Question

A brick of 203 x 102 x 57 mm in dimensions is being burned in a kiln to 1100°C, and then allowed to cool in a room with ambient air temperature of 30°C and convection heat transfer coefficient of 5 W/m2·K. If the brick has properties of rho=1920 kg/m3,Crho= 790 J/Kg·K, and k = 0.90 W/m·K, determine the time required to cool the brick to a temperature difference of 5°C from the ambient air temperature.

Answer 1

Answer:

407 minutes

Explanation:

Step 1: Calculate the volume of the brick

$V = 0.203 X 0.102 X 0.057$

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

$L_{C} =\frac{V}{A}$

$L_{C} = \frac{0.0012}{0.08}$ = 0.015 m

Step 4: calculate the biot number

$B_{i} = \frac{hL_{c} }{k}$

$B_{i} = \frac{5X0.015 }{0.9}$ = 0.083

⇒Since $B_{i}$ ∠ 0.1, the lumped system analysis is applicable. Then cooling time is determined from

$b = \frac{hA}{\rho c_{p}V } = \frac{h}{\rho c_{p}L_{c} }$

$b = \frac{h}{\rho c_{p}L_{c} }$

$b = \frac{5}{1920 X 790 X 0.015}$

b = 0.0002197 s⁻¹

$\frac{T(t) -T_{o}}{T_{i} - T_{o}} =e^{-bt}$

$\frac{5}{1100 - 30} =e^{-0.0002197t}$

Take natural log of both sides

-5.3659 = -0.0002197t

t = 24,424 seconds = 407 minutes

Answer 2

The required time "7 hours".

Air temperature:

Dimension of brick$= 203\times 102\times 57 \ mm$

kiln Temperature $\ T_t = 1100^{\circ}\ C$

Air temperature of Ambient $\ T_{\infty} = 30^{\circ}\ C$

Heat transmission coefficient by convection:

$\to h=5\ \frac{W}{m^2}\ K$

Properties of Bricks:  

$\to \rho = 1920\ \frac{kg}{m^3}\\\\ \to C_p = 790\ \frac{J}{kg-K}\\\\ \to k=0.9 \frac{W}{m-K}$

Calculating the temperature difference:

$\to T_t -T_{\infty} = 5^{\circ}\ C$

Where t represents the amount of time needed to cool the brick for a temperature of

$\to T_t = 35^{\circ}\ C$

We are familiar with the lumped system analysis for energy balance.

$\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = \frac{hA_s}{\rho V C_p} t$

Calculating the brick surface area:

$\to A_s = 2(ab + bc +ca)$

         $= 2(0.203\times 0.102 +0.102 \times 0.057 +0.057\times 0.203)\\ \\= 2(0.020706 +0.005814 +0.011571)\\\\ = 2 \times 0.038091 \\\\ =0.076182 \ m^2$

Calculating the volume:

$\to V = abc$

       $= 0.057 \times 0.203 \times 0.102 \\\\ = 0.00118\ m^3$

We know that:

$\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = - \frac{hA_s}{\rho V C_p} t \\\\\to \ln(\frac{5}{1100-30}) = \frac{5\times 0.076}{1920 \times 790 \times 0.00118 } t \\\\\to -5.3659=-2.128 \times 10^{-4}\ t\\\\\to t =\frac{-5.3659}{-2.128 \times 10^{-4}}\\\\\to t= 2.521423 \times 10^{4}\ s\\\\\to t=25214.23 \ s\\\\\to t=7.0039527778 \ h$

Therefore, the final answer is "7 hours".

Find out more information about air temperature here:

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