Mechanical Engineering Principles

HYDROSTATICS 231

Now try the following exercise

Exercise 106 Further problems on pres-

sure

1. A force of 280 N is applied to a piston
   of a hydraulic system of cross-sectional
   area 0.010 m^2. Determine the pressure
   produced by the piston in the hydraulic
   fluid. [28 kPa]


2. Find the force on the piston of question 1
   to produce a pressure of 450 kPa.
   [4.5 kN]


3. If the area of the piston in question
   1 is halved and the force applied is
   280 N, determine the new pressure in the
   hydraulic fluid. [56 kPa]

21.2 Fluid pressure

A fluid is either a liquid or a gas and there are
four basic factors governing the pressure within
fluids.

(a) The pressure at a given depth in a fluid is equal
in all directions, see Figure 21.1(a).

(b) The pressure at a given depth in a fluid is
independent of the shape of the container in
which the fluid is held. In Figure 21.1(b), the
pressure at X is the same as the pressure
atY.

F

P

P PP X

Y

A

B

C D

E

F

D

B

AC

(a) (b) (c) (d)

Figure 21.1

(c) Pressure acts at right angles to the surface con-
taining the fluid. In Figure 21.1(c), the pres-
sures at pointsAtoFall act at right angles to
the container.

(d) When a pressure is applied to a fluid, this
pressure is transmitted equally in all directions.
In Figure 21.1(d), if the mass of the fluid is
neglected, the pressures at pointsAtoDare
all the same.

The pressure,p, at any point in a fluid depends on

three factors:

(a) the density of the fluid,ρ, in kg/m^3 ,

(b) the gravitational acceleration, g, taken as

approximately 9.8 m/s^2 (or the gravitational

field force in N/kg), and

(c) the height of fluid vertically above the point,h

metres.

The relationship connecting these quantities is:

p=ρghPascal’s

When the container shown in Figure 21.2 is filled

with water of density 1000 kg/m^3 , the pressure due

to the water at a depth of 0.03 m below the surface

is given by:

p=ρgh=( 1000 × 9. 8 × 0. 03 )Pa=294 Pa

0.03 m

Figure 21.2

Problem 4. A tank contains water to a

depth of 600 mm. Calculate the water

pressure (a) at a depth of 350 mm, and (b) at

the base of the tank. Take the density of

water as 1000 kg/m^3 and the gravitational

acceleration as 9.8 m/s^2.

From above, pressurepat any point in a fluid is

given byp=ρghpascals, whereρis the density

in kg/m^3 ,gis the gravitational acceleration in m/s^2

andhis the height of fluid vertically above the point.

(a) At a depth of 350 mm, i.e. 0.35 m,

p= 1000 × 9. 8 × 0. 35 =3430 Pa= 3 .43 kPa

(b) At the base of the tank, the vertical height of

the water is 600 mm, i.e. 0.6 m. Hence,

p= 1000 × 9. 8 × 0. 6 =5880 Pa= 5 .88 kPa

Problem 5. A storage tank contains petrol

to a height of 4.7 m. If the pressure at the