fluid_mechanics

74 Chapter 2 ■ Fluid Statics
The pressure distribution
in a fluid
mass that is accelerating
along a
straight path is not
hydrostatic.
For the special circumstance in which a y 0, a z 0, which corresponds to the mass of
fluid accelerating in the vertical direction, Eq. 2.28 indicates that the fluid surface will be horizontal.
However, from Eq. 2.26 we see that the pressure distribution is not hydrostatic, but is
given by the equation
dp
dz r 1g a z 2
For fluids of constant density this equation shows that the pressure will vary linearly with depth,
but the variation is due to the combined effects of gravity and the externally induced acceleration,
r1g a z 2, rather than simply the specific weight rg. Thus, for example, the pressure along the bottom
of a liquid-filled tank which is resting on the floor of an elevator that is accelerating upward
will be increased over that which exists when the tank is at rest 1or moving with a constant velocity2.
It is to be noted that for a freely falling fluid mass 1a z g2, the pressure gradients in all
three coordinate directions are zero, which means that if the pressure surrounding the mass is zero,
the pressure throughout will be zero. The pressure throughout a “blob” of orange juice floating in
an orbiting space shuttle 1a form of free fall2 is zero. The only force holding the liquid together is
surface tension 1see Section 1.92.
E XAMPLE 2.11
Pressure Variation in an Accelerating Tank
GIVEN The cross section for the fuel tank of an experimental
vehicle is shown in Fig. E2.11. The rectangular tank is vented to
the atmosphere and the specific gravity of the fuel is SG 0.65.
A pressure transducer is located in its side as illustrated. During
testing of the vehicle, the tank is subjected to a constant linear acceleration,
a y .
FIND (a) Determine an expression that relates a y and the pressure
1in lbft 2 2 at the transducer. (b) What is the maximum acceleration
that can occur before the fuel level drops below the transducer?
a y Vent
Air
Fuel
(2)
0.75 ft 0.75 ft
F I G U R E E2.11
(1)
z
y
z 1
0.5 ft
Transducer
SOLUTION
(a) For a constant horizontal acceleration the fuel will move as
a rigid body, and from Eq. 2.28 the slope of the fuel surface can
be expressed as
since a z 0. Thus, for some arbitrary a y , the change in depth, z 1 , of
liquid on the right side of the tank can be found from the equation
or
dz
dy a y
g
z 1
0.75 ft a y
g
z 1 10.75 ft2 a a y
g b
Since there is no acceleration in the vertical, z, direction, the
pressure along the wall varies hydrostatically as shown by Eq.
2.26. Thus, the pressure at the transducer is given by the relationship
p gh
where h is the depth of fuel above the transducer, and therefore
p 10.652162.4 lbft 3 230.5 ft 10.75 ft21a yg24
a y
20.3 30.4
(Ans)
g
for z As written, p would be given in lbft 2 1 0.5 ft.
.
(b) The limiting value for 1a y 2 max 1when the fuel level reaches
the transducer2 can be found from the equation
or
0.5 ft 10.75 ft2 c 1a y2 max
d
g
1a y 2 max 2g
3
and for standard acceleration of gravity
1a y 2 max 2 3 132.2 ft s 2 2 21.5 fts 2
(Ans)
COMMENT Note that the pressure in horizontal layers is not
constant in this example since 0p0y ra y 0. Thus, for example,
p 1 p 2 .