Question

The Venturi tube discussed in Example 14.8 and shown in Figure P14.49 may be used as a fluid flow meter. Suppose the device is used at a service station to measure the flow rate of gasoline $\left(\rho =7.00\times {10}^{2}kg/m^3\right)$through a hose having an outlet radius of 1.20 cm. If the difference in pressure is measured to be${\mathit{P}}_{\mathbf{1}}\mathbf{-}{\mathit{P}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathit{k}\mathit{p}\mathit{a}$and the radius of the inlet tube to the meter is 2.40 cm, find (a) the speed of the gasoline as it leaves the hose and (b) the fluid flow rate in cubic meters per second.

Short answer

(a) The speed of the gasoline as it leaves the hose is$v_2=1.91m/s$

(b) The fluid flow rate in cubic meters per second is$R=8.6\times {10}^{-4}{m}^3/s$.

Step-by-step solution

Step 1:

The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume has the same value at all points along a streamline for an ideal fluid. This result is summarized in Bernoulli’s equation:

$P+\frac{1}{2}\rho v^2+\rho gy=cons\tan t$

Step 2:

Part(b):

Given:

Pressuredifference,$P_1-P_2=1.2kpa$

Liquid density,$\rho =7\times {10}^{2}kg/m^3$

Radius of in let tube$r_1=2.4cm=0.024m$

Radius of outer tube$r_2=1.2cm=0.012m$

Rate of flow is constant the means$Area\times velocity=cons\tan t$

Therefore;

$\begin{array}{rcl}\pi {r_1}^2\times v_1& =& \pi {r_2}^2\times v_2=cons\tan t\\ \frac{v_1}{v_2}& =& \left(\frac{{r_1}^2}{{r_2}^2}\right)={\left(\frac{0.024}{0.012}\right)}^2=4\\ v_2& =& 4v_1\\ & & \end{array}$

Applying Bernoulli’s equation, we get

$P_1+\frac{1}{2}\rho {v_1}^2+\rho g{h}_1=P_2+\frac{1}{2}\rho {v_1}^2+\rho g{h}_2$

$P_1=$Pressure of elevation (1)

$P_2=$Pressure of elevation (2)

$g=$Acceleration due to gravity

$\rho =$Density of water

$h_1=$Height at elevation (1)

$h_2=$Height at elevation (2)

As $h_1=h_2=same$

$\begin{array}{rcl}{P}_1-P_2& =& \frac{1}{2}\rho {v_2}^2-\frac{1}{2}\rho {v_1}^2\\ P_1-P_2& =& \frac{1}{2}\rho \left({v_2}^2-{v_1}^2\right)\\ \frac{2\left(P_1-P_2\right)}{\rho }& =& {\left(4v_1\right)}^2-{v_1}^2\\ & & \end{array}$

$\begin{array}{rcl}{P}_1-P_2& =& \frac{1}{2}\rho {v_2}^2-\frac{1}{2}\rho {v_1}^2\\ P_1-P_2& =& \frac{1}{2}\rho \left({v_2}^2-{v_1}^2\right)\\ \frac{2\left(P_1-P_2\right)}{\rho }& =& 15{v_1}^2\\ {v_1}^2& =& \frac{2\left(P_1-P_2\right)}{15\rho }\\ {v_1}^2& =& \frac{2\left(1.2\times {10}^3\right)}{15\times 7\times {10}^2}\\ v_1& =& 0.228\\ v_2& =& 4v_1\\ v_2& =& 4\times 0.228\\ v_2& =& 1.91m/s\\ & & \end{array}$

Step 3:

Part(b):

Rate of flow $=area\times velocity$

$\begin{array}{rcl}R& =& \pi {\left(r_2\right)}^2\times v_2\\ & =& 3.14\times {\left(0.012\right)}^2\times 1.91\\ & =& 8.6\times {10}^{-4}{m}^3/s\\ & & \end{array}$