Mechanical Engineering Principles

240 MECHANICAL ENGINEERING PRINCIPLES

Pirani gauge

The Pirani gauge measures the resistance and thus
the temperature of a wire through which current is
flowing. The thermal conductivity decreases with
the pressure in the range 10−^1 to 10−^4 mm of
mercury so that the increase in resistance can be
used to measure pressure in this region. The Pirani
gauge is calibrated by comparison with a McLeod
gauge.

21.11 Hydrostatic pressure on

submerged surfaces

From Section 21.2, it can be seen that hydro-
static pressure increases with depth according to the
formula:

p=ρgh

Problem 10. The deepest part of the oceans

is the Mariana’s trench, where its depth is

approximately 11.52 km (7.16 miles). What

is the gauge pressure at this depth, assuming

thatρ=1020 kg/m^3 andg= 9 .81 m/s^2

Gauge pressure,

p=ρgh

= 1020

kg

m^3

× 9. 81

m

s^2

× 11. 52 × 103 m

= 11. 527 × 107 N/m^2 ×

1bar

105 N/m^2

i.e. pressure,

p= 1152 .7bar

Note that from the above calculation, it can be
seen that a gauge pressure of 1 bar is approximately
equivalent to a depth of 10 m.

Problem 11. Determine an expression for

the thrust acting on a submerged plane

surface, which is inclined to the horizontal

by an angleθ, as shown in Figure 21.11.

q

y

Normal

view of

plane

Edge view

of plane

F

y

h

h

P(x

′, y

′)

dAC(x, y)

0 Free surface (atmosphere)

Figure 21.11

From Figure 21.11,

δF=elemental thrust ondA

=ρgh×dA

but h=ysinθ

Hence, δF=ρgysinθdA

Total thrust on plane surface

=F=

∫

dF=

∫

ρgysinθdA

or F=ρgsinθ

∫

ydA

However,

∫

ydA=Ah

whereA=area of the surface,

andh=distance of the centroid of the plane

from the free surface.

Problem 12. Determine an expression for

the position of the centre of pressure of the

plane surfaceP(x′,y′)of Figure 21.11; this

is also the position of the centre of thrust.

Taking moments aboutOgives:

Fy′=

∫

ρgysinθdA×y

However, F=ρgsinθ

∫

ydA