In: Civil Engineering
A 4 m-wide tank with height of 1.8 m filled with water is pulled with a cable inclined at 30o to the horizontal. The constant acceleration value in the cable direction is 4 m/s². The water depth is measured as 1.5 m before the motion starts. Determine the angle between the water surface and the horizontal. Compute the maximum and minimum pressure values on the bottom of the tank. How many volume of water are spilled?
Taking components of acceleration in horizontal and vertical directions- when taking an inertial frame of reference, accelerations will impose a load in the opposite direction. Thus in the vertical direction, effective gravitational acceleration will increase
geff = g+2 = 9.81+2 = 11.81 m/s²
And horizontal acceleration ax= 2√3 = 3.464 m/s²
(A) For constant linear acceleration case water forms linear surface inclined to horizontal such that its slope is in the ratio of Acceleration in horizontal and vertical direction i.e. if θ is the angle of the surface with horizontal, tanθ= ax/geff = 3.464/11.81
Thus θ = 16.347°
(B) There are two cases possible in the final arrangement of the tank-
Case 1: bottom of tank visible
Case 2: bottom of the tank not visible
At the farther end of the tank, final depth will be 1.8 m. With the known slope of water, we can check whether the corresponding horizontal width is less or greater than the available width(4m).
tanθ= 1.8/required width
Thus required width= 1.8/tan(16.347°) = 6.13m > 4m so it is case 2 i.e. bottom of tank not visible. So the arrangement of the tank is as follows
Writing slope equation for inclined height of water, tanθ = y/4
Thus y= 4*tan(16.347°) = 1.173 m
So depth of water at nearer end say A is 1.8-y=0.627 m. Since depth is minimum at A, pressure at bottom will also be minimum.
Thus min pressure= ρ*geff*h(A) = 1000*11.81*0.627 = 7.405 kPa
Also pressure at farther end say B will be max.
So max pressure = ρ*geff*h(B)= 1000*11.81*1.8= 21.258 kPa
(C) Considering unit length of tank, volume spilled= initial volume of water - final volume of water
Initial volume = 1.5*4*1 =6 m³
Final volume = volume of trapezoid= ½*4*(1.8+0.627) = 4.854 m³
Thus volume of water spilled= 1.146 m³