Chapter 2, Problem 2.34P

(a)

To determine

The formula for the pressure difference $p_a-p_b$ when the oil-water interface on the right rises $\Delta h<h$, for $d\ll D$.

Explanation of Solution

The following figure shows the two containers with water and oil.

Figure-(1)

The height of oil in container is $H$, the height of water in left container is $L$, the height of oil in limb is $h$ and the $\text{X}-\text{X}$ is the datum for analysis.

Write the hydrostatic equation for the left limb.

$p_l=p_a+{\gamma }_w\left(L+h\right)$

Here, the pressure at $a$ is $p_a$, and the specific weight of water is ${\gamma }_w$.

Write the hydrostatic equation for the right limb.

$p_r=p_b+{\gamma }_{oil}\left(H+h\right)$

Here, the pressure at $b$ is $p_b$ and the specific weight of oil is ${\gamma }_{oil}$.

The pressure of the right limb and the left limb is same at the datum $\text{X}-\text{X}$.

$p_l=p_r$ …… (I)

Substitute $p_b+{\gamma }_{oil}\left(H+h\right)$ for $p_r$ and $p_a+{\gamma }_w\left(L+h\right)$ for $p_l$ in Equation (I).

$\begin{array}{l}{p}_b+{\gamma }_{oil}\left(H+h\right)=p_a+{\gamma }_w\left(L+h\right)\\ p_a-p_b={\gamma }_{oil}\left(H+h\right)-{\gamma }_w\left(L+h\right)\end{array}$ …… (II)

The figure below shows the rise in limb $\Delta H$.

Figure-(2)

Both the containers are of equal diameter hence, the change in height of limb is $\Delta h$, the depression in container $a$ is equal to increase in other container $b$.

The volume rise and fall in both the containers is same.

Write the expression for balance the volume in both containers.

$\begin{array}{l}\frac{\pi }{4}{D}^2\Delta H=\frac{\pi }{4}{d}^2\times \Delta h\\ \Delta H={\left(\frac{d}{D}\right)}^2\Delta h\end{array}$

Here, the diameter of the limb is $d$.

Write the hydrostatic equation for the left limb at the datum $\text{X}-\text{X}$.

$p_l=p_a+{\gamma }_w\left(L-\Delta H+h-\Delta h\right)$

Write the hydrostatic equation for the right limb at the datum $\text{X}-\text{X}$.

$p_r=p_b+{\gamma }_{oil}\left(H+\Delta H+h-\Delta h\right)$

The pressure at both the limbs is same at the datum.

$p_l=p_r$ …… (III)

Substitute $\left(p_a+{\gamma }_w\left(L-\Delta H+h-\Delta h\right)\right)$ for $p_l$ and $\left(p_b+{\gamma }_{oil}\left(H+\Delta H+h-\Delta h\right)\right)$ for $p_r$ in Equation (III).

$\begin{array}{l}{p}_a+{\gamma }_w\left(L-\Delta H+h-\Delta h\right)=p_b+{\gamma }_{oil}\left(H+\Delta H+h-\Delta h\right)\\ p_a-p_b={\gamma }_{oil}\left(H+\Delta H+h-\Delta h\right)-{\gamma }_w\left(L-\Delta H+h-\Delta h\right)\\ p_a-p_b={\gamma }_{oil}\left(H+h\right)+{\gamma }_{oil}\left(\Delta H-\Delta h\right)-{\gamma }_w\left(L+h\right)+{\gamma }_w\left(\Delta H+\Delta h\right)\end{array}$ …. (IV)

From Equation (IV) and Equation (II).

$\begin{array}{c}{p}_a-p_b={\gamma }_{oil}\left(H+h\right)+{\gamma }_{oil}\left(H-\Delta h\right)-{\gamma }_{oil}\left(H+h\right)+{\gamma }_w\left(H+\Delta h\right)\\ ={\gamma }_{oil}\left(\Delta H-\Delta h\right)+{\gamma }_w\left(\Delta H+\Delta h\right)\end{array}$ …… (V)

Substitute ${\left(\frac{d}{D}\right)}^2\Delta h$ for $\Delta H$ in Equation (V).

$\begin{array}{c}{p}_a-p_b={\gamma }_{oil}\left[{\left(\frac{d}{D}\right)}^2\Delta h-\Delta h\right]+{\gamma }_w\left[{\left(\frac{d}{D}\right)}^2\Delta h+\Delta h\right]\\ =\Delta h\left\{{\gamma }_{oil}\left[{\left(\frac{d}{D}\right)}^2-1\right]+{\gamma }_w\left[{\left(\frac{d}{D}\right)}^2+1\right]\right\}\end{array}$ …… (VI)

When $d\ll D$ then $\frac{d}{D}\approx 0$.

Substitute $0$ for $\frac{d}{D}$ in Equation (VI).

$\begin{array}{c}{p}_a-p_b=\Delta h\left\{{\gamma }_{oil}\left[{\left(0\right)}^2-1\right]+{\gamma }_w\left[{\left(0\right)}^2+1\right]\right\}\\ =\Delta h\left[{\gamma }_w-{\gamma }_{oil}\right]\end{array}$ …… (VII)

Substitute $9790\text{N}/{\text{m}}^3$ for ${\gamma }_w$ and $8740.71\text{N}/{\text{m}}^3$ in Equation (VII).

$\begin{array}{c}{p}_a-p_b=\Delta h\left[9790\text{N}/{\text{m}}^3-8740.71\text{N}/{\text{m}}^3\right]\\ =1049.2\Delta h\text{N}/{\text{m}}^2\end{array}$

Thus, the formula for the pressure difference $p_a-p_b$ when the oil-water interface on the right rises $\Delta h<h$, for $d\ll D$ is $1049.2\Delta h\text{N}/{\text{m}}^2$.

(b)

To determine

The formula for the pressure difference $p_a-p_b$ when the oil-water interface on the right rises $\Delta h<h$, for $d=0.15D$.

The percentage change in pressure difference.

Answer to Problem 2.34P

The formula for the pressure difference $p_a-p_b$ when the oil-water interface on the right rises $\Delta h<h$, for $d=0.15D$ is $1466.231\Delta h\text{N}/{\text{m}}^2$.

The percentage change in pressure difference is $40\%$.

Explanation of Solution

Write the expression for diameter of limb.

$\begin{array}{l}d=0.15D\\ \frac{d}{D}=0.15\end{array}$

Substitute $0.15$ for $\frac{d}{D}$ in Equation (VI).

$\begin{array}{c}\Delta p=\Delta h\left\{{\gamma }_{oil}\left[{\left(0.15\right)}^2-1\right]+{\gamma }_w\left[{\left(0.15\right)}^2+1\right]\right\}\\ =\Delta h\left\{{\gamma }_{oil}\left[0.0225-1\right]+{\gamma }_w\left[0.0225+1\right]\right\}\\ =\Delta h\left\{{\gamma }_w\left[1.0225\right]-{\gamma }_{oil}\left[0.9775\right]\right\}\end{array}$ …… (VIII)

Write the percentage change in pressure.

$\Delta p\%=\frac{\Delta p_2-\Delta p_1}{\Delta p_1}\times 100$ …… (IX)

Here, the change in pressure in case 1 is $\Delta p_1$ and the change in pressure in case 2 is $\Delta p_2$.

Conclusion:

Substitute $9790\text{N}/{\text{m}}^3$ for ${\gamma }_w$ and $8740.71\text{N}/{\text{m}}^3$ in Equation (VIII).

$\begin{array}{c}\Delta p=\Delta h\left\{\left(9790\text{N}/{\text{m}}^3\right)\left[1.0225\right]-\left(8740.71\text{N}/{\text{m}}^3\right)\left[0.9775\right]\right\}\\ =1466.231\Delta h\text{N}/{\text{m}}^2\end{array}$

Thus, the formula for the pressure difference $p_a-p_b$ when the oil-water interface on the right rises $\Delta h<h$, for $d=0.15D$ is $1466.231\Delta h\text{N}/{\text{m}}^2$.

Substitute $1466.231\Delta h\text{N}/{\text{m}}^2$ for $\Delta p_2$ and $1049.2\Delta h\text{N}/{\text{m}}^2$ for $\Delta p_1$ in Equation (IX).

$\begin{array}{c}\Delta p\%=\frac{1466.231\Delta h\text{N}/{\text{m}}^2-1049.2\Delta h\text{N}/{\text{m}}^2}{1049.2\Delta h\text{N}/{\text{m}}^2}\times 100\\ =\frac{417.03}{1049.2}\times 100\\ \approx 40\%\end{array}$

Thus, the percentage change in pressure difference is $40\%$.