Question 3.SP.7: Water and oil in an open storage tank are in contact with th......

(a)            p_{\text {bott }}=\gamma_{\text {oil }} h_{\text {oil }}+\gamma_{\text {water }} h_{\text {water }}

  \begin{aligned} & =\left(0.8 \times 62.4 \mathrm{lb} / \mathrm{ft}^3\right)(1.5 \mathrm{ft})+\left(62.4 \mathrm{lb} / \mathrm{ft}^3\right)(1.5 \mathrm{ft}) \\ & =137.3 \mathrm{lb} / \mathrm{ft}^3=0.953 \mathrm{psi}  \end{aligned}

(b) The force acting on the end consists of three components, labeled A, B, and D, on the pressure distribution sketch. Note that component B has a uniform pressure distribution, due to the oil (A) above, which acts throughout the liquid below.

As a preliminary, we note for the semicircular end area (r=1 \mathrm{ft}) that

(i) A=\pi r^{2} / 2=\pi 1^{2} / 2=1.571 \mathrm{ft}^{2};

(ii) from Appendix A, Table A.7, the centroid is 4 r / 3 \pi=0.424 \mathrm{ft} from the center of the circle, i.e., below the water top surface.

For component A, i.e., the varying oil pressure distribution on the 1.5-\mathrm{ft} height of the end wall, the centroid C of the area of application is at

h_{c}=y_{c}=0.5(1.5 \mathrm{ft})=0.75 \mathrm{ft} \text { below the free oil surface, }

so, from Eq. (3.16),

For component B, the force F_{B} on the water-wetted area of the end wall due to the uniform pressure produced by the 1.5-ft depth of oil above is

For component D, i.e., the varying pressure distribution due to the water (only) on the water-wetted area of the end wall, the centroid C is at

\begin{gathered}h_{c}=y_{c}=0.424 \mathrm{ft} \text { below the water top surface, } \end{gathered}

so :      F_{D}=\gamma h_{c} A=62.4(0.424) \pi 1^{2} / 2=41.6 \mathrm{lb}

The total force F on the end of the tank is therefore

F=F_{A}+F_{B}+F_{D}=272 \mathrm{lb}

(c) As a preliminary to locating the center of pressure, we note that for the semicircular end area with D=2 \mathrm{ft},

(i) from Table A.7: I about the center of the circle, is I=\pi D^{4} / 128= \pi 2^{2} / 128=0.393 \mathrm{ft}^{4}, and

(ii) by the parallel axis theorem: I_{c} about the centroid, distance 0.424 \mathrm{ft} below the center of the circle, is I_{c}=I+A d^{2}=0.393+\left(\pi 1^{2} / 2\right)(0.424)^{2}=0.1098 \mathrm{ft}^{4}.

The locations of the centers of pressure, below the free oil surface, of the component forces are:

\left(y_{p}\right)_{A}=\frac{2}{3}(1.5 \mathrm{ft})=1.000 \mathrm{ft} for the varying oil pressure on the oil-wetted area, and \left.y_{p}\right)_{B}=y_{c}=1.5+0.424=1.924 \mathrm{ft} to the centroid of the water-wetted semicircular area, for the uniform pressure on this area due to 1.5 \mathrm{ft} of oil above the water; and

Eq. (3.18): \quad\left(y_{p}\right)_{D}=y_{c}+\frac{I_{c}}{y_{c} A}=0.424+\frac{0.1098}{0.424\left(\pi 1^{2} / 2\right)}=0.589 \mathrm{ft}

below the water top surface, for the varying water pressure on the water-wetted semicircular area,

=  1.5+0.589=2.09 ft below the free oil surface.

Finally,  F y_p=F_A\left(y_p\right)_A+F_B\left(y_p\right)_B+F_D\left(y_p\right)_D

\begin{aligned}& y_{p}=1.567 \mathrm{ft} \end{aligned}

F=\gamma h_c A        (3.16)

y_p=y_c+\frac{I_c}{y_c A}             (3.18)