🔄 Internal Flow — Circular Tubes
Analyze forced convection inside tubes with Dittus-Boelter and Gnielinski correlations, pressure drop, and outlet temperature.
📐 Calculation Domain & Boundary Layer Growth
📝 Configuration
Key Correlations:
Laminar (Re < 2300):
Nu = 3.66 (constant T_s)
Nu = 4.36 (constant q")
Turbulent:
Dittus-Boelter: Nu = 0.023 Re^0.8 Pr^n
Gnielinski: Nu = (f/8)(Re-1000)Pr / [1+12.7√(f/8)(Pr²ᐟ³-1)]
Pressure Drop:
ΔP = f(L/D)(ρU²/2)
Laminar (Re < 2300):
Nu = 3.66 (constant T_s)
Nu = 4.36 (constant q")
Turbulent:
Dittus-Boelter: Nu = 0.023 Re^0.8 Pr^n
Gnielinski: Nu = (f/8)(Re-1000)Pr / [1+12.7√(f/8)(Pr²ᐟ³-1)]
Pressure Drop:
ΔP = f(L/D)(ρU²/2)
📊 Results & Visualization
Configure the inputs and click Calculate to see results.
ℹ️ About Internal Flow Convection
Flow inside tubes is critical in many thermal systems: heat exchangers, HVAC, industrial piping. The flow transitions from laminar to turbulent at Re ≈ 2300.
Applications:
• Heat exchanger tube-side design
• Radiator and cooling system design
• Industrial process heating/cooling
• HVAC duct systems
Flow inside tubes is critical in many thermal systems: heat exchangers, HVAC, industrial piping. The flow transitions from laminar to turbulent at Re ≈ 2300.
Applications:
• Heat exchanger tube-side design
• Radiator and cooling system design
• Industrial process heating/cooling
• HVAC duct systems
📘 Calculation Methodology
Mathematical Model & Theory
For forced convection inside tubes, flow is laminar for $Re_D < 2300$ and fully turbulent for $Re_D > 10,000$. The Nusselt number is determined by Dittus-Boelter correlation for turbulent flow:
$$Re_D = \frac{\rho V D}{\mu} = \frac{4\dot{m}}{\pi D \mu}$$
$$\text{Turbulent (Dittus-Boelter): } Nu_D = 0.023 Re_D^{4/5} Pr^n$$
$$\text{Where } n=0.4 \text{ for heating } (T_s > T_m) \text{ and } n=0.3 \text{ for cooling } (T_s < T_m)$$
Worked Engineering Example
Problem Statement:
Water flows inside a 25 mm diameter tube ($k_f = 0.64$ W/m·K, $Pr = 3.5$, $\mu = 5.4 \times 10^{-4}$ Pa·s) at a mass flow rate of 0.5 kg/s. The tube wall is heated. Calculate the heat transfer coefficient.
Step-by-step Solution:
1. Calculate Reynolds number:
$$Re_D = \frac{4 \times 0.5}{\pi \times 0.025 \times 5.4 \times 10^{-4}} = 47,157 > 10,000 \quad \text{(Turbulent flow)}$$ 2. Calculate Nusselt number (heating: $n=0.4$):
$$Nu_D = 0.023 \times (47,157)^{0.8} \times (3.5)^{0.4} = 0.023 \times 5519.8 \times 1.65 = 209.4$$ 3. Calculate convection coefficient $h_c$:
$$h_c = \frac{Nu_D \cdot k_f}{D} = \frac{209.4 \times 0.64}{0.025} = 5360.6 \text{ W/m}^2\text{K}$$
Final Result:
The heat transfer coefficient is 5361 W/m²·K.
Water flows inside a 25 mm diameter tube ($k_f = 0.64$ W/m·K, $Pr = 3.5$, $\mu = 5.4 \times 10^{-4}$ Pa·s) at a mass flow rate of 0.5 kg/s. The tube wall is heated. Calculate the heat transfer coefficient.
Step-by-step Solution:
1. Calculate Reynolds number:
$$Re_D = \frac{4 \times 0.5}{\pi \times 0.025 \times 5.4 \times 10^{-4}} = 47,157 > 10,000 \quad \text{(Turbulent flow)}$$ 2. Calculate Nusselt number (heating: $n=0.4$):
$$Nu_D = 0.023 \times (47,157)^{0.8} \times (3.5)^{0.4} = 0.023 \times 5519.8 \times 1.65 = 209.4$$ 3. Calculate convection coefficient $h_c$:
$$h_c = \frac{Nu_D \cdot k_f}{D} = \frac{209.4 \times 0.64}{0.025} = 5360.6 \text{ W/m}^2\text{K}$$
Final Result:
The heat transfer coefficient is 5361 W/m²·K.