Extended Surfaces (Fins)

Abstract

Fins are widely used in various engineering equipments for increasing heat transfer rate from a surface. In this chapter, mathematical treatments of various types of uniform cross-section, triangular , trapezoidal and annular fins have been presented to determine the temperature variation along the fin length and heat transfer rate from the fin surface assuming one-dimensional steady-state heat flow condition. Fin performance parameters (effectiveness and efficiency) have been defined and discussed. Condition for enhancement of heat transfer from a fin has been deduced. A number of illustrative examples have been given.

Appendices

Review Questions

• 3.1 Write a short note on heat transfer enhancement from a surface.

• 3.2 Why fins are used? Discuss various types of fins used in engineering applications.

• 3.3 Derive the expression for temperature distribution and total heat flow rate under steady-state conditions for a fin of uniform cross-section which is so long that the temperature of the end of the fin can be assumed to be equal to the surrounding temperature.

• 3.4 Derive the expression for temperature distribution and total heat flow rate under steady-state conditions for a uniform cross-section fin of length L. The heat transfer from the free end of the fin may be neglected. Explain the different parameters that form a basis for assessing the utility of a fin.

• 3.5 Derive the equation for heat dissipation by a fin of uniform cross-section if the area of the fin end is a very small proportion of the total fin surface area and its contribution to the heat dissipation can be neglected.

• 3.6 Show that the temperature distribution for a fin of finite length, when the heat loss from the fin end by convection is taken into account, is given by

  $$\frac{{t - t_{\infty } }}{{t_{s} - t_{\infty } }} = \frac{{\cosh m(L - x) + (h_{L} /mk)\sinh m(L - x)}}{{(h_{L} /mk)\sinh mL + \cosh mL}}$$

  Hence, show that the heat transfer from this fin is given by

  $$q_{fin} = \sqrt {hPkA_{c} } \left( {t_{s} - t_{\infty } } \right)\frac{{(h_{L} /mk) + \tanh mL}}{{(h_{L} /mk)\tanh mL + 1}}$$

  where h_L is the heat transfer coefficient from the fin end.

• 3.7 How would you introduce correction in the fin length to take account of the heat transfer from the fin end? What will be the form of equation of the heat transfer from the fin?

• 3.8 Define the efficiency and effectiveness of a fin. Discuss the parameters that affect the fin efficiency and effectiveness.

• 3.9 Show that the total efficiency of a finned wall is given by

  $$\eta_{{total}} = 1 - \frac{{A_{f} }}{{A_{fw} }}(1 - \eta_{fin} )$$

  where

  A_f:

  = surface area of all fins attached to the wall

  A_fw:

  = total surface area of the finned wall rejecting heat

  η_fin:

  = fin efficiency.

• 3.10 Discuss the design of thermometer well to minimize the error in the measurement of the temperature of fluids flowing through a duct.

• 3.11 ‘Fins are more effective for surfaces with low value of surface heat transfer coefficient’. Justify the statement.

• 3.12 Show that fins provided on a surface will increase the heat transfer rate from the surface only if the Biot number is less than 1.

• 3.13 A thin rod of length L has its ends connected to two walls, which are maintained at temperatures T₁ and T₂, respectively. The rod loses heat to the surrounding fluid at T_∞ by convection. Derive the equations of temperature distribution in the rod and total heat rejection rate from the entire surface of the rod.

• 3.14 Present a scheme of experimental setup to measure heat transfer coefficient from a fin surface.

• 3.15 Show that for the triangular profile fin shown in Fig. 3.26, the temperature variation equation is

  $$\theta = \theta_{s} \frac{{I_{0} \left( {2\sqrt {\beta x} } \right)}}{{I_{0} \left( {2\sqrt {\beta L} } \right)}}$$

  where I_o(α) is Bessel’s function, β = (2hL/kδ), θ_s = temperature excess at the fin base and δ is thickness of the fin at the base.

  Determine the total heat rejected by the fin.

• 3.16 Figure 3.31 shows a fin of trapezoidal profile. Derive the equation for the temperature distribution. Neglect the heat transfer from the tip of the fin. θ₁ and θ₂ are the temperature excesses above the ambient. The width of the fin (perpendicular to the plane of the paper) is very large compared to the height L of the fin. Fin thicknesses at the base and tip are δ₁ and δ₂, respectively.

  Fig. 3.31

  

  Q. 3.16

Problems

• 3.1 A part of a very long 25-mm-diameter copper rod [k = 370 W/(m K)] is inserted into a furnace. Temperatures at two points 100 mm apart along its length are measured to be 150°C and 120°C when the steady state was achieved. If the ambient air is at 30°C, estimate the effective heat transfer coefficient from the rod surface.

  [Hint: Apply the fin equation for temperature distribution , i.e.

  $$\frac{{t_{1} - t_{\infty } }}{{t_{2} - t_{\infty } }} = \exp [m(x_{2} - x_{1} )]$$

  [Ans. m = 2.88 m⁻¹; h = 19.2 W/(m² K)]

• 3.2 One end of a long rod is inserted into a furnace, while the other end projects into outside air. Under steady state, the temperature of the rod is measured at two points 75 mm apart and found to be 125°C and 88.5°C when the ambient temperature is 20°C. The rod diameter is 250 mm, and the thermal conductivity of the rod material is 3.0 W/(m K). Find the convective heat transfer coefficient.

  [Ans. Given k = 3.0 W/(m K), P = πD = π × 0.25 m, A_c = (π/4) D² = (π/4) × (0.25)² m² and h is unknown. For a very long fin, $\dfrac{{t - t_{\infty } }}{{t_{s} - t_{\infty } }} = e^{ - mx}$, where $m = \sqrt {\frac{hP}{{kA_{c} }}} = \sqrt {\frac{h \times \pi \times 0.25 \times 4}{{\pi \times (0.25)^{2} \times 3}}} = 2.31\sqrt h ;$ Distance x₁ may be considered to be zero (refer to Example 3.23), then t_s = 125°C and t = 88.5°C at x = 0.075 m. Substitution gives $\frac{88.5 - 20}{125 - 20} = e^{{ - \left( {2.31\sqrt h \times 0.075} \right)}}$ or h = 6.08 W/(m² K).]

• 3.3 One end of a long rod is inserted into a furnace, while the other end projects into outside air. Under steady-state condition, the temperatures of the rod at 50 mm and 125 mm from the furnace wall measured with the help of thermocouples indicate 125°C and 88.5°C when the ambient temperature is 20°C. Find the temperature of the rod at the furnace wall.

  [Ans. For a very long fin, we have $\dfrac{{t - t_{\infty } }}{{t_{s} - t_{\infty } }} = e^{ - mx}$; for the given temperatures of t₁ = 125°C at x₁ = 0.05 m and t₂ = 88.5°C at x₂ = 0.125 m, we have $\dfrac{125 - 20}{{t_{s} - 20}} = e^{ - 0.05m}$ and $\dfrac{88.5 - 20}{{t_{s} - 20}} = e^{ - 0.125m}$, solution gives m =5.695 m⁻¹. With this value of m, $\dfrac{125 - 20}{{t_{s} - 20}} = e^{ - 0.05m}$ gives t_s = 159.6°C.]

• 3.4 Heat generated in a bearing causes the temperature of the end of a shaft to rise to 50°C above the ambient. The shaft is 50 mm in diameter and 800 mm long. Determine the equation of temperature distribution along the shaft if the heat transfer coefficient for the shaft surface is 10 W/(m² K) and the thermal conductivity of the shaft material is 45 W/(m K). What is the shaft temperature 100 mm away from the heated end?

  [Ans. m = 4.216 m⁻¹, mL = 3.373 > 3, hence, the equation of a very long fin can be used, which gives θ =50e^−4.216x. Temperature at 100 mm from the heated end, that is, at x = 100 mm is 32.8°C above the ambient.]

• 3.5 The head of a solid steel valve is at 600°C. The stem of the valve has a diameter of 10 mm and is cooled by water at 60°C. The heat transfer coefficient is 60 W/(m² K). The length of the stem is 200 mm. Determine temperature of the stem 50 mm from the heated end. k = 40 W/(m K).

  [Ans. m = 24.495, mL = 4.9 > 3. The temperature distribution equation of a very long fin gives T = 218.67°C. Equation (3.9) with end correction gives 218.75°C.]

• 3.6 Heat dissipation by convection from a metal tank containing cooling oil is to be increased by 50 per cent by addition of fins to the wall. The fins will be 10 mm thick and spaced 100 mm apart between the centres. The surface temperature of the tank is 100°C, and the surrounding air temperature is 20°C. Determine the height of each fin on the assumption that the convective heat transfer coefficient remains unchanged and the surface temperature of the tank is expected to drop to 95°C when fins are fitted. The thermal conductivity of the fins and tank material is 200 W/(m K). Take for simplicity 1 m × 1 m surface area of the tank. h = 20 W/(m² K).

  [Ans. Heat dissipation without fins, q = hA_c(t_s − t_∞) = 1600 W/m². Desired heat dissipation rate = 1.5 × 1600 = 2400 W. Fin width = width of the tank = 1 m; number of fins, n = (height of tank/fin pitch) = 1000/100 = 10; fin thickness, δ = 10 mm. Surface area not containing fins = 1 × 1 – 10 × 10/1000 = 0.9 m²; heat dissipation by this surface area = 20 × 0.9 × (95 – 20) = 1350 W; Heat to be dissipated by the fins = 2400–1350 = 1050 W; Heat to be dissipated by one fin = 1050/10 = 105 W; Fin perimeter, P = 2(W + δ) ≈ 2.0 m; A_c = 1 × 10/1000 = 1 × 10⁻²; m = 4.47. Considering fin end correction, q_fin = √(hPkA_c) × (t_s − t_∞) tanh(mL_c) gives tanh(mL_c) = 0.1565 and mL_c= 0.1578; L_c = 0.1578/4.47 = 35.3 mm; Required length of fin = L_c – ΔL = 35.3 − δ/2 = 30.3, say 31 mm.]

• 3.7 A steel well of 12 mm ID is placed in a duct of 75 mm ID for a thermometer. The fluid temperature in the duct is 300°C. Determine the length of the well so that the error in the temperature measurement is less than 1.5°C. The heat transfer coefficient is 100 W/(m² K). Wall thickness of the well is 1 mm. The temperature at the well base = 270°C. k = 50 W/(m K).

  [Ans. m = 44.72; (t_L − t_∞)/(t_s − t_∞) = 1/coshmL gives L = 82.5 mm for t_L − t_∞ = 1.5°C and t_s − t_∞ = 30°C. As the length of the well is greater than the diameter of the duct, the well is to be located inclined.]

• 3.8 For a rectangular section fin , the thickness and length are 5 mm and 100 mm, respectively. The width of the fin is 300 mm. The thermal conductivity of the fin material is 20 W/(m K). How will the effectiveness of the fin change as the heat transfer coefficient changes from 10 to 100 W/(m² K)?

  [Ans. First case, h = 10 W/(m² K): mL = 1.426, Bi = 1.25 × 10⁻³, and ε = tanhmL_c/√Bi = 25.4; Second case, h = 100 W/(m² K): mL = 4.509 > 3, Bi = 0.0125 and ε = 1/√Bi = 8.94; Fins are more effective when the heat transfer coefficient is low.]

• 3.9 20-mm-diameter steel rods [k = 50 W/(m K)] are to be used as fins on one side of a wall. The heat transfer coefficients for the two sides of the wall are 900 W/(m² K) and 25 W/(m² K), respectively. The length of the fins is to be restricted to 100 mm. On which side of the wall, would you install the fins? Also determine the heat transfer enhancement or the overall effectiveness if one fin is installed per 2000 mm² of the wall area.

  [Ans. For the high h side, Biot number Bi₁ = h₁D/4k = 900 × (0.02/4)/50 = 0.09; for the other side, Bi₂ = 0.0025, as Bi₂ < Bi₁, the fins must be installed on the low heat transfer coefficient side. m = √(hP/kA_c) = 10 and L_c = L + D/4 = 105 mm. mL_c = 1.05, q_fin = √(hPkA_c) (t_s − t_∞) tanhmL_c. Heat transfer from finned surface = h × (2000 × 10⁻⁶ – π/4 D²) (t_s − t_∞) + √(hPkA_c) (t_s − t_∞) tanhmL_c = 0.16495(t_s − t_∞). Heat transfer from 2000 mm² area without fins is h (2000 × 10⁻⁶) × (t_s − t_∞) = 0.05(t_s − t_∞). The ratio of these heat transfer rates = 3.3, which is the enhancement.]

• 3.10 The following data refer to a uniform section fin.

  Thickness = 6 mm; length = 60 mm; width = 1 m; temperature excess at the base = 100°C; h = 10 W/(m² K); thermal conductivity of the fin material = 60 W/(m K).

  Determine the temperature at the fin end and the efficiency of the fin.

  [Ans. m = 7.45 m⁻¹; mL_c = 0.47; q_fin = √(hPkA_c) (t_s − t_∞) tanhmL_c = 117.93 W; θ₂/θ_s = 1/coshmL_c gives θ₂ = 89.89°C; Fin efficiency = 117.93/[hPL_c(t_s − t_∞)] = 93.04%.]

• 3.11 Determine the end temperature and heat flow from a straight trapezoidal fin of length L = 75 mm, width = 1 m, δ₁ = 2 mm, δ₂ = 0.5 mm, h = 250 W/(m² K), k = 400 W/(m K) and θ₁= 100°C.

  [Ans. L₁ = 0.1 m, L₂ = 0.025 m, β = 2hL₁/(kδ₁) = 62.5, 2√(βL₂) = 2.5, 2√(βL₁) = 5, I₀[2√(βL₁)] = 27.24, I₀[2√(βL₂)] = 3.301, I₁[2√(βL₁)] = 24.336, I₁[2√(βL₂)] = 2.5265, K₀[2√(βL₁)] = 0.00369, K₀[2√(βL₂)] = 0.0628, K₁[2√(βL₁)] = 0.00404, K₁[2√(βL₂)] = 0.0745; From Eq. (3.44) for L₂ = 0.025 m, θ₂= 19.85°C; From Eq. (3.45), q = 1768.6 W.]

• 3.12 In order to increase the heat transfer rate from a cylindrical pipe (OD = 100 mm), circumferential fins of 4 mm thickness and 50 mm height are added at a pitch of 10 mm. If the heat transfer coefficient h = 40 W/(m² K) and the thermal conductivity of the fin and pipe material is 50 W/(m K), determine the heat transfer enhancement.

  [Ans. Without fins, q = 12.566 (t_s − t_∞); For the circumferential fins, efficiency ≈ 69% from Fig. 3.29b for L^3/2_c(h/kA_m)^1/2 ≈ 0.73 and r_2c/r₁ ≈ 2; Fin surface area = 2π(r²₂ − r²₁), q_fin = fin efficiency × fin surface area × h × (t_s − t_∞) = 1.3, 4.0(t_{s −} t_∞); From Eq. (3.47), q_fin = 1.3(t_s − t_∞) for β = √(2 h/kδ) = 20 (where δ = t), βr₁ = 1, βr₂ = 2, I_o(βr₁) = 1.2661, I₁(βr₁) = 0.5652, I₁(βr₂) = 1.591, K_o(βr₁) = 0.421, K₁(βr₁) = 0.602, K₁(βr₂) = 0.14. Number of fins/m = 100; Heat transfer from bare surface = 0.6 × 12.566(t_s − t_∞); Heat rejection from the finned tube = number of fins × 1.3 (t_{s −} t_∞) + 0.6 × 12.566(t_s − t_∞) = 137.54(t_s − t_∞); Enhancement = 137.54/12.566 = 10.95.]