Heat Transfer in Extended Surfaces

Why it matters

Extended surfaces, commonly known as fins, are crucial in enhancing heat transfer from surfaces to the surrounding environment. They are widely used in applications such as radiators, heat exchangers, and electronic cooling systems to improve efficiency and prevent overheating.

Key ideas

• Extended Surfaces (Fins): Structures that increase the surface area available for heat transfer, thereby enhancing the rate of heat dissipation.
• Types of Fins:
  • Straight Fins: Extend from a plane base; they may have a uniform section or a tapered profile.
  • Annular Fins: Circular fins used around cylindrical surfaces.
  • Pin Fins: Cylindrical rods extending from a surface.
• Heat Transfer Mechanism: Fins conduct heat from the base and dissipate it to the surrounding fluid through convection.
• Efficiency and Effectiveness:
  • Efficiency: Ratio of actual heat transfer to the maximum possible heat transfer if the entire fin were at the base temperature.
  • Effectiveness: Ratio of heat transfer with the fin to heat transfer without the fin.

Formulas

• Q = η·h·A_f·(T_b - T_∞)
  • Q: Heat transfer rate (W)
  • h: Convective heat transfer coefficient (W/m²·K)
  • A_f: Convecting fin surface area (m²)
  • T_b: Base temperature (K)
  • T_∞: Ambient temperature (K)
• η = (tanh(m·L))/(m·L)
  • η: Fin efficiency
  • m: Fin parameter, m = sqrt(h·P/(k·A_c)) (m⁻¹)
  • L: Length of the fin (m)
  • P: Perimeter of the fin (m)
  • k: Thermal conductivity of the fin material (W/m·K)
  • A_c: Cross-sectional area of the fin (m²)

The tanh efficiency below applies to a straight uniform-section fin with an adiabatic tip, constant k and h, steady one-dimensional conduction, no generation and negligible radiation. Then A_f = PL, Q = sqrt(hPkA_c)(T_b-T_∞)tanh(mL). A convecting tip requires a different boundary condition or an appropriate corrected-length approximation. Fin effectiveness compares fin heat flow with hA_c(T_b-T_∞), the heat that would leave the covered base area.

Worked example

Given: A straight uniform rectangular fin with an insulated tip with length L = 0.1 m, width w = 0.02 m, thickness t = 0.005 m, thermal conductivity k = 200 W/m·K, base temperature T_b = 373 K, ambient temperature T_∞ = 293 K, and convective heat transfer coefficient h = 25 W/m²·K.

1. Calculate the perimeter (P) and cross-sectional area (A_c):

   • P = 2·(w + t) = 2·(0.02 + 0.005) = 0.05 m
   • A_c = w·t = 0.02·0.005 = 0.0001 m²

2. Calculate the fin parameter (m):

   • m = sqrt(h·P/(k·A_c)) = sqrt(25·0.05/(200·0.0001)) = 7.90569 m⁻¹

3. Calculate the fin efficiency (η):

   • η = (tanh(m·L))/(m·L) = (tanh(11.18·0.1))/(11.18·0.1) = 0.83324

4. Calculate the heat transfer rate (Q):

   • A_f = PL = 0.05(0.1) = 0.005 m².
   • Q = 0.83324·25·0.005·(373-293) = 8.332 W

Final Answer: 8.332 W

Common mistakes

• Ignoring Fin Efficiency: Assuming the entire fin is at the base temperature, leading to overestimation of heat transfer.
• Incorrect Units: Not converting all measurements to SI units before calculations.
• Neglecting Convection: Forgetting that fins primarily enhance convective heat transfer.

For GATE ME

Questions often involve calculating the efficiency and effectiveness of fins, determining heat transfer rates, and analyzing the impact of different fin geometries. Practice problems involving various fin types and boundary conditions.

Quick check

1. What is the primary purpose of using fins in heat transfer?
2. How does fin efficiency differ from fin effectiveness?
3. What happens to the heat transfer rate if the fin length is doubled?

Answers: 1. To increase heat dissipation by increasing surface area. 2. Efficiency measures actual vs. maximum heat transfer; effectiveness compares fin vs. no fin. 3. It generally increases, but efficiency may decrease due to increased resistance.