Chapter 14, Problem 81AP

(a)

To determine

The height difference between the surface of both water and oil in the U tube.

Answer to Problem 81AP

The height difference between the surface of both water and oil in the U tube is $\underset{\_}{1.25\text{cm}}$.

Explanation of Solution

Three figures showing the three stages of the U tube is given below. The first figure shows water alone in the U tube. Figure 1 shows water in the U tube.

Consider Figure 2. The figure shows the oil and water in the tube. $L$ is the height of the oil column formed in the U tube and $h$ is the height difference between the surface of both liquids.

Consider the left tube of the U tube which contains water.

Write the expression for pressure in the left tube.

    $P_A=P_{\text{atm}}+{\rho }_{w}g\left(L-h\right)$                                                                                               (I)

Here, $P_A$ is the pressure in the left tube, $P_{\text{atm}}$ is the atmospheric pressure, ${\rho }_w$ is the density of water, $g$ is the acceleration due to gravity, $L$ is the height of oil column, and $h$ is the height difference between the surfaces of two liquids.

Consider the right hand tube which contains oil.

    $P_B=P_{\text{atm}}+{\rho }_{o}gL$                                                                                                       (II)

Here, $P_B$ is the pressure in the right tube, and ${\rho }_o$ is the density of oil.

Apply Pascal’s law. The change in pressure in any point of the fluid is equally transmitted to all points of the fluid. Here pressure at left tube equals to pressure at right tube.

    $P_A=P_B$                                                                                                                    (III)

Equate expressions (I) and (II) and solve for $h$.

    $\begin{array}{c}{P}_{\text{atm}}+{\rho }_{w}g\left(L-h\right)=P_{\text{atm}}+{\rho }_{o}gL\\ {\rho }_{w}h=\left({\rho }_w-{\rho }_o\right)L\\ h=\frac{\left({\rho }_w-{\rho }_o\right)}{{\rho }_w}L\end{array}$                                                                             (IV)

Conclusion:

Substitute $1000{\text{kg/m}}^3$ for ${\rho }_w$, $750{\text{kg/m}}^3$ for ${\rho }_o$, and $5.00\text{cm}$ for $L$ in equation (IV) to find $h$.

    $\begin{array}{c}h=\frac{\left(1000{\text{kg/m}}^3-750{\text{kg/m}}^3\right)}{1000{\text{kg/m}}^3}\left(5.00\text{cm}\right)\\ =1.25\text{cm}\end{array}$

Therefore, the height difference between the surface of both water and oil in the U tube is $\underset{\_}{1.25\text{cm}}$.

(b)

To determine

The speed of the air blown across the left arm.

Answer to Problem 81AP

The speed of the air blown across the left arm is $\underset{\_}{14.3\text{m/s}}$.

Explanation of Solution

Right arm of the U tube is shielded by blowing air from the left arm and the levels of liquids in both the arms of tube becomes equal.

Consider Figure 3. The fluid levels in both the arms is same.

Write Bernoulli’s equation for left part and right part of the U tube.

    $P_A+\frac{1}{2}{\rho }_a{v}_A^2+{\rho }_{a}g{y}_A=P_B+\frac{1}{2}{\rho }_a{v}_B^2+{\rho }_{a}g{y}_B$                                                            (V)

Here, ${\rho }_a$ is the atmospheric pressure, $P_A$ is the pressure at left arm, $v_A$ is the velocity of air in left arm, $g$ is the acceleration due to gravity, $y_A$ is the height of water in left arm, $v_B$ is the velocity of air in right arm, and $y_B$ is the height of oil in right arm.

Height of liquids in both right and left arm are equal. That is $y_A=y_B$. Also air is blown through the left arm of the tube. So velocity of air in right arm is zero, $v_B=0$. Let the velocity of air in left arm of tube be $v$.

Therefore rewrite expression (I).

    $P_A+\frac{1}{2}{\rho }_a{v}^2+{\rho }_{a}g{y}_A=P_B+\frac{1}{2}{\rho }_a{\left(0\right)}^2+{\rho }_{a}g{y}_A$                                                      (VI)

Solve expression (II) to find the difference of pressure between right and left arms of the tube.

    $P_B-P_A=\frac{1}{2}{\rho }_a{v}^2$                                                                                                     (VII)

Consider points C and D which are at the same level of the oil-water interface layer of right arm.

Write the expression for the pressure at point C of left arm.

    $P_C=P_A+{\rho }_{a}gH+{\rho }_{w}gL$                                                                                      (VIII)

Here, $P_C$ is the pressure at point C, $H$ is the height of air column, ${\rho }_w$ is the density of water, and $L$ is the height of water column from point C.

Write the expression for the pressure at point D of right arm.

    $P_D=P_B+{\rho }_{a}gH+{\rho }_{o}gL$                                                                                        (IX)

Here, $P_D$ is the pressure at point D, $H$ is the height of air column, ${\rho }_o$ is the density of oil, and $L$ is the height of oil column from point D.

By Pascal’s law, $P_C=P_D$.

    $P_A+{\rho }_{a}gH+{\rho }_{w}gL=P_B+{\rho }_{a}gH+{\rho }_{o}gL$                                                                 (X)

Solve expressions (VI).

    $\begin{array}{c}{P}_A+{\rho }_{a}gH+{\rho }_{w}gL=P_B+{\rho }_{a}gH+{\rho }_{o}gL\\ P_B-P_A=\left({\rho }_w-{\rho }_o\right)gL\end{array}$                                                                (XI)

Use expression (III) in (VII) to find $v$.

    $\begin{array}{c}\frac{1}{2}{\rho }_a{v}^2=\left({\rho }_w-{\rho }_o\right)gL\\ v=\sqrt{\frac{2gL\left({\rho }_w-{\rho }_o\right)}{{\rho }_a}}\end{array}$                                                                                    (VIII)

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $5.00\text{cm}$ for $L$, $1000{\text{kg/m}}^3$ for ${\rho }_w$, $750{\text{kg/m}}^3$ for ${\rho }_o$,and $1.20{\text{kg/m}}^3$ for ${\rho }_a$ in expression (VIII) to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{2\left(9.80{\text{m/s}}^2\right)\left(5.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1000{\text{kg/m}}^3-750{\text{kg/m}}^3\right)}{\left(1.20{\text{kg/m}}^3\right)}}\\ =14.3\text{m/s}\end{array}$

Therefore, the speed of the air blown across the left arm is $\underset{\_}{14.3\text{m/s}}$.