Fluid Mechanics: Principles and Engineering Calculations

Posted on Jun 14, 2026 in Chemical Engineering

Topic 1: Viscosity & Shear Stress

1. Find Clearance (dy): The gap is 30 mm, and the plate is 1.8 mm thick. The clearance on one side is: (30 mm – 1.8 mm) / 2 = 14.1 mm = 0.0141 m.
2. Shear Stress (τ):
τ = μ(dv/dy) = 3 × (0.12 / 0.0141) = 25.53 N/m²
3. Viscous Force (Fv): Acts on both sides of the plate (Area = 1.8 × 1.8 = 3.24 m²).
Fv = 2 × τ × A = 2 × 25.53 × 3.24 = 165.43 N
4. Total Force & Power:
F = Fv + Weight = 165.43 + 60 = 225.43 N
Power = F × v = 225.43 × 0.12 = 27.05 W

(Assuming a standard differential setup where H₁ and H₂ are measured from a common datum, and water is the manometric fluid).

3. 3D Continuity Equation:
- Definition: Mass can neither be created nor destroyed. Rate of mass in = rate of mass out + rate of accumulation.
- Derivation: Consider a small control volume dxdydz.
- Net mass entering: −(∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z) dxdydz
- Mass accumulation: (∂ρ/∂t) dxdydz
- Equating and dividing by volume yields:
  ∂ρ/∂t + ∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z = 0
4. Laminar Flow in Circular Pipe:
- (i) Shear Stress: Balance pressure force (ΔP · πr²) with shear force (τ · 2πrL).
  τ = (ΔP / 2L)r
- (ii) Pressure Drop: Let r = R (the wall), making τ = τ₀. Solve for ΔP:
  ΔP = (2τ₀L) / R = (4τ₀L) / D

1. Velocity & Flow Regime: v = Q / A = 0.18 / (π × 0.1²) = 5.73 m/s.
Re = (vD) / ν = (5.73 × 0.2) / (8.9 × 10⁻⁷) = 1.28 × 10⁶ (Turbulent)
2. Head Loss (Darcy-Weisbach):
hf = f(L/D)(v² / 2g) = 0.02 × (350 / 0.2) × (5.73² / (2 × 9.81)) = 58.55 m
3. Pumping Power (P):
P = (ρgQhf) / η = (997 × 9.81 × 0.18 × 58.55) / 0.75 = 137.4 kW

Venturimeter (Cd ≈ 0.98): High pressure recovery and almost no eddy formation due to the gradual converging/diverging cone. The vena contracta perfectly coincides with the throat.
Orifice Meter (Cd ≈ 0.60): Poor pressure recovery due to massive flow separation at the sudden restriction. Energy dissipates rapidly via turbulent eddies. The vena contracta forms loosely downstream, adding contraction losses.

(Note: v must dimensionally equal √(gH), not gH).

1. Variables: n = 6 (v, g, H, D, μ, ρ). Dimensions m = 3 (M, L, T). π-terms = 3.
2. Repeating Variables: H, g, ρ.
3. π-Terms: π₁ = v / √(gH) | π₂ = D / H | π₃ = μ / (ρH√(gH))
4. Relation: π₁ = Φ(π₂, π₃). Rearranging algebraically gives:
v = √(gH) Φ[D/H, μ / (ρvH)]

1. The Limit: Stokes’ law holds up to a Reynolds number of 1 (Re = 1).
2. Setup Equations:
vt = [gd²(ρs − ρ)] / 18μ and Re = (ρvt d) / μ = 1
3. Solve for d: Substitute vt into the Re equation:
d³ = (18μ²) / [ρg(ρs − ρ)] = [18 × (0.001)²] / [998 × 9.81 × (2650 − 998)] = 1.11 × 10⁻¹²
4. Result: d ≈ 1.03 × 10⁻⁴ m (or 0.103 mm). Plug this back into vt to get vt ≈ 0.0097 m/s.

1. Kozeny-Carman dominance: For tiny particles (0.5 mm) at minimum fluidization, the laminar Ergun term balances the bed weight:
150μUmf / (dp² · εmf³ / (1 − εmf)) = (ρs − ρ)g
2. Rearrange for Umf:
Umf = [dp²(ρs − ρ)g / 150μ] × [εmf³ / (1 − εmf)]
3. Calculate:
Umf = [(0.0005)² × (2650 − 1.18) × 9.81 / (150 × 1.8 × 10⁻⁵)] × [0.42³ / (1 − 0.42)] = 0.307 m/s

(i) Cavitation: The formation and violent collapse of vapor bubbles inside a pump when local pressure falls below the fluid’s vapor pressure, causing pitting and damage.
(ii) NPSH (Net Positive Suction Head): The total head of fluid at the pump inlet measured above the fluid’s vapor pressure. Sufficient NPSH is required to prevent cavitation.
(iii) Positive Displacement Pump: A pump that traps a fixed, discrete volume of fluid and physically forces it into the discharge line (e.g., piston or gear pumps). Ideal for highly viscous fluids.
(iv) Characteristics Curves: Graphical plots showing how a specific pump performs—typically mapping Head, Efficiency, and Power against the volumetric Flow Rate (Q).