💨 Forced Convection — Flow Over a Flat Plate
Analyze laminar, turbulent, and mixed boundary layer heat transfer with local and average Nusselt number correlations.
📐 Calculation Domain & Boundary Layer Growth
📝 Configuration
Key Correlations:
Laminar (Re_x < 5×10⁵):
Nu_x = 0.332 Re_x^0.5 Pr^(1/3)
δ = 5x / √Re_x
Turbulent (Re_x > 5×10⁵):
Nu_x = 0.0296 Re_x^0.8 Pr^(1/3)
δ = 0.37x / Re_x^0.2
Average (mixed):
Nu_L = (0.037 Re_L^0.8 - 871) Pr^(1/3)
Laminar (Re_x < 5×10⁵):
Nu_x = 0.332 Re_x^0.5 Pr^(1/3)
δ = 5x / √Re_x
Turbulent (Re_x > 5×10⁵):
Nu_x = 0.0296 Re_x^0.8 Pr^(1/3)
δ = 0.37x / Re_x^0.2
Average (mixed):
Nu_L = (0.037 Re_L^0.8 - 871) Pr^(1/3)
📊 Results & Visualization
Configure the inputs and click Calculate to see results.
ℹ️ About Flat Plate Convection
Flow over a flat plate is a fundamental problem in convective heat transfer. The boundary layer starts as laminar and transitions to turbulent at a critical Reynolds number (Re_crit ≈ 5×10⁵).
Applications:
• Solar collector plates
• Electronic cooling
• Aircraft wing surfaces
• Industrial drying processes
Flow over a flat plate is a fundamental problem in convective heat transfer. The boundary layer starts as laminar and transitions to turbulent at a critical Reynolds number (Re_crit ≈ 5×10⁵).
Applications:
• Solar collector plates
• Electronic cooling
• Aircraft wing surfaces
• Industrial drying processes
📘 Calculation Methodology
Mathematical Model & Theory
Convective heat transfer over a flat plate is analyzed using empirical correlations that relate the local and average Nusselt numbers ($Nu_L$) to the Reynolds ($Re_L$) and Prandtl ($Pr$) numbers. The transition from laminar to turbulent flow is assumed to occur at the critical Reynolds number ($Re_{crit} = 5 \times 10^5$).
$$Re_L = \frac{\rho U L}{\mu}$$
$$\text{Laminar (Re < } 5 \times 10^5\text{): } Nu_L = 0.664 Re_L^{1/2} Pr^{1/3}$$
$$\text{Turbulent (Re } \ge 5 \times 10^5\text{): } Nu_L = (0.037 Re_L^{4/5} - 871) Pr^{1/3}$$
$$h_c = \frac{Nu_L \cdot k}{L}, \quad Q = h_c A (T_s - T_{\infty})$$
Worked Engineering Example
Problem Statement:
Air at 20°C flows over a 1 m long, 0.5 m wide flat plate at 5 m/s. The plate surface temperature is kept at 80°C. Air properties: $\rho = 1.12$ kg/m³, $\mu = 1.9 \times 10^{-5}$ Pa·s, $k = 0.027$ W/m·K, $Pr = 0.71$. Calculate the heat transfer rate.
Step-by-step Solution:
1. Calculate plate Reynolds number:
$$Re_L = \frac{1.12 \times 5 \times 1.0}{1.9 \times 10^{-5}} = 2.947 \times 10^5 < 5 \times 10^5 \quad \text{(Laminar flow)}$$ 2. Calculate average Nusselt number:
$$Nu_L = 0.664 \times (2.947 \times 10^5)^{1/2} \times (0.71)^{1/3} = 0.664 \times 542.89 \times 0.892 = 321.57$$ 3. Calculate convection coefficient $h_c$:
$$h_c = \frac{321.57 \times 0.027}{1.0} = 8.68 \text{ W/m}^2\text{K}$$ 4. Calculate heat transfer rate ($A = 0.5$ m²):
$$Q = h_c A (T_s - T_{\infty}) = 8.68 \times 0.5 \times (80 - 20) = 260.47 \text{ W}$$
Final Result:
The convective heat transfer rate is 260.5 W.
Air at 20°C flows over a 1 m long, 0.5 m wide flat plate at 5 m/s. The plate surface temperature is kept at 80°C. Air properties: $\rho = 1.12$ kg/m³, $\mu = 1.9 \times 10^{-5}$ Pa·s, $k = 0.027$ W/m·K, $Pr = 0.71$. Calculate the heat transfer rate.
Step-by-step Solution:
1. Calculate plate Reynolds number:
$$Re_L = \frac{1.12 \times 5 \times 1.0}{1.9 \times 10^{-5}} = 2.947 \times 10^5 < 5 \times 10^5 \quad \text{(Laminar flow)}$$ 2. Calculate average Nusselt number:
$$Nu_L = 0.664 \times (2.947 \times 10^5)^{1/2} \times (0.71)^{1/3} = 0.664 \times 542.89 \times 0.892 = 321.57$$ 3. Calculate convection coefficient $h_c$:
$$h_c = \frac{321.57 \times 0.027}{1.0} = 8.68 \text{ W/m}^2\text{K}$$ 4. Calculate heat transfer rate ($A = 0.5$ m²):
$$Q = h_c A (T_s - T_{\infty}) = 8.68 \times 0.5 \times (80 - 20) = 260.47 \text{ W}$$
Final Result:
The convective heat transfer rate is 260.5 W.