Question

FIreclay brick (ρ=2050 kg/m3, cp=960 J/kg-K, k=1.1 W/m-K) with dimensions 0.06 m x 0.09 m x 0.20 m is removed from a kiln at 1600K and cooled in air at 40C with h = 30 W/m2-K. What is the temperature at the corner of the brick after 50 minutes of cooling? Give your answer in degrees C.

Answer 1

Solution

Given informations,

Dianmension of brick, (.06 m *.09 m *.2 m)

Density of brick,b=2050 kg/m^3

Thermal conductivity, K=1.1 W/m-K

Heat transfer ceofficient,h=30 W/m^2-K

Intial temprature , Ti=1600 K

cooled air temprature (surronding temprature), To=40 C

To=40+273

=313 K

specific heat,cp=960 J/kg-K

To find temprature after 50 minutes=3000sec?

.........................................................................................................

Assumption: Lump body analysis

we know that,

Temprature profile given as ,{(T - To)/(Ti - To)}=e^-(h*Acs*t)/(*V*cp)..............equation 1

where, Acs: cross sectional area of brick,m^2

t:time

V; volume of brick

Acs=2*(l*b+b*h+l*h), m^2

=2*(.2*.09+.09*.06+.2*.06)

Acs=.0708 m^2

V=l*b*h, m^3

=.2*.09*.06

v=1.08*10^-3 m^3

putting all values in equation 1

{(T -313)/(1600 - 313)}=e^-{(30*.0708*3000)/(2050*1.08*10^-3*960)}

(T-313)/(1287)=e^-(6372/2125.44)

(T-313)/(1287)=e^-2.998

T-313=1287*.0499

T-313=64.20

T=64.20+313

T=377.20 K

T=377.20-273

T=104.20 C

.................................................................................................................