Fluid Properties and Hydrostatics – Notes

Compressibility of water and basic fluid concepts

Myth vs reality: water is often treated as incompressible in everyday life, but it is compressible. A tap on a glass creates a compression wave in the water that travels at finite speed, not instantaneously. Sound in water is a longitudinal (compression) wave.
Importance of compressibility: essential for understanding phenomena like sound propagation, submarine sonar, and echolocation in whales. In everyday conditions, compressibility effects are tiny, but under high pressures or strong waves they become significant.
Quantitative example (compressibility under high pressure):
- Depth example: a glass of water at roughly sea level, pressure ~1 atm. If the same water is at a depth of about 5 km (≈ 16,400 ft) below the ocean surface, the pressure is about 500 times higher than at the surface.
- Resulting volume change: the water would compress by roughly $rac{ riangle V}{V} \, ext{(at }5 ext{ km)} \approx 2.3\%.$
- This illustrates that water compression is small but measurable under extreme pressures.
Everyday applications where compression matters:
- Water jet cutting uses high-pressure water pumps: typically $200\text{ to }600\ \text{MPa}$ (≈ 30,000 to 90,000 psi). Compare to surface glass pressure, this is about 2–6,000× higher, leading to significant compression effects on the jet fluid.
- At around $350\ \text{MPa}$ (~50,000 psi), water compresses by about $14.7\%$ , which is substantial.
- At similar pressures, the chamber pressure of a high-powered rifle underwater is ~ $350\ \text{MPa}$ ; forward-projectiles must compress water ahead of them in the barrel, drastically affecting performance.
Where the myth comes from:
- For most normal daily activities, water behaves nearly incompressibly due to its very large bulk modulus. Compressibility becomes relevant under high pressures or when strong waves/sonic disturbances propagate through the liquid.

Key fluid properties: density and specific volume

Density: $\rho = \frac{m}{V}$
- SI unit: $\text{kg/m}^3$
- Dimension: $[\text{M L}^{-3}]$ (mass per unit volume).
Specific volume: $v = \frac{V}{m} = \frac{1}{\rho}$
- SI unit: $\text{m}^3/\text{kg}$
Dependence on temperature and pressure:
- Liquids (e.g., water): density changes weakly with temperature/pressure. At a given pressure, density typically decreases with increasing temperature; there is a notable exception near four degrees Celsius where liquid water has maximum density at 1 bar (≈ $1000\ \text{kg/m}^3$ ).
- Gases (e.g., air): density changes strongly with temperature and pressure; heating lowers density, increasing volume at the same pressure.
- General rule: density variation is weak for liquids and strong for gases.

Specific weight and specific gravity

Specific weight: $\gamma = \rho g$
- Units: $\text{N/m}^3$
- Dimension: same as pressure per unit length; gamma is interchangeable with $\rho g$ .
Specific gravity: $SG = \frac{\rho}{\rho_{\text{ref}}}$
- Dimensionless; reference density typically the density of water at 20°C: $\rho_{\text{ref}} \approx 998 \ \text{kg/m}^3\;\text{(often rounded to }10^3\text{ kg/m}^3)$
- No units because the ratio cancels.

Pressure: definition and types

Pressure: $p = \frac{F}{A}$
- Force is normal to a surface; pressure is a scalar.
Pressure vs pressure force:
- Pressure is a scalar (magnitude only).
- Pressure force on a surface is a vector, normal to the surface, with magnitude $F = p A$ .
Gauge vs absolute (total) pressure:
- Gauge pressure: P{gauge} = p{total} - p{atm}
- Atmospheric pressure: $p_{atm} \approx 101.3\ \text{kPa}$
- Absolute (total) pressure: p{total} = p{atm} + p{gauge}
In fluid mechanics, unless stated otherwise, pressure often means gauge pressure; be mindful of the context.

Ideal gas law (useful in thermodynamics and some fluid problems)

Equation: $P = \rho R T$
- P: total pressure (Pa)
- ρ: density (kg/m^3)
- R: specific gas constant (J/(kg·K))
- T: absolute temperature (K)
Gas constant for air (example): at about 15°C, R \approx 286.9\ \text{J/(kg·K)}
Temperature scale:
- Absolute temperature: $T(K) = T(°C) + 273.15$
- Absolute zero: $T = 0\,\text{K} \quad\leftrightarrow\quad T = -273.15°\text{C}$
Note: R depends on the gas; for calculations in air at modest temperatures, the cited value is a typical reference.

Bulk modulus and compressibility

Concept: relationship between pressure change and volume change under compression/expansion.
Definition: $B = - V \frac{dP}{dV}$
- Rearranged: $\frac{dP}{dV} = -\frac{B}{V}$
- Unit: same as pressure (Pa).
Implications:
- Liquids (like water) have very large bulk modulus; compressibility is negligible for many civil engineering problems.
- Gases have small bulk modulus; they are easily compressible.

Viscosity and the nature of fluids

Viscosity concept: internal friction within a fluid; resistance to shear.
Newton's law of viscosity (dynamic/absolute viscosity):
- Consider two parallel plates with a fluid in between; top plate moves with velocity u, bottom plate fixed.
- Shear stress: $\tau = \mu \frac{du}{dy}$
- Here, $\mu$ is the dynamic (absolute) viscosity (Pa·s).
Units of viscosity: typically Pa·s, but also N·s/m^2 or kg/(m·s) through unit equivalences.
Velocity profile under simple shear (no-slip at boundaries): linear profile between plates for small gaps.
Kinematic viscosity: $\nu = \frac{\mu}{\rho}$
- Units: m^2/s
Newtonian vs non-Newtonian fluids:
- Newtonian fluid: viscosity is constant for given temperature/pressure; plot of shear stress vs. shear rate is linear through the origin: $\tau = \mu \frac{du}{dy}$ .
- Non-Newtonian fluids: viscosity varies with shear rate.
- Shear thinning (pseudoplastic): viscosity decreases as shear rate increases.
- Shear thickening (dilatant): viscosity increases as shear rate increases.
- Examples:
- Paint: shear thinning (easier to spread when stirred/painted).
- Cornstarch–water mixture: shear thickening in some regimes.
Temperature and density effects on viscosity:
- For liquids like water, increasing temperature generally decreases viscosity; density is relatively constant.
- For gases, increasing temperature can increase viscosity, while density decreases with temperature; kinematic viscosity may increase with temperature for gases.

Isothermal atmosphere and vertical density variation (example with ideal gas law)

Isothermal atmosphere problem (density not constant in height):
- Start with hydrostatic equation and ideal gas law; density ρ is not constant when height increases.
- Hydrostatic equation (density variable):
- General form: $\frac{dP}{dz} = -\rho(z) g$
- Using ideal gas law with isothermal assumption (constant T): ρ = P/(R T).
- Substitute into hydrostatic equation and separate variables:
 $\frac{dP}{P} = -\frac{g}{R T} dz$
- Integrate from sea level (z=0, pressure P0) to height h (z=h): $\ln\left(\frac{P(h)}{P0}\right) = -\frac{g h}{R T}$
- Solve for P(h):
 $P(h) = P_0 \exp\left(-\frac{g h}{R T}\right)$
Example numbers (isothermal atmosphere):
- Sea level pressure: $P_0 \approx 101.3\ \text{kPa}$ .
- At height $h = 1000\ \text{m}$ , using the isothermal relation, one gets roughly $P(h) \approx 89.96\ \text{kPa}$ (absolute).
- Gauge pressure at that height would be $P(h) - P_0 \approx -11.34\ \text{kPa}$ (lower than atmospheric pressure at the surface).

Hydrostatics: horizontal vs vertical pressure variation

Hydrostatic principles 1 Horizontal variation in static fluids is zero: in a static fluid, pressure does not vary with horizontal position along a horizontal plane.
- If you consider a small control volume, forces in the x- and y-directions balance, giving P{Left} = P{Right} $and$ P{front} = P{back} for the same fluid.
  2) Vertical variation (hydrostatic equation):
- For a small vertical element, include top/bottom pressures plus weight:
  $\frac{dP}{dz} = -\rho g$ , which is pressure change in the z direction / change in depth, and this is the hydrostatic equation. (and this is equation is correct regardless the density being constant or not) and this is principal 2 (and if what to integrate the equation need to keep in mind if density is constant or not)
- Sign convention on the RHS depends on the chosen axis; with the common choice of z positive upward, the derivative is negative.
- Principal 3 (pressure at a given Hight)
- If density is constant, integrate (the hydrostatic equation) to obtain the classic head relation: (had integrate both sides with lower limit 0 and upper limit as -h by dz)
  $P(z) = P_0 + \rho g h$ , this is general form z can be h also, also row g and be replaced by gama
- P(h) = $P_0$ + $\rho gh$ , which is Absolute Pressure
- and if $P_0$ is P{atm}, P(h)- $P_0$ = $\rho gh$ , Then its Gauge Pressure
- if density varies with height, integrate (the hydrostatic equation ) using the variable density:
Important remarks:
- The negative sign in the hydrostatic equation matches the chosen coordinate system and gravity direction.
- The hydrostatic equation holds regardless of whether density is constant or changing; the integration requires accounting for density variation if present.

Pressure measurement devices (basics)

Barometer:
- Uses a mercury column to measure atmospheric pressure; standard is 760 mm Hg, equivalent to 101.3 kPa at sea level.
- Pressure from the column: $P{atm} \approx \rho{Hg} g h{Hg}$ with h{Hg} ≈ 760 mm.
- P{atm}=P{vapor}+ $\rho gh$ , since P{vapor} very small we ignore it so,
- P{atm}= $\rho gh$
Piezometer:
- A simple tube open to the atmosphere used to indicate the pressure in a pipe.
- If the pipe fluid has density ρ1 and specific weight γ1, and the top is open to atmosphere, the reading is:
- Total pressure: PA = P{atm} + γ1h1
- Gauge pressure: p{g} = γ1h1
- Limitations: only measures moderate pressures and requires a tall column for high pressures.
U-tube manometer:
- Used to measure pressure difference between two points (A and B) in possibly different fluids or at different elevations.
- The relation depends on fluid density and column heights; a standard exercise is to derive the equation relating P at point A $and$ P at point B from the manometer readings.
- Practical hint: horizontal pressure variation in static fluids is zero for the same fluid; be careful with different fluids.

Practical example problems (two illustrations from the lecture)

Example 1: belt on the surface of a long water tank
- Setup: a belt moves on the surface with velocity $u = 10\ \text{m/s}$ ; fluid layer depth is given; viscosity is known; contact area A is known.
- Force and power: the friction force acts to resist motion; the required power is
  $P = F \cdot u$
- From Newton’s law of viscosity, friction (shear) stress is
  $\tau = \mu \frac{du}{dy}$
- Friction force is $F = \tau \cdot A$ and velocity gradient is approximated by $\frac{du}{dy} \approx \frac{u}{h}$ (for a layer of thickness h).
- Substitution gives
  $P \approx \mu \left(\frac{du}{dy}\right) A\, u = \mu \left(\frac{u}{h}\right) A \; u = \mu A \frac{u^2}{h}$
- With the provided numbers (viscosity, area, depth, velocity), the result given was approximately $P \approx 1.34\times 10^2\ \text{W}$ (134.4 W).
Example 2: isothermal atmosphere and the ideal gas law
- Question: what is the pressure at height h above the surface when ρ is not constant but P = ρ R T with T constant?
- Start with hydrostatic equation and ideal gas to derive the isothermal atmosphere relation:
- Isothermal hydrostatics leads to
  $\frac{dP}{P} = -\frac{g}{R T} dz$
- Integrating from 0 to h gives
  $P(h) = P_0 \exp\left(-\frac{g h}{R T}\right)$
- Example numbers (isothermal Earth scenario): at h = 1000 m, with typical surface conditions, one finds
- Absolute pressure: $P(h) \approx 89.96\ \text{kPa}$
- Gauge pressure: P(h){gauge} = P(h) - P{atm} = approx -11.34 kPa
- This demonstrates how pressure decreases with height in an isothermal atmosphere.

Connections to core principles and real-world relevance

Foundational concepts traced back to earlier lectures in fluid mechanics:
- Pressure is a scalar; pressure forces act normal to surfaces.
- The difference between gauge and absolute pressure and the role of atmospheric pressure in measurements.
- The distinction between density for liquids (weak sensitivity to T/P) and gases (strong sensitivity to T/P).
- The ideal gas law as a bridge between macroscopic properties (P, V, T) and microscopic behavior for gases.
- The hydrostatic equation as a fundamental tool for static fluids and the derivation of pressure with depth.
Practical engineering relevance:
- Designing piping, reservoirs, and hydrostatic structures requires accounting for pressure variation with depth and density changes.
- Viscosity and non-Newtonian behavior impact flow resistance and energy losses in pipes, coatings, and processes (e.g., paint, clumping materials).
- High-pressure processes (e.g., water-jet cutting) require awareness of compressibility effects and bulk modulus.
- Atmospheric pressure and buoyancy considerations in civil and environmental engineering rely on hydrostatics and the ideal gas law for air.

Summary of key equations (LaTeX)

Density and specific volume
$\rho = \frac{m}{V}$
$v = \frac{V}{m} = \frac{1}{\rho}$
Specific weight and specific gravity
$\gamma = \rho g$
$SG = \frac{\rho}{\rho_{\text{ref}}}$
Pressure concepts
$p = \frac{F}{A}$
$p{g} = p - p{atm}$
$p_{atm} \approx 101.3\ \text{kPa}$
Ideal gas law (isothermal context)
$P = \rho R T$
$T(K) = T(°C) + 273.15$
R \approx 286.9\ \text{J/(kg·K)}\;\text{(air at ~15°C)}
Bulk modulus
$B = - V \frac{dP}{dV}$
$\frac{dP}{dV} = -\frac{B}{V}$
Viscosity and shear (Newton’s law)
$\tau = \mu \frac{du}{dy}$
$\nu = \frac{\mu}{\rho}$
Newtonian vs non-Newtonian (conceptual)
- Newtonian: line through origin in $\tau$ vs. $\frac{du}{dy}$ plot.
- Shear thinning: viscosity decreases with shear rate.
- Shear thickening: viscosity increases with shear rate.
Hydrostatics $\frac{dP}{dz} = -\rho g$
- With constant density: $P(z) = P_0 + \rho g h$
- With variable density: $P(h) = P0 + \int0^h \rho(z) g \, dz$
Barometer relation
$P{atm} = \rho{Hg} g h{Hg} \quad (h{Hg} = 760\ \text{mm})$
Isothermal atmosphere pressure variant
$P(h) = P_0 \exp\left(-\frac{g h}{R T}\right)$
Pressure at depth (barometric-like relation)
$P(h) \;\text{and gauge} = P(h) - P_{atm}$
U-tube manometer (pressure difference, generic form)
- For same fluid: $PA - PB = \gamma h\,\text{(difference in column heights)}$