Question 1.5.3: A tank shown in Fig. E1.5.3 contains oil of specific gravity......
A tank shown in Fig. E1.5.3 contains oil of specific gravity 0.8. If it is given an acceleration of 5 m/s² along a 30° inclined plane in the upward direction, determine the slope of the free surface and pressure at b.
Components of the acceleration along the horizontal and vertical directions are
\mathrm{a_y = 5 \cos 30 = 4.33 and a_z = 5 \sin 30 = 2.5 m/s^2.}
Slope of the water surface is then \mathrm{\tan θ = a_y /(a_z + g) = 0.3519}
→ θ = 19.38°.
Depth at b = 2 – (3/2) tan θ = 1.472 m.
The pressure at b now can be determined using the equation for static fluid. However, there is acceleration in the z-direction too that would add to the pressure. We may use Eq. 1.5.2 without considering the first term in y as effect of [latex]\mathrm{a_y} has already been accounted while using the depth calculated above. Moreover, while going down, z is negative. With these comments, we get
\mathrm{p=-\rho [\alpha _y y+(\alpha _z+g)z]+p_0,} (1.5.2)
\mathrm{pb/w = (1 + a_z/g) × 1.472 = 1.847 m.}
or,
\mathrm{pb = 1.847 × 9787 × 0.8 = 14.463 kN/m^2} (gauge).