A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Figure); the cross-sectional area $\mathit{A}$of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed $\mathit{V}$and then through a narrow "throat" of cross-sectional area a with speed $\mathit{v}$. A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change $\mathit{\Delta }\mathit{p}$in the fluid's pressure, which causes a height difference $\mathit{h}$of the liquid in the two arms of the manometer. (Here $\mathit{\Delta }\mathit{p}$means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Figure, show that $\mathit{V}\mathbf{=}\sqrt{\frac{\mathbf{2}{\mathbf{a}}^{\mathbf{2}}\mathbf{\wedge }\mathbf{p}}{\mathbf{\rho }\left(a^2-A^2\right)}}$, where $\mathit{\rho }$is the density of the fluid.

(b) Suppose that the fluid is fresh water, that the cross-sectional areas are $\mathbf{64}{\mathbf{}\mathbf{c}\mathbf{m}}^{\mathbf{2}}$in the pipe and $\mathbf{32}{\mathbf{}\mathbf{c}\mathbf{m}}^{\mathbf{2}}$in the throat, and that the pressure is $\mathbf{55}\mathit{k}\mathit{P}\mathit{a}$in the pipe and $\mathbf{41}\mathit{k}\mathit{P}\mathit{a}$in the throat. What is the rate of water flow in cubic meters per second?

1) $\rho$is the density of fluid, and fluid is fresh water.

2) The cross-section area at the entrance,$A=64{\mathrm{cm}}^2=0.0064{\mathrm{m}}^2$

3) The cross-section area at the throat,$a=32{\mathrm{cm}}^2=0.0032{\mathrm{m}}^2$

4) Pressure in the pipe,$P_1=55\mathrm{kPa}=55000\mathrm{Pa}$

5) Pressure in the throat,$P_2=41\mathrm{kPa}=41000\mathrm{Pa}$

By using Bernoulli's equation and the equation of continuity to points 1 and 2 in given figure 14-50, prove the equation for the speed of the fluid $\mathit{V}$at the entrance and exit of the pipe. For the rate of flow of water $\mathit{Q}$, use the relation between the rate of flow of water $\mathit{Q}$and the speed of the fluid $\mathit{V}$at the entrance and exit of the pipe. According to Bernoulli's equation, the speed of a moving fluid increases, and the pressure within the fluid decreases.

Equations are as follows:

i) Bernoulli's equation

$pV+\frac{1}{2}\rho g^{2}y+-$constant

ii) Equation of continuity

$av=AV=$constant

iii) Rate of flow of water $Q$

$Q=VA$

Where, $p$is pressure, $v,V$are velocities, $y$is distance, $g$is an acceleration due to gravity, $h$is height, $A$is the area,$Q$is the rate of flow, and $\rho$is density.

Applying Bernoulli's equations, the total energy flow for the flow of fluid at point 1 of the pipe is,

$\rho V+\frac{1}{2}\rho g^{2}y+{}_1-$constant

The total energy flow for the flow of fluid at point 2 of the pipe is,

$\rho v+\rho g^{2}y+{}_2=$constant

The energy flow rate of the fluid in the pipe is,

localid="1657734863004" $$\begin{gathered} \rho V+\rho g^{2}y+\rho {}_1=\rho v+\rho g^{2}y+{}_2 \\ =cons\tan t \end{gathered}$$

But, the flow of fluid is on the same level as the ground. Thus,

$y_1=y_2$

It gives,

$pv+\frac{1}{2}{p}^2=\rho V+\frac{1}{2}={}^2$

localid="1657736361589" $\rho -vP=v\frac{1}{2}\left({}^2-^2\right)$

Now, applying the equation of continuity,

$VA=va=cons\tan t$

$\begin{array}{r}V=\frac{VA}{a}\\ V^2=\frac{V^2{A}^2}{a^2}\end{array}$

Also,

$p-P=\Delta P$

Putting these values in the above equation,

$$\begin{gathered} ∆p=\frac{1}{2}\rho \left(V^2-\frac{V^2{A}^2}{a^2}\right) \\ =\frac{1}{2}\rho \left(\frac{a^2{V}^2-V^2{A}^2}{a^2}\right) \\ =\frac{1}{2}\rho V\left(\frac{a^2-A^2}{a^2}\right) \\ {V}^2=\frac{2∆P}{\rho }\left(\frac{a^2}{a^2-A^2}\right) \\ V=\sqrt{\frac{2a^2∆p}{\rho \left(a^2-A^2\right)}} \end{gathered}$$

Hence, it proved.

Using the speed of the fluid$V$at the entrance and exit of the pipe,

$V=\sqrt{\frac{2a^2\Delta p}{\rho \left(a^2-A^2\right)}}$

Where the fluid is fresh water.

$\rho =1000\mathrm{kg}/{\mathrm{m}}^3$

Putting these values in the above equation,

$V=\sqrt{\frac{2(0.0032\mathrm{m}{)}^2(41000\mathrm{Pa}-55000\mathrm{Pa})}{1000\mathrm{kg}/{\mathrm{m}}^3\left({\left(0.0032{\mathrm{m}}^2\right)}^2-{\left(0.0064{\mathrm{m}}^2\right)}^2\right)}}$

$=\sqrt{\frac{2\times 0.00001024\times \left(-14000\right)}{1000\left(0.00001024-0.0004096\right)}}$

$=\sqrt{\frac{\left(-0.28672\right)}{\left(-0.03072\right)}}$

$=3.06m/s$

Rate of flow of water(Q),

$$\begin{gathered} Q=VA \\ =3.06m/s\times 0.0064m^2 \\ =0.01955m^3/s \\ =2.0\times {10}^{-2}{m}^3/s \end{gathered}$$

Hence, the rate of water flow $Q$in cubic meters per second is $2.0\times {10}^2{\mathrm{m}}^3/\mathrm{s}$