BIOT
Excel Usage
=BIOT(h, L, k)h(float, required): Heat transfer coefficient [W/m^2/K]L(float, required): Characteristic length [m]k(float, required): Thermal conductivity within the object [W/m/K]
Returns (float): Biot number [-]
Examples
Example 1: Biot number with h=1000, L=1.2, k=300
Inputs:
| h | L | k |
|---|---|---|
| 1000 | 1.2 | 300 |
Excel formula:
=BIOT(1000, 1.2, 300)Expected output:
4
Example 2: Biot number with h=10000, L=0.01, k=4000
Inputs:
| h | L | k |
|---|---|---|
| 10000 | 0.01 | 4000 |
Excel formula:
=BIOT(10000, 0.01, 4000)Expected output:
0.025
Example 3: Small Biot number (internal heat transfer dominant)
Inputs:
| h | L | k |
|---|---|---|
| 100 | 0.1 | 50 |
Excel formula:
=BIOT(100, 0.1, 50)Expected output:
0.2
Example 4: Large Biot number (surface heat transfer dominant)
Inputs:
| h | L | k |
|---|---|---|
| 5000 | 0.5 | 10 |
Excel formula:
=BIOT(5000, 0.5, 10)Expected output:
250
Python Code
from fluids.core import Biot as fluids_Biot
def biot(h, L, k):
"""
Calculate the Biot number for heat transfer.
See: https://fluids.readthedocs.io/fluids.core.html#fluids.core.Biot
This example function is provided as-is without any representation of accuracy.
Args:
h (float): Heat transfer coefficient [W/m^2/K]
L (float): Characteristic length [m]
k (float): Thermal conductivity within the object [W/m/K]
Returns:
float: Biot number [-]
"""
try:
h = float(h)
L = float(L)
k = float(k)
if h < 0 or L < 0 or k <= 0:
return "Error: h and L must be non-negative, k must be positive"
result = fluids_Biot(h, L, k)
return float(result)
except (TypeError, ValueError) as e:
return f"Error: Invalid input - {str(e)}"
except Exception as e:
return f"Error: {str(e)}"