CELE Hydraulics & Fluid Mechanics — Relative Equilibrium of LiquidsDetailed Explanation
A detailed, step-by-step explanation of Relative Equilibrium of Liquids for CELE aspirants. This page goes deeper than the summary and study notes, walking through the reasoning behind each concept so you understand why Professional Regulation Commission (PRC) — Board of Civil Engineering tests it the way it does in the CELE Hydraulics & Fluid Mechanics subtest.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Hydraulics & Fluid Mechanics section sits under a "Core" weighting, and Relative Equilibrium of Liquids is the 4th chapter in the 10-chapter CELE Hydraulics & Fluid Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Hydraulics & Fluid Mechanics.
Relative Equilibrium of Liquids - Detailed Explanation
Relative equilibrium describes the condition of a liquid that moves as a rigid body — no relative motion occurs between adjacent fluid particles, hence no shear stresses develop. The liquid behaves as if it were a solid block. This situation arises in three practical engineering scenarios: (1) a tank accelerating horizontally (e.g., a water truck on EDSA), (2) a tank accelerating vertically (e.g., an elevator tank or a bucket in a hoist), and (3) a vessel rotating about a vertical axis (e.g., a centrifuge or a spinning tank). In all three cases the free surface departs from the horizontal plane it occupies under true static equilibrium, and the pressure distribution shifts accordingly. Mastery of this topic is essential for the PRC Civil Engineer Licensure Examination — it consistently appears in the Hydraulics and Fluid Mechanics portion and tests both conceptual understanding and numerical computation. The governing equations are compact but their correct application — especially sign conventions — separates high scorers from average ones.
Concepts
Fundamental Principle of Relative Equilibrium
When a fluid mass accelerates uniformly (no rotation of the fluid itself relative to the container), Newton's second law applied to a fluid element yields the pressure gradient equations. In the x-direction (horizontal, direction of acceleration a): dp/dx = −ρa. In the z-direction (vertical, positive upward): dp/dz = −ρ(g ± a_z). Because there is no shear, the free surface must always be perpendicular to the resultant body-force vector (the vector sum of gravity and the pseudo-force −a). This is the physical key: the free surface orients itself so that it is always normal to the effective gravity vector. Pressure increases in the direction of effective gravity — that is, perpendicular to the free surface and into the fluid. The hydrostatic formula p = γh still holds, where h is the vertical depth measured below the (now tilted or curved) free surface.
Examples
Even though you cannot see the free surface (closed tank), the pressure distribution shifts. The back wall carries more hydrostatic load; the front wall carries less. Engineers must account for this in tank design — especially for fuel tanks in vehicles or water trucks.
Scenario
A closed rectangular tank is completely filled with water and accelerates horizontally. A technician asks: is the pressure distribution the same as when it was at rest?
Solution
No. When the tank accelerates to the right at a m/s², the pressure at any point depends on its position. At the left wall (back), the pressure is higher; at the right wall (front, in the direction of acceleration), the pressure is lower. The pressure gradient along the horizontal direction is dp/dx = −ρa ≠ 0.
Applications
- Design of water-carrying trucks and tankers — ensuring tanks do not rupture under braking or acceleration loads.
- Analysis of liquid cargo in ships during pitching and rolling motions.
- Design of fuel tanks in aircraft and spacecraft during maneuvers.
- Centrifugal pumps and separators — rotation-induced pressure gradients drive flow.
- Elevator water tanks in high-rise buildings — pressure surges during vertical acceleration.
Misconceptions
- MISCONCEPTION: 'The free surface always stays horizontal.' CORRECTION: It stays horizontal only for vertical acceleration. Horizontal acceleration tilts it; rotation curves it.
- MISCONCEPTION: 'p = γh cannot be used in relative equilibrium.' CORRECTION: It can — but h must be the vertical distance below the new (tilted or curved) free surface.
- MISCONCEPTION: 'Relative equilibrium means the fluid is at rest.' CORRECTION: The fluid moves, but as a rigid body — no internal relative motion between particles.
Related Concepts
- Hydrostatic pressure distribution
- Pascal's Law
- Body forces and surface forces in fluid mechanics
- Newton's second law applied to fluid elements
Common Exam Questions
Example
A tank of water is accelerated horizontally. Describe qualitatively the pressure distribution on the front and rear walls.
Approach
Identify what type of acceleration is present and which formula applies. Draw the free body diagram of the fluid element, resolve the body force, and determine free surface orientation.
Question Type
Conceptual identification
Example
Which of the following correctly expresses the pressure at the bottom of a tank of water depth h accelerating upward at a? Answer: p = γh(1 + a/g).
Approach
Recognize that relative equilibrium problems always start with identifying the direction and magnitude of acceleration, then selecting the correct formula variant.
Question Type
Formula selection
Key Points To Remember
- No shear stress → no relative motion between particles → the fluid moves as a rigid body.
- The free surface is always perpendicular to the resultant of gravity and the applied acceleration.
- p = γh is still valid, but h is the vertical depth below the new free surface.
- Horizontal acceleration tilts the free surface; vertical acceleration keeps it horizontal but changes its effective weight; rotation curves it into a paraboloid.
- The volume of liquid is conserved — the tilted or curved surface cannot add or remove fluid.
Horizontal Acceleration
Consider an open rectangular tank of water accelerating horizontally to the right at acceleration a (m/s²). The effective body force on any fluid particle is the vector sum of gravity g (downward) and the inertial pseudo-force a (backward, opposite to motion). The resultant makes an angle θ with the vertical. The free surface, being perpendicular to this resultant, tilts at angle θ from the horizontal — sloping downward in the direction of acceleration (the front of the tank is lower; the rear is higher). The governing relation is: tan θ = a/g. For pressure at any point: since no shear exists, p = γh where h is the vertical depth below the tilted free surface at that location. For a tank of length L accelerating at a, the difference in water depth between the rear and front is: Δh = L · tan θ = La/g. Conservation of volume means the average depth remains unchanged — the surface pivots about its original centerline.
Examples
Step 1: Compute the surface angle using tan θ = a/g. Step 2: The surface pivots about the midpoint of the original surface, so each end rises or falls by (L/2) tan θ. Step 3: Pressure at any bottom point equals unit weight times the depth of water directly above it. The rear corner has the greatest depth, hence the greatest pressure.
Scenario
BOARD-TYPE PROBLEM: An open rectangular tank is 4.0 m long, 1.5 m wide, and initially contains water to a depth of 1.2 m. The tank accelerates horizontally at 3.0 m/s² in the direction of its length. Find: (a) the angle of the free surface, (b) the depth at the rear and front walls, (c) the pressure at the bottom rear corner.
Solution
(a) tan θ = a/g = 3.0/9.81 = 0.3058 → θ = 17.0° (b) Rise/fall at each end = (L/2) × tan θ = (4.0/2)(0.3058) = 0.6116 m Rear depth = 1.2 + 0.6116 = 1.8116 m ≈ 1.81 m Front depth = 1.2 − 0.6116 = 0.5884 m ≈ 0.59 m (Check: tank height not given as a limit, so no spillage assumed.) (c) Pressure at bottom rear corner = γ × h_rear = 9.81 × 1.8116 = 17.77 kPa
The spillage scenario is a classic board exam extension. Always check whether the computed rear depth exceeds the tank wall height. If it does, set the rear depth = wall height and recompute the wetted length from the volume constraint.
Scenario
SPILLAGE CHECK PROBLEM: The same tank (4.0 m long, 1.2 m deep, tank wall height = 1.5 m) now accelerates at 6.0 m/s². Does water spill?
Solution
tan θ = 6.0/9.81 = 0.6116 Rear rise = (4.0/2)(0.6116) = 1.2232 m → Rear depth would be 1.2 + 1.2232 = 2.4232 m But tank wall height = 1.5 m → 2.4232 m > 1.5 m → SPILL occurs from the rear. After spilling: the surface must pass through the rear top corner (1.5 m deep at x = 0). The surface still has slope tan θ = 0.6116. Front depth = 1.5 − (4.0)(0.6116) = 1.5 − 2.4464 = −0.9464 m → Negative means the front is dry (bottom exposed). Recalculate using the triangular water volume: Triangular prism volume = (1/2)(x)(1.5)(1.5) where x is the wetted length from the rear. x = 1.5/tan θ = 1.5/0.6116 = 2.452 m Volume check: Initial volume = 4.0 × 1.5 × 1.2 = 7.2 m³ (per unit width) Triangular volume retained = (1/2)(2.452)(1.5)(1.5) = 2.754 m² per unit width — not equal. Full solution requires equating the retained triangular volume to the original volume to find the exact geometry, then computing pressures at specific points.
Applications
- Water trucks and tankers braking on EDSA — computing maximum braking deceleration without spillage.
- Pressure surges on the rear bulkhead of a tank truck during sudden braking.
- Analysis of partially filled tanks on cargo ships during pitching.
- Design of sloshing baffles in liquid storage vessels subjected to earthquake (seismic) excitation.
- Fuel tank design for vehicles — ensuring adequate fuel supply to engine under acceleration.
Misconceptions
- MISCONCEPTION: 'The free surface slopes UP in the direction of acceleration.' CORRECTION: It slopes DOWN in the direction of acceleration — the liquid lags behind, piling up at the rear.
- MISCONCEPTION: 'θ is measured from the vertical.' CORRECTION: θ is measured from the horizontal. tan θ = a/g gives the angle from horizontal.
- MISCONCEPTION: 'The pivot point is at the front or rear wall.' CORRECTION: The surface pivots about the midpoint (centroid) of the original free surface area.
- MISCONCEPTION: 'Pressure at the bottom front corner is zero if depth is small.' CORRECTION: Pressure is γ × actual depth at that point — it approaches zero only if the free surface reaches that point.
Related Concepts
- Hydrostatic force on plane surfaces
- Volume conservation
- Seismic sloshing analysis
- Pressure on curved and plane walls
Common Exam Questions
Example
A tank accelerates at 4.905 m/s². Find θ. Answer: tan θ = 4.905/9.81 = 0.5, θ = 26.57°.
Approach
Direct application of tan θ = a/g. Convert a to m/s² if given in other units. The answer is the angle from the horizontal.
Question Type
Find the angle of the free surface
Example
Find the pressure at the midpoint of the bottom of a 6 m long tank, depth 1.5 m, a = 2 m/s². The surface is horizontal at the midpoint (pivot point), so depth there = 1.5 m. p = 9.81 × 1.5 = 14.72 kPa.
Approach
Compute the tilted free surface position at the horizontal location of the point. Then p = γ × (vertical depth below the free surface at that x-location).
Question Type
Find pressure at a specific point
Example
A tank 3 m long, 1 m deep, wall height 1.2 m, a = 5 m/s². tan θ = 5/9.81 = 0.5097. Rear depth = 1 + 1.5(0.5097) = 1.765 m > 1.2 m → spill occurs.
Approach
Check if rear depth exceeds tank height. If yes, set up the geometry using the retained water volume equal to the original volume.
Question Type
Spillage / volume conservation
Key Points To Remember
- tan θ = a/g — memorize this; θ is measured from the horizontal.
- Free surface slopes DOWN in the direction of acceleration.
- The surface pivots about the centroid of the original free surface — average depth is unchanged.
- For a tank of length L: rear depth increase = front depth decrease = (L/2)(a/g).
- If the computed rear depth exceeds the tank height, liquid spills from the back — recalculate using the remaining (spilled) volume.
- Pressure at any point = γ × (vertical depth below the tilted free surface at that horizontal position).
Vertical Acceleration
When a tank accelerates vertically, the free surface remains horizontal (because the acceleration has no horizontal component to tilt it). However, the effective gravity changes. For upward acceleration at a_z = a: effective gravity = g + a (the fluid feels heavier). For downward acceleration at a_z = a: effective gravity = g − a (the fluid feels lighter). The pressure at depth h below the free surface is: p = γh(1 + a/g) [upward acceleration] p = γh(1 − a/g) [downward acceleration] Alternatively: p = ρh(g ± a), where + is upward and − is downward. Special case — FREE FALL (a = g downward): p_gauge = γh(1 − 1) = 0. The gauge pressure throughout is zero — the liquid exerts no net force on the container walls in the vertical direction (astronaut's water bottle in freefall). Absolute pressure remains approximately atmospheric throughout.
Examples
The three answers bracket the static case: upward gives the highest pressure (25.62 kPa > static 19.62 kPa), downward gives lower pressure (13.62 kPa), and free fall gives zero gauge pressure. Note that 'accelerates downward' does NOT mean the elevator moves downward — it means the acceleration vector points downward (it could be slowing down while moving upward).
Scenario
BOARD-TYPE PROBLEM: A rectangular tank 2.0 m deep contains water and is attached to an elevator. Find the pressure at the bottom when the elevator: (a) accelerates upward at 3.0 m/s², (b) accelerates downward at 3.0 m/s², (c) is in free fall.
Solution
Given: h = 2.0 m, γ = 9.81 kN/m³, g = 9.81 m/s² (a) Upward a = 3.0 m/s²: p = γh(1 + a/g) = 9.81(2.0)(1 + 3.0/9.81) = 19.62(1.3058) = 25.62 kPa (b) Downward a = 3.0 m/s²: p = γh(1 − a/g) = 9.81(2.0)(1 − 3.0/9.81) = 19.62(0.6942) = 13.62 kPa (c) Free fall (a = g = 9.81 m/s² downward): p = γh(1 − 9.81/9.81) = 19.62(0) = 0 kPa (gauge)
This is a straightforward application. The bucket is being lifted (upward acceleration), so pressure increases above the hydrostatic value. Static pressure would have been 9.81 × 0.40 = 3.924 kPa. The 25.48% increase corresponds exactly to the ratio a/g.
Scenario
BOARD-TYPE PROBLEM: A bucket of water 0.40 m deep hangs from a crane hook and is lifted upward with an acceleration of 2.5 m/s². Find the pressure at the bottom of the bucket.
Solution
p = γh(1 + a/g) = 9.81(0.40)(1 + 2.5/9.81) = 3.924(1.2548) = 4.924 kPa ≈ 4.92 kPa
Applications
- Elevator water tanks in high-rise buildings — surge analysis during acceleration/deceleration of elevator cars.
- Hydraulic testing of pipes and vessels aboard ships under pitch and heave motions.
- Space launch vehicles — fuel tank pressurization during liftoff (upward acceleration increases effective hydrostatic head).
- Drop-weight impact tests — understanding fluid behavior in free-fall conditions.
- Seismic analysis — vertical ground acceleration component affects fluid pressure in storage tanks.
Misconceptions
- MISCONCEPTION: 'Accelerating downward always means the tank moves downward.' CORRECTION: Acceleration direction and velocity direction are independent. A tank moving up but slowing down has downward acceleration.
- MISCONCEPTION: 'In free fall, pressure equals γh because gravity still acts.' CORRECTION: The inertial pseudo-force exactly cancels gravity in the reference frame of the container, giving zero gauge pressure.
- MISCONCEPTION: 'The free surface tilts during vertical acceleration.' CORRECTION: No — it remains horizontal. Only the pressure magnitude changes.
- MISCONCEPTION: 'The formula with ± applies to the entire pressure, not just the modification.' CORRECTION: The formula p = γh(1 ± a/g) gives the total gauge pressure. The factor (1 ± a/g) multiplies the entire γh term.
Related Concepts
- Hydrostatic pressure
- Effective gravity in non-inertial reference frames
- Cavitation — occurs when absolute pressure drops to vapor pressure
- Seismic vertical acceleration effects on liquid storage tanks
Common Exam Questions
Example
Water tank 1.5 m deep, downward a = 4.9 m/s². p_bottom = 9.81(1.5)(1 − 4.9/9.81) = 14.715(0.5) = 7.36 kPa.
Approach
Identify direction of acceleration (up or down), apply p = γh(1 ± a/g) with correct sign. Compute numerically.
Question Type
Compute pressure for given vertical acceleration
Example
Pressure at bottom of 1.0 m deep tank is 12.26 kPa. Is this upward or downward acceleration and what is a? Static p = 9.81 kPa. Since 12.26 > 9.81 → upward. 12.26 = 9.81(1 + a/9.81) → a = (12.26/9.81 − 1)(9.81) = 2.45 m/s² upward.
Approach
Set up p = γh(1 ± a/g) and solve for a. Be careful with the sign — determine from context whether a is up or down.
Question Type
Find the acceleration given pressure reading
Example
A tank falls freely from a crane. What is the gauge pressure 0.8 m below the free surface? Answer: 0 kPa.
Approach
Recognize free fall means a = g downward. Gauge pressure = 0 throughout regardless of depth.
Question Type
Free fall gauge pressure
Key Points To Remember
- Vertical acceleration keeps the free surface horizontal — only the effective pressure changes.
- Upward acceleration → pressure INCREASES: p = γh(1 + a/g).
- Downward acceleration → pressure DECREASES: p = γh(1 − a/g).
- Free fall (a = g downward) → gauge pressure = 0 everywhere.
- If a > g downward (hypothetically), gauge pressure would be negative — cavitation could occur.
- The sign convention '+' for upward, '−' for downward must be applied consistently.
- Units: a and g must be in the same units (m/s²); γ in kN/m³; h in m → p in kPa.
Rotation of a Liquid Mass (Rotating Vessels)
When an open cylindrical vessel containing liquid rotates about its vertical axis at constant angular velocity ω (rad/s), the liquid eventually reaches rigid-body rotation — every particle rotates at the same ω with no relative motion. The centrifugal acceleration at radius r is ω²r directed outward. The free surface forms a paraboloid of revolution. The equation of the free surface (height z above the vertex, the lowest point of the paraboloid) at radius r is: z = ω²r²/(2g) The total rise from the center (r = 0) to the rim (r = R) is: Δz = ω²R²/(2g) Key geometric property: The volume of the paraboloid equals exactly half the volume of its enclosing cylinder (πR² × Δz)/2. This means: if the vertex of the paraboloid is at the same level as the original free surface, then Δz above original = Δz/2 at the rim (the center drops by Δz/2 and the rim rises by Δz/2 from the original level). More precisely: - Center drops by: Δz/2 = ω²R²/(4g) - Rim rises by: Δz/2 = ω²R²/(4g) (Only valid when the vertex stays above the bottom of the tank.) Pressure at any interior point: p = γ × (vertical depth below the paraboloid surface directly above that point). Alternatively, using the full pressure equation at a point (r, z) inside the liquid: p = p_0 + ρ(ω²r²/2 − gz') where p_0 and z' are reference values. Conversion: ω = 2πN/60 where N is rotation speed in rpm.
Examples
This comprehensive problem illustrates the full analysis: (1) convert rpm to rad/s, (2) compute the paraboloid geometry, (3) check for spillage, (4) check if the vertex hits the bottom, and (5) use volume conservation if either condition is violated. In board exams, Steps 3 and 4 are the most commonly tested traps.
Scenario
BOARD-TYPE PROBLEM: An open cylindrical tank of radius R = 0.5 m and height 2.0 m contains water initially at a depth of 1.2 m. It rotates about its vertical axis at 120 rpm. Find: (a) ω in rad/s, (b) rise from center to rim of the paraboloid, (c) the depth at the rim and at the center after rotation, (d) whether water spills.
Solution
(a) ω = 2π(120)/60 = 4π = 12.566 rad/s (b) Total rise (vertex to rim) = ω²R²/(2g) = (12.566)²(0.5)²/(2×9.81) = (157.91)(0.25)/19.62 = 39.478/19.62 = 2.012 m (c) Center drops by: 2.012/2 = 1.006 m from original Rim rises by: 2.012/2 = 1.006 m from original Depth at center = 1.2 − 1.006 = 0.194 m (vertex above bottom — OK) Depth at rim = 1.2 + 1.006 = 2.206 m (d) Tank height = 2.0 m. Rim depth = 2.206 m > 2.0 m → SPILL occurs. After spilling, recalculate using volume conservation: Set depth at rim = 2.0 m (liquid level at rim = tank top). Let h_c = depth at center. Paraboloid height = 2.0 − h_c. Also, paraboloid height = ω²R²/(2g) = 2.012 m (this is fixed by ω and R). So h_c = 2.0 − 2.012 = −0.012 m → negative → center is DRY (vertex hits bottom). Recalculate with vertex at bottom: Volume of liquid retained = volume of paraboloid = (1/2)πR²(Δz) where Δz is the paraboloid height above the bottom. Original volume = π(0.5)²(1.2) = 0.9425 m³ For vertex at bottom: V_retained = (1/2)π(0.5)²(Δz) Set equal to original volume (if no spill yet) or solve for actual Δz after spill. Since spill occurs: V_retained = (1/2)π(0.5)²(2.0) = (1/2)(0.7854)(2.0) = 0.7854 m³ Volume spilled = 0.9425 − 0.7854 = 0.1571 m³
The key insight: when the vertex just touches the bottom, the drop from the original level at center = h_0 (original depth). By the symmetry property, this drop = Δz/2, so Δz = 2h_0. This is a frequently tested concept in Philippine board exams.
Scenario
BOARD-TYPE PROBLEM: An open cylinder of radius 0.30 m contains water at rest at a depth of 0.80 m. Find the angular velocity ω at which the water just starts to uncover the bottom (vertex just reaches the bottom).
Solution
When the vertex just touches the bottom, the depth at the center = 0, and the original depth = 0.80 m. The paraboloid height from center (bottom) to rim = total height Δz. By volume conservation (paraboloid volume = original volume before any spill): Paraboloid volume = (1/2)πR²(Δz) = πR²(h_0) (1/2)(Δz) = h_0 Δz = 2h_0 = 2(0.80) = 1.60 m But Δz = ω²R²/(2g): 1.60 = ω²(0.30)²/(2×9.81) 1.60 = ω²(0.09)/19.62 ω² = 1.60 × 19.62/0.09 = 31.392/0.09 = 348.8 ω = √348.8 = 18.68 rad/s N = 60ω/(2π) = 60(18.68)/(6.283) = 178.5 rpm
Applications
- Centrifugal pumps and turbines — the rotating liquid creates the pressure gradient that drives flow.
- Centrifuge machines in laboratories and industries — separation of mixtures by density difference.
- Mixing tanks with rotating impellers — understanding the free surface profile during operation.
- Rotating biological contactors in wastewater treatment.
- Design of storage tanks subject to seismic ground rotation.
- Blood centrifuges in medical laboratories — determining rotation speeds for blood component separation.
Misconceptions
- MISCONCEPTION: 'The water level rises everywhere during rotation.' CORRECTION: The water level rises at the rim but drops at the center. The average level remains the same (conservation of volume).
- MISCONCEPTION: 'Δz = ω²R²/(2g) is the rise above the original level at the rim.' CORRECTION: Δz is the total rise from the vertex (center, lowest point) to the rim. The rim rises by only Δz/2 above the original level.
- MISCONCEPTION: 'ω in rpm can be directly substituted into z = ω²r²/(2g).' CORRECTION: ω must be in rad/s. Always convert: ω = 2πN/60.
- MISCONCEPTION: 'The paraboloid volume equals the enclosing cylinder volume.' CORRECTION: The paraboloid volume equals HALF the enclosing cylinder volume — this is critical for volume conservation problems.
- MISCONCEPTION: 'If the vertex hits the bottom, all calculations break down.' CORRECTION: When the vertex hits the bottom, use the annular volume of the remaining liquid (which forms a truncated paraboloid bowl) and apply volume conservation to find the new geometry.
Related Concepts
- Centrifugal force and centripetal acceleration
- Volume of a paraboloid (= 1/2 × base × height)
- Centrifugal pump hydraulics
- Forced vortex vs. free vortex
Common Exam Questions
Example
Cylinder R = 0.4 m, initial depth 0.9 m, ω = 8 rad/s. Δz = (64)(0.16)/19.62 = 0.522 m. Center drop = 0.261 m, depth at center = 0.9 − 0.261 = 0.639 m (OK). Rim rise = 0.261 m, depth at rim = 0.9 + 0.261 = 1.161 m.
Approach
Compute ω (convert from rpm if necessary), then Δz = ω²R²/(2g). Check if vertex stays above bottom and if rim stays below tank top.
Question Type
Find the rim rise or vertex depth
Example
R = 0.3 m, h_0 = 0.5 m. Δz = 1.0 m. ω² = 1.0 × 2 × 9.81/0.09 = 218. ω = 14.77 rad/s.
Approach
Use Δz = 2h_0 (from volume conservation when vertex just touches bottom, no spill). Then solve ω²R²/(2g) = 2h_0.
Question Type
Find ω for vertex to just reach the bottom
Example
Find pressure at r = 0.2 m, 0.3 m below original surface level in a rotating tank (ω, R given). First find z at r = 0.2 m on the paraboloid, determine depth at that radius, then p = γ × depth.
Approach
Identify the paraboloid surface height at the radial position of the point. Compute vertical depth from free surface to the point. p = γ × depth.
Question Type
Pressure at a specific point inside the rotating liquid
Key Points To Remember
- ω must be in rad/s. Convert from rpm: ω = 2πN/60.
- Free surface equation: z = ω²r²/(2g) — z measured from the vertex (lowest point).
- Total rim rise from vertex: Δz = ω²R²/(2g).
- The paraboloid volume = (1/2) × volume of enclosing cylinder.
- Center drops by ω²R²/(4g); rim rises by ω²R²/(4g) from original horizontal surface.
- If the paraboloid vertex reaches the bottom → a dry center occurs; recalculate using volume conservation.
- Pressure = γ × vertical depth below the paraboloid surface at that radial position.
- The paraboloid is symmetric about the axis of rotation.
Practice Problems
Always perform the spillage check by comparing the computed rear depth to the tank wall height. When water forms a triangle (front bottom exposed), the wetted length is 2.0/tan θ. The volume conservation check confirms spillage. After spillage, the rear depth equals the wall height (2.0 m).
Problem
PROBLEM 1 (Horizontal Acceleration): An open rectangular tank is 5.0 m long, 2.0 m wide, and has water initially at a depth of 1.4 m. The tank wall height is 2.0 m. The tank accelerates horizontally at 4.0 m/s² along its length. Find: (a) the angle of the free surface, (b) the depth at the rear wall, (c) the pressure at the bottom of the rear wall, and (d) whether any water spills.
Solution
(a) tan θ = a/g = 4.0/9.81 = 0.4077 → θ = arctan(0.4077) = 22.17° (b) Surface pivots about the midpoint. Rise at rear = (L/2) × tan θ = (5.0/2)(0.4077) = 1.019 m Depth at rear wall = 1.4 + 1.019 = 2.419 m (c) p_rear_bottom = γ × h_rear = 9.81 × 2.419 = 23.73 kPa (d) Depth at rear = 2.419 m > wall height 2.0 m → SPILL OCCURS. After spill: rear water level = 2.0 m (at the tank rim). Front depth = 2.0 − (5.0)(tan θ) = 2.0 − 5.0(0.4077) = 2.0 − 2.039 = −0.039 m → negative → front bottom is exposed. Water forms a triangle in cross-section with maximum depth 2.0 m at the rear. Wetted length from rear = 2.0/tan θ = 2.0/0.4077 = 4.906 m (< 5.0 m → consistent with exposed front). Volume retained per unit width = (1/2)(4.906)(2.0) = 4.906 m² Per unit width, original volume = 5.0 × 1.4 = 7.0 m² Since retained volume (4.906) < original (7.0), spill has occurred. Actual volume spilled per unit width = 7.0 − 4.906 = 2.094 m² Total volume spilled = 2.094 × 2.0 = 4.188 m³ Rear wall pressure (after spill) = 9.81 × 2.0 = 19.62 kPa
Part (b) is a classic trap: decelerating while moving upward means the velocity and acceleration vectors point in OPPOSITE directions. The object slows down while going up, so the acceleration is downward — use the minus sign. Always determine the DIRECTION of the acceleration vector, not the direction of motion.
Problem
PROBLEM 2 (Vertical Acceleration): A 1.5 m × 1.5 m square tank contains oil (SG = 0.85) to a depth of 1.8 m. Find the pressure at the bottom when the tank: (a) accelerates upward at 5.0 m/s², (b) decelerates while moving upward at 3.0 m/s², (c) is in free fall.
Solution
γ_oil = 0.85 × 9.81 = 8.3385 kN/m³, h = 1.8 m (a) Upward acceleration a = 5.0 m/s²: p = γh(1 + a/g) = 8.3385(1.8)(1 + 5.0/9.81) = 15.009(1.5097) = 22.66 kPa (b) Decelerating while moving upward → acceleration is DOWNWARD at 3.0 m/s²: p = γh(1 − a/g) = 8.3385(1.8)(1 − 3.0/9.81) = 15.009(0.6942) = 10.42 kPa (c) Free fall (a = g = 9.81 m/s² downward): p = γh(1 − g/g) = 15.009(0) = 0 kPa (gauge)
When the vertex just touches the bottom, the geometric center of the paraboloid base is at the tank floor. The water depth at any radius r is exactly z(r) = ω²r²/(2g) measured from the bottom. At the rim, this equals Δz = 2h_0 = 2.00 m, giving a pressure of 19.62 kPa. The critical formula Δz = 2h_0 comes directly from the property that the paraboloid volume = half its enclosing cylinder's volume.
Problem
PROBLEM 3 (Rotation): An open cylindrical tank has a diameter of 1.20 m and contains water at a depth of 1.00 m. The tank is rotated about its vertical axis. Find: (a) the rotational speed (rpm) at which the water just begins to uncover the bottom (vertex reaches the bottom), (b) the depth at the rim at that speed, and (c) the pressure at the bottom at the rim when rotating at this speed.
Solution
R = 0.60 m, h_0 = 1.00 m (a) Vertex just touches bottom → Δz = 2h_0 = 2(1.00) = 2.00 m Δz = ω²R²/(2g) 2.00 = ω²(0.60)²/(2 × 9.81) 2.00 = ω²(0.36)/19.62 ω² = 2.00 × 19.62/0.36 = 109.0 ω = 10.44 rad/s N = 60ω/(2π) = 60(10.44)/6.2832 = 99.72 rpm ≈ 99.7 rpm (b) Depth at rim: Height of paraboloid from bottom = Δz = 2.00 m Depth at rim = Δz = 2.00 m (since the center is at the bottom, z = 0 at center) The water surface at the rim is 2.00 m above the bottom. Depth at the rim = 2.00 m (c) Pressure at the bottom at the rim: The water surface is 2.00 m above the bottom at the rim, so the vertical depth at the rim bottom = 2.00 m. But the pressure at the BOTTOM of the tank at radius R: The free surface at r = R is at height z = ω²R²/(2g) = 2.00 m above the vertex (at the bottom). Depth below free surface = 2.00 m p = γ × depth = 9.81 × 2.00 = 19.62 kPa
This is the most complex scenario: both vertex-at-bottom and rim-at-top occur simultaneously. The retained volume is a paraboloid with height equal to the tank height (from bottom to rim). Volume conservation gives the spilled volume. In board exams, this level of problem is typically worth the most points and is where most examinees lose marks due to not checking BOTH conditions.
Problem
PROBLEM 4 (Combined Concept — Rotation with Spill): An open cylinder of radius 0.40 m and height 1.0 m is filled with water to a depth of 0.70 m. It is then rotated at 15 rad/s. Determine: (a) whether water spills, (b) the volume of water spilled (if any), and (c) the depth at the rim after equilibrium.
Solution
(a) Paraboloid height: Δz = ω²R²/(2g) = (15)²(0.40)²/(2×9.81) = (225)(0.16)/19.62 = 36/19.62 = 1.835 m Expected center drop = Δz/2 = 0.917 m Expected center depth = 0.70 − 0.917 = −0.217 m → Negative → vertex hits the bottom first. Recalculate with vertex at bottom: Expected rim depth = h_0 + Δz/2 = 0.70 + 0.917 = 1.617 m > 1.0 m (tank height) → SPILL also occurs. Both conditions triggered: vertex at bottom AND rim at tank top. Set rim depth = 1.0 m (at tank top), vertex at bottom. Volume retained = (1/2)πR²(Δz_new), where Δz_new = 1.0 m (rim − bottom). Volume retained = (1/2)π(0.40)²(1.0) = (1/2)(0.5027) = 0.2513 m³ (b) Original volume = πR²h_0 = π(0.40)²(0.70) = (0.5027)(0.70) = 0.3519 m³ Volume spilled = 0.3519 − 0.2513 = 0.1006 m³ ≈ 0.101 m³ (c) The rim depth after spilling = 1.0 m (the tank is full at the rim, vertex is at the bottom). Verify ω: Δz = ω²R²/(2g) → 1.0 = (15)²(0.40)²/(2×9.81) = 1.835 m ≠ 1.0 m. This means the actual paraboloid height from vertex to rim = 1.0 m, but ω²R²/(2g) = 1.835 m. The rim depth = 1.0 m is confirmed (the tank rim constrains the liquid level); the rest has spilled.
When the vertex is at the bottom and volume conservation is applied: H = 2h_0 = 1.20 m. The free surface at the rim is 1.20 m above the bottom. The pressure at the bottom rim = γ × H = 9.81 × 1.20 = 11.77 kPa. Note: The high ω (300 rpm) causes severe paraboloid formation — in practice, this high speed would cause centrifugal effects that dominate.
Problem
PROBLEM 5 (RPM Conversion and Pressure): A cylindrical container of radius 0.25 m contains water at a depth of 0.60 m. It rotates at 300 rpm. Find the pressure at the bottom of the container at the rim (r = R).
Solution
ω = 2π(300)/60 = 10π = 31.416 rad/s Paraboloid height: Δz = ω²R²/(2g) = (31.416)²(0.25)²/(2×9.81) = (987.0)(0.0625)/19.62 = 61.69/19.62 = 3.144 m Center drop = Δz/2 = 1.572 m Depth at center = 0.60 − 1.572 = −0.972 m → vertex hits bottom. With vertex at bottom: Rim rise = Δz = 3.144 m from the vertex (bottom). But original depth = 0.60 m → the paraboloid height Δz = 2h_0 for no-spill vertex-at-bottom case. Δz_critical = 2(0.60) = 1.20 m → Since actual Δz = 3.144 m > 1.20 m, water has spilled beyond the vertex-at-bottom scenario. With vertex at bottom and some water spilled: Let actual paraboloid height = H (from bottom to rim surface). Assume tank is open-topped (no height limit given — so no rim spillage constraint from tank walls). Volume conservation: (1/2)πR²H = πR²h_0 H = 2h_0 = 2(0.60) = 1.20 m So the paraboloid has its vertex at the bottom and rim water level at H = 1.20 m. The water surface at r = R is at height H = 1.20 m above the bottom. Depth below free surface at r = R (which is the water surface): depth = 0 at the free surface. Pressure at the BOTTOM at r = R: Height of water surface above the bottom at r = R = H = 1.20 m. Depth of bottom below surface at r = R = 1.20 m. p = γ × 1.20 = 9.81 × 1.20 = 11.77 kPa
Exam Preparation Tips
- FORMULA CARD: Write the three master formulas on a card and memorize them cold: (1) tan θ = a/g [horizontal], (2) p = γh(1 ± a/g) [vertical, + up, − down], (3) z = ω²r²/(2g) and Δz = ω²R²/(2g) [rotation]. These three cover 95% of board exam problems in this topic.
- SIGN CONVENTION DRILL: Before every vertical acceleration problem, write 'UP = +, DOWN = −' at the top of your solution. Deceleration while moving upward = downward acceleration (negative sign). This is the single most common source of error in this topic.
- RPM TO RAD/S: Drill the conversion ω = 2πN/60 until it is automatic. Never substitute N (rpm) directly into the paraboloid formula — the units will be wrong and your answer will be off by a factor of (2π/60)² ≈ 0.011.
- SPILLAGE CHECKLIST: For horizontal acceleration — compare computed rear depth to wall height. For rotation — (Step 1) check if center drops below tank bottom (Δz > 2h_0 means vertex hits bottom for a no-spill scenario); (Step 2) check if rim exceeds tank height. Check BOTH conditions.
- VOLUME CONSERVATION IS YOUR FRIEND: In all spill and vertex-at-bottom problems, volume conservation is the rescue equation. Original volume = retained volume. For horizontal: triangular cross-section if front is dry. For rotation: paraboloid volume = (1/2)πR²Δz.
- PARABOLOID GEOMETRY PROPERTY: Memorize that the paraboloid volume = (1/2) × enclosing cylinder volume. This gives you the critical result: when vertex just touches the bottom without spilling, Δz = 2h_0.
- UNIT CONSISTENCY: Always use γ in kN/m³, h in m → answer in kPa. Or use γ in N/m³ (9810), h in m → answer in Pa. Never mix units.
- CLOSED TANK PROBLEMS: For a completely filled closed tank accelerating horizontally, there is no free surface. The pressure at any point is determined by integrating the pressure gradient dp/dx = −ρa from a known reference pressure. Identify a reference point (usually at a corner) and compute pressure at the point of interest.
- PRACTICE WITH NUMBERS: The board exam often uses 'nice' values of a — such as 4.905 m/s² (= g/2), 9.81 m/s² (= g), or 2.0 m/s². When you see a/g = 0.5, recognize immediately that the effective gravity modification factor is 1.5 (up) or 0.5 (down). This speeds up computation.
- DRAW THE DIAGRAM FIRST: Before writing any equation, sketch the tank, mark the direction of acceleration, draw the expected free surface (tilted down toward acceleration for horizontal; horizontal for vertical; paraboloid for rotation). This visual check catches conceptual errors before they propagate into calculations.
- TIME MANAGEMENT: Horizontal and vertical acceleration problems should take 3–5 minutes each. Rotation problems with spill/vertex-at-bottom take 7–10 minutes. Allocate exam time accordingly — do the simpler parts first to secure partial credit.
- BOARD EXAM FREQUENCY: This topic appears in virtually every Civil Engineer board exam. Based on historical patterns (2015–2024), expect 2–4 problems from this topic, typically one from each subtopic. The rotation subtype with spillage is most frequently appearing in recent years.
In summary
Relative equilibrium of liquids is a compact but highly testable topic in the PRC Civil Engineer Licensure Examination. The core insight — that a liquid accelerating as a rigid body behaves like a modified hydrostatic system with an altered effective gravity and a repositioned free surface — reduces all three cases to the application of just three master equations: tan θ = a/g for horizontal acceleration, p = γh(1 ± a/g) for vertical acceleration, and z = ω²r²/(2g) for rotation. Mastery requires not just memorizing formulas but understanding the physical reasoning behind each: the free surface is always perpendicular to the resultant body force; pressure always equals γ times the vertical depth below that new free surface; and volume of liquid is always conserved. The critical differentiators in board exam performance are: (1) correct sign convention in vertical acceleration problems, (2) proper unit conversion (rpm to rad/s) in rotation problems, (3) systematic spillage checks in both horizontal and rotation cases, and (4) application of the paraboloid volume property (equal to half the enclosing cylinder) in rotation problems with vertex-at-bottom conditions. Practice with the worked examples and board-type problems in this chapter, apply the decision flowcharts until the logic is automatic, and you will consistently score full marks on this topic. As a future licensed Civil Engineer in the Philippines under RA 544, the ability to analyze fluid behavior under dynamic conditions is essential for designing safe water supply systems, tanker vehicles, elevated tanks subject to seismic loads, and centrifugal pump installations — all core competencies expected of a professional engineer.
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