Mechanical Engineering Principles

232 MECHANICAL ENGINEERING PRINCIPLES

base of the tank is 32.3 kPa, determine the

density of the petrol. Take the gravitational

field force as 9.8 m/s^2.

From above, pressure p = ρghPascal’s, where
ρis the density in kg/m^3 ,g is the gravitational
acceleration in m/s^2 andhis the vertical height of
the petrol.

Transposing gives: ρ=

p

gh

The pressurepis 32.2kPa=32200 Pa, hence,

density,ρ=

32200

9. 8 × 4. 7

=699 kg/m^3

That is,the density of the petrol is 699 kg/m^3.

Problem 6. A vertical tube is partly filled

with mercury of density 13600 kg/m^3 .Find

the height, in millimetres, of the column of

mercury, when the pressure at the base of the

tube is 101 kPa. Take the gravitational field

force as 9.8 m/s^2.

From above, pressure p = ρgh, hence vertical
heighthis given by:

h=

p

ρg

Pressure p=101 kPa=101000 Pa, thus,

h=

101000

13600 × 9. 8

= 0 .758 m

That is,the height of the column of mercury is
758 mm.

Now try the following exercise

Exercise 107 Further problems on fluid

pressure

Take the gravitational acceleration as 9.8 m/s^2

1. Determine the pressure acting at the base
   of a dam, when the surface of the water is
   35 m above base level. Take the density
   of water as 1000 kg/m^3. [343 kPa]


2. An uncorked bottle is full of sea water
   of density 1030 kg/m^3. Calculate, correct
   to 3 significant figures, the pressures on

the side wall of the bottle at depths of

(a) 30 mm, and (b) 70 mm below the

top of the bottle.

[(a) 303 Pa (b) 707 Pa]

3. A U-tube manometer is used to determine
   the pressure at a depth of 500 mm below
   the free surface of a fluid. If the pressure
   at this depth is 6.86 kPa, calculate the
   density of the liquid used in the manome-
   ter. [1400 kg/m^3 ]

21.3 Atmospheric pressure

The air above the Earth’s surface is a fluid, hav-

ing a density,ρ, which varies from approximately

1.225 kg/m^3 at sea level to zero in outer space.

Sincep= ρgh, where heighthis several thou-

sands of metres, the air exerts a pressure on all

points on the earth’s surface. This pressure, called

atmospheric pressure, has a value of approxi-

mately 100 kilopascals. Two terms are commonly

used when measuring pressures:

(a) absolute pressure, meaning the pressure above

that of an absolute vacuum (i.e. zero pres-

sure), and

(b) gauge pressure, meaning the pressure above

that normally present due to the atmosphere.

Thus,absolute pressure=atmospheric pressure+

gauge pressure

Thus, a gauge pressure of 50 kPa is equivalent to an

absolute pressure of( 100 + 50 )kPa, i.e. 150 kPa,

since the atmospheric pressure is approximately

100 kPa.

Problem 7. Calculate the absolute pressure

at a point on a submarine, at a depth of 30 m

below the surface of the sea, when the

atmospheric pressure is 101 kPa. Take the

density of sea water as 1030 kg/m^3 and the

gravitational acceleration as 9.8 m/s^2.

From Section 21.2, the pressure due to the sea, that

is, the gauge pressure (pg) is given by:

pg=ρghPascal’s, i.e.

pg= 1030 × 9. 8 × 30 =302820 Pa= 302 .82 kPa