Heat Transfer and Lumped Capacitance Method Lecture

Introduction

• In heat transfer problems, we often need to determine cooling time or temperature after a period of time.
• Temperature is a function of both time and position, making the problem complex.

Methods to Solve Heat Transfer Problems

1. Lumped Capacitance Method
2. Exact Solution
   • Solving heat equation (second order in space, first order in time).
   • Beyond scope for this lecture.
3. Numerical Methods

Lumped Capacitance Method

Concept

• Simplification of heat conduction problems by assuming uniform temperature throughout the object.
• Typically, temperature varies with position (x, y, z) and time.
• Assumption reduces the complexity significantly at a small expense of accuracy.

Assumptions and Example

• Assumes negligible temperature difference between object's surface and center.
• Example: A small copper ball with high thermal conductivity.
• Temperature is a function of time only, not position.

Mathematical Formulation

• Object initially at uniform temperature T₀ is immersed in a fluid at temperature T∞.
• Heat transfer occurs between object and fluid.
• Energy balance: rate of change of internal energy = rate of convective heat transfer.

Specific Heat and Energy Equations

• Specific Heat (C) is the energy required to raise the temperature per unit mass per degree.
• Energy lost by the object = convective heat transfer from surface to surrounding fluid.

Derivation and Equations

• Simplification: θ = T - T∞
• Integration from initial to final temperature results in:
  • θ/θ₀ = exp(-t/τ)
  • Where τ (Tau) = (ρVC) / (hA) is the time constant

Using the Equations

• Determining object temperature at any time t, or the time required to reach a certain temperature.

Time Constant and Its Importance

• Time constant (τ) signifies how quickly an object responds to temperature changes.
• Small τ: Quick response
• Large τ: Slow response

Heating and Cooling Graphs

• Heating: Temperature increases to T∞.
• Cooling: Temperature decreases to T∞.
• Dimensionless temperature approaches zero over time.

Practical Application

• Identify valid use of lumped capacitance model via Biot number (Bi).

Validating the Lumped Capacitance Model

Biot Number (Bi)

• Defined as: Bi = (hL)/k
• Ratio of conduction resistance to convection resistance.
• Criteria: Bi ≤ 0.1 suggests valid use of lumped capacitance method.

Factors Influencing the Biot Number

• Small objects: Smaller characteristic length (L).
• High thermal conductivity: Solid's capacity to conduct heat internally.
• Low convection coefficient: Surrounding fluid's ability to transfer heat.

Example Analysis

• Metal plate submerged in a fluid at 800°C.
• Using given properties to check if Bi ≤ 0.1.
• Calculation steps demonstrate the lumped capacitance model's applicability.

Practical Implications

• Use the derived equation to solve practical problems.
• Depending on initial and surrounding temperatures, calculate either time or temperature at given conditions.

Conclusion

• Lumped capacitance method offers a simplified approach to complex heat transfer problems.
• Important to verify assumptions using Biot number for accurate results.

Thank you.