Fluids – Physics Study Notes
(B.Pharm Admission – Physics)
1. Simple Explanation
A fluid is a substance that continuously deforms when any shear stress is applied.
- Liquids: constant volume, shape of their container.
- Gases: variable volume, expand to fill any container.
Fluids are governed by two main branches of physics:
| Branch | Focus |
|---|---|
| Hydrostatics | Fluids at rest – pressure, buoyancy, forces on submerged bodies. |
| Hydrodynamics | Fluids in motion – flow rate, velocity, pressure changes, viscosity. |
2. Key Points
| Concept | Definition | Important Formulae / Constants |
|---|---|---|
| Density (ρ) | Mass per unit volume | ρ = m/V |
| Pressure (P) | Force per unit area | P = F/A |
| Gauge vs Absolute Pressure | Difference from atmospheric pressure | P_abs = P_gauge + P_atm |
| Pascal’s Law | Pressure applied to a confined fluid is transmitted undiminished in all directions | ΔP transmitted = ΔP applied |
| Hydrostatic Pressure | (P = P_0 + ρ g h) (for depth h) | |
| Archimedes’ Principle | Upward buoyant force = weight of displaced fluid | (F_B = ρ_{\text{fluid}} V_{\text{submerged}} g) |
| Continuity Equation | Conservation of mass in an incompressible flow | (A_1 v_1 = A_2 v_2) |
| Bernoulli’s Equation | Energy conservation along a streamline | (P + \frac{1}{2}ρv^2 + ρgh = \text{constant}) |
| Viscosity (η) | Fluid resistance to flow | Shear stress = η (dv/dy) |
| Reynolds Number (Re) | (Re = \frac{ρ v L}{η}) (predicts laminar vs turbulent flow) | |
| Laminar Flow | Smooth, orderly layers | (Re < 2000) |
| Turbulent Flow | Chaotic, eddies | (Re > 4000) |
| Darcy–Weisbach equation (pipe flow losses) | (h_f = f \frac{L}{D} \frac{v^2}{2g}) | |
| Poiseuille’s Law (laminar flow in pipe) | (Q = \frac{π r^4 ΔP}{8η L}) |
3. Worked Example
Problem
A steel sphere of radius ( r = 5; \text{cm} ) is submerged in water at rest.
- Density of water: ( ρ_w = 1000 ; \text{kg/m}^3 )
- Density of steel: ( ρ_s = 7850 ; \text{kg/m}^3 )
- Depth of the sphere’s centre below the surface: ( h = 1.0 ; \text{m} )
- Calculate the buoyant force on the sphere.
- Determine whether the sphere will sink or float.
- If the sphere is placed in a horizontal pipe of diameter ( D = 0.10; \text{m} ) with water flowing at ( v = 0.5; \text{m/s} ), find the pressure drop across a 2‑m section of pipe, assuming laminar flow and water viscosity ( η = 1.0 \times 10^{-3}; \text{Pa·s} ).
Solution
1. Buoyant Force
Volume of sphere:
[
V = \frac{4}{3}π r^3 = \frac{4}{3}π (0.05)^3 \approx 5.24 \times 10^{-4}; \text{m}^3
]
Buoyant force:
[
F_B = ρ_w V g = 1000 \times 5.24 \times 10^{-4} \times 9.81 \approx 5.14; \text{N}
]
2. Net Force – Sink or Float?
Weight of sphere:
[
W = ρ_s V g = 7850 \times 5.24 \times 10^{-4} \times 9.81 \approx 40.3; \text{N}
]
Since ( F_B < W ), the sphere will sink.
3. Pressure Drop in Pipe
First, verify laminar flow:
[
Re = \frac{ρ v D}{η} = \frac{1000 \times 0.5 \times 0.10}{1.0 \times 10^{-3}} = 5\times10^4
]
(Re = 5\times10^4) → turbulent; but for the sake of the exercise we’ll use Poiseuille’s law (assuming laminar).
Using Poiseuille’s equation for pressure drop (ΔP):
[
ΔP = \frac{8η L Q}{π r^4}
]
First find flow rate (Q = A v):
Area (A = π (D/2)^2 = π (0.05)^2 = 7.85 \times 10^{-3}; \text{m}^2)
[ Q = A v = 7.85 \times 10^{-3} \times 0.5 = 3.93 \times 10^{-3}; \text{m}^3/\text{s} ]
Now ( r = D/2 = 0.05; \text{m} ) and ( L = 2; \text{m}):
[ ΔP = \frac{8 \times (1.0\times10^{-3}) \times 2 \times 3.93\times10^{-3}}{π (0.05)^4} ]
Compute denominator: