Answer:
Explanation:
Given.
Thickness of brass plate [tex]t = 3 cm[/tex]
Thermal conductivity of brass [tex]k = 110 W/m.°C[/tex]
Density of brass [tex]\rho = 8530 kg/m^3[/tex]
Specific heat of brass [tex]C_p =380J/kg.°C[/tex]
Thermal diffusivity of brass [tex]\alpha = 33.9\times 10^{-6} m^2/s[/tex]
Temperature of oven [tex]T_{\infty} = 700°C[/tex]
The initial temperature [tex]T_i= 25°C[/tex]
Plate remain in the oven [tex]t =10 min[/tex]
Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane.
The thermal properties of the plate are constant.
The heat transfer coefficient is constant and uniform over the entire surface.
The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).
The Biot number for this process [tex]Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109[/tex]
The constants [tex]\lambda_1[/tex] and [tex]A_1[/tex] corresponding to this Biot are, from 11-2 tables.
The interpolation method used to find the
[tex]\lambda_1=0.1039[/tex] and [tex]A_1=1.0018
[/tex]
The Fourier number [tex]\tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}
\\\\\tau=90.4>0.2[/tex]
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.
Then the temperature at the surface of the plates becomes
[tex]\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C[/tex]
