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PHYSICS
Pressure and flow rate
Part 1
The two fundamental physical quantities of pneumatics are:
pressure
and
flow rate
.
The flow rate represents the volume of fluid that crosses the section of aduct ina certainunit of time. The flow rate
Q
is defined as the ratio between the volume
∆
V
of the fluid that passes through the duct and the time interval
∆
t
employed to pass it through or as the product of flow section
S
and fluid velocity
ν
, if they are known.
Q=
∆
V
Q= S *
ν
∆
t
In the International System it ismeasured in
m
3
/s
.
The flow rate of a valve is influenced by two factors:
• the flow section;
• theweight of the column of liquidwhich acts on the valve (remember that awater column of height 10,33
m
exerts a pressure of 1
bar
).
Example
:
A tank has an outlet valve
A
and an inlet valve
B
with the same section
(S
A
= S
B
)
both with an adjustable flow.
This can bemapped using a Cartesian axis diagram, inwhich the
x-
axis represents the time
t
necessary to empty
the tank and the
y-
axis thewater level
L
. Observe the following situations.
Figure 15
Pos. 1
: both valves are closed. The diagram has a point at 0,0.
Pos. 2
: drain valve
A
is closed and the inlet valve
B
open. The tank is filled up to the level
L
p
(full tank level).
In the diagram,
t
= 0 and
L
=
L
p
.
Pos. 3
: inlet valve
B
is closed and the exhaust
A
valve open. The water level is lowered until the tank is
completely empty (the emptying time
t
s
depends on
S
A
). In the diagramwe obtain a segment that goes from
L
p
to
t
s
(time required for emptying the tank).
Pos. 4
: both valves are open. There’s an equal amount of water entering and exiting, the water level in the tank
remains constant and the diagram shows a linewith
L
=
L
p
.
Tests have verified that from a holewith a section
S
=1
cm²
(1*10
-4
m
2
) with a pressure
p
=1
bar
, water leaks
at about 84
l/min
= 1,4
l/s
.
In the following exercise calculate the level
L
e
so that the amount of water inside the tank is constant, knowing
the flow rate of the valve
B
has been adjusted to provide 40
l/min
, i.e. 0,66
l/s
.
We proceed by assuming
Q
A
=
Q
B
and therefore
ν
A
=
ν
B
, as the sections of the two valves are identical. Through
Q
B
it is possible to calculate the velocity of thewater passing through valve
B
with the inverse formula.
ν
B
=
Q
B
=
0,66
[l ⁄ s]
=
0,66 * 10
-3
[m
3
⁄ s]
ν
B
=
6,6
m ⁄ s
S
1 * 10
-4
[m
2
]
1 * 10
-4
[m
2
]
To obtain the height of equilibrium
L
e
, we apply Torricelli’s law: the speedwithwhich thewater exits from a hole
of a tank is equal to the speed in a vacuumwhich a stonewould possess if it were dropped from a height equal to
the distance from the surface of thewater to the hole.
2
g
*
L
e
ν
A
=
L
e
=
ν
2
A
6,6
2
[m ⁄ s]
2
L
e
=
2,22
m
2
g
2 * 9,81
[m ⁄ s
2
]
To get an output flow equal to the input flow rate
Q
A
=
Q
B
, we set the speed
ν
A
equal to
ν
B
(
ν
A
=
ν
B
) and using
Torricelli’s lawwe find the height of the liquid required to verify these equalities; in the example just shown we
have:
L
e
=
2,22
m
Q
A
=
Q
B
=
0,66
l/s
1
25
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