Chapter 15, Problem 67P

A U-tube open at both ends is partially filled with water (Fig. P15.67a). Oil having a density 750 kg/m³ is then poured into the right arm and forms a column L = 5.00 cm high (Fig. P15.67b). (a) Determine the difference h in the heights of the two liquid surfaces. (b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Fig. P15.67c). Determine the speed of the air being blown across the left arm. Take the density of air as constant at 1.20 kg/m³.

To determine

The difference $h$ in the heights of the two liquid surfaces.

Answer to Problem 67P

The difference $h$ in the heights of the two liquid surfaces is $1.25\text{ cm}$.

Explanation of Solution

The initial condition of the U-tube is shown in figure 1.

The representation of the tube after the oil is poured is shown in figure 2.

Consider two points A and B in figure 2.

According to the Pascal’s principle, the pressure at points A and B must be equal.

Write the relationship between the pressure at points A and B.

  $P_A=P_B$        (I)

Here, $P_A$ is the pressure at point A and $P_B$ is the pressure at point B.

Write the equation for $P_A$ using the left tube.

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

Here, $P_{\text{atm}}$ is the atmospheric pressure, ${\rho }_w$ is the density of water, $g$ is the acceleration due to gravity and $L$ is the length of the oil column.

Write the equation for $P_B$ using the right tube.

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

Here, ${\rho }_o$ is the density of oil.

Put equations (II) and (III) in equation (I) and rewrite for $h$ .

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

Conclusion:

The density of water is $1000{\text{ kg/m}}^3$ .

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 difference $h$ in the heights of the two liquid surfaces is $1.25\text{ cm}$.

To determine

The speed of the air being blown across the left arm.

Answer to Problem 67P

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

Explanation of Solution

The situation when the air flow over the left tube stabilizes the fluid levels in the two tubes is shown in figure 3.

Write the Bernoulli’s equation for the two points A and B.

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

Here, ${\rho }_a$ is the density of air, $v_A$ is the speed of the air at point A, $y_A$ is the height of the point A above the reference position, $v_B$ is the speed of the liquid at point B and $y_B$ is the height of the point B above the reference position.

The value of $v_B$ is zero and the value of $y_A$ and $y_B$ are the same. Take $v_A$ to be $v$ .

Replace $y_B$ by $y_A$ , $v_A$ by $v$ and substitute $0$ for $v_B$ in equation (V).

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

Consider two points C and D which are at the level of oil-water interface in the right tube.

According to the Pascal’s principle, the pressure at points C and D must be equal.

Write the relationship between the pressure at points C and D.

  $P_C=P_D$        (VII)

Here, $P_C$ is the pressure at point C and $P_D$ is the pressure at point D.

Write the equation for $P_C$ .

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

Here, $H$ is the distance as shown in figure 3.

Write the equation for $P_D$ .

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

Put equations (VIII) and (IX) in equation (VII).

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

Put equation (VI) in the above equation.

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

Conclusion:

The value of $g$ is $9.80{\text{ m/s}}^2$.

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

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

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