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Variation of Pressure with Depth

🚁 Overview

Overview: In a static (non-moving) fluid of uniform density ρ, the pressure at a depth h below the free surface increases linearly with h. By performing a force-balance on a fluid element, one arrives at the hydrostatic formula p = p0 + ρgh, where p0 is the pressure at the free surface (often atmospheric).

Course Hub

Prep4Uni Fluid Mechanics 1
Central course portal outlining all core topics, syllabus milestones, and foundational fluid statics modules.

Chapter 1: Pressure Fundamentals

Chapter 1: Pressure
Establishes basic definitions of normal force per unit area, unit systems, and absolute versus gauge pressure concepts.

Chapter 3: Pressure Measurement

Chapter 3: Pressure Measurement
Applies depth-variation principles to laboratory manometers, barometers, and differential pressure devices.

Chapter 5: Dynamics

Chapter 5: Fluid Dynamics
Extends pressure concepts into moving fluids governed by mass conservation and momentum principles.

📖 Contents

  1. Concept of Hydrostatic Pressure
  2. Derivation by Force Balance on a Fluid Element
  3. Differential Derivation
  4. Free-Surface vs. Gauge Pressure
  5. Worked Examples
  6. Practice Problems & Solutions
  7. Key Formula Recap

🎯 Learning Outcomes

  • Explain the concept of hydrostatic pressure and how it develops in a static fluid.
  • Derive the hydrostatic pressure equation p = p0 + ρgh using both a force-balance on a fluid element and the differential form dp/dz = ρg.
  • Apply the formula p = p0 + ρgh to compute absolute and gauge pressures at a given depth in fluids of uniform density.
  • Account for non-atmospheric free-surface pressure p0 when calculating pressure in enclosed or pressurized containers.
  • Solve hydrostatic-pressure problems, including finding the depth corresponding to a specified pressure and determining pressure differences between two depths.
  • Use the differential relation dp/dz = ρg to generalize pressure variation in fluids with depth (including variable density cases).

2.1 Concept of Hydrostatic Pressure

  • In a fluid at rest, any small element feels force on its top and bottom faces due to the surrounding pressure, plus its own weight.
  • Because pressure transmits equally in all directions at a point, a simple vertical slice (column) of fluid suffices to derive the depth-dependence.
Hydrostatic Pressure on Water
Hydrostatic Pressure on Water

2.2 Derivation by Force Balance on a Fluid Element

Consider a right-prism of fluid of cross-sectional area A and height h:

  • Top face pressure: p0 → downward force on prism: Ftop = p0A
  • Bottom face pressure: p → upward reaction on prism: Fbottom = pA
  • Weight of fluid: W = ρgAh acting downward

Static equilibrium requirement: Fbottom − Ftop − W = 0.

pA − p0A − ρgAh = 0
⇒ p = p0 + ρgh
Fbottom=Ftop+Weight
F at bottom = F at top + Weight

2.3 Differential Derivation

Alternatively, consider an infinitesimal layer of thickness dz at depth z:

  • Pressure on bottom: p(z + dz)
  • Pressure on top: p(z)
  • Weight of layer: dW = ρgA dz
Infinitesimal Layer of Water
Infinitesimal Layer of Water

Force balance yields: p(z + dz)A − p(z)A − ρgA dz = 0.

Dividing by A and taking the limit dz → 0 gives the differential hydrostatic gradient:

dp / dz = ρg

Integrating from the surface (z = 0, p = p0) down to depth h:

∫p0p dp = ∫0h ρg dz → p − p0 = ρgh → p = p0 + ρgh

2.4 Free-Surface vs. Gauge Pressure

  • Absolute pressure: p = p0 + ρgh, with p0 = absolute free-surface pressure (often 1 atm).
  • Gauge pressure: pgauge ≡ p − p0 = ρgh.

In many practical engineering applications, using pgauge = ρgh is sufficient since free-surface atmospheric pressure acts ambiently on surrounding structures.

2.5 Worked Examples

  1. Example 1: Depth = 5 m in fresh water (ρ = 1000 kg/m3, g = 9.81 m/s2). Find absolute pressure.
    Solution:
    p = 1.013 × 105 Pa + (1000)(9.81)(5) ≈ 1.062 × 105 Pa (106.2 kPa).
  2. Example 2: Oil of density ρ = 850 kg/m3 is used in a hydraulic press system. What gauge pressure corresponds to 3 m depth?
    Solution:
    pgauge = ρgh = 850 × 9.81 × 3 ≈ 2.50 × 104 Pa (≈ 25.0 kPa).

2.6 Practice Problems & Solutions

  1. Problem 1: Calculate the depth at which pgauge = 150 kPa in seawater (ρ = 1025 kg/m3).
    Solution:
    Gauge pressure in a fluid: pgauge = ρgh
    ⇒ h = pgauge / (ρg) = (1.50 × 105 Pa) / (1025 kg/m3 × 9.81 m/s2) ≈ 14.9 m.
  2. Problem 2: Find the absolute pressure 12 m below the surface if p0 = 120 kPa (e.g. inside a pressurized tank).
    Solution:
    Absolute pressure at depth: p = p0 + ρgh
    Assuming water (ρ = 1000 kg/m3, g = 9.81 m/s2):
    p = 1.20 × 105 Pa + (1000 × 9.81 × 12) Pa = 1.20 × 105 + 1.18 × 105 Pa ≈ 2.38 × 105 Pa.
  3. Problem 3: A dam wall at depth h experiences force F = pA. Express F as a function of h, ρ, and A.
    Solution:
    At depth h, absolute pressure is p = p0 + ρgh. Evaluating net gauge force on the wall (ignoring ambient atmosphere), p = ρgh. Thus F = pA = (ρgh)A.
  4. Problem 4: Derive the pressure difference between two depths in a fluid with variable density ρ(z) by integrating dp/dz = ρ(z)g.
    Solution:
    Starting from the differential relation: dp = ρ(z)g dz.
    Integrate from depth z1 (pressure p1) to z2 (pressure p2):
    ∫p1p2 dp = ∫z1z2 ρ(z)g dz ⇒ p2 − p1 = g ∫z1z2 ρ(z) dz.

2.7 Key Formula Recap

FormulaDescription
p − p0 = ρghIncrease in pressure below free surface
p = p0 + ρghTotal (absolute) pressure at depth
pgauge = ρghGauge pressure (relative to surface reference)
dp / dz = ρgDifferential form for hydrostatic equilibrium

Interactive Hydrostatic Profile Simulator

Dynamically adjust fluid density (ρ), target depth (h), surface pressure (p0), and gravitational acceleration (g) to observe hydrostatic pressure progression over time and depth.

Gauge Pressure at Max Depth: 196.20 kPa
Absolute Pressure at Max Depth: 297.50 kPa

Interactive Hydrostatic Concept Toggles

Click to expand each conceptual section for detailed derivations and fluid statics insights.

Why is hydrostatic pressure independent of container cross-sectional geometry?
Hydrostatic pressure depends solely on vertical depth h, fluid density ρ, and acceleration due to gravity g. When performing a force balance on a fluid prism, cross-sectional area A cancels out entirely from both side walls and face interactions, demonstrating that container shape does not alter static pressure at a given depth (Hydrostatic Paradox).
How does non-uniform density affect the vertical pressure gradient?
When fluid density varies with elevation or temperature ρ(z), pressure variation is non-linear. Calculating total pressure requires integrating the fundamental differential equation dp = ρ(z)g dz across the height boundaries.
What is the significance of the piezometric head concept in fluid statics?
Piezometric head represents the total mechanical energy potential per unit weight in static fluids, expressed as z + p / (ρg) = constant. It combines elevation head and static pressure head into a unified parameter across continuous static fluid bodies.

Frequently Asked Questions

Does hydrostatic pressure act horizontally as well as vertically?

Yes. Pressure at any given point in a static fluid is isotropic, meaning it acts with equal magnitude in all directions against surrounding fluid elements or solid container boundaries.

How do multi-layer fluids affect pressure calculations at the container base?

In stratified immiscible fluids, hydrostatic pressure at the base equals the surface pressure plus the sum of individual hydrostatic contributions: p = p0 + ∑ (ρighi).

Can hydrostatic pressure formulas be applied to compressible gases?

The simplified formula p = p0 + ρgh assumes constant density. For large gas columns where density changes significantly with pressure and temperature, differential integration using ideal gas equations is required.

📝 Comprehensive Practice Module

Section 1: Foundational Core Assessment

  1. What is hydrostatic pressure?
    Answer: Pressure exerted by a fluid at rest due to the gravitational weight of the fluid column above; mathematically p = p0 + ρgh.
  2. State the hydrostatic pressure equation.
    Answer: p = p0 + ρgh, where p0 is free-surface pressure, ρ is fluid density, g is gravitational acceleration, and h is depth.
  3. How is gauge pressure defined in a static fluid column?
    Answer: Gauge pressure is the hydrostatic pressure relative to ambient atmospheric pressure: pgauge = ρgh.
  4. How is absolute pressure in a fluid expressed?
    Answer: Absolute pressure incorporates total free-surface pressure: pabs = p0 + ρgh.
  5. What are the SI units of pressure?
    Answer: The pascal (Pa), defined as 1 N/m2.
  6. How does hydrostatic pressure scale with fluid depth?
    Answer: It increases linearly with depth h.
  7. What does ρ represent in the hydrostatic equation?
    Answer: The mass density of the fluid expressed in kg/m3.
  8. What is the local gauge pressure at the free fluid surface (h = 0)?
    Answer: Zero gauge pressure (0 Pa gauge).
  9. What differential equation governs hydrostatic equilibrium?
    Answer: dp / dz = ρg, where z represents vertical depth.
  10. What fundamental assumption underlies the standard hydrostatic formula p = p0 + ρgh?
    Answer: The fluid is incompressible (constant density ρ) and static under constant gravitational acceleration g.

Section 2: Analytical & Conceptual Analysis

  1. Why is hydrostatic pressure at a specific depth isotropic in all directions?
    Answer: Static fluids cannot support shear stress without undergoing deformation; thus normal forces transmit equally across all orientation planes at a point.
  2. How is fluid density variation with elevation ρ(z) mathematically integrated?
    Answer: By performing definite integration on the differential relation: p2 − p1 = g ∫ ρ(z) dz.
  3. How do liquid-filled manometers measure pressure differences using column height?
    Answer: Manometers equate unknown pressure forces to the hydrostatic head weight generated by manometric fluid columns (Δp = ρgΔh).
  4. Why can free-surface pressure p0 be omitted when calculating net hydrostatic force on submerged structures open to atmosphere on both sides?
    Answer: Atmospheric pressure acts equally on both opposing faces of the structural boundary, canceling out in net force calculations.
  5. How are multi-layered immiscible fluid columns evaluated for total bottom pressure?
    Answer: By summing the individual hydrostatic pressure heads of each fluid layer: ptotal = p0 + ∑ ρighi.
  6. What role does hydrostatic pressure play in structural dam wall engineering?
    Answer: It establishes a triangular force loading distribution that increases linearly toward the reservoir bottom, determining required dam thickness.
  7. How does fluid temperature alter hydrostatic pressure gradients in deep reservoirs?
    Answer: Temperature shifts liquid density ρ(T); thermal expansion reduces density, slightly lessening the hydrostatic pressure gradient dp/dz.
  8. State Pascal's Law and its relationship to hydrostatic systems.
    Answer: Pascal's Law states that pressure applied to an enclosed static fluid is transmitted undiminished throughout the fluid and container walls.
  9. Why must submarine pressure hulls be designed specifically around hydrostatic depth curves?
    Answer: External hydrostatic compressive stress grows continuously with depth, requiring hull materials capable of preventing buckling failure.

Section 3: Computational & Numerical Exercises

  1. Calculate the gauge pressure at an 8 m depth in fresh water (ρ = 1000 kg/m3, g = 9.81 m/s2).
    Solution:
    pgauge = ρgh = 1000 × 9.81 × 8 = 78 480 Pa ≈ 78.5 kPa.
  2. Find the depth in hydraulic oil (ρ = 850 kg/m3) required to produce a gauge pressure of 200 kPa.
    Solution:
    h = pgauge / (ρg) = 200 000 / (850 × 9.81) ≈ 24.0 m.
  3. What is the absolute pressure 15 m below the ocean surface if atmospheric pressure is 101 kPa and seawater density is 1025 kg/m3?
    Solution:
    pabs = p0 + ρgh = 101 000 + (1025 × 9.81 × 15) ≈ 101 000 + 150 828 = 251 828 Pa ≈ 251.8 kPa.
  4. A storage tank is pressurized at p0 = 150 kPa. Find the absolute pressure at 5 m depth in water (ρ = 1000 kg/m3).
    Solution:
    p = 1.50 × 105 + (1000 × 9.81 × 5) = 150 000 + 49 050 = 199 050 Pa ≈ 199.1 kPa.
  5. In a U-tube manometer with mercury (ρ = 13 600 kg/m3), a fluid pressure causes a column deflection Δh = 0.05 m. Calculate the gauge pressure.
    Solution:
    pgauge = ρgΔh = 13 600 × 9.81 × 0.05 ≈ 6670.8 Pa ≈ 6.67 kPa.
  6. A container holds a 10 m layer of water (ρ = 1000 kg/m3) over a 2 m layer of dense oil (ρ = 850 kg/m3). Find total gauge pressure at the tank bottom.
    Solution:
    pgauge = (1000 × 9.81 × 10) + (850 × 9.81 × 2) = 98 100 + 16 677 = 114 777 Pa ≈ 114.8 kPa.
  7. Determine the absolute pressure at a depth of 20 m in seawater (ρ = 1025 kg/m3, patm = 101.3 kPa).
    Solution:
    pabs = 101 300 + (1025 × 9.81 × 20) = 101 300 + 201 105 = 302 405 Pa ≈ 302.4 kPa.
  8. Calculate the required column depth in mercury (ρ = 13 600 kg/m3) to develop a gauge pressure of 50 kPa.
    Solution:
    h = pgauge / (ρg) = 50 000 / (13 600 × 9.81) ≈ 0.375 m = 37.5 cm.
  9. Calculate the hydrostatic pressure difference between depths z1 = 2 m and z2 = 7 m in pure water (ρ = 1000 kg/m3).
    Solution:
    Δp = ρg(z2 − z1) = 1000 × 9.81 × (7 − 2) = 1000 × 9.81 × 5 = 49 050 Pa ≈ 49.1 kPa.
  10. If a fluid's density varies linearly as ρ(z) = 1000 + z (kg/m3), calculate pressure rise Δp between depth z1 = 0 and z2 = 2 m.
    Solution:
    Δp = g ∫02 (1000 + z) dz = 9.81 × [1000z + 0.5z2]02 = 9.81 × (2000 + 2) = 9.81 × 2002 ≈ 19 639.6 Pa ≈ 19.64 kPa.
Last updated: 05 Aug 2026