it has several disadvantages. It is only suitable if the pressure in the container (pipe or vessel) is greater than the atmospheric pressure (otherwise air would be sucked into system), and the pressure to be measured must be relatively small so that the required height of column is reasonable. Also, the fluid in the container in which the pressure is to be measured must be a liquid rather than a gas.

**Example 3.1 **

A pressure tube is used to measure the pressure of oil ( mass density,_{640}_{kg}_{/}* _{m}*3

_{) in a }

pipeline. If the oil rises to a height of 1.2 above the centre of the pipe, what is the
gauge pressure in *N/ m*2*at that point? (gravity = 9.81 m/s2*)

**Solution to Example 3.1 **
Putting _{ρ} _{=}_{640}_{kg}_{/}* _{m}*3

_{ }and

*h*=1.2

*m*We know that,

*p*=ρ

*gh*So,

*p*=640×9.81×1.2 2 / 55 . 7

*kN*

*m*

*p*=

**3.2 BAROMETERS **

Figure 3.2: Mercury Barometer

A Barometer is a device used for measuring atmospheric pressure. A
*simple Barometer consists of a tube of more than 30 inch (760 mm) long *
inserted into an open container of mercury with a closed and evacuated end at
the top and open tube end at the bottom and with mercury extending from the
container up into the tube. Strictly, the space above the liquid cannot be a true
vacuum. It contains mercury vapour at its saturated vapour pressure, but this is
*extremely small at room temperatures (e.g. 0.173 Pa at 20*o*C). The atmospheric *

*pressure is calculated from the relation patm = ρgh where ρ is the density of *

fluid in the barometer. There are two types of Barometer; Mercury Barometer and Aneroid Barometer.

**Example 3.2 **

Describe with a sketch, one method of measuring atmospheric pressure.

**Solution to Example 3.2 **

Vacuum (zero pressure)

Mercury sp.wg, ωm
* p p *
A
*pA*
Figure 3.3

*A Mercury Barometer in its simplest form consists of a glass tube, about 1 m *
long and closed at one end, which is completely filled with mercury and inverted in a
bowl of mercury (Figure 3.3). A vacuum forms at the top of the tube and the
atmospheric pressure acting on the surface of the mercury in the bowl supports a
*column of mercury in the tube height, h. *

**Example 3.3 **

*What is the atmospheric pressure in N/m2 *if the level of mercury in a Barometer
*(Figure 3.3) tube is 760 mm above the level of the mercury in the bowl? Given the *
specific gravity of mercury is 13.6 and specific weight of water is _{9}_{.}_{81}_{×}_{10}3_{N}_{/}* _{m}*3

_{. }

**Solution to Example 3.3 **

*If A is a point in the tube at the same level as the free surface outside, the pressure pA*

*at A is equal to the atmospheric pressure p at the surface because, if the fluid is at rest, *
pressure is the same at all points at the same level.

The column of mercury in the tube is in equilibrium under the action of the force due
*to pA* acting upwards and its weight acting downwards; there is no pressure on the top

of the column as there is a vacuum at the top of the tube.

So,

*A*
*column*
*of*

*area*

*p _{A}* × =

*specificweight*

*of*

*mercury*×

*specificweightof*

*water*

*p _{A}*×

*A*=ω

*×*

_{m}*ah*or

*p*=ω

_{A}*×*

_{m}*h*Putting

*h = 760 mm = 0.76 mm*While

*water*

*of*

*weight*

*specific*

*mercury*

*of*

*gravity*

*specific*

*m*= × ω

**ω**

*=13.6×9.81×103*

_{m}*N*/

*m*2 From

*h*

*p*=ω

_{A}*× So*

_{m}

_{p}_{13}

_{.}

_{6}

_{9}

_{.}

_{81}

_{10}3

_{0}

_{.}

_{76}

_{N}_{/}

*2*

_{m}*A*= × × ×

_{=}

_{101}

_{.}

_{3}

_{kN}_{/}

*2*

_{m}**3.3 BOURDON GAUGE **

Figure 3.4

*Bourdon Gauge is used to measure pressure differences that are more than 1.2 bar. *
The pressure to be measured is applied to a curved tube, oval in cross section. Pressure
applied to the tube tends to cause the tube to straighten out, and the deflection of the
end of the tube is communicated through a system of levers to a recording needle. This
gauge is widely used for steam and compressed gases. The pressure indicated is the
difference between that communicated by the system to the external (ambient)
pressure, and is usually referred to as the gauge pressure.

The pressure sensing element is a tube of oval cross-section bent to a circular shape.

One end of the tube is fixed to the gauge case and is connected to the fluid whose pressure is to be measured.

The other end is closed and is free to move as it is connected via mechanical linkage and gear sector to a pointer.

As measured fluid pressure increases above the surroundings, the tube cross-section tends to become circular and causes the tube to deflect at the second end.

This motion is transmitted via linkage to the pointer, which would directly indicate on the calibrated scale or dial on the gauge pressure.

**ACTIVITY 3A **

**TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT **

INPUT…!

**3.1 What is the maximum gauge pressure of water of height 1.5 m that can be measured **

by a Piezometer? If the liquid had a relative density of 0.85 what would the maximum measurable gauge pressure?

**FEEDBACK ON ACTIVITY 3A **

**3.1**

*Gauge pressure, p =*ρ

*gh*

**For water,**

*p =*ρ

*watergh*

*p = 1000 x 9.81x 1.5*

*p = 14715 N/m2*

_{ (or Pa) }*p = 14.715 kN/m2 (or kPa)*

**For Liquid,**

*p =*ρ

*liquidgh*

ρ*liquid = *ρ*water x relative density *

*p = 1000 x 0.85 x 9.81 x 1.5 *

*p = 12507.75 N/m2 (or Pa) *

** **

**3.4 MANOMETERS **

The relationship between pressure and head is utilized for pressure measurement in the manometer or liquid gauge. We can measure comparatively high pressures and negative pressures with the manometer. The following are a few types of manometers: a) Simple manometer,

b) Differential manometer and c) Inverted differential manometer.

**3.4.1 SIMPLE MANOMETER **

A simple manometer is a tube bent in U-shape. One end of which is attached to the gauge point and the other is open to the atmosphere as shown in (Figure 3.5)

The liquid used in the bent tube or simple manometer is generally mercury which is 13.6 times heavier than water. Hence, it is also suitable for measuring high pressure.

Now consider a simple manometer connected to a pipe containing a light liquid under high pressure. The high pressure in the pipe will force the heavy liquid, in the left-hand limb of the U-tube, to move downward. This downward movement of the heavy liquid in the left-hand limb will cause a corresponding rise of the heavy liquid in the right-hand limb. The horizontal surface, at which the heavy and light liquid meet in the left-hand limb is known as a common surface or datum line. Let B-C be the datum line, as shown in Figure 3.5.

*Let h1* = Height of the light liquid in the left-hand limb above the common

surface in cm.

*h2*= Height of the heavy liquid in the right-hand limb above the common

surface in cm.

*p _{A}*= Pressure in the pipe, expressed in terms of head of water in cm.

* *ω* _{P}* = Specific weight of the light liquid

*s = Specific gravity of the heavy liquid.*

_{Q}The pressure in the left-hand limb and the right-hand limb above the datum line is equal.

** Pressure **

*p*

*B*

** at B = Pressure **

**p at C **

**p at C**

*C*

### ω

**P**

**s**

**s**

**Q**

**Pressure in the left-hand limb above the datum line **

*B*

*p* = Pressure, *p _{A}* at A + Pressure due to depth,

*h*

_{1}of fluid P =

*p*+ω

_{A}

_{P}h_{1}=

*p*+ ρ

_{A}

_{P}gh_{1}

### Specific weight,

**ω = **

### ρ

** g **

**g**

### ω

**P**Imagine that the right limb is hidden.

**Thus pressure in the right-hand limb above the datum line; **

*C*

*p = Pressure p _{D}* at D + Pressure due to depth

*h*

_{2}of liquid Q

But *p _{D}* = Atmospheric pressure = Zero gauge pressure
And so,

*p =*0+ω

_{C}

_{Q}h_{2}= 0+ ρ

_{Q}gh_{2}Since

*p*=

_{B}*p*, 2 1

_{C}*gh*

*gh*

*p*+ρ

_{A}*=ρ*

_{P}*so,*

_{Q}*p*=ρ

_{A}

_{Q}gh_{2}−ρ

_{P}gh_{1}

**sQ**

Imagine that
**sQ**

the left limb is hidden.

**Example 3.4 **

A U-tube manometer similar to that shown in Figure 3.6 is used to measure the gauge
pressure of water (mass density *ρ = 1000 kg /m3*). If the density of mercury is 13.6 ×

103*kg /m3*, what will be the gauge pressure at A if *h1* = 0.45 *m and D is 0.7 m above *

BC.
** **
** **
**Solution to Example 3.4 **
Considering
3
3 _{/}
10
6
.
13 *kg* *m*
*Q* = ×
ρ
3
3 _{/}
10
0
.
1 *kg* *m*
*P* = ×
ρ
*m*
*h*_{1} =0.45
*m*
*h*_{2} =0.7

the pressure at left-hand limb;

*B*

*p* = Pressure, *p _{A}* at A + Pressure due to depth,

*h*

_{1}of fluid P =

*p*+ω

_{A}

_{P}h_{1}

= *p _{A}* + ρ

_{P}gh_{1}

### ω

water### ω

mercurythe pressure at right-hand limb;

*C*

*p = Pressure p _{D}* at D + Pressure due to depth

*h*

_{2}of liquid Q

*C*
*p = *0+ω_{Q}h_{2}
= 0+ ρ_{Q}gh_{2}
Since *p _{B}* =

*p*2 1

_{C}*gh*

*gh*

*p*+ρ

_{A}*=ρ*

_{P}*1 2*

_{Q}*gh*

*gh*

*p*= ρ

_{A}*−ρ*

_{Q}

_{P}_{=}13.6

_{×}103

_{×}9.81

_{×}0.7

_{−}1.0

_{×}103

_{×}9.81

_{×}0.45 =88976.7

*N*/

*m*2 =88.97×103

*N*/

*m*2

If negative pressure is to be measured by a simple manometer, this can be measured easily as discussed below:

In this case, the negative pressure in the pipe will suck the light liquid which will pull up the heavy liquid in the left-hand limb of the U-tube. This upward movement of the heavy liquid, in the left-hand limb will cause a corresponding fall of the liquid in the right-hand limb as shown in Figure 3.7.

In this case, the datum line B-C may be considered to correspond with the top level of the heavy liquid in the right column as shown in the Figure 3.7.

Now to calculate the pressure in the left- hand limb above the datum line.

*Let h1* = Height of the light liquid in the left-hand limb above the common

surface in cm.

*h2*= Height of the heavy liquid in the left-hand limb above the common

* surface in cm *

* pA* = Pressure in the pipe, expressed in terms of head of water in cm.

* sP* = Specific gravity of the light liquid

* sQ* = Specific gravity of the heavy liquid.

Pressure in the left-hand limb above the datum line;

*B*

*p* = Pressure *p _{A}* at A + Pressure due to depth

*h*

_{1}of fluid P + Pressure due to depth

*h*

_{2}of liquid Q

= *p _{A}* +ω

_{P}h_{1}+ω

_{Q}h_{2}

= *p _{A}* + ρ

_{P}gh_{1}+ρ

_{Q}gh_{2}

Pressure in the right-hand limb above the datum line;

*C*

*p = Pressure p _{D}* at D

But *p _{D}* = Atmospheric pressure
And so,

*p =*

_{C}*p*

_{atm}Since *p _{B}* =

*p*

_{C}*D*

*Q*

*P*

*A*

*gh*

*gh*

*p*

*p*+ρ

_{1}+ ρ

_{2}=

*p*=

_{A}*p*−

_{B}## (

ρ

_{P}gh_{1}+ρ

_{Q}gh_{2}

## )

**Example 3.5 **

A U-tube manometer similar to that shown in Figure 3.8 is used to measure the gauge
*pressure of a fluid P of density ρ = 1000 kg/m3*. If the density of the liquid Q is 13.6 ×
103* kg/m3, what will be the gauge pressure at A if h1 = 0.15 m and h*2* = 0.25 m above *

*BC. Take into consideration p*atm* = 101.3 kN/m2*.

** ** ** **
** Solution to Example 3.5 **
Putting ,
*ρ*Q = 13.6 ×103
*ρ*P* = 1000 kg/m3 *
*h*1* = 0.15 m *
*h*2* = 0.25 m *

pressure at left-hand limb;

*B*

*p* = Pressure *p _{A}* at A + Pressure due to depth

*h*

_{1}of fluid P + Pressure due to depth

*h*

_{2}of liquid Q

= *p _{A}* +ω

_{P}h_{1}+ω

_{Q}h_{2}

= *p _{A}* + ρ

_{P}gh_{1}+ρ

_{Q}gh_{2}

pressure at right-hand limb;

*C*
*p = Pressure p _{D}* at D

*D*

*p*= Atmospheric pressure

*C*

*p = p*Figure 3.8

_{atm}Since *p _{B}* =

*p*,

_{C}*D*

*Q*

*P*

*A*

*gh*

*gh*

*p*

*p*+ρ

_{1}+ ρ

_{2}=

*p*=

_{A}*p*−

_{B}## (

ρ

_{P}gh_{1}+ρ

_{Q}gh_{2}

## )

### (

13.6 103 9.81 0.15 1000 9.81 0.25### )

3 . 101 − × × × + × × = 2 / 1 . 70835*N*

*m*= 2 / 84 . 70

*kN*

*m*=

**ACTIVITY 3B **

**3.2 ** A U-tube manometer is used to measure the pressure which is more than the

atmospheric pressure in a pipe, the water being in contact with the mercury in
*the left-hand limb. The mercury is 20 cm below A in the left-hand limb and 25 *

**3.3 ** The U-tube manometer measures the pressure of water at A which is below the

atmospheric pressure. If the specific weight of mercury is 13.6 times that of
*water and the atmospheric pressure is 101.3 kN/m2*, find what is the absolute
*pressure at A when h*1* = 10 cm, h*2* = 25 cm and the specific weight of water is *

**FEEDBACK ON ACTIVITY 3B **

**3.2**

**3.3 **
2
1 *h*
*h*
*p*
*p _{B}* =

*+ω*

_{A}*+ω*

_{air}*2 / 3 . 101*

_{mercury}*kN*

*m*

*p*

*p*=

_{C}*=*

_{atm}*C*

*B*

*p*

*p*=

*atm*

*mercury*

*air*

*A*

*h*

*h*

*p*

*p*+ω

_{1}+ω

_{2}= 2 1

*h*

*h*

*p*

*p _{A}* =

*−ω*

_{atm}*−ω*

_{air}

_{mercury}### ( )

0.1 9810### (

13.6### )(

0.25### )

9810 10 3 . 101_{×}3

_{−}

_{−}=

*A*

*p*

**=66965**

*N*/

*m*2

**=66.965**

*kN*/

*m*2

** **

**3.4.2 DIFFERENTIAL MANOMETER **

It is a device used for measuring the difference of pressures, between two points in a pipe, or in two different pipes.

A differential manometer consists of a U-tube, containing a heavy liquid with two ends connected to two different points. We are required to find the difference of pressure at these two points, as shown in Figure 3.9.

*A differential manometer is connected to two different points A and B. A little *
*consideration will show that the greater pressure at A will force the heavy liquid in the *
U-tube to move downwards. This downward movement of the heavy liquid, in the
left-hand limb, will cause a corresponding rise of the heavy liquid in the right-left-hand limb as
shown in Figure 3.9.

*The horizontal surface C-D, at which the heavy liquid meet in the left-hand limb, is *
the datum line.

**INPUT **

**INPUT**

*Let h = Height of the light liquid in the left-hand limb above the datum line. *

* h*1 = Height of the heavy liquid in the right-hand limb above the datum line

* h*2 = Height of the light liquid in the right-hand limb above the datum line

* p*A = Pressure in the pipe A, expressed in term of head of the liquid in cm

* p*B = Pressure in the pipe B, expressed in term of head of the liquid in cm

* ω*P = Specific weight of the light liquid

* ω*Q = Specific weight of the heavy liquid

We know that the pressures in the left-hand limb and right-hand limb , above the datum line are equal.

**Pressure in the left-hand limb above the datum line **

*C*

*p = Pressure p _{A} at A + Pressure due to depth h of fluid P *

*h*
*p*
*p _{C}* =

*+ω*

_{A}

_{P}*gh*

*p*

*p*=

_{C}*+ρ*

_{A}

_{P}** Pressure **

*p*

*C*

** at C = Pressure **

*p*

*D*

** at D **

**Pressure in the right-hand limb above the datum line **

*D*

*p* = Pressure *p _{A}* at A + Pressure due to depth

*h*

_{1}of fluid P + Pressure due to depth

2
*h* of liquid Q
2
1 *h*
*h*
*p*
*p _{D}* =

*+ω*

_{B}*+ω*

_{Q}

_{P}**=**

*p*+ρ

_{B}

_{Q}gh_{1}+ρ

_{P}gh_{2}Since,

*p*=

_{C}*p*2 1

_{D}*gh*

*gh*

*p*

*gh*

*p*+ρ

_{A}*=*

_{P}*+ρ*

_{B}*+ρ*

_{Q}

_{P}

*p*−

_{A}*p*=ρ

_{B}

_{Q}gh_{1}+ρ

_{P}gh_{2}−ρ

_{P}gh**Example 3.6 **

A U tube manometer measures the pressure difference between two points A and B in
*a liquid. The U tube contains mercury. Calculate the difference in pressure if h =1.5 m, *

*h2 = 0.75 m and h1 = 0.5 m. The liquid at A and B is water ( ω = 9.81 × 10*3* N/m*2) and

the specific gravity of mercury is 13.6.

**Solution to Example 3.6 **

Since C and D are at the same level in the same liquid at rest
Pressure *p _{P}* at C = Pressure

*p at D*

_{Q}For the left hand limb

*h*
*p*

*p _{C}* =

*+ω*

_{A}For the right hand limb

### (

*h*2

*h*1

### )

*s*

*h*1

*p*

*p*=

_{D}*+ω − + ω =*

_{B}*p*+ω

_{B}*h*

_{2}−ω

*h*

_{1}+

*s*ω

*h*

_{1}since

*p*=

_{C}*p*1 1 2

_{D}*h*

*s*

*h*

*h*

*p*

*h*

*p*+ω =

_{A}*+ω −ω + ω Pressure difference*

_{B}*p*−

_{A}*p*

_{B}*h*

*h*

*s*

*h*

*h*ω ω ω ω − + − =

_{2}

_{1}

_{1}1 1 2

*h*

*s*

*h*

*h*

*h*ω ω ω ω − + − =

### (

_{2}−

### )

+_{1}

### (

−1### )

=ω*h*

*h*ω

*h*

*s*

### (

0.75 1.5### )

9.81 10### ( )(

0.5 13.6 1### )

10 81 . 9_{×}3

_{−}

_{+}

_{×}3

_{−}= 2 / 5 . 54445

*N*

*m*=

_{=}

_{54}

_{.}

_{44}

_{kN}_{/}

*2*

_{m}

**Figure 3.10**

**ACTIVITY 3C **

**3.4 ** A U tube manometer measures the pressure difference between two points A

and B in a liquid. The U tube contains mercury. Calculate the difference in
*pressure if h = 2.0 m, h2 = 0.35 m and h1 = 0.5 m. The liquid at A and B is oil *

**FEEDBACK ON ACTIVITY 3C **

**3.4 **

Since C and D are at the same level in the same liquid at rest
Pressure *p _{C}* at C = Pressure

*p*at D

_{D}For the left hand limb

*h*
*p*

*p _{C}* =

*+ω*

_{A}For the right hand limb

1
2 *s* *h*
*h*
*p*
*p _{D}* =

*+ω + ω since*

_{B}*p*=

_{P}*p*1 2

_{Q}*s*

*h*

*h*

*p*

*h*

*p*+ω =

_{A}*+ω + ω Pressure difference*

_{B}*p*−

_{A}*p*

_{B}*h*

*h*

*s*

*h*

_{oil}*oil*ω ω ω + − =

_{2}

_{1}=0.85

### (

9810### )(

0.35### )

+13.6### (

9810### )( )

0.5 +0.85### (

9810### )( )

2.0 2 / 475 . 69626*N*

*m*= 2 / 626 . 69

*kN*

*m*= mercury oil

** **

**3.4.3 INVERTED DIFFERENTIAL MANOMETER **

It is a particular type of differential manometer, in which an inverted U-tube is
used. An inverted differential manometer is used for measuring the difference
of low pressure, where accuracy is the prime consideration. It consists of an
*inverted U-tube, containing a light liquid. The two ends of the U-tube are *
connected to the points where the difference of pressure is to be found out as
shown in Figure 3.10.

Now consider an inverted differential manometer whose two ends are
*connected to two different points A and B. Let us assume that the pressure at *
*point A is more than that at point B, a greater pressure at A will force the light *
liquid in the inverted U-tube to move upwards. This upward movement of
liquid in the left limb will cause a corresponding fall of the light liquid in the
*right limb as shown in Figure 3.10. Let us take C-D as the datum line in this *
case.

**INPUT **

**INPUT**

**ω**

**ω**

**P**

**ωQ**

Figure 3.11
**ωQ**

Let *h = Height of the heavy liquid in the left-hand limb below the datum line, *
*h1*= Height of the light liquid in the left-hand limb below the datum line ,

*h2*= Height of the light liquid in the right-hand limb below the datum line,

*ω*P*= Specific weight of the light liquid *

*ω*Q*= Specific weight of the heavy liquid *

We know that pressures in the left limb and right limb below the datum line are equal.

**Example 3.7 **

The top of an inverted U tube manometer is filled with oil of specific gravity,

*s*oil=0.98 and the remainder of the tube with water whose specific weight of water,

ω= 9.81×103* N/m2. Find the pressure difference in N/m2* between two points A and B
*at the same level at the base of the legs where the difference in water level h is 75 mm.*

** Pressure **

**p at C = Pressure **

**p at C = Pressure**

*C*

*p*

*D*

** at D **

**Solution to Example 3.7 **

For the left hand limb

1

2 *s* *h*

*h*
*p*

*p _{D}* =

*−ω −*

_{A}*ω*

_{o}for the right hand limb

### (

*h*1

*h*2

### )

*p*

*p*=

_{C}*−ω − =*

_{B}*p*−ω

_{B}*h*

_{1}−ω

*h*

_{2}since,

*p*=

_{C}*p*1 2 2 1

_{D}*h*

*p*

*h*

*s*

*h*

*h*

*p*−ω −ω =

_{B}*−ω − ω 2 1 1 2*

_{A}*s*

*h*

*h*

*h*

*h*

*p*

*p*−

_{B}*=−ω − ω +ω +ω =ω*

_{A}*h*

_{1}−

*s*ω

*h*

_{1}=ω

*h*

_{1}

### (

1−*s*

### )

_{=}9.81

_{×}103

### (

0.075### )

### [

1_{−}0.98

### ]

_{=}

_{14}

_{.}

_{715}

_{N}_{/}

*2*

_{m}**ACTIVITY 3D **

**3.5 ** An inverted U tube as shown in the figure below is used to measure the pressure

difference between two points A and B which has water flowing. The difference in
*level h = 0.3 m, a = 0.25 m and b = 0.15 m. Calculate the pressure difference p*B* – p*A if

the top of the manometer is filled with: (a) air

(b) oil of relative density 0.8.

**FEEDBACK ON ACTIVITY 3D **

** 3.5 **

In either case, the pressure at X-X will be the same in both limbs, so that

### (

*b*

*h*

### )

*g*

*p*

*gh*

*ga*

*p*

*p*=

_{XX}*−ρ −ρ*

_{A}*=*

_{mano}*−ρ +*

_{B}### (

### )

### (

*mano*

### )

*A*

*B*

*p*

*g*

*b*

*a*

*gh*

*p*− =ρ − + ρ−ρ

*(a) if the top is filled with air ρmano is negligible compared with ρ. Therefore, *

### (

*b*

*a*

### )

*gh*

*g*

*p*

*p*−

_{B}*=ρ − +ρ = ρ*

_{A}*g*

### (

*b*−

*a*+

*h*

### )

putting 103 / 3, 0.15 , 0.25 , 0.3 : 2*O*

*kg*

*m*

*b*

*m*

*a*

*m*

*h*

*m*

*H*= = = = = ρ ρ

### (

0.15 0.25 0.3### )

81 . 9 103_{×}

_{−}

_{+}= −

_{A}*B*

*p*

*p*

_{=}

_{1}

_{.}

_{962}

_{×}

_{10}3

_{N}_{/}

*2*

_{m}*(b) if the top is filled with oil of relative density 0.8, ρmano *= 0.8 ρ_{H}_{2}* _{O}*,

*p _{B}* −

*p*=ρ

_{A}*g*

### (

*b*−

*a*

### )

+*gh*

### (

ρ−ρ

_{mano}### )

_{=}103_{×}9.81

### (

0.15_{−}0.25

### )

_{+}9.81

_{×}0.3

_{×}103

### (

1_{−}0.8

### )

_{=}103

_{×}9.81

### (

_{−}0.1

_{+}0.06

### )

**SELF-ASSESSMENT **

**You are approaching success. Try all the questions in this self-assessment section and check **
your answers with those given in the Feedback on Self-Assessment. If you face any problems,
discuss it with your lecturer. Good luck.

**3.1 ** *What is the gauge pressure of the water at A if h*1* = 0.6 m and the mercury in *

the right hand limb, h2* = 0.9 m as shown in the figure below? *

**3.2 ** *In the figure below, fluid at A is water and fluid B is mercury (s =13.6). What *

*will be the difference in level h if the pressure at X is 140 kN/m2 _{ and a =1.5 m? }*

**3.3 ** *Assuming that the atmospheric pressure is 101.3 kN/m2* find the absolute

*pressure at A in the figure below when *

*a) fluid P is water, fluid Q is mercury ω = 13.6, a = 1 m and h = 0.4 m. *
*b) fluid P is oil ω = 0.82, fluid Q is brine ω = 1.10, a = 20 cm and h = 55 cm. *

**3.4 ** In the figure below, fluid P is water and fluid Q is mercury (specific

*gravity=13.6). If the pressure difference between A and B is 35 kN/m*2*, a = 1 m *
*and b = 30 cm, what is the difference in level h? *

**3.5 ** According to the figure in question 3.4, fluid P is oil (specific gravity = 0.85)

*and fluid Q is water. If a = 120 cm, b = 60 cm and h = 45 cm, what is the *
*difference in pressure in kN/m*2_{ between A and B? }

**3.6 ** In the figure below, fluid Q is water and fluid P is oil (specific gravity = 0.9).

*If h = 69 cm and z = 23 cm, what is the difference in pressure in kN/m*2
between A and B?

**3.7 ** In question 6, fluid Q is water and fluid P is air. Assuming that the specific

*weight of air is negligible, what is the pressure difference in kN/m*2 between A
and B?

** **

**FEEDBACK ON SELF-ASSESSMENT **

**Answers:**1.

*114.188 kN/m2*2. 1.164

*m*3. a) 38.2

*kN/m2*b) 93.8

*kN/m2*4. 30.7 cm 5. -5.23

*kN/m2*6. -1.57

*kN/m2*

*7. 4.51 kN/m2*