Chapter 3, Problem 91P

To determine

The ratio of $\frac{h}{H}$.

Answer to Problem 91P

The ratio of $\frac{h}{H}$ is $0.636$.

Explanation of Solution

Given information:

Specific gravity of the fluid 1 is $0.86$, specific gravity of the fluid 2 is $1.25$ and the ratio of forces applied on it is $1.70$.

The below figure shows the free body diagram of the partially submerged gate.

    

    Figure-(1)

Write the expression for the density of the fluid 1.

    ${\rho }_1={\left(SG\right)}_1\times {\rho }_{water}$   ..... (I)

Here, the specific gravity of the fluid 1 is ${\left(SG\right)}_1$ and the density of water is ${\rho }_{water}$.

Write the expression for the density of the fluid 2.

    ${\rho }_2={\left(SG\right)}_2\times {\rho }_{water}$   ..... (II)

Here, the specific gravity of the fluid 2 is ${\left(SG\right)}_2$.

Write the expression for the force on wall due to fluid 1.

    $F_1=P_1{A}_1$   ..... (III)

Here, the pressure exerted by fluid 1 is $p_1$ and the area of the gate resisting to the pressure to fluid 1 is $A_1$.

Write the expression for the pressure acting on the wall due to fluid 1.

    $p_1=p_0+{\rho }_{1}g\left(\frac{b_1}{2}\right)$   ..... (IV)

Here, the atmospheric pressure is $p_0$, acceleration due to gravity is $g$ and the slant height of the gate in contact with the fluid 1 is $b_1$.

Write the expression for the slant height of the gate in contact with the fluid 1.

    $\begin{array}{l}\cos \left(90-\alpha \right)=\frac{H}{b_1}\\ b_1=\frac{H}{\sin \alpha }\end{array}$   ..... (V)

Here, the angle made by the gate is $\alpha$.

Write the expression for the area of the gate resisting to the pressure to fluid 1.

    $A_1=a{b}_1$   ..... (VI)

Here, the width of the gate is $a$.

Write the expression for the force on wall due to fluid 2.

    $F_2=P_2{A}_2$   ..... (VII)

Here, the pressure exerted by fluid 2 is $p_2$ and the area of the gate resisting to the pressure to fluid 2 is $A_2$.

Write the expression for the pressure acting on the wall due to fluid 2.

    $p_2=p_0+{\rho }_{2}g\left(\frac{b_2}{2}\right)$   ..... (VIII)

Here, and the slant height of the gate in contact with the fluid 2 is $b_2$.

Write the expression for the slant height of the gate in contact with the fluid 2.

    $\begin{array}{l}\sin \alpha =\frac{h}{b_2}\\ b_2=\frac{h}{\sin \alpha }\end{array}$   ..... (IX)

Write the expression for the area of the gate resisting to the pressure to fluid 2.

    $A_2=a{b}_2$   ..... (X)

Write the expression for the ratio of the forces.

    $\frac{F_1}{F_2}=1.70$   ..... (XI)

Calculation:

Substitute $0.86$ for ${\left(SG\right)}_1$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$ in Equation (I).

    $\begin{array}{c}{\rho }_1=0.86\times 1000\text{kg}/{\text{m}}^{\text{3}}\\ =860\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.25$ for ${\left(SG\right)}_2$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$ in Equation (II).

    $\begin{array}{c}{\rho }_2=1.25\times 1000\text{kg}/{\text{m}}^{\text{3}}\\ =1250\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $p_0+{\rho }_{1}g\left(\frac{b_1}{2}\right)$ for $p_1$ and $a{b}_1$ for $A_1$ in Equation (III).

    $F_1=\left(p_0+{\rho }_{1}g\left(\frac{b_1}{2}\right)\right)a{b}_1$   ..... (XII)

Substitute $0$ for $p_0$, $860\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_1$ and $\frac{H}{\sin \alpha }$ for $b_1$ in Equation (XII).

    $\begin{array}{c}{F}_1=\left(0+\left(860\text{kg}/{\text{m}}^{\text{3}}\right)g\left(\frac{\frac{H}{\sin \alpha }}{2}\right)\right)a\left(\frac{H}{\sin \alpha }\right)\\ =\left(860\text{kg}/{\text{m}}^{\text{3}}\right)\left(\frac{ag{H}^2}{2{\sin }^2\alpha }\right)\\ =\left(\left(860\text{kg}/{\text{m}}^{\text{3}}\right)ag\right)\left(\frac{H^2}{2{\sin }^2\alpha }\right)\end{array}$

Substitute $p_0+{\rho }_{2}g\left(\frac{b_2}{2}\right)$ for $p_2$ and $a{b}_2$ for $A_2$ in Equation (VII).

    $F_2=\left(p_0+{\rho }_{2}g\left(\frac{b_2}{2}\right)\right)a{b}_2$   ..... (XIII)

Substitute $0$ for $p_0$, $1250\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_2$ and $\frac{h}{\sin \alpha }$ for $b_2$ in Equation (XII).

    $\begin{array}{c}{F}_1=\left(0+\left(1250\text{kg}/{\text{m}}^{\text{3}}\right)g\left(\frac{\frac{h}{\sin \alpha }}{2}\right)\right)a\left(\frac{h}{\sin \alpha }\right)\\ =\left(1250\text{kg}/{\text{m}}^{\text{3}}\right)\left(\frac{ag{h}^2}{2{\sin }^2\alpha }\right)\\ =\left(\left(1250\text{kg}/{\text{m}}^{\text{3}}\right)ag\right)\left(\frac{h^2}{2{\sin }^2\alpha }\right)\end{array}$

Substitute $\left(\left(860\text{kg}/{\text{m}}^{\text{3}}\right)ag\right)\left(\frac{H^2}{2{\sin }^2\alpha }\right)$ for $F_1$ and $\left(\left(1250\text{kg}/{\text{m}}^{\text{3}}\right)ag\right)\left(\frac{h^2}{2{\sin }^2\alpha }\right)$ for $F_2$ in Equation (XI).

    $\begin{array}{l}\frac{\left(\left(860\text{kg}/{\text{m}}^{\text{3}}\right)ag\right)\left(\frac{H^2}{2{\sin }^2\alpha }\right)}{\left(\left(1250\text{kg}/{\text{m}}^{\text{3}}\right)ag\right)\left(\frac{h^2}{2{\sin }^2\alpha }\right)}=1.70\\ \frac{h^2}{H^2}=\frac{860\text{kg}/{\text{m}}^{\text{3}}}{\left(1250\text{kg}/{\text{m}}^{\text{3}}\right)1.7}\\ \frac{h}{H}=0.636\end{array}$

Conclusion:

The ratio of $\frac{h}{H}$ is $0.636$.