Chapter 3, Problem 165P

To determine

The density of the fluid in the left arm.

Answer to Problem 165P

The density of the fluid in the left arm is $841.60\text{kg}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Given information:

The fluid is incompressible and stay separate and do not mix with each other, the angular velocity of rotation of the U-tube is $50\text{rpm}$, the distance of the centre of rotation from the left arm is $5\text{cm}$, the distance of the centre of rotation from the left arm is $15\text{cm}$ and the height of both the liquids in the tube is $18\text{cm}$.

The figure shows the different pressure points.

Figure-(1)

Write the expression for converting the angular velocity from $\text{rpm}$ to $\text{rad}/\text{s}$.

$\omega =\frac{2\pi n}{60}$ …… (I)

Here, the angular velocity in $\text{rpm}$ is $n$.

Write the expression for the pressure difference for the fluid at point 2.

$P_2-P_1=\frac{{\rho }_{fluid}{\omega }^2}{2}\left(0-r_1^2\right)-{\rho }_{fluid}g\left(h_2-h_1\right)$ …… (II)

Here, the pressure of the fluid at point 1 is $P_1$, is the pressure of the fluid at point 2 is $P_2$, the density of the fluid is ${\rho }_{fluid}$, the angular velocity is ${\omega }^{}$, the distance of the centre of rotation from the left arm $r_1$, the acceleration due to gravity is $g$, the height of the fluid at point 1 is $h_1$ and the height of the fluid at point 2 is $h_2$.

Write the expression for the pressure difference for the water at point 3.

$P_2-P_3=\frac{{\rho }_{water}{\omega }^2}{2}\left(0-r_2^2\right)-{\rho }_{water}g\left(h_2-h_1\right)$ …… (III)

Here, the pressure of the water at point 2 is $P_2$, the pressure of the water at point 3 is $P_3$, the density of water is ${\rho }_{water}$, the distance of the centre of rotation from the right arm $r_2$, the height of the water at point 1 is $h_1$ and the height of the water at point 2 is $h_2$.

Write the expression for the different points at atmospheric pressure.

$\begin{array}{c}{P}_2-P_1=P_{atm}\\ P_2-P_3=P_{atm}\\ P_2-P_1=P_2-P_3\end{array}$ …… (IV)

Calculation:

Substitute $50\text{rpm}$ for $n$ in the Equation (I).

$\begin{array}{c}\omega =\frac{2\pi \left(50\text{rpm}\right)}{60}\\ =0.1047\left(50\text{rpm}\right)\\ =5.235\text{rad}/\text{s}\end{array}$

Substitute $5.235\text{rad}/\text{s}$ for $\omega$, $5\text{cm}$ for $r_1$, $0$ for $h_2$, $18\text{cm}$ for $h_1$ and $9.81\text{m}/{\text{sec}}^{\text{2}}$ for $g$ in the Equation (II).

$\begin{array}{c}{P}_2-P_1=\frac{{\rho }_{fluid}\left(5.235\right)}{2}\left(0-{\left(5\text{cm}\right)}^2\right)-{\rho }_{fluid}9.81\text{m}/{\text{sec}}^{\text{2}}\left(0-18\text{cm}\right)\sqrt{a^2+b^2}\\ =\frac{{\rho }_{fluid}\left(5.235\right)}{2}\left(0-{\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)-{\rho }_{fluid}9.81\text{m}/{\text{sec}}^{\text{2}}\left(0-18\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\sqrt{a^2+b^2}\\ =\frac{{\rho }_{fluid}\left(5.235\right)}{2}\left(0-{0.05}^2{\text{m}}^2\right)-{\rho }_{fluid}9.81\text{m}/{\text{sec}}^{\text{2}}\left(0.18\text{m}\right)\\ =-0.034{\rho }_{fluid}+1.7658{\rho }_{fluid}\end{array}$ $P_2-P_1=1.7318{\rho }_{fluid}$

Substitute $5.235\text{rad}/\text{s}$ for $\omega$, $15\text{cm}$ for $r_2$, $0$ for $h_2$, $18\text{cm}$ for $h_1$ and $9.81\text{m}/{\text{sec}}^{\text{2}}$ in the Equation (III).

$\begin{array}{c}{P}_2-P_3=\frac{{\rho }_{water}\left(5.235\right)}{2}\left(0-{\left(15\text{cm}\right)}^2\right)-{\rho }_{water}9.81\text{m}/{\text{sec}}^{\text{2}}\left(0-18\text{cm}\right)\\ =\left[\begin{array}{c}\frac{{\rho }_{water}\left(5.235\right)}{2}\left(0-{\left(15\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)-{\rho }_{water}9.81\text{m}/{\text{sec}}^{\text{2}}\\ \left(0-18\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\end{array}\right]\\ =\left[\begin{array}{c}\frac{1000\text{kg}/{\text{m}}^{\text{3}}\left(5.235\right)}{2}\left(0-{0.15}^2{\text{m}}^2\right)-1000\text{kg}/{\text{m}}^{\text{3}}9.81\text{m}/{\text{sec}}^{\text{2}}\\ \left(0.18\text{m}\right)\end{array}\right]\\ =-308.30+1765.8\end{array}$

$$

   $P_2-P_3=1457.5\text{kg}/{\text{m}}^{\text{2}}$

Substitute $1.7318{\rho }_{fluid}$ for $P_2-P_1$ and $1457.5$ for $P_2-P_3$ in the Equation (V).

$\begin{array}{c}1.7318{\rho }_{fluid}=1457.5\text{kg}/{\text{m}}^{\text{2}}\\ {\rho }_{fluid}=\frac{1457.5\text{kg}/{\text{m}}^{\text{2}}}{1.7318}\\ =841.60\text{kg}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

The density of the fluid in the left arm is $841.60\text{kg}/{\text{m}}^{\text{3}}$.