Consider the venturi tube of Figure without the manometer. Let, $A$equal localid="1658313222020" $\mathbf{5}\mathit{a}$. Suppose the pressure localid="1658313216049" ${\mathit{p}}_{\mathbf{1}}$at localid="1658313234980" $\mathit{A}$is localid="1658313228211" $\mathbf{2}\mathbf{.}\mathbf{0}\mathit{a}\mathit{t}\mathit{m}$.

(a) Compute the values of the speed localid="1658313241250" $\mathit{v}$at A.

(b) Compute the values of the speed localid="1658313246652" $\mathit{v}$at a that make the pressure localid="1658313253283" ${\mathit{p}}_{\mathbf{2}}$at an equal to zero.

(c) Compute the corresponding volume flow rate if the diameter at A is localid="1658313260347" $\mathbf{5}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$. The phenomenon that occurs at localid="1658313273085" $\mathit{a}$when localid="1658313267168" ${\mathit{p}}_{\mathbf{2}}$ falls to nearly zero is known as cavitation. The water vaporizes into small bubbles.

$$\begin{gathered} A=5a \\ {P}_1=2.0atm=202650pa \\ {P}_2=0atm=0pa \\ Diameter=5.0cm=0.05m \end{gathered}$$

By using Bernoulli’s equation and the equation of continuity to points$\mathbf{1}$and$\mathbf{2}$in the given figure$\mathbf{14}\mathbf{-}\mathbf{50}\mathbf{,}$find the speed$\mathit{V}$at $\mathit{A}$and the speed$\mathit{v}$at$\mathit{a}$. 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 fluid $\mathit{V}\mathbf{}$at the entrance and exit of the pipe. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

1. Bernoulli’s equation, $pv+\frac{1}{2}\rho g^{2}y+=cons\tan t$

2. Equation of continuity, $av=AV$

3. The speed $V$at $A$, $V=\sqrt{\frac{2a^2∆p}{\rho \left(a^2-A^2\right)}}$

4. Rate of flow of water$Q$, $Q=VA$

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

Here,

$A=5a$

Putting this value in the equation,

$$\begin{gathered} v=\sqrt{\frac{2a^2∆p}{\rho \left(a^2-A^2\right)}} \\ =\sqrt{\frac{2a^2∆p}{\rho \left(a^2-{\left(5a\right)}^2\right)}} \\ =\sqrt{\frac{∆p}{\left(-12\right)\rho }} \\ =\sqrt{\frac{p_1-p_2}{12\rho }} \end{gathered}$$

Substitute the given values.

$v=\sqrt{\frac{\left(202650pa-0\right)}{12\times 1000kg/m^3}}$

$\left\{Densityofwateris\rho =1000\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.\right\}$

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{v}=4.1094\mathrm{m}/\mathrm{s} \\ \approx 4.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed $v$at $A$ is $4.1m/s$.

Now,

$A=5a$

Putting in the equation of continuity,

$av=AV$

Substitute the value ofA

$$\begin{gathered} av=5aV \\ v=5V \\ =5\times 4.1094 \\ =20.5472m/s \\ \approx 21m/s \end{gathered}$$

Hence, the speed $v$at a that makes pressure $p_2$at $a$equal to zero is $21m/s$.

Rate of flow of water$Q$,

Here,

$Q=VA$$$\begin{gathered} A=\mathrm{\pi }{\left(\frac{\mathrm{Diameter}}{2}\right)}^2 \\ =\mathrm{\pi }{\left(\frac{0.05\mathrm{m}}{2}\right)}^2 \\ =0.0019635{\mathrm{m}}^2 \end{gathered}$$

Putting in the above equation of rate of flow of water (Q),

$$\begin{gathered} Q=4.1094\times 0.0019635 \\ =8.0688\times {10}^{-3}{m}^3/s \\ \approx 8.1\times {10}^{-3}{m}^3/s \end{gathered}$$

Hence, the corresponding volume flow rate if the diameter at $A$is $5.0cm$is$8.1\times {10}^{-3}{m}^3/s$.