Answer to Question #191134 in Mechanical Engineering for Srn

Question

A fin in the shape of an upright cone with a fin root diameter of 2 cm and a length of 30 cm. The convection heat transfer coefficient is 100 W/ m² and can be considered constant and uniform across the surface of the fin. The conductivity, density and density of the fin material are 200 W/m.K, 900 J/kg.K and 2700 kg/m², respectively. The temperature of the fin roots and ambient air were 100 and 25 ° C, respectively.

a. Make a meshing of temperature distribution calculations by placing the nodal points at the root of the fin, tip of the fin, and along the length of the fin with a uniform distance between. Make the number of your elements as much as 15 nodes and place the boundary between elementsxactly in the middle plane between

b. Derive the discretization equation around the ith nodal point of the fin root. You can start thew twodjacent nodes. analysis the Fourier equation. If ther processs stationary.

c. Calculate the discretization equation around the nodes to nomor (11), and (7)

Answer 1

$q_x=q_{x+dx}+q_{convential}$

$q_x=q_{x}+ \frac{\partial}{\partial x}(q_x)dx+h(Pdx)(T-T_{\infin})$

$0= \frac{\partial}{\partial x}(-kA \frac{dT}{dx})dx+h(Pdx)(T-T_{\infin})$

$\frac{d^2T}{dx^2}-\frac{hP}{kA}(T-T_{\infin})=0$

Let, $(T-T_{\infin})=\theta=f(x)$

$\frac{dT}{dx}=\frac{d \theta}{dx}$

$\frac{d^2T}{dx^2}=\frac{d ^2\theta}{dx^2}$

$m^2=(\frac{hP}{kA})$

$\frac{d ^2\theta}{dx^2}-m^2 \theta =0 \implies \theta= C_1e^{-mx}+C_2e^{mx}$

$m=\sqrt{\frac{hP}{kA}} \implies x=0 , T=T_0 \space and \space \theta=\theta_0 = T_0-T_{\infin}$

$x =\infin \implies T=T_{\infin}, \theta=0$

$\frac{\theta}{\theta_0}=\frac{T-T_{\infin}}{T_0-T_{\infin}}=e^{-mx}$

$\frac{T-T_{\infin}}{T_0-T_{\infin}}=e^{-mx}$

$\frac{dT}{dx}=(T_o-T_{\infin}) e^{-mx}$

Therefore, $q_{fin}=\sqrt{hPkA}(T_0-T_{\infin}) Watts$

b)$m= \sqrt{\frac{hP}{kA}}= \sqrt{\frac{100*6.28*0.01}{200*3.14*0.0001}}=10$

c) $\frac{T-25}{100-25}=e^{-10*0.30} \implies T=28.73^0C$