Overflow from a Water Tank During Acceleration
An 80-cm-high fish tank of cross section 2 m × 0.6 m that is partially filled with water is to be transported on the back of a truck (Fig. 3–62). The truck accelerates from 0 to 90 km/h in 10 s. If it is desired that no water spills during acceleration, determine the allowable initial water height in the tank. Would you recommend the tank to be aligned with the long or short side parallel to the direction of motion?
A fish tank is to be transported on a truck. The allowable water height to avoid spill of water during acceleration and the proper orientation are to be determined.
Assumptions 1 The road is horizontal during acceleration so that acceleration has no vertical component \left(a_{z}=0\right). 2 Effects of splashing, braking, shifting gears, driving over bumps, climbing hills, etc., are assumed to be secondary and are not considered. 3 The acceleration remains constant.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction, and the origin to be the lower left corner of the tank. Noting that the truck goes from 0 to 90 km/h in 10 s, the acceleration of the truck is
a_{x}=\frac{\Delta V}{\Delta t}=\frac{(90-0) km / h }{10 s }\left(\frac{1 m / s }{3.6 km / h }\right)=2.5 m / s ^{2}
The tangent of the angle the free surface makes with the horizontal is
\tan \theta=\frac{a_{x}}{g+a_{z}}=\frac{2.5}{9.81+0}=0.255 \quad \text { (and thus } \theta=14.3^{\circ} \text { ) }
The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midplane experiences no rise or drop during acceleration since it is a plane of symmetry. Then the vertical rise at the back of the tank relative to the midplane for the two possible orientations becomes
Case 1: The long side is parallel to the direction of motion:
\Delta z_{s 1}=\left(b_{1} / 2\right) \tan \theta=[(2 m ) / 2] \times 0.255=0.255 m =25.5 cm
Case 2: The short side is parallel to the direction of motion:
\Delta z_{s 2}=\left(b_{2} / 2\right) \tan \theta=[(0.6 m ) / 2] \times 0.255=0.076 m =7.6 cm
Therefore, assuming tipping is not a problem, the tank should definitely be oriented such that its short side is parallel to the direction of motion. Emptying the tank such that its free surface level drops just 7.6 cm in this case will be adequate to avoid spilling during acceleration.
Discussion Note that the orientation of the tank is important in controlling the vertical rise. Also, the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution.