Abrick 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.
Step 1: Calculate the volume of the brick
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
= 0.015 m
Step 4: calculate the biot number
⇒Since ∠ 0.1, the lumped system analysis is applicable. Then cooling time is determined from
b = 0.0002197 s⁻¹
Take natural log of both sides
-5.3659 = -0.0002197t
t = 24,424 seconds = 407 minutes
the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.