Question

A U-tube open at both ends is partially filled with water (Fig. P14.81a). Oil having a densit$750kg/m^3$ is then poured into the right arm and forms a column $L=5.00cm$high (Fig. P14.81b). (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. P14.81c). Determine the speed of the air being blown across the left arm. Take the density of air as constant at localid="1663657678687" $1.20kg/m^3$.

Short answer

(a) The difference $h$ in the heights of the two liquid surfaces is $h=1.25cm$.

(b) The speed of the air being blown across the left arm is $v=14.3m/s$.

Step-by-step solution

Pascal’s law:

Pascal's law states that when pressure is applied to a confined fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container.

The pressure in a fluid at rest varies with depth h in the fluid according to the expression

$P=P_o+\rho gh$

Where, $P_o$is the pressure at h = 0 and is the density of the fluid, assumed uniform.

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

P+role="math" localid="1663653234946" $P+\frac{1}{2}\rho v^2+\rho gy=cons\tan t$

Here, $P$is the pressure, $\rho$ is the density, $g$ is the gravity, and $y$ is the distance.

(a) The difference h in the heights of the two liquid surfaces:

Consider the pressure at points A and B in FIG. (a) P . 14 . 81.

Using the left tube: $P_A=P_{atm}+{\rho }_{w}g(L-h)$

Using the right tube: $P_B=P_{atm}+{\rho }_{o}gL$

Here,

Water density , ${\rho }_w=1000kg/m^3$

Oil density , ${\rho }_o=750kg/m^3$

Height , $L=5.00cm=0.05m$

Acceleration due to gravity , $g=9.8m/s^2$

From Pascal's principle,

$$\begin{gathered} P_{atm}+{\rho }_{w}g(L-h)=P_{atm}+{\rho }_{o}gL \\ {\rho }_{w}h=({\rho }_o-{\rho }_w)L \end{gathered}$$

This giving the height as below.

$$\begin{gathered} h=\frac{({\rho }_w-{\rho }_o)}{{\rho }_w}L \\ =\frac{(1000kg/m^3-750kg/m^3)}{1000kg/m^3}(5.00cm) \end{gathered}$$

Hence , the difference $h$in the heights of the two liquid surfaces is$1.25cm$.

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

Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid levels in the two tubes. First, apply Bernoulli’s equation to points A and B $\left(y_A=y_B,v_A=v,and{v}_B=0\right)$.

This gives,

$P_A+\frac{1}{2}{\rho }_a{v}^2+{\rho }_{a}g{y}_A=P_B+\frac{1}{2}{\rho }_a{0}^2+{\rho }_{a}g{y}_B$

And since $y_A=y_B,$this reduces to

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

….. (1)

Now consider points C and D, both at the level of the oil-water interface in the right tube. Using the variation of pressure with depth in static fluids, we have

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

And

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

But Pascal’s principle says that,

$$\begin{gathered} P_A+{\rho }_{a}gH+{\rho }_{o}gL=P_B+{\rho }_{a}gH+{\rho }_{o}gL \\ {P}_B-P_A=({\rho }_w-{\rho }_o)gL \end{gathered}$$

Substitute equation (1) for $P_B-P_A,$into (2) to obtain

$$\begin{gathered} \frac{1}{2}{\rho }_a{v}^2=\left({\rho }_w-{\rho }_0\right)gL \\ {v}^2=\frac{2\left({\rho }_w-{\rho }_0\right)}{{\rho }_a}gL \\ v=\sqrt{\frac{2gL\left({\rho }_w-{\rho }_0\right)}{{\rho }_a}} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} v=\sqrt{\frac{2(9.8m/s^s)(0.0500m)(10000kg/m^3-750kg/m^3)}{1.20kg/m^3}} \\ =14.3m/s \end{gathered}$$

Hence, the speed of the air being blown across the left arm is $14.3m/s$.