Fluid Statics Notes

Fluid Statics

Subtopics

• Hydrostatic equilibrium
• Hydrostatic equilibrium in a Centrifugal field
• Manometers
  • Vertical Manometer
  • Inclined Manometer
• Continuous Gravity Decanter
• Centrifugal Decanter

Introduction

• Statics is the area of fluid mechanics that studies fluid at rest.
• Application: designing liquid storage tanks.

Importance and Applications of Fluid Statics

• Applications in Engineering:
  • Water supply systems (e.g., water towers)
  • Hydraulic presses (brake systems)
  • Dam design (pressure on dam walls)
  • Design of storage tanks

Pressure

• Pressure is defined as P = \frac{F}{A}
• Pressure = Total static force exerted normal to the area / Area on which the force is applied

Pressure Details

• Force is a vector.
• Area is a scalar.
• Pressure is a scalar with units N/m2 or Pascal or mm Hg, bar, psi.

Basic Properties of Pressure

• Definition: Pressure (P) = Force (F) / Area (A)
• Units: Pascals (Pa), N/m²
• Pressure is a scalar and acts equally in all directions in a fluid at rest.

Hydrostatic Equilibrium

• In a stationary mass of a single static fluid, the pressure is constant in any cross section parallel to the earth's surface but varies from height to height.

Variation of Pressure with Depth

• Pressure increases linearly with depth in a static fluid: P = P_0 + \rho g h
  • \rho = density of fluid
  • g = gravity
  • h = depth

Pascal's Law

• Pascal's Law: A pressure change at any point in a confined fluid is transmitted equally in all directions.
• Applications: Hydraulic lifts, braking systems

Vertical Variation Of Pressure In A Fluid Under Gravity

• Forces involved:
  • Force due to p1 on A (upward) = p1A
  • Force due to p2 on A (downward) = p2A
  • Force due to weight of element (downward) = mg = mass density x volume x g = \rho g A(z2 - z1)

Hydrostatic Equilibrium Equation

• Taking upward as positive:
  • \Sigma F = 0
  • P1A - P2A - \rho gA(z2 - z1) = 0
  • P2 - P1 = -\rho g (z2 - z1)
  • \frac{dp}{dz} = -\rho g

Hydrostatic Pressure Change

• In a fluid under gravity, pressure decreases linearly with an increase in height.
• This is the hydrostatic pressure change.

Incompressible Fluid

• \rho is constant.
• p2 –p1 = -\rho g(Z2-Z1) = \rho g(Z1-Z2)
• Can be written as: dp + \rho g dZ = 0
• Pressure is commonly expressed as a head, which is the height of the liquid column that would exert that amount of pressure at its base.
• The equation can be rewritten as: p/\rho g + Z = constant
  • p/\rho g is the pressure head.
  • Z is the static head.

Compressible Fluid

• \rho = f(p), density is not constant, but a function of pressure.
• \rho = \frac{pM}{RT}
• dp + \rho g dZ = 0
• \frac{dp}{p} + \frac{gM}{RT} dZ = 0
• Integrating gives: \ln(\frac{p2}{p1}) = -(\frac{gM}{RT})(Z2 - Z1)
• OR: \frac{p2}{p1} = \exp[-(\frac{gM}{RT})(Z2 - Z1)] (barometric equation)

Activity Questions & Answers

• Question 1: Why does pressure increase with depth in a fluid but not in a solid?
  • Answer: In fluids, particles move freely and transmit forces in all directions. As depth increases, more fluid is stacked above, increasing the weight exerted downward, thus pressure increases with depth. In solids, pressure isn't transmitted equally, and the internal structure supports the load without uniform distribution.
• Question 2: How would you calculate the pressure at the bottom of a 10-meter water tank?
  • Answer: Use the hydrostatic pressure formula: P = \rho g h
    • \rho (density of water) = 1000 kg/m³
    • g (gravitational acceleration) = 9.81 m/s²
    • h = 10 m
    • P = 1000 \times 9.81 \times 10 = 98,100 Pa = 98.1 kPa
    • So, the pressure at the bottom of the tank is 98.1 kPa (excluding atmospheric pressure).
• Question 3: Explain how hydraulic braking in cars utilizes Pascal’s Law.
  • Answer: Pascal’s Law states that a pressure applied to a confined fluid is transmitted undiminished in all directions. In hydraulic brakes: A small force on the brake pedal applies pressure to a fluid (brake fluid). This pressure is transmitted through the fluid to brake pads at all wheels. Because pressure is consistent throughout the fluid, even a small pedal force can generate enough pressure to stop the vehicle. This allows the use of small input forces to create large braking forces efficiently.
• Question 4: What are the advantages of using a manometer for pressure measurements?
  • Answer: Accuracy (especially for low-pressure measurements), simplicity (no moving parts, easy to use), cost-effectiveness (inexpensive compared to electronic pressure sensors), visual clarity (fluid height difference provides direct, readable result), and reliability (good for calibration and educational demonstrations).

Fluid Static Applications

• Manometer: Device for measuring pressure difference.

Manometers: U-Tube or Vertical Manometer

• A manometer is a device for measuring pressure differences (or gradients).

• A simple vertical U-tube manometer is used.

• The pressure on the left arm of the manometer, pa, is higher compared to the pressure on the right arm of the manometer,pb.

• pa > pb

• The total pressure at point 2 is a product of Pa plus the pressure exerted by the column of fluid B: Pressure at point 2 = Pa + \rhoB g (Zm + R_m)

• Within a column of fluid in hydrostatic equilibrium, at a certain elevation, the pressure in a cross section parallel to the earth's surface is constant. Therefore, the pressure at point 2 is equivalent to the pressure at point 3. P2 = P3

• The pressure at point 4 is less than at point 3 because point 4 is positioned atop the liquid column of fluid A. The pressure exerted by column 3-4 can be given by: Pressure exerted by column 3-4 = \rhoA g Rm

• The pressure at point 5 is Pb. The pressure exerted by column 4-5 of fluid B is: Pressure exerted by column 4-5 = \rhoB g Z_m

• The total pressure at point 3 is the total of column 3-4, plus column 4-5, plus Pb. Thus, Pressure at point 3 = Pb + \rhoB g Zm + \rhoA g Rm

• Expressed mathematically:

  • Pa + \rhoB g(Zm + Rm) = Pb + \rhoB g Zm + \rhoA g R_m
  • Pa - Pb = \rhoB g Zm + \rhoA g Rm - \rhoB g(Zm + R_m)
  • Pa - Pb = g Rm (\rhoA - \rho_B)

Inclined Manometer

• Used for measuring small differences in pressure.
• For a small height magnitude of Rm, the meniscus in the inclined tube must move a considerable distance, given by \frac{Rm}{\sin \alpha}, where \alpha is the angle of inclination.
• The pressure drop Pa - Pb can be given by:
  • Pa - Pb = g Rl (\rhoA - \rho_B) \sin \alpha

Continuous Gravity Decanter

• A continuous gravity decanter is used for continuously separating two immiscible liquids of differing densities.
• The two liquids are flowing continuously and slowly.
• The two liquids would overflow at the end of the decanter.

Continuous Gravity Decanter Function

• a.) The feed, which consists of a mixture of two liquids, enters at one end of the decanter.
• b.) The two liquids flow through the separator and form two layers: light liquid B on top and heavy liquid A at the bottom.
• c.) The two liquids are then discharged through overflow lines at the other end of the decanter.
• The density of the heavier liquid is \rhoA, while the density of the lighter liquid is \rhoB.
• The depth of layer of heavy liquid is Z{A1}, while the depth of the layer of light liquid is ZB. The total depth of the liquid within the vessel is Z_T.
• ZB + Z{A1} = Z_T
• The value of ZT is fixed by the position of the overflow line for the light liquid (liquid B). The heavy liquid (liquid A) is discharged through an overflow leg connected to the bottom of the vessel and rises to a height of Z{A2} above the vessel floor. Executing a hydrostatic balance for the columns of liquid yields:
  • ZB \rhoB + Z{A1} \rhoA = Z{A2} \rhoA
  • Solving for Z_{A1} gives:
    • Z{A1} = Z{A2} - \frac{ZB \rhoB}{\rho_A}
    • Z{A1} = Z{A2} - \frac{(ZT - Z{A1})\rhoB}{\rhoA}
    • Z{A1}(\rhoA - \rhoB) = Z{A2}\rhoA - ZT\rho_B
    • So, Z{A1} = \frac{Z{A2} - ZT\frac{\rhoB}{\rhoA}}{(1-\frac{\rhoB}{\rho_A})}
• The size of a decanter is determined by the time required for the separation of the two liquids. The time required for separation is dependent upon the densities of the two liquids and the viscosity (\mu) of the continuous phase. The separation time may be estimated from the empirical equation:
  • t = \frac{100\mu}{(\rhoA - \rhoB)}
    • t = separation time, hour (h)
    • \rhoA, \rhoB = densities of A and B, kg/m³
    • \mu = viscosity of continuous phase, centipoises (cP)

Fluid statics: Sizing the decanter

• Depends on the time required for separation of the two liquids.
• Separation time in hours (h), t = \frac{100 \mu}{\rhoA – \rhoB}
  • \mu = viscosity of continuous phase, cP
  • \rho density in kg/m3

Centrifugal Decanter

• Used for separation between two liquids which have small densities difference
• In a centrifuge that rotates at a very high angular velocity, the centrifugal force is so much larger than the force of gravity that the fluid surface is virtually cylindrical and coaxial with the axis of rotation.

Hydrostatic Equilibrium in a Centrifugal Field

• The centrifugal force is much larger than gravity, resulting in a cylindrical fluid surface coaxial with the rotation axis.
  • dF = \rho \omega^2 (2 \pi r b dr)

Fluid statics: Centrifugal decanter

• A centrifugal decanter is used if the difference densities of the two liquids are small.
• A hydrostatic pressure balance exists at the neutral interface.
  • Pi – PB = Pi – PA
    • where P_i = pressure at the neutral interface

Fluid statics: Centrifugal decanter Equation

• At steady state, the position of the liquid-liquid interface is established by hydrostatic balance. The pressure difference in the light liquid between r and ri must equal the pressure difference in the heavy liquid between rA and r_i. Thus:
  • Pi - PB = Pi - PA
  • \frac{\omega^2 \rhoB}{2}(ri^2 - rB^2) = \frac{\omega^2 \rhoA}{2} (rA^2 - ri^2)
  • \rhoB (ri^2 - rB^2) = \rhoA(rA^2 - ri^2)
  • Solving for r_i yields:
    • ri^2 = \frac{\rhoBrB^2 - \rhoA rA}{\rhoB - \rho_A}

Conditions For Centrifugal Decanter

• Equations can only be applied where the difference between \rhoA and \rhoB is not less than 3% to ensure stable operation.
• If the densities of the fluids are too nearly alike, the neutral zone may be unstable even if the speed of the rotation is sufficient to separate the liquids quickly.
• Percentage difference = \frac{|E1 - E2|}{\frac{1}{2}(E1 + E2)} \times 100

Example 1 (Simple U-tube manometer)

• A simple U-tube manometer is installed across an orifice meter.
• The manometer is filled with mercury (spec. gravity 13.6), and the liquid above the mercury is carbon tetrachloride (spec. gravity 1.6).
• The manometer reads 200 mm.
• What is the pressure difference over the manometer in newtons per square meter?

Example 1: U-Tube Manometer (Step-by-Step Solution)

• Given:
  • SG(Hg) = 13.6
  • SG(CCl₄) = 1.6
  • h = 200 mm = 0.2 m
  • g = 9.81 m/s²
• Step 1: Convert SG to density:
  • \rho_{Hg} = 13.6 \times 1000 = 13,600 kg/m³
  • \rho_{CCl₄} = 1.6 \times 1000 = 1,600 kg/m³
• Step 2: Use \Delta P = h \times g \times (\rho{Hg} - \rho{CCl₄}):
  • \Delta P = 0.2 \times 9.81 \times (13,600 - 1,600)
  • \Delta P = 0.2 \times 9.81 \times 12,000 = 23,544 Pa
• Final Answer:
  • \Delta P = 23.54 kPa

Example 2: U tube Manometer

• A simple U-tube vertical manometer is used to measure the pressure drop across an orifice. Liquid A is mercury (density = 13 590 kg/m³) and liquid B is brine (density = 1260 kg/m³), which flows through the orifice and fills the manometer leads. Under operating conditions, the pressure of the gauge at the upstream tap is 0.14 bar, while the pressure at the downstream tap is 250 mm Hg below atmosphere. What is the reading of the manometer in millimetres?
• Answer:
  • P_a = 0.14 bar
  • P_b = -250 mm Hg

Example 2 Solution

• Convert units:
  • P_a = 0.14 bar \times \frac{10^5 N/m^2}{1 bar} = 14000 N/m^2
  • P_b = -[13 590 \frac{kg}{m^3} \times 9.80665 \frac{m}{s^2} \times (\frac{250}{1000})m] = -33 318 \frac{kg}{ms^2} = -33 318 N/m^2
• From Equation (41): Pa-Pb = gRm (\rhoA-\rho_B)
  • 14000 \frac{kg}{ms^2} - (-33318 \frac{kg}{ms^2}) = 9.80665 \frac{m}{s^2} \times R_m \times (13590 \frac{kg}{m^3} – 1260 \frac{kg}{m^3})
  • 47318 \frac{kg}{ms^2} = 120 916 \frac{kg}{m^2s^2} \times R_m
  • R_m = 0.391 m = 391 mm

Example 3

• What is the height in m, of a column of water at 10oC equivalent to a pressure of 1 atm?
• What is this height in ft?

Solution to Example 3: Water Column Height from Pressure

• Given:
  • Pressure = 1 atm = 101325 Pa
  • \rho (water @10°C) = 999.7 kg/m³
  • g = 9.81 m/s²
• Step 1: Use h = \frac{P}{\rho \times g}
• Step 2: h = \frac{101325}{999.7 \times 9.81}
• Step 3: h ≈ \frac{101325}{9806.657} ≈ 10.33 m
• Step 4: Convert to feet: h = 10.33 \times 3.281 ≈ 33.89 ft
• Final Answers: h = 10.33 m = 33.89 ft

Example 4: Continuous Decanter

• A horizontal cylindrical continuous decanter is to separate 1500 bbl/d of a liquid petroleum fraction from an equal volume of wash acid. The oil is the continuous phase and, at the operating temperature, has a viscosity of 1.1 cP and a density of 54 lb/ft3 (865 kg/m3). The density of the acid is 72 lb/ft³ (1153 kg/m³).
• Determine:
  • a.) the length and diameter of the vessel (in ft³). The length of the cylindrical tank should be 5 times its diameter.
  • b.) the height of the acid overflow above the vessel floor.
  • barrels of crude oil per day (bbl/d) = 42 gal

Viscosity Units

• cP (centipoise): A unit of measure of dynamic viscosity in the centimeter-gram-second system of units equal to one one-hundredth of a poise (1 P = 100 cP = 1 g x cm-1 x s-1).

Horizontal Cylinder Filling

• Nonlinear filling: When volume = 95%, the corresponding liquid height ≈ 90% of diameter. This is based on pre-tabulated geometrical relationships or numerical integration of the area of a circular segment.

Summary: Fluid Statics Overview

• Fluid Statics deals with fluids at rest and pressure variation due to depth or motion.
• Key Concepts:
  • Pressure = Force / Area (scalar, acts in all directions)
  • Hydrostatic Pressure: Increases linearly with depth (P = \rho g h)
  • Pascal’s Law: Pressure applied to a fluid is transmitted equally
  • Manometers: Devices to measure pressure differences using fluid columns
  • Absolute, Gauge & Atmospheric Pressure:P{abs} = P{gauge} + P_{atm}
• Applications:
  • Water tanks and dams
  • Hydraulic brakes and lifts
  • Pressure measurement in pipelines (U-tube or inclined manometers)
  • Separation of immiscible liquids using gravity decanters
  • Centrifugal decanters for low-density-difference separations

Fluid Statics – Main Concepts

Fluid Statics – More Concepts

Test Information

• TEST 1 (7.5%) 17/4/2023; 9.00 -11.00 AM
• MCQ (1 HOUR)
• DKA COVERING UNIT AND DIMENSIONS ONLY
• 20-30 QUESTIONS
• PLEASE BRING LAPTOP/TABLET

Attendance

• 10/4/2025