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Chapter-2---One-dimensional--steady-state-heat-co_2021_Heat-Transfer-Enginee

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	2.10 Analysis of fin heat transfer		55
2.10.2 Case 2: Long fin
The fin equation, for this case, is given by
d2θ	2
dx2	− m θ = 0	(2.199)
The general solution is
θ = c1emx + c2e−mx		(2.200)

Boundary condition 1: At x = 0; T = Tb, θ = θb From Eq. (2.200) c1 = 0; c2 = θb

θ = θbe−mx

The heat transfer from a long fin is given by

Q = −kAc dθ dx x=0

Q = hpkAcθb

The efficiency of a long fin is given by

η =	hpkAc
	h( pL)θb

η = mL1

The effectiveness of the fin is given by

ε =	pk

	hA
	c

(2.201)

(2.202)

(2.203)

(2.204)

(2.205)

(2.206)

Eq. (2.205) can be reduced for the specific case of a rectangular long fin to be ε = √(2k/ht).

2.10.3 Case 3: Convecting tip

The more general case is; where the fin tip is convective. The governing equation is

d2θ − m2θ = 0 dx2

The general solution is

θ = c1cosh(mx) + c2 sinh(mx)

Boundary condition 1: At x = 0; θ = θb From Eq. (2.208)

c1 = θb

(2.207)

(2.208)

(2.209)

56 CHAPTER 2 One-dimensional, steady state heat conduction

θ = θbcosh(mx) + c2 sinh(mx)

Boundary condition 2: −k dθ

= h θ |

x=L

tip

x=L

dθ

= θbmsinh(mx) + c2mcosh(mx)

From boundary condition 2, we get

−km (θb sinh(mL) + c2cosh(mL)) = htip (θbcosh(mL) + c2 sinh(mL))

htip

= −θb sinh(mL) −

htip

θbcosh(mL)

sinhmL + cosh(mL)

htip

sinh(mL) +

cosh(mL)

= −θb

htip

cosh(mL) +

sinh(mL)

The temperature distribution in the fin is given by

(2.210)

(2.211)

(2.212)

(2.213)

(2.214)

htip

sinh(mL) +

cosh(mL)

θ = θb cosh(mx) − θb

sinh(mx)

(2.215)

htip

cosh(mL) +

sinh(mL)

cosh(m(L − x)) +

htip

sinh(m(L − x))

θ = θb

(2.216)

cosh(mL) +

htip

sinhmL

It is instructive to see that the expression reduces to the following for an adiabatic fin tip case, when we set htip to be 0.

θ	= cosh(m(L − x))	(2.217)
θ0
	cosh(mL)

Eqn. (2.217) is the same as the equation obtained for the insulated fin tip case (Eqn. 2.182)

The heat transfer from the fin with convecting fin tip is given by

Q = −kAc	dθ					(2.218)
Q = −kAc	dθ					(2.218)
	dx		x=0
	dx		x=0
		sinh(mL) +		htip		cosh(mL)
Q = kAcθbm		sinh(mL) +		mk		cosh(mL)
Q = kAcθbm				mk			(2.219)
		cosh(mL) +			htip	sinh(mL)
		cosh(mL) +			mk	sinh(mL)
					mk


			2.10 Analysis of fin heat transfer				57
	tanh(mL) +				htip
Q = hpkAcθb	tanh(mL) +				mk	(2.220)
					mk
		+	htip
	1			tanh(mL)
	1		mk	tanh(mL)
			mk

Again when htip →0 Eq. (2.220) reduces to Eq. (2.186); i.e., it reduces to the case of a fin with adiabatic tip.

We now look at an engineering approach to get rid of the mathematical tedium associated with this case

Fig. 2.29 represents the idea behind the use of a corrected length in the analysis of fin heat transfer. Fig. 2.29 shows that we are looking for an equivalent length Lc of a similar fin as the original, which transfers the same heat but has an adiabatic tip. Assuming (Lc − L) to be at temperature TL, we equate the heat transfer at the tip from Fig. 2.29A to the convection across (Lc − L) in Fig. 2.29B.

hAcθL = hp(Lc − L)θL

(2.221)

In the above equation θL is the temperature excess at the tip given by (TL – T∞).

Lc = L + Ap

Here Lc is the corrected length.

The heat transfer from the fin is given by

Q = hpkAcθb tanh mLc

The efficiency of the fin is given by

η = tanh mLc mLc

(2.222)

(2.223)

(2.224)

The above-corrected length was conceptualized in Harper and Brown (1923) and must be considered as a smart simplification, considering the fact that it is nearly 100 years since the original work was published, is still frequently used.

FIGURE 2.29

Schematic representation of a fin to demonstrate the corrected fin length concept. (A) Actual length with convection at the tip; (B) increased length with an adiabatic tip.

58CHAPTER 2 One-dimensional, steady state heat conduction

Example 2.9: Consider a long rod that is placed vertically. The rod cools in ambient air at 30 °C. Thermocouples at 30 mm and 60 mm from the base indicate temperatures of 90 °C and 74 °C respectively. Determine the base temperature.

Solution:

At x = x1 = 30 mm, T1 = 90 °C At x = x2 = 60 mm, T2 = 74 °C

θ1 = T1 − T∞ = 60 °C θ2 = T2 − T∞ = 44 °C

For a long fin:


			θ	= e−mx
				= e−mx
		θb
		θ			= −mx
ln
ln
	θb
	θ1				= −mx1
ln
ln	θb
	θb
	θ2				= −mx2
ln
ln	θb
	θb

Dividing Eq. (2.226) by Eq. (2.227) results in Eq. (2.228).

ln(θ1 /θb )

30 =

ln(θ2 /θb )

60 2

θ1

θ2

θb

θ1

θ2 1/2

θb

= 81.82 °C

= 111.82 °C

(2.225)

(2.226)

(2.227)

(2.228)

(2.229)

(2.230)

Please note that the above example is again an inverse problem, wherein from temperatures at two locations we were able to obtain the base temperature. In all of the preceding discussions, θb was given or known. Hence this problem represents a ‘surprising’ example. One can go one step further and obtain m, from either Eq. (2.225) or Eq. (2.226). Proceeding even further, if the geometrical details and the thermal conductivity are known, making use of the definition of m as √(hp/kAc), we can actually “pull out” the heat transfer coefficient in the situation. If more measurements


2.10 Analysis of fin heat transfer	59

are available, one can find the “best” value of the heat transfer coefficient, h that minimizes the difference between the measure temperatures and those obtained with given values of ‘h.’ This minimization is invariably done in a least square sense and opens up new vistas in the application of heat transfer principles in the solution of practical inverse problems in thermal science and engineering.

2.10.4 Variable area fins

For variable area fins with trapezoidal or parabolic profile, the solution leads to Bessel functions. These can be concisely presented as a chart for fin efficiency as a function of the fin parameter and are shown in Fig. 2.30. A similar chart can also be made for radial fins across a cylinder. These charts can be used to estimate the heat transfer from such variable area fins quickly. Analytical solutions to these are available in many advanced texts on conduction (see, for example, Poulikakos, 1994).

Fig. 2.30 shows the fin efficiency chart of rectangular, triangular, and parabolic fins with and rectangular cross-section as a function of fin parameters. Please note that the corrected length should be used here along with specific expressions for the profile area, Ap and length parameter ξ which are embedded in the figure.

Fig. 2.31 shows efficiency curves for annular fins placed circumferentially over a pipe or tube. Please note that here too corrected fin length should be used.

FIGURE 2.30

Fin efficiency of rectangular, triangular, and parabolic profiles as a function of key parameters.

60 CHAPTER 2 One-dimensional, steady state heat conduction

FIGURE 2.31

Efficiency of an annular fin as a function of key parameters.

For a detailed presentation of the analytical results for these, please refer to Poulikakos (1994).

Problems

2.1Acompositewallismadeofa30mmstainlesssteelplatewithathermalconductivity of 16 W/mK, 50 mm of fiberglass insulation with a thermal conductivity of 0.04 W/mK, and 20 mm of plastic insulation with thermal conductivity of 0.03 W/mK. The temperature on the stainless steel side of the wall is 600 °C. The right side is exposed to ambient air with a heat transfer coefficient of 6 W/m2K and T∞ = 30 °C.

a.Draw a resistance diagram for this problem.

b.Determine the heat transfer across the composite wall.

c.Calculate all the intermediate temperatures.

2.2Consider a plane wall, 15 cm in thickness, made of stainless steel (SS304) and very long such that the heat transfer across the wall is one dimensional (axial direction-x, with the origin at the left wall). The left end of the wall is maintained at 573 K, and the right end is at 373 K. The thermal conductivity of the material is known to vary as k = 16.5 + 0.0175(T − 373), where T is the temperature in Kelvin and k is in W/mK. Steady state conditions prevail, and there is no heat generation in the material.


2.10 Analysis of fin heat transfer	61

a.Write down the equation governing the temperature distribution in the plane wall.

b.Solve (a) for the boundary conditions given.

c.Determine the value of dT /dx at both ends of the wall.

d.Determine the heat transfer rate across the wall.

2.3A long heat-generating wall is bathed by a cold fluid on both sides. The thickness of the wall is 12 cm, and the thermal conductivity of the wall material is 48 W/mK. A thermocouple located in the midplane shows a temperature of 88 °C. The cold fluid is at a temperature of 30 °C. The heat transfer coefficient on both sides of the wall is 200 W/m2K.

a.Identify the governing equation for the problem, recognizing that a heat generation (which may be assumed to be uniform) is present in the problem.

b.Solve the governing equation assuming that the temperatures are symmetric about the midplane.

c.From the data given and the solution obtained, determine the volumetric heat generation in the slab.

d.Determine the temperature at the two ends of the slab.

2.4A very long copper wire of specification AWG 4/0 (AWG, American wire gauge) has a diameter of 11.68 mm. It is capable of carrying a current of 200 A. The thermal conductivity of copper is 400 W/mK, and its electrical resistivity is 1.72 × 10−8 Ωm. The surroundings are at 30 °C with a free convection heat transfer coefficient of 6 W/m2K.

a.Write down the governing equation for temperature distribution in the wire under the conditions of steady state, together with boundary conditions.

b.Solve (a) to obtain the temperature distribution in the wire.

c.Determine the maximum temperature in the wire.

d.Determine the minimum temperature in the wire. Where does it occur?

e.From the solution to this problem, reason out why a maximum currentcarrying capacity is specified on a wire.

2.5Revisit Problem 2.4. However, now the goal is to insulate the wire to reduce losses. It is preferable to use Bakelite (k = 0.2 W/mK) to insulate the wire.

a.Determine the critical radius of insulation.

b.Determine the total heat rate with the Bakelite insulation for a thickness equal to the critical thickness of insulation.

c.Is the result obtained in (b) baffling to you? Why?

d.Determine the temperature at the interface between the copper wire and Bakelite.

e.Determine the maximum temperature inside the copper wire with the insulation.

f.Has the insulation helped or hurt the heat transfer from the wire?

62CHAPTER 2 One-dimensional, steady state heat conduction

2.6One possibility of having a nuclear fuel element is to have a spherical capsule of fissionable material and a spherical shell of aluminum cladding. Coolant flows outside of the cladding and removes the heat generated due to nuclear

fission. Consider a homogeneous fissionable material of radius r1 = 5 cm, with uniform heat generation of qv = 4.5 × 107 W/m3. The thermal conductivity of the fissionable material is 10 W/m/K. The thermal conductivity of the aluminum cladding is 200 W/mK and has a thickness of 1 cm. The heat transfer coefficient on the coolant side is 4000 W/m2K, and the bulk temperature of the coolant is 90 °C. Assume steady state, one-dimensional heat transfer.

a.Write down the governing equations for the temperature distribution within the fissionable material along with the boundary conditions.

b.Solve the governing equation and obtain an expression for temperature distribution in the sphere.

c.From the data given, determine the temperature at the outer end of the aluminum cladding.

d.Determine the temperature at the interface between the cladding and the fuel.

e.Determine the temperature at the center of the sphere.

2.7Consider two long rods (or long fins) made of aluminum (k = 205 W/mK) that have diameter d = 12 mm. They are soldered together end to end with the solder having a melting point of 700 °C. The rods are kept in quiescent air at 30 °C, with a convection coefficient of 10 W/m2K. Determine the minimum power input that is required to accomplish the soldering.

2.8An inverse problem in heat transfer concerns the estimation of properties like k,

Cp, and α, to name a few, from heat flux or temperature measurements. A long fin is a heat transfer device that can be ingeniously used to estimate the thermal conductivity of material if simple temperature measurements are made. One approach would be to use two identical, very long rods (or long fins) that are geometrically the same in all aspects and are subject to the same environment

and base temperatures but are made of different materials, k1 and k2. Consider, a situation where thermocouple measurements with a fin of thermal conductiv-

ity k1 show a temperature of 72 °C at an axial location x, when Tb = 100 °C and T∞ = 25 °C. For the fin of thermal conductivity k2, the temperature is measured to be 85 °C at the same location x. If k2 = 205 W/mK corresponding to aluminum, what is k1?

2.9Consider an aluminum rectangular fin (k = 205 W/m K) of length L = 10 mm,

thickness t = 1 mm, and width w»t. The base temperature of the fin is Tb = l00 °C, and the fin is exposed to a fluid of temperature = 30 °C. (a) Assuming a uniform convection coefficient of h = 100 W/m2K over the entire fin surface, determine the fin heat transfer rate per unit width, efficiency, effectiveness, thermal resistance per unit width, and tip temperature for the following variants of the problem:

a.Convection at the tip


2.10 Analysis of fin heat transfer	63

b.Adiabatic tip

c.Adiabatic tip with length corrected for tip convection and compare these with case (a)

d.Infinite fin (long fin) approximation

2.10Consider a horizontally oriented triangular fin made of aluminum (k = 205 W/mK) having a rectangular cross-section placed on a base at T = 100 °C. The thickness of the fin is 3.5 mm at the base, and the length of the fin is 16 mm. The fin

is exposed to the ambient at T∞ = 30 °C that affords a convection heat transfer coefficient of h = 70 W/m2K. For unit width of the fin, determine the following:

a.Fin tip temperature

b.Heat transfer rate from the fin

c.Fin efficiency

d.Fin effectiveness

2.11Revisit problem 2.10. Now consider two variants (1) rectangular fin with the same thickness of 3.5 mm throughout, and (2) trapezoidal fin with a tip thickness of 1.5 mm. All other conditions remain the same as before. For these two cases, determine the fin tip temperature, fin heat transfer rate, fin efficiency, and fin effectiveness, and compare their values along with those of the fin considered in problem 2.10. The tip may be assumed to be insulated for both the cases.

2.12Annular aluminum fins of rectangular profile (k = 205 W/mK) are attached to a circular tube The outside diameter of the tube 50 mm and the outer surface temperature of the tube is 200 °C. The fins are 3 mm thick and 15 mm long (in the radial direction). The system is in ambient air at 30 °C, and the associated convection heat transfer coefficient is 40 W/m2K.

a.Determine fin efficiency.

b.Determine the heat transfer from one fin.

c.Determine the effectiveness of one fin.

d.If there are 120 fins per meter length of the tube, what is the overall heat transfer rate per unit length of the system?

e.Determine the overall surface effectiveness of the fin array.

2.13A heat sink consists of a horizontal base with an array of vertical aluminum fins of a rectangular cross-section. The base of the heat sink is 15 cm long and 8 cm deep. There are 20 equispaced (spaced equally along the length) fins, each of 3 mm thickness and 3 cm height. The heat transfer coefficient under these conditions for the finned and unfinned portions of the heat sink is 6.5 W/

m2K with a T∞ of 30 °C. The heat sink is placed on heat-generating electronic equipment, and it can be assumed that the electronic equipment and the base are at the same temperature. If the electronic component dissipates 6W of heat and the heat transfer from the underside of the base and the equipment can be neglected, what is the steady state temperature of the electronic equipment?

(Note: you may assume the tip of the fins to be insulated, and properties of aluminum are k = 205 W/mK, ρ = 2700 kg/m3, and Cp = 900 J/kg K.)

64 CHAPTER 2 One-dimensional, steady state heat conduction

FIGURE 2.32

Schematic representation of heat sink with aluminium fins described in Problem 2.14.

2.14A silicon chip of width W = 24.2 mm on a side is soldered to an aluminum heat sink (k = 205 W/m K) of the same width as shown in Fig. 2.32. The heat sink has

a base thickness of tb = 3 mm and consists of an array of vertical rectangular fins, each of length L = 14 mm. Air flows with T∞ = 30 °C through channels formed between the fins and the cover plate. The convection coefficient h is 150 W/m2K, and a minimum fin spacing of 2 mm is required considering the restrictions on the pressure drop. The solder joint offers a thermal resistance of Rt,c = 3 × 10−6 m2 K/W. The silicon chip may be assumed to be isothermal.

The array has 13 fins, the fin thickness t = 0.2 mm, and pitch S = 2 mm. If the

maximum permissible temperature of the chip is Tc = 80 °C, determine the chip power. Assume adiabatic fin tip condition (adapted from Incropera et al.,

Fundamentals of Heat Transfer, 2013).

References

Harper, R.R., and Brown, W.B., 1923. Mathematical equations for heat conduction in the fins of air-cooled engines. 1923, NACA-TR-158, NACA Annual Report 8, pp. 677–708.

Incropera F.P., Lavine A.S., Bergman T.L., DeWitt D.P., 2013. Principles of Heat and Mass Transfer. John Wiley and sons, New York.

Kraus, A.D., Aziz, A., Welty, J., 2002. Extended Surface Heat Transfer. John Wiley & Sons Inc., New York.

Poulikakos, D., 1994. Conduction Heat Transfer. Prentice Hall, Englewood Cliffs, NJ.

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