Eight numericals covering laminar, transitional and turbulent flow, dynamic and kinematic viscosity, across water, oil, gas, and viscous fluids.
Practice Problems
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Problem 1 — Water in a garden hoseEasy
Given: ρ = 998 kg/m³, v = 5 m/s, D = 0.019 m, μ = 0.001 Pa·s
Step 1 — Formula: Re = ρvD/μ
Step 2 — Substitute: (998 × 5 × 0.019) / 0.001
Step 3 — Compute: Re = 94,810
Step 4 — Classify: Re > 4,000, so the flow is turbulent.
Answer: Re ≈ 94,810 (turbulent)
Problem 2 — Crude oil in a transfer lineMedium
Given: ρ = 870 kg/m³, v = 1.2 m/s, D = 0.15 m, μ = 0.012 Pa·s
Step 1 — Formula: Re = ρvD/μ
Step 2 — Substitute: (870 × 1.2 × 0.15) / 0.012
Step 3 — Compute: Re = 13,050
Step 4 — Classify: Re > 4,000, so the flow is turbulent.
Answer: Re ≈ 13,050 (turbulent)
Problem 3 — Air in an HVAC ductMedium
Given: ρ = 1.2 kg/m³, v = 8 m/s, D = 0.3 m, μ = 0.000018 Pa·s
Step 1 — Formula: Re = ρvD/μ
Step 2 — Substitute: (1.2 × 8 × 0.3) / 0.000018
Step 3 — Compute: Re = 160,000
Step 4 — Classify: Re > 4,000, so the flow is turbulent.
Answer: Re ≈ 160,000 (turbulent)
Problem 4 — Glycerin in a lab tubeEasy
Given: ρ = 1260 kg/m³, v = 0.05 m/s, D = 0.01 m, μ = 1.5 Pa·s
Step 1 — Formula: Re = ρvD/μ
Step 2 — Substitute: (1260 × 0.05 × 0.01) / 1.5
Step 3 — Compute: Re = 0.4
Step 4 — Classify: Re < 2,300, so the flow is laminar.
Answer: Re ≈ 0.4 (laminar)
Problem 5 — Seawater in an intake pipeMedium
Given: ρ = 1025 kg/m³, v = 3 m/s, D = 0.2 m, μ = 0.00108 Pa·s
Step 1 — Formula: Re = ρvD/μ
Step 2 — Substitute: (1025 × 3 × 0.2) / 0.00108
Step 3 — Compute: Re = 569,444.4
Step 4 — Classify: Re > 4,000, so the flow is turbulent.
Answer: Re ≈ 569,444.4 (turbulent)
Problem 6 — Molten sulfur in a process lineHard
Given: ρ = 1800 kg/m³, v = 0.8 m/s, D = 0.08 m, μ = 0.01 Pa·s
Step 1 — Formula: Re = ρvD/μ
Step 2 — Substitute: (1800 × 0.8 × 0.08) / 0.01
Step 3 — Compute: Re = 11,520
Step 4 — Classify: Re > 4,000, so the flow is turbulent.
Answer: Re ≈ 11,520 (turbulent)
Problem 7 — Milk in a dairy pipelineMedium
Given: ρ = 1030 kg/m³, v = 1.5 m/s, D = 0.05 m, μ = 0.002 Pa·s
Step 1 — Formula: Re = ρvD/μ
Step 2 — Substitute: (1030 × 1.5 × 0.05) / 0.002
Step 3 — Compute: Re = 38,625
Step 4 — Classify: Re > 4,000, so the flow is turbulent.
Answer: Re ≈ 38,625 (turbulent)
Problem 8 — Natural gas in a pipeline (kinematic)Hard
Given: v = 10 m/s, D = 0.25 m, ν = 0.0000015 m²/s (kinematic viscosity given)
Step 1 — Formula (kinematic form): Re = vD/ν
Step 2 — Substitute: (10 × 0.25) / 0.0000015
Step 3 — Compute: Re = 1,666,666.7
Step 4 — Classify: Re > 4,000, so the flow is turbulent.
Answer: Re ≈ 1,666,666.7 (turbulent)
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