Mechanical Engineering Principles

HYDROSTATICS 233

From above, absolute pressure

=atmospheric pressure+gauge pressure

=( 101 + 302. 82 )kPa= 403 .82 kPa

That is,the absolute pressure at a depth of 30 m
is 403.82 kPa.

Now try the following exercise

Exercise 108 Further problems on atmo-

spheric pressure

Take the gravitational acceleration as 9.8 m/s^2 ,

the density of water as 1000 kg/m^3 ,andthe

density of mercury as 13600 kg/m^3.

1. The height of a column of mercury in
   a barometer is 750 mm. Determine the
   atmospheric pressure, correct to 3 signif-
   icant figures. [100 kPa]


2. A U-tube manometer containing mercury
   gives a height reading of 250 mm of mer-
   cury when connected to a gas cylinder. If
   the barometer reading at the same time is
   756 mm of mercury, calculate the abso-
   lute pressure of the gas in the cylinder,
   correct to 3 significant figures.
   [134 kPa]


3. A water manometer connected to a
   condenser shows that the pressure in the
   condenser is 350 mm below atmospheric
   pressure. If the barometer is reading
   760 mm of mercury, determine the
   absolute pressure in the condenser, correct
   to 3 significant figures. [97.9 kPa]


4. A Bourdon pressure gauge shows a pres-
   sure of 1.151 MPa. If the absolute pres-
   sure is 1.25 MPa, find the atmospheric
   pressure in millimetres of mercury.
   [743 mm]

21.4 Archimedes’ principle

Archimedes’ principle states that:

If a solid body floats, or is submerged, in a liquid,

the liquid exerts an upthrust on the body equal to the

gravitational force on the liquid displaced by the body.

In other words, if a solid body is immersed in a

liquid, the apparent loss of weight is equal to the

weight of liquid displaced.

IfVis the volume of the body below the surface

of the liquid, then the apparent loss of weightWis

given by:

W=Vω=Vρg

whereωis the specific weight (i.e. weight per unit

volume) andρis the density.

If a body floats on the surface of a liquid all of

its weight appears to have been lost. The weight

of liquid displaced is equal to the weight of the

floating body.

Problem 8. A body weighs 2.760 N in air

and 1.925 N when completely immersed in

water of density 1000 kg/m^3. Calculate

(a) the volume of the body, (b) the density of

the body and (c) the relative density of the

body. Take the gravitational acceleration as

9.81 m/s^2.

(a) The apparent loss of weight is 2.760 N−

1 .925 N = 0 .835 N. This is the weight of

water displaced, i.e. Vρg,whereV is the

volume of the body andρis the density of

water, i.e.

0 .835 N=V×1000 kg/m^3 × 9 .81 m/s^2

=V× 9 .81 kN/m^3

Hence, V=

0. 835

9. 81 × 103

m^3

= 8. 512 × 10 −^5 m^3

= 8. 512 × 104 mm^3

(b) The density of the body

=

mass

volume

=

weight

g×V

=

2 .760 N

9 .81 m/s^2 × 8. 512 × 10 −^5 m^3

=

2. 760

9. 81

kg× 105

8 .512 m^3

=3305 kg/m^3

= 3 .305 tonne/m^3

(c) Relative density=

density

density of water