Table of Contents
What is Biot Number?
Biot Number (Bi) is a dimensionless quantity used in heat transfer calculations. It is defined as internal thermal resistance within a body (conduction) to the external thermal resistance due to convection.
The Biot number provides a measure of the temperature drop in the solid relative to the temperature difference between the surface and the fluid.
Mathematically, it is defined as,
\[Bi = \frac{\text{Internal Thermal Resistance}}{\text{External convective Resistance}}\]
\[Bi = \frac{R_{\text{Conduction}}}{R_{\text{convection}}} = \frac{hL_c}{k}\]
where:
- \( h \) is the convective heat transfer coefficient (W/m2·K)
- \(L_c \) is the characteristic length (m)
- \( k \) is the thermal conductivity of the material (W/m·K)
Related: Heat Transfer through Conduction Calculator – Fourier’s law
Biot Number Calculator
This Calculator / web tool helps user to the Biot number. User inputs the values for the convective heat transfer coefficient \(h\), the characteristic length \(L_c \), and the thermal conductivity \(k\).
Related: Hydraulic Diameter Calculator for Circular and Non-Circular cross-section
Related: Inverse Square Law for Radiation
Physical Significance of Biot Number
Bi Number is significant in Heat Transfer Calculations, it compares internal thermal resistance within a solid to the external thermal resistance due to convection.
Consider a steady-state temperature distribution in plane wall with surface convection as shown in figure below.
Low Biot Number (\( Bi \ll 1 \))
The resistance to conduction within the solid is much less than the resistance to convection. The temperature within the solid is relatively uniform because heat conduction inside the solid is very effective compared to the convective heat transfer at the surface.
High Biot Number (\( Bi \gg 1 \))
The resistance due to conduction is much larger than the resistance due to convection. The temperature difference across the solid is much larger than that between the surface and the fluid.
The Bi number is important in determining the mechanism of heat transfer and helps in simplifying the analysis of thermal systems.
Example: Consider a small hot metal sphere, when immersed in a pool of water, conduction within a sphere and convection at the sphere surface will be influenced by the water.
If the Bi number is low, the temperature sphere is assumed to be uniform. If the Bi number is high it means that temperature gradient exists.
Python Code for Biot Number calculation
This Python code helps user to find the Bi number for different materials and convective heat transfer coefficient of 100 W/m²·K. It calculates the values of Bi at different Characteristic length for Aluminum, Copper and Steel.
Note: This Python code solves the specified problem. Users can copy the code and run it in a suitable Python environment. By adjusting the input parameters, and observe how the output changes accordingly.
import numpy as np
import matplotlib.pyplot as plt
def calculate_biot_number(h, L_c, k):
"""
Calculate the Biot_number for a plane wall.
Parameters:
h (float): Convective heat transfer coefficient (W/m²·K)
L_c (float): Characteristic length (m)
k (float): Thermal conductivity of the material (W/m·K)
Returns:
float: The Biot number
"""
return (h * L_c) / k
# Example data
h = 100 # W/m²·K, convective heat transfer coefficient
# Thermal conductivities for different materials
thermal_conductivities = {
'Aluminum': 237, # W/m·K
'Copper': 398, # W/m·K
'Steel': 50 # W/m·K
}
# Range of characteristic lengths
L_c_values = np.linspace(0.01, 0.1, 100) # Characteristic lengths from 0.01m to 0.1m
# Plotting
plt.figure(figsize=(12, 6))
for material, k in thermal_conductivities.items():
biot_numbers = calculate_biot_number(h, L_c_values, k)
plt.plot(L_c_values, biot_numbers, label=f'{material} (k = {k} W/m·K)')
plt.xlabel('Characteristic Length ($L_c$) [m]')
plt.ylabel('Biot Number (Bi)')
plt.title('Biot Number as a Function of Characteristic Length for Different Materials for plane wall')
plt.grid(True)
plt.legend()
plt.show()
# Display Biot numbers for specific characteristic lengths
for material, k in thermal_conductivities.items():
for example_L_c in [0.02, 0.05, 0.08]:
biot_number_example = calculate_biot_number(h, example_L_c, k)
print(f"Biot Number for {material} with L_c = {example_L_c} m: {biot_number_example:.4f}")
Output:
Resources
- “Fundamentals of Heat and Mass Transfer” by Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera, and David P. DeWitt.
- “Heat Transfer Book” by David W. Hahn.
- “Conduction of Heat in Solids” by H.S. Carslaw and J.C. Jaeger.
- “Introduction to Heat Transfer” by Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, and Adrienne S. Lavine.
Disclaimer: The Solver provided here is for educational purposes. While efforts ensure accuracy, results may not always reflect real-world scenarios. Verify results with other sources and consult professionals for critical applications. Contact us for any suggestions or corrections.