Chapter 3: Thermal Resistance and Heat Transfer

Objectives

• Learn the analogy between thermal resistance and electrical resistance
• Analyze heat transfer from extended surfaces (fins)

Thermal Resistance Concept

• Conduction: Analysis of heat transfer through a plane wall.
• Temperature Distribution:
  • Left surface: 20°C
  • Mid plane: 11°C
  • Right surface: 3°C
• Temperature Gradient:
  • No gradient in Y and Z directions (no heat transfer in Y and Z)
  • Significant gradient in X direction (heat flows from higher to lower temperatures)
  • Steady State Condition: Temperature at a given location does not change with time.
  • One-dimensional Heat Conduction: Heat flows only in X direction.

Analogy between Electrical and Thermal Resistance

• Ohm’s Law: Electric current (I) is proportional to voltage difference (ΔV) and inversely proportional to electrical resistance (R).
• Heat Conduction (Fourier's Law):
  • Q = kA ΔT / L (k - thermal conductivity, A - area, L - distance)
  • Thermal Resistance (R_th): R_th = L / kA
  • Comparison: Electrical current (I) ~ Heat transfer rate (Q)
  • Potential Difference (ΔV) ~ Temperature Difference (ΔT)
  • Electrical Resistance (R) ~ Thermal Resistance (R_th)

Types of Thermal Resistance

• Conduction: R_th (conduction) = L / kA
• Convection: R_th (convection) = 1 / hA (h - convection heat transfer coefficient)
• Example: Heat transfer from surface to fluid via convection

Thermal Resistance Network (Circuit)

• Analysis with thermal resistance circuits to simplify heat conduction problems.
• Single-layer Wall Example: Thermal resistance network for heat transfer through a wall surrounded by fluid.
  • Temperature nodes: T∞1, T1, T2, T∞2
  • Resistances in series: Convection, Conduction, Convection
  • Heat Transfer Rate (Q): Convection into wall = Conduction through wall = Convection from wall
  • Total Resistance (R_total): Sum of individual resistances

Multi-layer Heat Transfer

• Composite Medium: Multiple thermal resistances in series/parallel (e.g., two layers).
• Total Thermal Resistance: Sum of all resistances
• Temperature Calculation: Using thermal resistance network to find interface temperatures (T1, T2, T3, T4)

Example Problem: Double Pane Glass

• Objective: Determine interface temperatures (T1, T2, T3, T4)
• Assumptions: One-dimensional steady-state heat conduction with no heat generation.
• Setup: Thermal resistance network with given parameters (convective coefficients, thermal conductivities, dimensions)
• Calculation Steps:
  1. Draw thermal resistance network
  2. Calculate individual thermal resistances
  3. Find total thermal resistance (R_total)
  4. Determine heat transfer rate (Q)
  5. Calculate temperatures at interfaces (T1, T2, T3, T4)

Example Solution Details

• Given:
  • Left fluid temperature (T∞1): 20°C, h1 = 10 W/m²·°C
  • Thermal conductivities: Glass (k_G = 0.78 W/m·°C), Air (k_A = 0.026 W/m·°C)
  • Thicknesses: Glass (L_G = 4 mm), Air (L_A = 10 mm)
  • Right fluid temperature (T∞2): -10°C, h2 = 40 W/m²·°C
  • Area (A): 1.2 m²
• Thermal Resistance Calculations:
  • R_conv1 (left fluid): 1 / (h1 * A)
  • R_cond1 (glass): L_G / (k_G * A)
  • R_cond2 (air): L_A / (k_A * A)
  • R_conv2 (right fluid): 1 / (h2 * A)
  • R_total: Sum of all thermal resistances
• Heat Transfer Rate (Q): Calculate using T∞1 and T∞2 with R_total
• Interface Temperatures (T1, T2, T3, T4): Using Q and thermal resistances

Key Conditions for Thermal Resistance Application

• Steady-state heat conduction
• One-dimensional heat conduction
• No heat generation

Summary

• Thermal resistance simplifies analysis of heat conduction problems.
• Analogous to electrical resistance, it allows for easier calculation of temperature distributions and heat transfer rates in a system.
• Application extends to both conduction and convection scenarios.