1.3 Fluid Dynamics
13
Exercise 1.4 The diagram represents a Pitot tube inserted into a duct through which
air circulates. The difference in height of mercury measured on the Pitot was 10 mm.
You are asked to determine the speed at which air moves inside the duct. Data: Air
density: 1.2 kg m
−3 , mercury density: 13.60 g cm
−3 .
10 mm
B
A
1
2
Solution
Considering the arrangement at the inlet A, both static and dynamic pressure enter
the Pitot tube. On the other hand, at inlet B, only static pressure enters the tube. The
pressure in points at the same high (1 and 2) must be the same in the Pitot tube. Thus:
P + ρ Hg gh =
1
2
ρ a v
2
+ P
You finally get that:
v =
2gh
ρ Hg
ρ a
=
2 · 9.81
m
s 2 0.01 m
13,600
1.2
= 47.2
m
s
1.3.6 Conservation of Linear Momentum
To immobilize a tennis ball travelling at 200 km h
−1 in a time t, is much simpler than
to stop, at that same time, a vehicle that moves at the same speed. To evaluate the
effects of motion, it is, therefore, necessary to know not only the velocity of bodies
but also their mass. This can be done using the kinetic energy equation; however,
what cannot be considered is the time it takes for the body to stop. It is, therefore,
necessary to have an expression to explain this phenomenon.
Thus, for two interacting bodies, if F 1 is the force exerted by m A in m B and F 2 by
m B in m A we have by Newton’s second law that:
F 1 = m B
dv B
t
= m B
v
B − v B
t
13
Exercise 1.4 The diagram represents a Pitot tube inserted into a duct through which
air circulates. The difference in height of mercury measured on the Pitot was 10 mm.
You are asked to determine the speed at which air moves inside the duct. Data: Air
density: 1.2 kg m
−3 , mercury density: 13.60 g cm
−3 .
10 mm
B
A
1
2
Solution
Considering the arrangement at the inlet A, both static and dynamic pressure enter
the Pitot tube. On the other hand, at inlet B, only static pressure enters the tube. The
pressure in points at the same high (1 and 2) must be the same in the Pitot tube. Thus:
P + ρ Hg gh =
1
2
ρ a v
2
+ P
You finally get that:
v =
2gh
ρ Hg
ρ a
=
2 · 9.81
m
s 2 0.01 m
13,600
1.2
= 47.2
m
s
1.3.6 Conservation of Linear Momentum
To immobilize a tennis ball travelling at 200 km h
−1 in a time t, is much simpler than
to stop, at that same time, a vehicle that moves at the same speed. To evaluate the
effects of motion, it is, therefore, necessary to know not only the velocity of bodies
but also their mass. This can be done using the kinetic energy equation; however,
what cannot be considered is the time it takes for the body to stop. It is, therefore,
necessary to have an expression to explain this phenomenon.
Thus, for two interacting bodies, if F 1 is the force exerted by m A in m B and F 2 by
m B in m A we have by Newton’s second law that:
F 1 = m B
dv B
t
= m B
v
B − v B
t