Consider a flat fin of thickness 1 mm, length 4 cm, and made of aluminum of thermal conductivity equal to 237W/m^◦C. The fin is losing heat to an environment by convection subject to a heat transfer coefficient of 150W/m^2^◦C. Calculate the fin heat loss per unit width and per unit temperature diff

Question

Consider a flat fin of thickness 1 mm, length 4 cm, and made of aluminum of thermal conductivity equal to 237 W/m°C. The fin is losing heat to an environment by convection subject to a heat transfer coefficient of [latex]150W/m^{2}°C[/latex]. Calculate the fin heat loss per unit width and per unit temperature difference between the base and the ambient. Rework the problem with an iron fin with a thermal conductivity of 45W/m°C, all other parameters being held fixed. Show the temperature profiles in the two cases on a common plot for a base temperature of 100 ◦C and an ambient temperature of 30°C.

Accepted Answer

Step 1 Given data is listed:

Fin thickness: [latex]2t = 1 mm = 0.001 m[/latex]
and hence [latex]t = 0.0005 m[/latex]
Fin length: [latex]L = 4 cm = 0.04 m[/latex]
Heat transfer coefficient: [latex]h = 150 W/m^{2}°C[/latex]

Case 1: Aluminum fin

Step 2 With thermal conductivity of k = 237W/m°C the fin parameter is calculated based on Table 4.1 as

[latex]\mu =L\sqrt{\frac{h}{kt} }=0.04 imes \sqrt{\frac{150}{237 imes 0.0005} } =1.423[/latex]

Table 4.1 Fin parameter of pin fins of elliptic cross section

AR	m	AR	m	AR	m
2	[latex]\sqrt{\frac{1.542h}{kb} }[/latex]	1.5	[latex]\sqrt{\frac{1.683h}{kb} }[/latex]	1.25	[latex]\sqrt{\frac{1.806h}{kb} }[/latex]
1.125	[latex]\sqrt{\frac{1.891h}{kb} }[/latex]	1†	[latex]\sqrt{\frac{2h}{kb} }[/latex]

†Cross section is a circle with b = D/2

The fin efficiency is calculated using Eq. 4.22 as

[latex]\mu =\frac{Q_{b} }{Q_{max} } =\frac{mk heta _{b} tanh(mL) }{PLh heta _{b} } =\frac{tanh(mL)}{mL} =\frac{tanh(\mu )}{\mu } [/latex] (4.22)

[latex]\mu=\frac{tanh(1.423)}{1.423} =0.626[/latex]

The heat loss per unit width and unit temperature difference [latex] heta _{b} = 1°C [/latex] is then calculated as

[latex]Q_{b} = 2hLη = 2 × 150 × 0.04 × 0.626 = 7.512W/m°C[/latex]

Step 3 We nowgenerate the desired data for making a plot of temperature variation along the fin. The temperatures are taken as

[latex]T_{b} = 100°C; T_{amb} = 30°C[/latex]

We rewrite Eq. 4.15 in dimensional form as follows:

[latex]\phi =\frac{ heta }{ heta_{b} } ;\xi =\frac{x}{L} and \mu =mL[/latex] (4.15)

[latex]T(x)=30+(100-30)\frac{cosh\left[1.423(1-\frac{x}{0.04} ) ight] }{cosh(1.423)}[/latex]

[latex]=30+31.89cosh\left[1.423\left(1-\frac{x}{0.04} ight) ight][/latex]

Case 2: Iron fin

Step 4 With thermal conductivity of k = 45W/m°C, the fin parameter is calculated based on Table 4.1 as

[latex]\mu =L\sqrt{\frac{h}{kt} }=0.04 imes \sqrt{\frac{150}{45 imes 0.0005} } =3.266[/latex]

The fin efficiency is calculated using Eq. 4.22 as

[latex]\eta =\frac{tanh(3.266)}{3.266} =0.305[/latex]

The heat loss per unit width and unit temperature difference [latex] heta _{b} = 1°C [/latex] is then calculated as

[latex]Q _{b}= 2hLη = 2 × 150 × 0.04 × 0.305 = 3.66W/m°C[/latex]

We nowgenerate the desired data for making a plot of temperature variation along the fin. The temperatures are taken as

[latex]T _{b} = 100 °C; T _{amb} = 30 °C[/latex]

We rewrite Eq. 4.15 in dimensional form as follows:

[latex]T(x)=30+(100-30)\frac{cosh\left[3.266(1-\frac{x}{0.04} ) ight] }{cosh(3.266)}[/latex]

[latex]=30+5.33cosh\left[3.266\left(1-\frac{x}{0.04} ight) ight][/latex]

Step 5 The temperature data generated above is plotted in Fig. 4.5.

The figure compares the temperature variations along the fins of the two different materials. The figure indicates that the temperature variation is strongly affected by the thermal conductivity of the material. Temperature variation is more pronounced in iron (lower thermal conductivity material) as compared to that in aluminum (higher thermal conductivity material). It is as though the iron fin is thermally “longer” than the aluminum fin.

Question 4.1: Consider a flat fin of thickness 1 mm, length 4 cm, and made...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions