⏱️ Lumped System Analysis
Analyze transient heat conduction using the lumped capacitance method for objects with uniform temperature (Bi < 0.1).
📝 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]
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.