An industrial kiln is designed to operate at a very high temperature of 2100 deg
ID: 1297889 • Letter: A
Question
An industrial kiln is designed to operate at a very high temperature of 2100 degrees Celsius. It is 1.03 m long, 0.82 m wide, and 0.55 m high; and it will be insulated using 19 cm thick firebrick with a thermal conductivity of 0.096 W m-1 K-1. (Remember that the kiln is three dimensional with six faces: it might be a good idea to draw at least a crude 3D diagram of it.)
a) If the temperature outside the kiln is 36 degrees Celsius when it begins operating, find the total heat current coming from the kiln.
b) Eventually the outside surface of the kiln reaches a temperature of 137 degrees Celsius. Don't stand too close! Now recalculate the heat current coming from the kiln, taking the new outside temperature into consideration.
c) The kiln is then turned off, and with its outside temperature still at 137 degrees Celsius, cooler air is blown in at 20 degrees Celsius. Now find the power radiated from the hot walls into the cooler air. Assume the emissivity of the firebrick is unity, and ignore any convection that will occur.
Explanation / Answer
a)
heat current=dQ/dt=K*A(T2-T1)/t
here
heat flow throuh a fire brick of thickness t=19cm
and K=0.096 W/m*K
T1=2100 C
T2=36 C
now
kiln dimensional are
length l=1.03 m
width b=0.82 m
height h=0.55 m
kiln has six faces and all faces are coverd with a fire brick
hence we need to calclate the heat current throuh all six faces
let cross-sectinal area are
A1=l*b, A2=b*h, A3=l*h
and
total heat current through all six faces is
dQ/dt=(K*2A1*(T2-T1)/t+K*2A2*(T2-T1)/t +K*2A3*(T2-T1)/t)
=(K*(T2-T1)/t)*2(A1+A2+A3)
=(K*(T2-T1)/t)*2(l*b+b*h+l*h)
=(0.096*(36-2100)/0.19)*2(1.03*0.82+0.82*0.55+1.03*0.55)
=-3723.036 J/sec
negative sign indicting direction of heat flow
b)
now,
T1=137 C
total heat current
=(K*(T2-T1)/t)*2(l*b+b*h+l*h)
=(0.096*(36-137)/0.19)*2(1.03*0.82+0.82*0.55+1.03*0.55)
=-182.183 J/sec
c)
from the Stefan- Biltzmann law,
thermal power radiated is,
u=e*sigma*A*(T1^4-T2^4)
here
emisivity e=1
and
T1=137 C=411 K
T2=36 K=310 K
Stefan's constant sigma=5.67*10^-8 watt/m^2*K^4
power radiated from the all six hot wall into cooler air is
u=e*sigma*2(A1+A2+A3)*(T1^4-T2^4)
=1*5.67*10^-8*2(1.03*0.82+0.82*0.55+1.03*0.55)*(411^4-310^4)
=21.116214*19299094241
=4075.23 J/sec