Relative Equilibrium of Liquids - Revision Notes
Relative equilibrium occurs when a liquid moves as a rigid body — no relative motion between fluid particles, hence no shear stress develops. The liquid behaves as if it were a solid mass subjected to modified gravity. This topic is a consistent fixture in PRC Civil Engineer board examinations (Hydraulics & Fluid Mechanics component) and typically appears as 2–4 problems per examination. Three scenarios are tested: (1) horizontal acceleration, (2) vertical acceleration, and (3) rotation about a vertical axis. Mastery requires confident manipulation of the governing equations, correct sign conventions, and awareness of spill/no-spill conditions in rotating vessels.
Fundamentals of Relative Equilibrium
Formulas
p = γh
Example
Water (γ = 9 810 N/m³) at 1.5 m vertical depth below a tilted free surface: p = 9 810 × 1.5 = 14 715 Pa = 14.72 kPa
Variables
p = pressure (Pa or kPa); γ = specific weight of liquid (N/m³); h = vertical depth below the free surface (m)
Application
Valid in all relative-equilibrium cases; h is measured vertically from the tilted or curved free surface to the point of interest.
Exam tips
- Draw a clear sketch of the tank, mark the new free surface, and label all relevant dimensions before applying any formula.
- Always confirm whether the problem asks for gauge pressure or absolute pressure; board problems typically ask for gauge pressure.
- When the problem gives γ in kN/m³ (e.g., 9.81 kN/m³), the resulting pressure will be in kPa directly — a common computational shortcut.
Key points
- Relative equilibrium is a state where a fluid body accelerates as a unit — no particle moves relative to another, so viscous shear is absent.
- Because there is no shear, the equations of static fluid pressure (p = γh) still apply, but measured from the NEW free surface (tilted or curved).
- The free surface is always perpendicular to the resultant of all body forces acting on the fluid (gravity + inertial body force).
- Pressure at any point equals γ × vertical depth below the free surface, regardless of the direction of acceleration.
- Three standard cases appear in PRC board exams: horizontal rectilinear acceleration, vertical rectilinear acceleration, and uniform rotation about a vertical axis.
Definitions
Relative Equilibrium
A condition in which the entire fluid mass accelerates uniformly so that there is no relative motion between adjacent fluid layers; shear stress is zero throughout.
Importance
Justifies using hydrostatic pressure formulas (p = γh) even for accelerating fluids — a foundational concept for all three board-exam cases.
Free Surface
The boundary between the liquid and the atmosphere (or vapor space) where gauge pressure equals zero. In relative equilibrium, this surface tilts or curves depending on the type of acceleration.
Importance
The free surface is the datum from which vertical depth h is always measured.
Effective Gravity (g_eff)
The vector resultant of true gravitational acceleration g and the negative of the imposed translational acceleration a. The free surface is perpendicular to g_eff.
Importance
Provides the physical explanation for why the free surface tilts or why pressure increases/decreases with vertical acceleration.
Common mistakes
- Measuring depth h along the slant of the free surface instead of vertically — always use vertical depth.
- Forgetting that the free surface is the zero-pressure datum; pressure at the bottom of a tilted tank must use the vertical distance from the (tilted) surface to the bottom.
- Assuming the pressure distribution is the same as static when the tank is accelerating — it is NOT; only the formula form p = γh is preserved, but h changes with the new surface geometry.
Horizontal Acceleration
Formulas
tan θ = a / g
Example
Tank accelerates at a = 3 m/s²: tan θ = 3/9.81 = 0.3058, θ = arctan(0.3058) = 17.0°
Variables
θ = angle of free surface below horizontal (degrees or radians); a = horizontal acceleration (m/s²); g = 9.81 m/s²
Application
Directly gives the slope of the free surface. Used to find how much liquid spills or the new depth at any point along the tank.
Δh = L × tan θ / 2 = L × a / (2g)
Example
L = 4 m, a = 3 m/s²: Δh = 4 × 3/(2 × 9.81) = 0.612 m rise at rear, 0.612 m fall at front
Variables
Δh = rise/fall of liquid surface at each end from the undisturbed level (m); L = length of tank in direction of acceleration (m)
Application
Determines how much the surface rises at the rear and falls at the front; used for spill-check problems.
Exam tips
- Memorize: 'Surface tilts down toward the acceleration.' This is your quick sanity check.
- For a rectangular tank of width W perpendicular to acceleration: pressure at the bottom is uniform across W (no variation in that direction).
- Spill-check: compute Δh; if Δh > (tank height − initial depth at front), liquid spills from the front. If spill occurs, recompute with the condition that the surface just touches the front top edge.
- In board problems, a = g tan θ is frequently needed in reverse: given θ, find a.
Key points
- The free surface tilts — it slopes downward in the direction of acceleration and upward on the trailing side.
- The angle θ of the free surface with the horizontal is given by tan θ = a/g.
- Pressure still follows p = γh where h is vertical depth below the (now tilted) free surface.
- For a closed, fully-filled tank, no surface tilt is visible but pressure increases at the back and decreases at the front; the imaginary free surface concept (piezometric surface) is used.
- Horizontal acceleration does NOT change the total volume of liquid; it merely redistributes it.
Definitions
Piezometric (Imaginary Free) Surface
For a closed tank, the imaginary surface where the pressure would be zero (atmospheric). It tilts at angle θ even though no actual surface is visible.
Importance
Allows use of the same tan θ = a/g formula for closed tanks; pressure at any point is γ × vertical depth below this imaginary surface.
Common mistakes
- Using tan θ = g/a instead of a/g — the acceleration is in the numerator because it is the inertial 'horizontal component' of effective gravity.
- Not checking whether the tilted surface goes below the tank bottom on the front side — if it does, the front portion has zero pressure (air pocket), and a separate analysis is needed.
- Applying Δh = L tan θ / 2 when the tank is NOT initially full; only valid if the original surface is at mid-level or you account for the actual initial depth.
- Forgetting the direction: surface goes DOWN at the FRONT (direction of a), UP at the REAR.
Vertical Acceleration
Formulas
p = γ h (1 + a/g) [upward acceleration]
Example
h = 2 m, a = 4 m/s² upward: p = 9 810 × 2 × (1 + 4/9.81) = 19 620 × 1.408 = 27 621 Pa ≈ 27.62 kPa
Variables
p = gauge pressure at depth h (Pa); γ = specific weight (N/m³); h = vertical depth below free surface (m); a = magnitude of upward acceleration (m/s²)
Application
Use when the tank moves upward (elevator going up, crane lifting, rocket ascending). Pressure INCREASES relative to static.
p = γ h (1 − a/g) [downward acceleration]
Example
h = 0.4 m, a = 2 m/s² downward: p = 9 810 × 0.4 × (1 − 2/9.81) = 3 924 × 0.796 = 3 123 Pa ≈ 3.12 kPa
Variables
p = gauge pressure at depth h (Pa); γ = specific weight (N/m³); h = vertical depth below free surface (m); a = magnitude of downward acceleration (m/s²)
Application
Use when the tank moves downward (elevator going down, free fall). Pressure DECREASES relative to static. At a = g, p = 0 (free fall).
γ_eff = γ (1 ± a/g)
Example
γ_water = 9 810 N/m³, a = 3 m/s² upward: γ_eff = 9 810(1 + 3/9.81) = 12 810 N/m³
Variables
γ_eff = effective specific weight (N/m³); + for upward acceleration; − for downward acceleration
Application
Treat the accelerating liquid as a static liquid of modified specific weight γ_eff. Simplifies multi-point pressure calculations.
Exam tips
- Memory aid: 'Going up → pressure goes up; going down → pressure goes down.'
- When a problem says 'decelerating at X m/s² while moving upward,' the net acceleration is DOWNWARD at X m/s² — use the minus formula.
- When a problem says 'decelerating at X m/s² while moving downward,' the net acceleration is UPWARD at X m/s² — use the plus formula.
- Board exam favorite: find the acceleration required so that the pressure at the bottom equals zero — solve (1 − a/g) = 0, giving a = g = 9.81 m/s² downward (free fall).
Key points
- The free surface remains horizontal — only the magnitude of effective pressure changes.
- Upward acceleration increases pressure (equivalent to higher g); downward acceleration decreases pressure.
- Free fall (a = g downward) produces zero gauge pressure throughout — weightlessness condition for the liquid.
- This case directly models elevators, rockets at launch, and crane-lifted tanks of water.
- The specific weight effectively becomes γ_eff = γ(1 ± a/g).
Definitions
Free Fall Condition
When a = g downward (container in free fall), the gauge pressure throughout the liquid is zero. The liquid exerts no net force on the container walls (beyond atmospheric).
Importance
A classical board exam conceptual question; also used to explain why astronauts see liquid float in free-fall environments.
Common mistakes
- Using + sign for downward acceleration (the most common sign-error in board exams for this topic).
- Computing a = g for free fall and getting a negative pressure — gauge pressure cannot be negative for normal liquids (cavitation would occur first). The answer should be zero, not negative.
- Confusing 'decelerating upward' with 'accelerating downward' — both produce effective downward acceleration and reduce pressure. Always resolve the net acceleration direction first.
- Neglecting to add atmospheric pressure when absolute pressure is asked.
Rotation About a Vertical Axis (Rotating Vessel)
Formulas
z = ω² r² / (2g)
Example
ω = 10 rad/s, r = R = 0.5 m: z = (10²)(0.5²)/(2×9.81) = 25/19.62 = 1.274 m
Variables
z = height of free surface above the vertex at radius r (m); ω = angular velocity (rad/s); r = radial distance from axis of rotation (m); g = 9.81 m/s²
Application
Defines the paraboloid shape. Set r = R (radius of cylinder) to find the maximum rise at the rim.
h_rise = ω² R² / (2g)
Example
N = 120 rpm → ω = 2π(120)/60 = 4π rad/s; R = 0.3 m: h_rise = (4π)²(0.3)²/(2×9.81) = 157.91×0.09/19.62 = 0.725 m
Variables
h_rise = total rise of surface from center vertex to rim (m); R = radius of cylinder (m)
Application
Directly gives the height difference between the lowest point (center) and the highest point (rim) of the free surface.
ω = 2πN / 60
Example
N = 120 rpm: ω = 2π(120)/60 = 4π = 12.566 rad/s
Variables
ω = angular velocity (rad/s); N = rotational speed (rpm)
Application
Unit conversion from rpm (given in most problems) to rad/s (required in formula).
V_paraboloid = (1/2) × π R² × h_rise
Example
R = 0.3 m, h_rise = 0.725 m: V_para = 0.5 × π(0.3²)(0.725) = 0.5 × 0.2054 = 0.1027 m³
Variables
V_paraboloid = volume of the paraboloid solid of revolution (m³); R = cylinder radius (m); h_rise = paraboloid height (m)
Application
Used in spill problems: if the paraboloid volume exceeds the air space above the original liquid level, liquid spills.
p = γ (z_surface − z_point) at radius r
Example
At r = 0.2 m on the bottom (z_point = 0, vertex at elevation z_0 above bottom): first find z_surface above vertex = ω²(0.2)²/(2g), then add the vertex elevation to get surface elevation, then subtract z_point.
Variables
z_surface = elevation of free surface at radius r (from vertex); z_point = elevation of point of interest; both measured from the same datum
Application
Pressure at any interior point in the rotating liquid.
Exam tips
- Always convert N (rpm) to ω (rad/s) as the very first step — write it explicitly to avoid errors.
- No-spill check: Is h_rise/2 ≤ (tank height − initial liquid depth)? If yes, no spill (vertex rises, volume redistributes within tank). If no, spill occurs.
- More precisely: The vertex drops by h_rise/2 and the rim rises by h_rise/2 from the undisturbed level. So check if h_rise/2 ≤ freeboard.
- For closed, fully-filled rotating tanks, the paraboloid concept applies to the pressure distribution — no actual free surface, but piezometric surface is a paraboloid.
- Memorize: Volume of paraboloid = ½ × base area × height. This is identical to: V_para = ½ × V_enclosing_cylinder.
Key points
- A liquid in an open cylinder rotating at constant angular velocity ω forms a paraboloid free surface.
- The shape is a paraboloid of revolution: z = ω²r²/(2g), where z is measured upward from the vertex of the paraboloid.
- The rise from the center (vertex) to the rim (radius R) is h_rise = ω²R²/(2g).
- Volume of the paraboloid = (1/2) × volume of its bounding cylinder — a key property for spill-check problems.
- ω must be in rad/s; convert from rpm: ω = 2πN/60.
- Pressure at any interior point (r, z_point) is p = γ × (vertical depth from the paraboloid surface directly above that point to the point itself).
Definitions
Paraboloid of Revolution
The 3-D surface formed by rotating a parabola about its axis. In rotating liquid problems, it is the shape of the free surface described by z = ω²r²/(2g).
Importance
Recognition of this shape is critical for setting up volume calculations and spill-check problems.
Vertex of the Paraboloid
The lowest point of the curved free surface, located at the center (r = 0) of the rotating cylinder. The elevation of the vertex changes if liquid spills or if the initial fill level changes.
Importance
All z-measurements for the paraboloid are referenced from the vertex; locating the vertex correctly is the key step in any rotation problem.
Spill Condition
Occurs when the paraboloid, if the cylinder were infinitely tall, would require a vertex below the bottom of the tank — meaning the rim rises higher than the tank wall and liquid overflows.
Importance
Triggers a redesign of the problem: the surface passes through the top rim of the tank, the vertex goes below the floor, and pressure at the bottom center becomes non-zero.
Common mistakes
- Using N (rpm) directly in the formula instead of converting to ω (rad/s).
- Confusing h_rise (vertex-to-rim) with the absolute elevation of the free surface.
- In spill problems, forgetting that the volume of liquid is conserved — the paraboloid volume above the original level equals the volume removed from below it.
- Assuming the vertex stays at the bottom of the tank — the vertex can drop below the floor if rotation is fast enough (indicating dry zone at center bottom).
- Using r = diameter instead of r = radius in the formula.
Spill Analysis and Special Cases
Formulas
Volume spilled (horizontal) = (1/2) × W × L × (Δh − freeboard)
Example
L = 3 m, W = 1 m, Δh = 0.8 m, freeboard = 0.5 m: Excess Δh = 0.3 m; Volume spilled ≈ 0.5 × 1 × 3 × (0.8 − 0.5) = 0.45 m³ [approximate; exact calc requires geometry]
Variables
W = tank width perpendicular to acceleration (m); L = tank length in direction of acceleration (m); Δh = computed surface rise at rear (m); freeboard = (tank height − initial depth) (m)
Application
Estimates how much liquid spills when Δh exceeds the freeboard. More commonly, the board exam asks for depth at a specific location after spilling.
r_dry² = R² − (4g × V_liquid)/(π ω² R²) × (2g/ω²) [simplified: solve from V_cone = V_liquid]
Example
Set up from first principles: total liquid volume = volume of paraboloid from r_dry to R, then solve for r_dry.
Variables
r_dry = radius of dry zone at bottom when vertex is below floor (m); V_liquid = volume of liquid remaining (m³)
Application
Used in advanced board problems where the vertex falls below the tank floor; the bottom is dry for r < r_dry.
Exam tips
- In board exams, 'spill problems' are usually flagged by phrases like 'the tank is initially full' or 'find the rotational speed that just causes the liquid to spill from the rim.'
- The critical (just-about-to-spill) angular velocity for a full tank: h_rise = H (full tank height): ω_critical = √(2gH/R²).
- For partial-fill rotation (liquid depth d, tank height H): no-spill condition is h_rise/2 ≤ H − d. The vertex drops by h_rise/2.
Key points
- Spill analysis is required whenever computed Δh (horizontal case) or h_rise/2 (rotation case) exceeds the available freeboard.
- For horizontal acceleration with spill: Set the tilted surface to pass through the top-rear corner; recompute effective length and volume constraints.
- For rotation with spill: The vertex of the paraboloid descends below the tank bottom, creating a dry annular zone at the bottom. The problem becomes: find the dry radius.
- Conservation of liquid volume is always enforced — the volume of paraboloid above original level equals volume 'vacated' below.
- Closed-tank rotation: use the imaginary free surface (piezometric paraboloid) with pressure = γ × vertical distance to the imaginary surface.
Definitions
Freeboard
The vertical distance from the undisturbed liquid surface to the top rim of the tank (= tank height − initial liquid depth). Represents the margin before spilling.
Importance
Determines whether a spill analysis is needed. Always compute freeboard before solving horizontal acceleration or rotation problems.
Common mistakes
- Skipping the spill check and using the no-spill formula when the tank has actually spilled — leads to gross errors in computed pressures.
- In rotation-with-spill problems, forgetting that the vertex may descend below the bottom, creating a dry ring zone — treating the bottom as fully wetted when it is not.
Connections
- Hydrostatics (p = γh) — Relative equilibrium extends hydrostatics to accelerating frames; the core pressure formula is unchanged, only the geometry of the free surface changes.
- Kinematics/Dynamics — Angular velocity (ω), linear acceleration (a), and free fall are mechanical concepts that directly enter the fluid equations.
- Centrifugal Force / Pseudo-force — Rotation problems are solved in the rotating reference frame where centrifugal body force balances the pressure gradient; identical approach to non-inertial frame mechanics.
- Bernoulli's Equation — Relative equilibrium is the limiting case where velocity is zero relative to the fluid; Bernoulli reduces to the hydrostatic equation in the effective gravity field.
- Fluid Statics — Forces on submerged curved surfaces and plane surfaces use p = γh; the same formula applies here with the modified (tilted or curved) surface.
- Structural Engineering (Tank Design) — Understanding pressure distribution under acceleration is essential for designing liquid storage tanks (e.g., water towers, fuel tanks) subject to seismic or vehicular loading — relevant to NSCP 2015 seismic provisions for non-structural components.
- Unit Conversions — rpm to rad/s, kPa to Pa, kN/m³ to N/m³ — these conversions are tested implicitly in every rotation problem.
- Volume of Solids of Revolution (Mathematics) — The paraboloid volume formula (½πR²h) comes from calculus (Pappus' theorem / integration); understanding this prevents sign and factor errors.
Quick review questions
A tank of water accelerates horizontally at 4.905 m/s². What is the angle of the free surface with the horizontal?
Answer
θ = 26.57° (approximately 26.6°)
tan θ = a/g = 4.905/9.81 = 0.5; θ = arctan(0.5) = 26.57°. Note that 4.905 = g/2, so tan θ = 0.5 exactly — a common board-exam value.
A water tank 3 m deep accelerates downward at 3 m/s². Find the gauge pressure at the bottom.
Answer
p = 19.62 kPa × (1 − 3/9.81) = 19.62 × 0.6942 = 13.62 kPa
Use p = γh(1 − a/g). γ = 9.81 kN/m³, h = 3 m: p = 9.81 × 3 × (1 − 3/9.81) = 29.43 × 0.6942 = 20.43 kPa. (Recompute: 9.81×3 = 29.43 kN/m²×(1−0.3058) = 29.43×0.6942 = 20.43 kPa.) Downward acceleration → use minus sign.
An open cylinder of radius 0.4 m and height 1.5 m, initially half-full of water, rotates at ω = 8 rad/s. Does the water spill? Find h_rise.
Answer
h_rise = 1.307 m; half of this is 0.653 m. Freeboard = 1.5 − 0.75 = 0.75 m. Since 0.653 m < 0.75 m, no spill occurs.
h_rise = ω²R²/(2g) = 64 × 0.16/(19.62) = 10.24/19.62 = 0.522 m. Wait — recompute: ω=8, R=0.4: h_rise = (8²)(0.4²)/(2×9.81) = 64×0.16/19.62 = 10.24/19.62 = 0.522 m. Half-rise = 0.261 m. Freeboard = 0.75 m. 0.261 < 0.75, so no spill. (Vertex drops 0.261 m; rim rises 0.261 m from mid-level.)
At what downward acceleration does the pressure at the bottom of a water tank become zero?
Answer
a = g = 9.81 m/s² (free fall)
Set p = γh(1 − a/g) = 0. Since γh ≠ 0, we need (1 − a/g) = 0, which gives a = g = 9.81 m/s². This is free fall — the liquid is weightless relative to the container.
A cylinder of diameter 0.6 m rotates at 120 rpm. Find (a) ω in rad/s and (b) the rise of the paraboloid from center to rim.
Answer
(a) ω = 4π ≈ 12.566 rad/s; (b) h_rise ≈ 1.157 m
(a) ω = 2π×120/60 = 4π = 12.566 rad/s. (b) R = 0.3 m; h_rise = ω²R²/(2g) = (4π)²(0.3)²/(2×9.81) = 157.91×0.09/19.62 = 14.21/19.62 = 0.724 m. (Exact: 16π²×0.09/19.62 = 14.21/19.62 = 0.724 m.)
A rectangular open tank is 2 m long, 1 m wide, and 1.2 m deep, initially filled with water to 1 m depth. It accelerates horizontally at a = 6 m/s² along its 2-m length. Does water spill?
Answer
Δh = 0.612 m; freeboard = 0.2 m. Since 0.612 > 0.2, water DOES spill from the front-leading edge.
Δh = L×a/(2g) = 2×6/(2×9.81) = 12/19.62 = 0.612 m. Freeboard = 1.2 − 1.0 = 0.2 m. The surface would rise 0.612 m at the rear and drop 0.612 m at the front, but the front depth would be 1.0 − 0.612 = 0.388 m (still positive, so the rear spills over the top). Since 0.612 > 0.2, water spills from the REAR (the side that rises).
What is the effective specific weight of water (γ = 9.81 kN/m³) when the container accelerates upward at a = 4.905 m/s²?
Answer
γ_eff = 14.715 kN/m³
γ_eff = γ(1 + a/g) = 9.81 × (1 + 4.905/9.81) = 9.81 × 1.5 = 14.715 kN/m³. The liquid behaves as a heavier fluid.
State the property of a paraboloid's volume relative to its bounding cylinder.
Answer
The volume of a paraboloid of revolution is exactly one-half the volume of its bounding cylinder (same base radius and same height).
V_paraboloid = (1/2)πR²h_rise. This result is used directly in spill problems: when the paraboloid forms, the liquid that 'rises' on the outside equals the volume 'vacated' at the center, consistent with this half-volume property.
Diagrams
Relative Equilibrium of Liquids — Chapter Overview
Mind map summarizing all three cases of relative equilibrium (horizontal acceleration, vertical acceleration, rotation), key formulas, and common pitfalls. Use this as a quick orientation before diving into each section.
Step-by-Step Problem-Solving Flowchart for Relative Equilibrium
Decision-tree flowchart guiding the solver through identifying the case, performing a spill check, and applying the correct formula. Follow this flowchart on every board exam problem to avoid common errors.
Sign Convention for Vertical Acceleration
Flowchart clarifying the critical sign convention for vertical acceleration problems, including the special case of free fall. This addresses the most common error (wrong sign) in board exam answers for this topic.
States of the Paraboloid in a Rotating Cylinder
State diagram showing the progression of the paraboloid shape as angular velocity increases from zero to very high values, including the critical transitions (no-spill, spill onset, vertex below floor). Helps visualize the four distinct regimes tested in board problems.
Distribution of Relative Equilibrium Topics in PRC Board Exams (Typical)
Approximate distribution of question types on this chapter in recent PRC board examinations. Horizontal acceleration and rotation (no-spill) dominate; allocate review time proportionally.
Exam strategy
For PRC board exam problems on Relative Equilibrium of Liquids: (1) Identify the case (horizontal, vertical, or rotation) from the problem statement. (2) Draw a labeled sketch immediately — mark tank dimensions, initial liquid level, acceleration arrow, and the new free surface. (3) Write down the governing formula for that case. (4) Perform the spill check BEFORE computing pressures. (5) Convert units first (rpm → rad/s; kN/m³ → N/m³ if needed for Pa). (6) Solve step-by-step, showing substitution clearly. (7) Check reasonableness: upward accel → higher pressure than static; downward → lower; horizontal → one side higher, other lower. Budget approximately 4–6 minutes per numerical problem. For conceptual/multiple-choice items, apply the memory aids: 'surface down toward a' (horizontal), 'up = up, down = down' (vertical), 'paraboloid = half-cylinder volume' (rotation). Common trap: sign errors in vertical acceleration — always explicitly state the direction of net acceleration before substituting.