I have forgotten
my Password

Or login with:

  • Facebookhttp://facebook.com/
  • Googlehttps://www.google.com/accounts/o8/id
  • Yahoohttps://me.yahoo.com

Venturi Meters

The measurement of incompressible liquids flowing in a pipe using a Venturi meter.
+ View other versions (5)

Key Facts

Gyroscopic Couple: The rate of change of angular momentum (\tau) = I\omega\Omega (In the limit).
  • I = Moment of Inertia.
  • \omega = Angular velocity
  • \Omega = Angular velocity of precession.

Introduction

Venturi Meter are used to measure the velocity of flow of fluids in a pipe. They consist of a short length of pipe shaped like a vena contracta, or the portion with the least cross-sectional area, which fits into a normal pipe-line. The obstruction caused to the flow of liquid at the throat of the venturi produces a local pressure drop in the region that is proportional to the rate of discharge. This phenomenon, using Bernoullis equation, is used to calculate the rate of flow of the fluid flowing through the pipe.
23287/Venturi-Meters-017.png
+

Venturi Meters have the following characteristics:
  • Theoretically there is no restriction to the flow down the pipe.
  • They can be manufactured to fit any required pipe size.
  • The temperature and pressure within the pipe does not affect the meter or its accuracy.
  • There are no moving parts.
  • Unfortunately the accurate shape required of the inside of the meter makes them relatively expensive to manufacture.

Measurement Of Flow.

23287/venturi-meters-001.png
+

For a meter with the above arrangements of manometers, the quantity flowing is given by:

Q=K\sqrt{H}
a_1\times v_1 = a_2\times v_2

Applying Bernoulli's equation at stations 1 and 2:

\frac{p_1}{w}+\frac{v_1^2}{2g}=\frac{p_2}{w}+\frac{v_2^2}{2g}
\therefore\;\;\;\;\frac{v_2^2-v_1^2}{2g}=\frac{p_1-p_2}{w}
\therefore\;\:\;\frac{v_2^2-(\displaystyle\frac{a_2\,v_2}{a_1})^2}{2g}=\frac{p_1-p_2}{w}
\therefore\;\;\;v_2=\frac{1}{\sqrt[]{1\,-\,(\displaystyle\frac{A_2}{A_1})^2}}x\sqrt[]{2g\:(\frac{P_1\,-\,P_2}{W})}
\therefore\:\:\;v_2= Const\times \sqrt[]{2g\,H}
where
H=\left( \frac{p_1-p_2}{w} \right)
Q=a_2v_2=a_2\timesC\;\sqrt[]{2g\:H}

Which can be written as:
Q =K\sqrt[]{H}

In practice, because of fluid resistance, the actual velocity and consequently actual discharge is LESS than that given by the above equations. A coefficient of discharge is therefore introduced, which usually lies between 0.96 to 0.99.
23287/Venturi-Meters-002.png
+

In an actual meter it is not be practical for the tubes to be taken straight up as shown, since the pressures would require the use of long tubes. A more practical arrangement is to measure the difference in pressure rather than the absolute values. This is achieved as shown in the following diagram.

For the above arrangement the Quantity flowing is given by:
\text{Q}=K\sqrt{h_1-h_2}
where the constant K is specific to a particular meter, and will include an allowance for a coefficient of discharge.

p_1+w\,h_1=p_2+wh_2+w_u(h_1-h_2)
where w_u is the specific weight of the liquid in the U tube.
\therefore\;\;\;p_1-p_2=wh_2-wh_1+w_u(h_1-h_2)
\therefore\;\:\;p_1-p_2=(h_1-h_2)(w_u-w)
Q= K\sqrt[]{2\,g\,(\frac{w_u-w}{w}})\times \sqrt[]{h_1-h_2}

For any given meter this can be written as:
Q=Constant\times\sqrt[]{h_1-h_2}

Example:

[imperial]
Example - Example 1
Problem
A venturi meter with a 3 in. diameter throat is installed in a 6 in. pipe-line. The pressure at the entrance to the meter is 10\;lb./in^2 gauge and it is undesirable that the pressure should at any point, fall below 8\;lb./in^2 absolute.



Assuming that C_d for the meter is 0.96, find the maximum flow for which it may be used. Take the specific weight of the liquid as 60\;lb./ft.^3 and atmospheric pressure 14.7\;lb./in^2

23287/venturi-meters-003.png
+
Workings
Applying Bernoulli to an ideal horizontal venturi eter, i.e. one with no losses :

\frac{p_1}{w}+\frac{{v_{1}}^{2}}{2g}=\frac{p_2}{w}+\frac{{v_{2}}^{2}}{2g}

Re-writing the equation:

The quantity of fluid flowing along the pipe Q is given by:

Q=v\times a

From equations (1) and (2)

h=15\frac{{v_{1}}^{2}}{2g}
And:
{v_{1}}^{2}=\frac{2gh}{15}
\therefore \;\;\;\;\;v_1=2.07h^{\displaystyle\frac{1}{2}}

Thus for an Ideal meter:

Q=\frac{\pi\times \left (  \displaystyle\frac{6}{12}  \right )^2}{4}\times 2.07h^{\displaystyle\frac{1}{2}}=0.407\times h^{\displaystyle\frac{1}{2}}

For the actual meter taking into account the C_d of 0.96:

Q_{actual}=0.96\times 0.407h^{\displaystyle\frac{1}{2}}=0.39\times h^{\displaystyle\frac{1}{2}}

Solution
The maximum flow is 0.39\times h^{\displaystyle\frac{1}{2}}

Vertical Venturi Meters.

All the examples above and the theory have examined horizontal meters.

23287/Venturi-Meters-018.png
+
The following section considers a meter mounted in the vertical. It will be found that the formulae which have already been proved are equally applicable to vertical meters.

Applying Bernoulli and ignoring losses:

\frac{p_1}{w}+\frac{{v_{1}}^{2}}{2g}+Z_1=\frac{p_2}{w}+\frac{{v_{2}}^{2}}{2g}+Z_2

\therefore \;\;\;\;\;\frac{{v_{2}}^{2}-{v_{1}}^{1}}{2g}=\left ( \frac{p_1-p_2}{w} +(Z_1-Z_2)\right )

A note is required too to explain the following equation!

p_1+w(Z_1-Z_2)+wh_1=p_2+wh_2+w_u(h_1-h_2)

\therefore \;\;\;\;\;\;p_1-p_2=wh_2+w_u(h_1-h_2)-w(Z_1-Z_2)
=(h_1-h_2)(w_u-w)-(Z_1-Z_2)

\therefore \;\;\;\;\frac{p_1-p_2}{w}+(Z_1-Z_2)=(h_1-h_2)\left ( \frac{w_u-w}{w} \right )

Which is the same as equation (2) which was proved for a horizontal meter. It is worth noting however that the pressures are different.

The following example is of a non-horizontal meter.

Example:

Example - Example 1
Problem
A venturi meter is connected at the main and throat sections by tubes filled with the fluid being metered by a differential mercury manometer. Prove that for any flow the reading is unaffected by the slope of the meter.

If the mains diameter is 3\displaystyle\frac{1}{2} in. and the throat diameter 1\displaystyle\frac{1}{4} in., calculate the flow of fuel oil in gals./hr. if its relative density relative to water is 0.8 and the difference of level of the mercury columns is 7 in.

Use a direct application of Bernoulli's theorem taking the relative density of mercury to water as 13.6 and the meter coefficient as 0.96.

23287/Venturi-Meters-008.png
+
Workings
Applying Bernoulli:

\frac{p_1}{w}+\frac{{v_{1}}^{2}}{2g}=\frac{p_2}{w}+\frac{{v_{2}}^{2}}{2g}+Z_2

\therefore \;\;\;\;\;\frac{{v_{2}}^{2}-{v_{1}}^{2}}{2g}=\frac{p_1-p_2}{w}-Z_2
But for a given flow v_1 and v_2 are constant:

is constant

Now the pressures at level XX , in the U-tube are equal and if the subscript m refers to mercury, then:

\frac{p_1}{w}+h_1=\frac{p_2}{w}+Z_2+h_2+H_m
or:

From equations (3) and (4) where C= Constant

Now:

\frac{{v_{2}}^{2}-{v_{1}}^{2}}{2g}= \frac{{v_{1}}^{2}}{2g}\left ( \frac{{v_{2}}^{2}}{{v_{1}}^{2}}-1 \right )

From equations (5) and (6) and substituting values:

\frac{{v_{1}}^{1}}{2g}\left \{ \left ( \frac{3.5}{1.25} \right )^4-1 \right \}=\frac{7}{12}\left (  \frac{13.6}{0/8}-1  \right )\;ft.\;of\;oil

From which:
v_1=3.14\;ft./sec.

Q=C_d\times a_1v_1=0.96\times \frac{\pi}{4}\times\left ( \frac{3.5}{12} \right )^2\times 3.14=0.201\;ft.^3/sec.
Thus the flow of oil in gallons/hour
=0.201\times 6.24\times 3600=4515
Solution
The flow of oil in gallons/hour is 4515.