Compare the heat loss per unit mass for the following two configurations, viz., (a) Uniform area flat plate type fin, (b) Triangular fin. Both fins are 10 cm long. The thickness of the flat plate type fin and the base thickness of the triangular fin are both equal to 1.2 mm. Compare on the basis of

Question

Compare the heat loss per unit mass for the following two configurations, viz., (a) Uniform area flat plate type fin, (b) Triangular fin. Both fins are 10 cm long. The thickness of the flat plate type fin and the base thickness of the triangular fin are both equal to 1.2 mm. Compare on the basis of unit width of fin in both the cases. The material of the fin is pure aluminum with thermal conductivity equal to 237W/m°C and density of [latex]2700 kg/m^{3}[/latex]. The base and ambient temperatures are, respectively, 60 and 15°C. The heat transfer coefficient has been determined to be [latex] 26W/m^{ 2}°C[/latex] in both cases.

Accepted Answer

Data that is common to the two cases are

Fin length: [latex]L = 0.1 m[/latex]
Thermal conductivity: [latex]k = 237 W/m°C[/latex]
Heat transfer coefficient: [latex]h = 26 W/m^{ 2} °C[/latex]
Base temperature: [latex]T_{b} = 60°C[/latex]
Ambient temperature: [latex]T_{amb} = 15 °C[/latex]
Density of material of fin: [latex]ρ = 2700 kg/m^{ 3}[/latex]

Case (a) Uniform area fin:

Step 1 The data specific to this case is

Fin thickness: 2t = 1.2mm or t = 0.0006m

The fin parameter may be calculated as

[latex]m_{a} =\sqrt{h} kt=\frac{26}{237 imes 0.0006} =13.522m^{-1}[/latex]

where the subscript “a” represents Case (a). The non dimensional fin parameter is then calculated as

[latex]μ_{a} = m_{a}L = 13.522 × 0.1 = 1.352[/latex]

Step 2 The fin efficiency is calculated as

[latex]\eta _{a}=\frac{tanh(\mu _{a})}{\mu_{a}} =\frac{tanh(1.352)}{1.352}=0.647[/latex]

Alternately the efficiency may be read off Fig. 4.4b.

Step 3 The heat loss from the fin may be calculated as

[latex]Q_{a} = 2hLη_{a}(T_{b} − T_{amb}) = 2 × 26 × 0.1 × 0.647 × (60 − 15) = 151.4W/m[/latex]

Step 4 Per unit width the volume of the fin is

[latex]V_{a} = 2t L = 2 × 0.0006 × 0.1 = 0.00012m^{ 3}[/latex]

Mass of fin per unit width is
[latex]M_{a} = ρV_{a}= 2700 × 0.00012 = 0.324 kg/m[/latex]

Heat loss per unit mass of fin material is then given by

[latex]q_{a}=\frac{Q_{a}}{M_{a}} =\frac{151.4}{0.324} =467.3W/kg[/latex]

Case (b) Triangular fin:

Step 5 The data specific to this case is

Base thickness: [latex]2y_{b} = 1.2mm or y_{b} = 0.0006m[/latex]

The fin parameter may be calculated as

[latex]p_{b} = \sqrt{ \frac{hL}{ky_{b} }\sqrt{1+\left(\frac{y_{b}}{L} ight)^{2} }}=\sqrt{\frac{26 imes 0.1}{237 imes 0.0006 } \sqrt{1+\left(\frac{0.0006}{0.1} ight)^{2} }} =4.276m^{-\frac{1}{2} }[/latex]

where the subscript “b” represents Case (b). The non dimensional fin parameter is then calculated as

[latex]\mu_{b}=p_{b}\sqrt{L} =4.276 imes \sqrt{0.1} =1.352[/latex]

Step 6 The fin efficiency is read off Fig. 4.9b as [latex]η_{b} = 0.58[/latex]. The heat loss from the fin may be calculated as

[latex]Q_{b}=2hL\sqrt{1+\left(\frac{y_{b} }{L} ight)^{2} } \eta _{b} (T_{b} -T_{amb} )[/latex]

[latex]=2 imes 26 imes 0.1\sqrt{1+\left(\frac{0.0006}{0.1} ight)^{2} } imes 0.85 imes (60-15)=135.9W/m[/latex]

Step 7 Per unit width the volume of the fin is

[latex]V_b = t L = 2 × 0.0006 × 0.1 = 0.00006m^{3}[/latex]

Mass of fin per unit width is

[latex]M_{a} = ρV_{b} = 2700 × 0.00006 = 0.162 kg/m[/latex]

Heat loss per unit mass of fin material is then given by

[latex]q_{b}=\frac{Q_{b}}{M_{b}}=\frac{135.9}{0.162} =839.1W/kg[/latex]

Step 8 Even though the heat loss is a bit lower in the case of the triangular fin, it is seen that per unit mass it loses 839.1/467.2 = 1.796 ≈ 1.8 times the heat loss by the uniform area fin.

Question 4.4: Compare the heat loss per unit mass for the following two co...

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