Relative equilibrium means the liquid has no motion relative to its container, but the whole liquid mass may be accelerating. Replace gravity with an effective acceleration field.
The free surface is perpendicular to the resultant effective gravity.
When the tank accelerates vertically, use an effective unit weight. Upward acceleration increases pressure; downward acceleration reduces it.
An open tank 2 m long accelerates horizontally at $2.5\text{ m/s}^2$. Find the difference in liquid level between the ends.
Answer: The liquid level differs by about 0.51 m, rising opposite the direction of acceleration.
Sketch the pseudo-force direction before writing equations. For open tanks, conserve volume when finding spill or final depth; for closed tanks, include trapped air pressure when given.
An open rectangular tank 3.0 m long (in direction of motion) and 1.5 m wide is mounted on a truck. The tank is filled with water to a depth of 0.90 m. The truck accelerates forward at 3.0 m/s². Determine: (a) the slope of the water surface, (b) the rise of water at the rear wall and the drop at the front wall, and (c) whether any water spills.
Tank depth is not given, but if the tank walls are at least 1.36 m tall, no spill occurs. If tank wall height is 1.20 m (common), the rear overflows.
Answer: The water surface tilts at $\tan\theta = 0.306$. Water rises 0.459 m at the rear and drops 0.459 m at the front. The rear level reaches 1.359 m — water would spill if the tank walls are less than 1.36 m tall.
A closed tank 1.5 m tall is completely filled with water and is accelerating upward at 4.0 m/s². Determine the pressure in kPa at the bottom of the tank if the pressure at the top is zero gage. Compare this to the static case.
For static case (a = 0):
Answer: Bottom pressure under upward acceleration is 20.72 kPa vs. 14.72 kPa static — an increase of 6.0 kPa. Downward acceleration at 4 m/s² would give only 8.72 kPa, confirming that upward acceleration amplifies hydrostatic pressure.
A rectangular open tank is 4.0 m long, 2.0 m wide, and has walls 1.20 m tall. It is initially filled to a depth of 1.00 m and accelerates horizontally at 5.0 m/s². Determine whether water spills and, if so, compute the volume of water lost.
Since $h_{rear} = 2.02 \text{ m} > h_{wall} = 1.20 \text{ m}$, spill occurs at the rear.
After spilling, the rear water surface is at the wall top (1.20 m). Let $x$ = depth at front wall:
Since $x < 0$, the front is dry — the water surface intersects the bottom. The water surface is a triangle in the rear portion. Let $L_w$ = length from rear where water contacts floor:
Answer: Water spills. The volume lost is approximately 5.18 m³ — the remaining water occupies only a triangular wedge against the rear wall during acceleration.
A closed rectangular tank 2.0 m long and 1.0 m tall is completely filled with water and accelerates to the right along an incline at 30° above horizontal. The magnitude of acceleration is 5.0 m/s². Determine: (a) the slope of the pressure isobars (equipressure planes) inside the tank, and (b) the gage pressure at the rear-bottom corner if the front-top corner has gage pressure of 10 kPa.
Resolve the acceleration: $a_x = 5\cos30° = 4.33 \text{ m/s}^2$ (horizontal), $a_z = 5\sin30° = 2.50 \text{ m/s}^2$ (upward).
Pressure gradient: along horizontal (x-direction): $\frac{\partial p}{\partial x} = -\rho a_x = -1000(4.33) = -4330 \text{ Pa/m}$. Along vertical (z-direction): $\frac{\partial p}{\partial z} = -\rho(g + a_z) = -1000(12.31) = -12{,}310 \text{ Pa/m}$.
Answer: Pressure isobars slope at 19.4° from horizontal. The rear-bottom corner has gage pressure 30.97 kPa.
A pump draws water from a tank through a 100 mm pipe. The pump is 3.5 m above the water surface in the tank. The pump-suction assembly accelerates downward (momentarily) at 3.0 m/s² due to vibration. The velocity in the pipe is 2.0 m/s and there are no friction losses during this instant. Determine the absolute pressure at the pump inlet. Does cavitation risk increase? Atmospheric pressure = 101.3 kPa, vapor pressure of water = 2.34 kPa absolute.
Applying Bernoulli with effective gravity $g_{eff} = g - a = 9.81 - 3.0 = 6.81 \text{ m/s}^2$ from the free surface (point 1) to pump inlet (point 2):
With downward acceleration — the effective head available is reduced further. Using effective weight $\gamma_{eff} = \rho(g-a)$, the static head contribution decreases.
Answer: Pump inlet absolute pressure drops to about 64.97 kPa, well above vapor pressure of 2.34 kPa in this case. However, during downward acceleration, effective gravity is reduced, making suction head more critical. In deep-suction pumps, vibration-induced downward acceleration can trigger cavitation.
Additional board-style practice items for this topic.
An open rectangular tank mounted on a moving vehicle 6m long, 2m wide and 3m high is filled with water to a depth of 2.5 m.
What is the maximum horizontal acceleration in m/sec2 that can be imposed on the tank without spilling any water?
Determine the accelerating force on the liquid mass.
If the acceleration is increased to 6m/sec2, how much water is spilled out?
Part 1.
For a horizontally accelerating tank, the free-surface difference is:Part 2.
Volume of water in the tank:Part 3.
At $a=6$ m/s2, the free-surface slope gives a height difference:A bucket with water 0.55 m deep is carried inside an elevator that accelerates at 3 m/s2. Evaluate the pressure in kPa exerted by the water at the base of the bucket if the elevator is travelling downward.
Inside a vertically accelerating container the water behaves with an effective unit weight $\gamma_{eff}=\gamma\left(1\pm\dfrac{a}{g}\right)$ — the sign is minus for downward acceleration. The pressure at the base is
$$p=\gamma_{eff}\,h=\gamma h\left(1-\frac{a}{g}\right).$$An open rectangular tank 1.6 m wide, 6 m long, and 2.2 m high is mounted on a truck. The still-water depth is 1.5 m.
Determine the maximum horizontal acceleration without spillage, in m/s2.
Using that maximum acceleration, determine the force required to accelerate the water mass, in kN.
If the acceleration is increased to 5.1 m/s2, how much water is spilled, in cubic meters?
Part 1 — Maximum acceleration. Under horizontal acceleration the free surface tilts at $\tan\theta=\dfrac{a}{g}$. Impending spillage occurs when the back rises by $H_{tank}-h_0$ over the half-length:
$$H_{tank}-h_0=\frac{L}{2}\cdot\frac{a}{g}\ \Rightarrow\ a=\frac{2g\,(H_{tank}-h_0)}{L}.$$Part 2 — Accelerating force. With water volume $\mathcal V=W L h_0$ and mass $m=\rho\mathcal V$,
$$F=m\,a=\rho\,(WLh_0)\,a.$$Part 3 — Spillage at $a_2$. The surface slope is $a_2/g$. After spilling, the water is a triangular wedge of base $x=\dfrac{H_{tank}}{a_2/g}$, so
$$V_{left}=\tfrac12 H_{tank}\,x\,W,\qquad V_{spill}=WLh_0-V_{left}.$$A container holds water 1.1 m deep in an elevator accelerating upward at 1.5 m/s2. Determine the base pressure in kPa.