A pin fin of brass of diameter D = 3mm = 0.003m and length L = 60mm = 0.06m is used to remove heat from a parent surface maintained at Tb = 70^◦C. Fin loses heat to an ambient at a temperature of Tamb = 20^◦C via convection with a uniform heat transfer coefficient of h = 26W/m^2^◦C. What is the hea

Question

A pin fin of brass of diameter D = 3mm = 0.003m and length L = 60mm = 0.06m is used to remove heat from a parent surface maintained at [latex] T_{b} = 70°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 = 26W/m^{2}°C[/latex] . What is the heat transfer from the parent surface via the fin? What is the temperature at the fin tip? Take into account heat transfer from the fin tip assuming [latex] h_{t} = h[/latex] .

Accepted Answer

Step 1 Fin parameter is calculated as [latex] m=\sqrt{\frac{4h}{kD} } =\sqrt{\frac{4 imes 26}{116 imes 0.003} } =17.287[/latex]. Hence the non-dimensional fin parameter is [latex]μ = mL = 17.287 × 0.06 = 1.037[/latex] . Parameter that accounts for tip heat loss is calculated as [latex] \frac{h_{t} }{km} =\frac{h }{km}={\frac{26}{116 imes 17.287} }=0.013[/latex] .

Step 2 Temperature gradient at the base of pin fin is obtained by differentiating with respect to x expression 4.27 and substituting x = 0 in the resulting expression. Thus we have

[latex]\frac{ heta }{ heta _{b} } =\frac{cosh\left\{m(L-x) ight\}+\left(\frac{h_{t}}{km} ight) sinh\left\{m(L-x) ight\} }{cosh(mL)+\left( \frac{h_{t}}{km} ight) sinh(mL)}[/latex] (4.27)

[latex]\frac{1}{ heta _{b} } \frac{\partial heta }{\partial x} \mid _{x=0} =-m\left\lgroup\frac{sinh(\mu )+\frac{h_{t}}{km}cosh(\mu )}{cosh(\mu )+\frac{h_{t}}{km}sinh(\mu )} ight group[/latex]

Substituting the numerical values we get

[latex]\frac{1}{ heta _{b} } \frac{\partial heta }{\partial x} \mid _{x=0} =-m\left\lgroup\frac{sinh(1.037)+0.013cosh(1.037)}{cosh(1.037)+0.013sinh(1.037)} ight group =-13.603 m^{-1}[/latex]

Step 3 Heat loss from the fin may then be calculated as

[latex]Q= -k \left(\frac{\pi D^{2} }{4} ight) (T_{b}-T_{amb} )\frac{1}{ heta _{b} } \frac{\partial heta }{\partial x} \mid _{x=0}[/latex]

[latex]=-116 \left(\frac{ \pi imes 0.003^{2} }{4} ight)(70-20) (-13.603)=0.558W[/latex]

Step 4 Tip temperature is determined by letting x = L in Eq. 4.27 as

[latex]\frac{ heta_{t} }{ heta _{b} } =\frac{cosh(0)+\frac{h_{t}}{km} sinh(0)}{cosh(\mu )+\frac{h_{t}}{km} sinh(\mu )}=\frac{1}{cosh(\mu )+\frac{h_{t}}{km} sinh(\mu )}[/latex]

We thus have

[latex]\frac{ heta_{t} }{ heta _{b} }=\frac{1}{cosh(1.037)+0.013sinh(1.037)} =0.617[/latex]

Hence the fin tip temperature is

[latex]T_{t} = T_{ amb} + θ_{t} = 20 + (70 − 20) × 0.617 = 50.86 °C[/latex]

Question 4.2: A pin fin of brass of diameter D = 3mm = 0.003m and length L...

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