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1 Fundamental Concepts of Fluid Mechanics for Mine Ventilation
Wherein the superscript (
) represents the final velocities (v).
Also, by Newton’s third law: F 2 = −F 1 , so:
−F 1 = m A
dv A
t
= m A
v
A − v A
t
Then:
m B
v
B − v B
t
= −m A
v
A − v A
t
And therefore we get Eq. 1.15:
m B v B + m A v A = m B v
B + m A v
A
(1.15)
With what one finally obtains the law of conservation of the linear moment for
the system (Eq. 1.16):
p before = p after
(1.16)
Exercise 1.5 The figure represents a regulator with a 0.75 m diameter orifice, which
is located in a 3 m diameter circular gallery in which the air circulates at 8 m s
−1 .
Determine the force (F) acting on the walls of the regulator if the pressure difference
on both sides is 200 Pa. Assume that there is no vena contracta.
F/2
P1, v1
P2, v2
A
a , v
F/2
1
2
Resolve the conservation of the linear momentum for the indicated control volume.
Assume the air is incompressible and do not consider its friction on the walls of the
system.
Note: The indication of
F
2
is by symmetry in the figure.
Solution
By the continuity equation, the mass flow ( ˙
m) is the same on both sides of the control
volume, so since the cross section (A) is the same, the speed (v i ) also coincides.
Applying then the momentum conservation equation, one obtains:
1 Fundamental Concepts of Fluid Mechanics for Mine Ventilation
Wherein the superscript (
) represents the final velocities (v).
Also, by Newton’s third law: F 2 = −F 1 , so:
−F 1 = m A
dv A
t
= m A
v
A − v A
t
Then:
m B
v
B − v B
t
= −m A
v
A − v A
t
And therefore we get Eq. 1.15:
m B v B + m A v A = m B v
B + m A v
A
(1.15)
With what one finally obtains the law of conservation of the linear moment for
the system (Eq. 1.16):
p before = p after
(1.16)
Exercise 1.5 The figure represents a regulator with a 0.75 m diameter orifice, which
is located in a 3 m diameter circular gallery in which the air circulates at 8 m s
−1 .
Determine the force (F) acting on the walls of the regulator if the pressure difference
on both sides is 200 Pa. Assume that there is no vena contracta.
F/2
P1, v1
P2, v2
A
a , v
F/2
1
2
Resolve the conservation of the linear momentum for the indicated control volume.
Assume the air is incompressible and do not consider its friction on the walls of the
system.
Note: The indication of
F
2
is by symmetry in the figure.
Solution
By the continuity equation, the mass flow ( ˙
m) is the same on both sides of the control
volume, so since the cross section (A) is the same, the speed (v i ) also coincides.
Applying then the momentum conservation equation, one obtains: