Respuesta :
Answer:
407 minutes
Explanation:
Step 1: Calculate the volume of the brick
[tex]V = 0.203 X 0.102 X 0.057[/tex]
V = 0.0012 m³
Step 2: Calculate the surface area of the brick
A= 2[(0.203 X 0.102) +(0.203 X 0.057) +(0.102 X0.057)] = 0.08 m²
Step 3: calculate the characteristic length
[tex]L_{C} =\frac{V}{A}[/tex]
[tex]L_{C} = \frac{0.0012}{0.08}[/tex] = 0.015 m
Step 4: calculate the biot number
[tex]B_{i} = \frac{hL_{c} }{k}[/tex]
[tex]B_{i} = \frac{5X0.015 }{0.9}[/tex] = 0.083
⇒Since [tex]B_{i}[/tex] ∠ 0.1, the lumped system analysis is applicable. Then cooling time is determined from
[tex]b = \frac{hA}{\rho c_{p}V } = \frac{h}{\rho c_{p}L_{c} }[/tex]
[tex]b = \frac{h}{\rho c_{p}L_{c} }[/tex]
[tex]b = \frac{5}{1920 X 790 X 0.015}[/tex]
b = 0.0002197 s⁻¹
[tex]\frac{T(t) -T_{o}}{T_{i} - T_{o}} =e^{-bt}[/tex]
[tex]\frac{5}{1100 - 30} =e^{-0.0002197t}[/tex]
Take natural log of both sides
-5.3659 = -0.0002197t
t = 24,424 seconds = 407 minutes
The required time "7 hours".
Air temperature:
Dimension of brick[tex]= 203\times 102\times 57 \ mm\\\\[/tex]
kiln Temperature [tex]\ T_t = 1100^{\circ}\ C \\\\[/tex]
Air temperature of Ambient [tex]\ T_{\infty} = 30^{\circ}\ C\\\\[/tex]
Heat transmission coefficient by convection:
[tex]\to h=5\ \frac{W}{m^2}\ K\\\\[/tex]
Properties of Bricks:
[tex]\to \rho = 1920\ \frac{kg}{m^3}\\\\ \to C_p = 790\ \frac{J}{kg-K}\\\\ \to k=0.9 \frac{W}{m-K}\\\\[/tex]
Calculating the temperature difference:
[tex]\to T_t -T_{\infty} = 5^{\circ}\ C\\\\[/tex]
Where t represents the amount of time needed to cool the brick for a temperature of
[tex]\to T_t = 35^{\circ}\ C\\\\[/tex]
We are familiar with the lumped system analysis for energy balance.
[tex]\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = \frac{hA_s}{\rho V C_p} t \\\\\\[/tex]
Calculating the brick surface area:
[tex]\to A_s = 2(ab + bc +ca)[/tex]
[tex]= 2(0.203\times 0.102 +0.102 \times 0.057 +0.057\times 0.203)\\ \\= 2(0.020706 +0.005814 +0.011571)\\\\ = 2 \times 0.038091 \\\\ =0.076182 \ m^2\\\\[/tex]
Calculating the volume:
[tex]\to V = abc[/tex]
[tex]= 0.057 \times 0.203 \times 0.102 \\\\ = 0.00118\ m^3\\\\[/tex]
We know that:
[tex]\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = - \frac{hA_s}{\rho V C_p} t \\\\\to \ln(\frac{5}{1100-30}) = \frac{5\times 0.076}{1920 \times 790 \times 0.00118 } t \\\\\to -5.3659=-2.128 \times 10^{-4}\ t\\\\\to t =\frac{-5.3659}{-2.128 \times 10^{-4}}\\\\\to t= 2.521423 \times 10^{4}\ s\\\\\to t=25214.23 \ s\\\\\to t=7.0039527778 \ h\\\\[/tex]
Therefore, the final answer is "7 hours".
Find out more information about air temperature here:
brainly.com/question/11329440
