Question
What temperature gradient must exist in aluminium rod for it to transmit 8.0 cal per second per cm.
Solution;
P="\\frac {k\u00d7\u2206T\u00d7A}{L}"
p is power
k= thermal conductivity of Aluminium
Take k=205W/mK
∆T is temperature gradient.
L is the length.
Given:
"\\frac P A" =8.0cal/scm2=33.4864×104W/sm2
"\\frac {\u2206T}{L}=\\frac PA\\times \\frac 1 k"
"\\frac {\u2206T}{L}=\\frac {33.4944\u00d710^4}{205}"=1633.87K/m
Answer
∆T/L=1633.87K/m or 1360.87°C/m
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