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The measurement, using a Venturi meter, of incompressible liquids flowing down a pipe

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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, which fits into a normal pipe-line.

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.

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'\b{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.

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. $\therefore\;\;\;p_1-p_2=wh_2-wh_1+w_u(h_1-h_2)$ $\therefore\;\:\;p_1-p_2=(h_1-h_2)(w_u-w)$

$\therefore\;\;\;\frac{p_1-p_2}{w}=(h_1-h_2)(\frac{w_u-w}{w})$

(1)

$Q= K\sqrt[]{2\,g\,(\frac{w_u-w}{w}})\times \sqrt[]{h_1-h_2}$

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

Example 1 [imperial]

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 $14.7\;lb./in^2$

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:

$\frac{p_1-p_2}{w}=h=\frac{1}{2g}({v_{2}}^{2}-{v_{1}}^{2})=\frac{{v_{1}}^{2}}{2g}\left ( \frac{{v_{2}}^{2}}{{v_{1}}^{2}}-1 \right )$

(2)

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

$Q=v\times a$

$\therefore \;\;\;\;\frac{{v_{2}}^{2}}{{v_{1}}^{2}}=\frac{{a_{1}}^{2}}{{a_{2}}^{2}}=\frac{{d_{1}}^{4}}{{d_{2}}^{2}}=\frac{6^4}{3^4}=16$

(3)

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.

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 (#1) 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 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.

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:

$\frac{p_1-p_2}{w}-Z_2$

(4)

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:

$\frac{p_1-p_2}{w}-Z_2=h_2-h_1+H_m=H_m-H$

(5)

From equations (#1) and (#2)

$H_m-H=\frac{{v_{2}}^{2}-{v_{1}}^{2}}{2g}= C$

(6)

where $C=$ Constant

Now:

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

$= \frac{{v_{1}}^{2}}{2g}\left ( \frac{{d_{1}}^{4}}{{d_{2}}^{4}}-1 \right )$

(7)

From equations (#3) and (#4) 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 $4515$ .

Venturi Meters

Introduction

Measurement of Flow.

Example 1 [imperial]

Example 2 [imperial]

Vertical Venturi Meters.

Example 1