A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy&nbsp;(k=160Wm⋅K). The fin diameter is&nbsp;D=4&nbsp;mm, and the fin is exposed to convective conditions characterized by&nbsp;h=220Wm2⋅K. It is reported that the fin efficiency is&nbsp;ηf=0.65. Determine the fin length L and the fin effectiveness&nbsp;εf&nbsp;Account for tip convection.

Question

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy&amp;nbsp;(k=160Wm⋅K). The fin diameter is&amp;nbsp;D=4&amp;nbsp;mm, and the fin is exposed to convective conditions characterized by&amp;nbsp;h=220Wm2⋅K. It is reported that the fin efficiency is&amp;nbsp;ηf=0.65. Determine the fin length L and the fin effectiveness&amp;nbsp;εf&amp;nbsp;Account for tip convection.

Deufemiak7 · Accepted Answer

Step 1  Given: D=4mm, k=160Wm⋅K, h=220Wm2⋅K, ηf=0.65  Calculate the parameter m as follows:  m=hPkA  =h×πDk×πD24  =4hkD  =4×220160×0.004  m=37.08  Step 2  The expression for the efficiency of the after taking account for convection transfer from the tip of the fin as follows:  ηf∈=tanh(mLC)mLC  0.65=tan⁡h(37.08LC)37.08LC  Let 37.08LC=x  0.65=tanh(x)x  0.65=1xex−exxex+e−x  0.65x=e2x−1e2x+1  e2x(0.65x−1)+0.65x+1=0  By solving the equation, we get  x=1.34164  37.08LC=1.34164  LC=0.03618m  LC=36.18mm  Hence, the length of the fin is 36.18mm  Step 3  Calculate the fin effectiveness  εf=qfhAθb  =Mtanh(mLC)hAθb  =hPktanh(mLC}}{hA}  =2hDktanh(mLC)  =2220×4×10−3160tanh(37.08×36.18×10−3)

SlabydouluS62 · Accepted Answer

Step 1  Given that  D=4mm  K=160Wm−K  h=h=220Wm2−K  ηf=0.65  We know that  m=hPKA  For circular fin  m=4hKD  m=4×220160×0.004  m=37.08  ηf=tanhmLmL  0.65=tanh37.08L37.08L  By solving above equation we get  L=36.18mm  The effectiveness for circular fin given as  ε=2tanhmLhDK  ε=2tanh(37.08×0.03618)220×0.004160  ε=23.52

Don Sumner · Accepted Answer

Step 1 As per the question: Thermal conductivity, k=160Wm−K Diameter of the fin, D=4mm=4×10−3m Coefficient of convective heat transfer, h=220Wm2−K Efficiency of the fin, ηf=0.65 Now, To calculate the length if the fin: 1) ηf=tanh⁡mLCmLC where LC=L+D4=(L+1)mm Also, m=PhACk where P=Perimeter of the fin=πD AC=Convection Area AC=πD24 Now, m=πDhπD24k m=4hDk=4×2204×10−3×160=37.08m−3 Now, using eqn (1) 0.65=tanh⁡37.08LC37.08LC tanh⁡(37.08LC)=24.10LC Now, by trial-error: LC=0.0362m=36.2mm Also, we know that: LC=L+1 36.2=L+1 L=35.2mm Now, the fin effectiveness is given by: εf=PkhAC εf=πD×160220×πD24tanh⁡mLC εf4×160220×4×10−3tanh⁡(37.08×0.0362)=23.5

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy PS

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