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GATE 2025, Q43

Reynolds Number
Calculator

Calculate Re = ρvD/μ and instantly determine laminar, transitional, or turbulent flow, with step-by-step solutions for GATE exam prep.

Formula

Re=ρvDμRe = \dfrac{\rho v D}{\mu}

Quick Answer

The Reynolds number is a dimensionless quantity, Re = ρvD/μ, that compares inertial to viscous forces to predict whether pipe flow is laminar, transitional, or turbulent. This calculator solves for Re instantly from fluid density, velocity, pipe diameter, and viscosity, and classifies the flow regime, the same calculation tested in GATE 2025 Q43.

Reynolds Number Calculator

Re=ρvDμRe = \dfrac{\rho v D}{\mu} · Re=vDνRe = \dfrac{vD}{\nu}

Flow in Pipe

┌────────────────────────────────┐
→→ →→ →→ v = 1.5 m/s →→ →→ →→
D = 0.05 m
└────────────────────────────────┘

How to Use the Reynolds Number Calculator

  1. 1

    Enter the fluid density

    Enter the fluid density ρ (kg/m³).

  2. 2

    Enter the mean velocity and pipe diameter

    Enter the mean velocity v (m/s) and the pipe diameter D (m).

  3. 3

    Enter the dynamic viscosity

    Enter the dynamic viscosity μ (Pa·s), then click Calculate.

  4. 4

    Read the Reynolds number result

    Read the calculated Reynolds number Re and its flow regime classification (laminar, transitional, or turbulent).

The GATE 2025 Q43 pipe flow problem, solved with this calculator, is below, or browse the full GATE ChemE previous year questions collection.

What Is the Reynolds Number?

The Reynolds number (Re) is a dimensionless quantity that predicts whether fluid flow is laminar (smooth, ordered layers) or turbulent (chaotic, mixing eddies). It represents the ratio of inertial forces to viscous forces in a flowing fluid: Re=ρvDμRe = \dfrac{\rho v D}{\mu}, or equivalently Re=vDνRe = \dfrac{vD}{\nu} using kinematic viscosity ν=μ/ρ\nu = \mu/\rho.

At low Re, viscous forces dominate and damp out disturbances, keeping the flow in smooth parallel layers. At high Re, inertial forces dominate and small disturbances grow into the chaotic eddies of turbulence. For flow in a circular pipe, Re < 2100 is laminar, 2100 ≤ Re ≤ 4000 is transitional, and Re > 4000 is turbulent.

This calculator solves Re=ρvDμRe = \dfrac{\rho v D}{\mu} instantly from fluid density, velocity, pipe diameter, and viscosity, and classifies the resulting flow regime, the same calculation tested in GATE 2025 Q43. For non-circular ducts, replace D with the hydraulic diameter Dh=4A/PD_h = 4A/P.

Workers laying a large-diameter steel pipeline in a trench, 1899
Real pipe, real diameter — the D that goes straight into Re = ρvD/μ, whatever era it's laid in. Anders Beer Wilse, Public domain, via Wikimedia Commons.

Derivation: Non-Dimensionalizing the Flow Equations

The Reynolds number falls out of non-dimensionalizing the Navier-Stokes momentum equation for incompressible flow:

ρ(vt+vv)=p+μ2v\rho\left(\dfrac{\partial v}{\partial t} + v\cdot\nabla v\right) = -\nabla p + \mu \nabla^2 v

Introduce a characteristic length L (say, the pipe diameter) and characteristic velocity U, and define dimensionless variables x=x/L, v=v/U, t=tU/L, p=p/(ρU2)x^{*}=x/L,\ v^{*}=v/U,\ t^{*}=tU/L,\ p^{*}=p/(\rho U^2). Substituting these into the momentum equation and simplifying, every term collapses to order-1 dimensionless quantities except for a single leftover group multiplying the viscous term:

vt+vv=p+μρUL2v\dfrac{\partial v^{*}}{\partial t^{*}} + v^{*}\cdot\nabla^{*} v^{*} = -\nabla^{*} p^{*} + \dfrac{\mu}{\rho U L}\nabla^{*2} v^{*}

That coefficient, μ/(ρUL), is 1/Re. Equivalently, comparing the order of magnitude of the inertial term (~ρU²/L) to the viscous term (~μU/L²) gives the same ratio:

InertialViscous=ρU2/LμU/L2=ρULμ=Re\dfrac{\text{Inertial}}{\text{Viscous}} = \dfrac{\rho U^2/L}{\mu U/L^2} = \dfrac{\rho U L}{\mu} = Re

So Re isn't just a convenient curve-fit number, it falls straight out of the governing equations as a measure of which physics actually dominates the flow. That's why it correctly predicts the switch between two genuinely different flow regimes, not just a rough correlation.

Laminar vs Turbulent Velocity Profiles

Re doesn't just flip a label, it changes the actual shape of the velocity field across the pipe, computed here from the two profiles' governing equations, not drawn freehand.

Laminaru(r) = umax(1 − (r/R)²)Turbulentu(r) ≈ umax(1 − |r|/R)^(1/7)
Laminar flow's sharp parabola versus turbulent flow's flatter, fuller profile — turbulent mixing drags the core velocity down and the near-wall velocity up compared to the laminar case.

When You'll Need It in GATE Chemical Engineering

Re is almost never the final answer on GATE, it's the gatekeeper that tells you which formula to use next:

  • Friction factor and pressure drop, laminar flow uses f=64/Ref = 64/Re (from Hagen-Poiseuille); turbulent flow needs the Moody chart or a Colebrook-type correlation. Using the wrong one is a common trap.
  • Hagen-Poiseuille pressure drop, ΔP=128μLQπD4\Delta P = \dfrac{128\mu L Q}{\pi D^4}is valid only for laminar flow (Re < 2100); GATE regularly checks whether you verify Re before applying it.
  • Heat and mass transfer correlations, many Nusselt and Sherwood number correlations are piecewise in Re, so classifying the flow regime is a prerequisite step before picking the right correlation.
  • Pump and pipe network design, sizing problems often require iterating between an assumed velocity, the resulting Re, and a friction factor until the system is self-consistent.

Three Fully Worked Examples

Example 1: Basic Reynolds Number and Regime

Problem: Water at 25°C flows through a horizontal pipe of inner diameter 2 cm at a mean velocity of 0.05 m/s. Find the Reynolds number and identify the flow regime.

Given: ρ = 1000 kg/m³, v = 0.05 m/s, D = 0.02 m, μ = 1×10⁻³ Pa·s.

Re=ρvDμ=1000×0.05×0.021×103Re = \dfrac{\rho v D}{\mu} = \dfrac{1000 \times 0.05 \times 0.02}{1\times10^{-3}}
Re=1.00.001Re = \dfrac{1.0}{0.001}

Answer: Re = 1000, Laminar flow (Re < 2100), so Hagen-Poiseuille applies.

Example 2: Maximum Velocity for Laminar Flow

Problem: Water at 20°C flows through a 5 cm diameter pipe. What is the maximum mean velocity for which the flow remains laminar?

Given: ρ = 998 kg/m³, μ = 1.002×10⁻³ Pa·s, D = 0.05 m, Re_critical = 2100.

Approach: Rearrange Re=ρvD/μRe = \rho v D/\mu to solve for v at the laminar limit.

vmax=ReμρD=2100×1.002×103998×0.05v_{max} = \dfrac{Re \cdot \mu}{\rho \cdot D} = \dfrac{2100 \times 1.002\times10^{-3}}{998 \times 0.05}
vmax=2.104249.9v_{max} = \dfrac{2.1042}{49.9}

Answer: v_max ≈ 0.0422 m/s (42.2 mm/s), above this, the flow transitions out of the laminar regime.

Example 3: Verifying Laminar Flow, Then Finding Pressure Drop

Problem: Oil flows through a 3 cm diameter, 10 m long pipe at 0.5 m/s. Verify the flow is laminar, then find the pressure drop.

Given: ρ = 880 kg/m³, μ = 0.29 Pa·s, D = 0.03 m, L = 10 m, v = 0.5 m/s.

Re=ρvDμ=880×0.5×0.030.29=13.20.2945.5 (2100, laminar confirmed)Re = \dfrac{\rho v D}{\mu} = \dfrac{880 \times 0.5 \times 0.03}{0.29} = \dfrac{13.2}{0.29} \approx 45.5\ (\ll 2100,\ \text{laminar confirmed})
Hagen-Poiseuille (mean-velocity form): ΔP=32μLvD2\Delta P = \dfrac{32\mu L v}{D^2}
ΔP=32×0.29×10×0.5(0.03)2=46.40.0009\Delta P = \dfrac{32 \times 0.29 \times 10 \times 0.5}{(0.03)^2} = \dfrac{46.4}{0.0009}

Answer: Re ≈ 45.5 (laminar); ΔP ≈ 51.6 kPa (0.516 bar)

Common Mistakes GATE Students Make

  • Using the pipe-flow threshold for a different geometry.Re = 2100–4000 applies specifically to flow inside a circular pipe. External flow over a flat plate, packed beds, and stirred tanks each have their own modified Reynolds number and critical value , applying the pipe-flow numbers there is a common conceptual slip.
  • Unit mismatches, especially diameter. Forgetting to convert a diameter given in cm or mm to meters before plugging into the SI-based Re=ρvD/μRe = \rho v D/\mu formula throws the answer off by a factor of 100 or 1000.
  • Applying Hagen-Poiseuille without checking Re first. ΔP=128μLQπD4\Delta P = \dfrac{128\mu L Q}{\pi D^4}assumes a parabolic, laminar velocity profile , using it when Re > 2100 gives a confidently wrong answer with no warning.
  • Mixing up dynamic and kinematic viscosity. μ (Pa·s) and ν = μ/ρ (m²/s) are related but not interchangeable, plugging one into the formula meant for the other is off by a factor of density.

Key Takeaways

  • Re=ρvD/μRe = \rho v D/\mu compares inertial to viscous forces and falls directly out of non-dimensionalizing the Navier-Stokes equation.
  • For pipe flow: Re < 2100 laminar, 2100–4000 transitional, Re > 4000 turbulent, but this range is specific to pipes.
  • Always check Re before choosing a friction factor correlation or applying Hagen-Poiseuille.
  • Use the hydraulic diameter Dh=4A/PD_h = 4A/P for non-circular ducts.
  • Keep dynamic viscosity μ and kinematic viscosity ν straight, they differ by a factor of density.

Common Fluid Properties for GATE

FluidT (°C)ρ (kg/m³)μ (Pa·s)ν (m²/s)
Water209981.002×10⁻³1.004×10⁻⁶
Water259978.90×10⁻⁴8.93×10⁻⁷
Water609834.67×10⁻⁴4.75×10⁻⁷
Air201.2041.81×10⁻⁵1.51×10⁻⁵
Benzene258746.00×10⁻⁴6.87×10⁻⁷

Source: Perry's Chemical Engineers' Handbook, 9th ed.

Frequently Asked Questions

Re < 2100 for flow in a circular pipe. That number isn't a hard physical wall, real pipes have run laminar a bit past it and turbulent a bit before it depending on disturbances and pipe entry conditions, but it's the design threshold worth treating as firm unless you have a specific reason not to.

Further reading: Reynolds Number in Real Pipelines: Laminar vs Turbulent in Practice

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