A conical spine of a material of thermal conductivity k = 45W/m^◦C, base diameter of db = 3mm and length L = 100mm is used to remove heat from a parent surface maintained at Tb = 120^◦C. Fin loses heat to an ambient at a temperature of Tamb = 20^◦C via convection with a uniform heat transfer coeffi

Question

A conical spine of a material of thermal conductivity k = 45W/m°C, base diameter of db = 3mm and length L = 100 mm is used to remove heat from a parent surface maintained at [latex] T_{b} = 120°C[/latex] . Fin loses heat to an ambient at a temperature of [latex] T_{amb} = 20 °C[/latex] via convection with a uniform heat transfer coefficient of [latex] h = 15W/m^{2} °C[/latex] . Make a plot of temperature variation along the fin. Determine the efficiency of the spine. What is the heat loss from the spine?

Accepted Answer

Step 1 Radius of cross section of spine at its base is [latex] r_{b} = \frac{d_{b}}{2}=\frac{0.003} {2}=0.0015m[/latex] . Length of spine is L = 100mm = 0.1m. Base radius to length ratio is hence given by [latex] \frac{r_{b}}{L} = \frac{0.0015}{0.1} = 0.015[/latex] . Hence parameter p may now be calculated, neglecting the factor involving this ratio as

[latex]p=\sqrt{\frac{2hL}{kr_{b} } } =\sqrt{\frac{2 imes 15 imes 0.1}{45 imes 0.0015} } =44.4444[/latex]

We then have [latex]p\sqrt{L} = 2 × 44.4444 \sqrt{0.1}= 2.1082.[/latex]

Step 2 We shall introduce non-dimensional location along the spine by the relation [latex]ξ=1-\frac{x}{L}[/latex] such that ξ = 0 corresponds to the base of the spine and ξ = 1 corresponds to the tip of the spine. With [latex]\phi =\frac{ heta }{ heta _{b} }[/latex] where [latex]θ_{b} = T_{b} − T_{amb} = 120 − 20 = 100 °C[/latex] the non dimensional temperature variation with ξ is given, using Eq. 4.46, as

[latex]\frac{ heta_{x}}{ heta _{b} } =\sqrt{\frac{L}{x} } \frac{I_{1}(2p\sqrt{x} ) }{I_{1}(2p\sqrt{L} )} =\frac{2p\sqrt{L} }{X} \frac{I_{1}(X)}{I_{1}(2p\sqrt{L} )}[/latex] (4.46)

[latex]\phi(ξ)=\frac{1}{\sqrt{1-ξ} } \frac{I_{1}(2p\sqrt{L(1-ξ)} )}{I_{1}(2p\sqrt{L} )} =\frac{1}{\sqrt{1-ξ} } \frac{I_{1}(4.2164\sqrt{1-ξ} )}{I_{1}(4.2164)}[/latex]

Step 3 A plot of [latex]\phi vs ξ[/latex] is shown in Fig. 4.11. The tip temperature is obtained by letting ξ = 1 in the expression for [latex]\phi[/latex]. It is noted that the ratio [latex]\frac{I_{1}(4.2164\sqrt{t} )}{\sqrt{t} }[/latex] as t → 0 is required. This ratio is given by [latex]p\sqrt{L}[/latex] and hence the tip temperature is obtained as

[latex]\phi ( \xi =1)=\frac{2.1082}{I_{1}(4.2164)}=0.1774[/latex]

Hence the dimensional tip temperature is T (x = 0) = 0.1774 × 100 = 17.74 °C.

Step 4 We calculate the fin efficiency using expression 4.47 as

[latex] \eta =\frac{45 imes 0.0015 imes 44.4444}{15 imes 0.1^{3/2} }\cdot \frac{I_{2}(4.2164) }{I_{2}(4.2164)} =0.6387[/latex]

Maximum possible heat transfer by the spine is given by

[latex]Q_{max} = 15 × π × 0.0015 × 0.1 × 100 = 0.707W[/latex]

Hence the heat loss from the spine is given by

[latex]Q_{b} = Q_{max}η = 0.707 × 0.6387 = 0.452W[/latex]

Question 4.5: A conical spine of a material of thermal conductivity k = 45...

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