ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-05-30 00:14
Lumped System Analysis
Analyze transient heat conduction using the lumped capacitance method for objects with uniform temperature (Bi < 0.1).
Calculation Domain Inputs
Specify geometry parameters, material presets, and boundary conditions to calculate the transient thermal history of the lumped system:
- Geometry: Choose between a Sphere, Cylinder, Plane Wall, or Cube.
- Properties ($\rho, c_p, k$): Density, specific heat, and thermal conductivity.
- Convection ($h$): Heat transfer coefficient of the surrounding fluid.
- Temperatures ($T_i, T_\infty, T_{target}$): Initial, ambient fluid, and target temperature thresholds.
Configuration
Lumped System Analysis:
Biot Number: $Bi = \frac{h L_c}{k} \lt 0.1$
Time Constant: $\tau = \frac{\rho c_p V}{h A_s}$
Temperature: $T(t) = T_\infty + (T_i - T_\infty) e^{-t/\tau}$
• $L_c = V/A_s$ (characteristic length)
• Valid when $Bi \lt 0.1$ (uniform $T$ inside solid)
Biot Number: $Bi = \frac{h L_c}{k} \lt 0.1$
Time Constant: $\tau = \frac{\rho c_p V}{h A_s}$
Temperature: $T(t) = T_\infty + (T_i - T_\infty) e^{-t/\tau}$
• $L_c = V/A_s$ (characteristic length)
• Valid when $Bi \lt 0.1$ (uniform $T$ inside solid)
Results & Visualization
Results and visualizations will appear here after calculation.
ℹ️ About Lumped System Analysis
The lumped capacitance method assumes the temperature within an object is spatially uniform at any given time.
When is it valid?
• When the Biot number (Bi = hLc/k) is less than 0.1
• This means internal conduction is much faster than surface convection
Applications:
• Quenching of small metal parts
• Thermocouple response time estimation
• Food processing temperature calculations
• Electronic component thermal analysis
The lumped capacitance method assumes the temperature within an object is spatially uniform at any given time.
When is it valid?
• When the Biot number (Bi = hLc/k) is less than 0.1
• This means internal conduction is much faster than surface convection
Applications:
• Quenching of small metal parts
• Thermocouple response time estimation
• Food processing temperature calculations
• Electronic component thermal analysis
Calculation Methodology
Mathematical Model & Theory
Lumped system analysis simplifies transient heat transfer by assuming that the solid is at a uniform temperature at any instant during the process. This is valid when internal conduction resistance is much smaller than external convection resistance:
$$Bi = \frac{h L_c}{k} < 0.1, \quad L_c = \frac{V}{A_s}$$
$$T(t) = T_{\infty} + (T_i - T_{\infty}) e^{-t/\tau}, \quad \tau = \frac{\rho V C_p}{h A_s}$$
Variable Definitions & Units:
- $Bi$: Biot number [-]
- $L_c$: Characteristic length of the solid [m]
- $V$: Volume [m³], $A_s$: Surface area [m²]
- $T(t)$: Temperature at time $t$ [°C]
- $\tau$: Time constant of the system [s]
Assumptions & Boundary Conditions:
- Uniform temperature distribution inside the solid at any instant ($Bi < 0.1$).
- Constant material density $\rho$, specific heat $C_p$, and thermal conductivity $k$.
- Uniform and constant convection heat transfer coefficient $h$ over the entire boundary surface.
- Constant ambient fluid temperature $T_\infty$.
- Uniform initial temperature $T_i$ inside the solid at time $t = 0$.
- No internal heat generation.
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer.
- Çengel, Y. A. (2015). Heat and Mass Transfer: Fundamentals and Applications.
Worked Engineering Example
Problem Statement:
A copper sphere ($D = 10$ mm, $\rho = 8933$ kg/m³, $C_p = 385$ J/kg·K, $k = 401$ W/m·K) initially at 150°C is quenched in air at 25°C with $h = 20$ W/m²·K. Calculate the Biot number and the time required for the center to reach 50°C.
Step-by-step Solution:
1. Calculate characteristic length $L_c$ for a sphere:
$$L_c = \frac{V}{A_s} = \frac{D}{6} = \frac{0.01}{6} = 0.001667 \text{ m}$$ 2. Calculate Biot number:
$$Bi = \frac{h L_c}{k} = \frac{20 \times 0.001667}{401} = 8.31 \times 10^{-5} < 0.1 \quad \text{(Lumped system is valid)}$$ 3. Calculate time constant $\tau$:
$$\tau = \frac{\rho C_p L_c}{h} = \frac{8933 \times 385 \times 0.001667}{20} = 286.65 \text{ s}$$ 4. Calculate cooling time $t$:
$$\frac{T(t) - T_{\infty}}{T_i - T_{\infty}} = e^{-t/\tau} \Rightarrow \frac{50 - 25}{150 - 25} = 0.2$$ $$t = -\tau \ln(0.2) = -286.65 \times (-1.6094) = 461.35 \text{ s}$$
Final Result:
Biot number is $8.31 \times 10^{-5}$ and time to cool is 461.4 s.
A copper sphere ($D = 10$ mm, $\rho = 8933$ kg/m³, $C_p = 385$ J/kg·K, $k = 401$ W/m·K) initially at 150°C is quenched in air at 25°C with $h = 20$ W/m²·K. Calculate the Biot number and the time required for the center to reach 50°C.
Step-by-step Solution:
1. Calculate characteristic length $L_c$ for a sphere:
$$L_c = \frac{V}{A_s} = \frac{D}{6} = \frac{0.01}{6} = 0.001667 \text{ m}$$ 2. Calculate Biot number:
$$Bi = \frac{h L_c}{k} = \frac{20 \times 0.001667}{401} = 8.31 \times 10^{-5} < 0.1 \quad \text{(Lumped system is valid)}$$ 3. Calculate time constant $\tau$:
$$\tau = \frac{\rho C_p L_c}{h} = \frac{8933 \times 385 \times 0.001667}{20} = 286.65 \text{ s}$$ 4. Calculate cooling time $t$:
$$\frac{T(t) - T_{\infty}}{T_i - T_{\infty}} = e^{-t/\tau} \Rightarrow \frac{50 - 25}{150 - 25} = 0.2$$ $$t = -\tau \ln(0.2) = -286.65 \times (-1.6094) = 461.35 \text{ s}$$
Final Result:
Biot number is $8.31 \times 10^{-5}$ and time to cool is 461.4 s.