Example 11-15 A Venturi Meter
Figure 11-30 shows a simple device called a Venturi meter for measuring fluid velocity in a gas such as air. When gas
passes from left to right through the horizontal pipe, it speeds up as it passes through the constriction at point 2.
­Bernoulli’s principle tells us that the gas pressure must be lower at point 2 than at
point 1, and the pressure difference causes the liquid in the U-tube to drop on the
v2
A1
left-hand side and rise on the right-hand side. Suppose the gas is air ­(density
A2
1.20 kg>m3) that enters the left-hand side of the Venturi meter at 25.0 m>s. The
v1
horizontal tube has cross-­sectional area 2.00 cm2 at point 1 and cross-­sectional
2
1
area 1.00 cm2 at point 2. If the liquid in the U-tube is water, what is the difference
in height between the water columns on the left-hand and right-hand sides of the
tube?
Figure 11-30 ​A Venturi meter A Venturi meter, or flow meter, can be used to measure the flow
speed of a gas.
Set Up
In this problem there are two fluids, the air
that flows through the horizontal pipe and the
water in the U-tube. Hence we’ll use Bernoulli’s
equation twice: once to relate the moving air at
point 1 to the moving air at point 2, and once
to relate the heights of the water on the two
sides of the U-tube (which is what we’re trying
to find). We’ll also use the equation of continuity to relate the speeds of the air at points 1
and 2.
Solve
We need to find the difference in air pressure
at points 1 and 2, since this is what causes the
difference in height of the water on the two
sides of the U-tube. Find this difference using
Bernoulli’s equation and the equation of continuity applied to the air in the horizontal pipe,
keeping in mind that we know the values of v1,
A1, and A2.
Bernoulli’s equation:
1
p + rv 2 + rgy = a constant
2
(11-27)
relate properties of air
here and here
1
2
Equation of continuity for steady
flow of an incompressible fluid:
A1v1 = A2v2
(11-19)
relate properties
of water
here and
here
Bernoulli’s equation for the air at points 1 and 2:
p1 +
1
1
r v 21 + rair gy1 = p2 + rair v 22 + rair gy2
2 air
2
At the center of the horizontal pipe, y1 = y2, so
1
1
rair v 21 = p2 + rair v 22
2
2
1
1
p1 - p2 = rair v 22 - rair v 21
2
2
p1 +
From the equation of continuity,
v2 =
A1
v so
A2 1
p1 - p2 =
=
=
A1 2
1
1
rair 1v 22 - v 21 2 = rair c a b v 21 - v 21 d
2
2
A2
A1 2
1
rair v 21 c a b - 1 d
2
A2
1
2.00 cm2 2
11.20 kg>m3 2 125.0 m>s2 2 c a
b - 1d
2
1.00 cm2
= 1.13 * 103 kg> 1ms 2 2 = 1.13 * 103 N>m2
The pressure difference between points 1
and 2 is also the pressure difference between
the water at two points: the top of the water
column on the left-hand side of the U-tube and
the water column on the right-hand side. The
water is at rest (v = 0) at both points. Use this
technique to find the height difference between
the two water columns.
Bernoulli’s principle for the water at the tops of the two
columns:
1
rwater v 2water,1 + rwater gywater,1
2
1
= p2 + rwater v 2water, 2 + rwater gywater, 2
2
p1 +
Water is at rest: vwater,1 = vwater, 2 = 0
Solve for the height difference of the two water columns:
p1 + rwater gywater, 1 = p2 + rwater gywater, 2
rwater gywater, 2 - rwater gywater, 1 = p1 - p2
1.13 * 103 kg> 1m # s 2 2
p1 - p2
=
rwater g
11.000 * 103 kg>m3 2 19.80 m>s 2 2
= 0.115 m = 11.5 cm
ywater, 2 - ywater, 1 =
Reflect
This height difference is large enough to easily
measure, so the Venturi meter is a practical
device.
We made an approximation in our ­solution
that the pressure difference between points 1
and 2 within the horizontal pipe is the same as
the pressure difference between the two water
columns. This isn’t exactly correct, because
there is a greater weight of air above the column on the left-hand side. However, the resulting additional pressure difference is so small
that we can ignore it.
Additional pressure difference between the tops of the two water
columns due to the extra weight of air on the left-hand side:
rair g1ywater, 2 - ywater, 1 2 = 11.20 kg>m3 2 19.80 m>s 2 2 10.115 m2
= 1.35 kg> 1m # s 2 2 = 1.35 N>m2
This is 0.12% of the pressure difference calculated above
(p1 - p2 = 1.13 * 103 N>m2), so we can neglect it.