1temp at anyy juction is found out by the princliple
taht
heat current reived at any junction is equal to the heat current going out from that junc.
just like elec current.
we have, Q/t = - kA(T2 - T1)/x
Q/t is the heat current ...
can any1 tell me da logic bhind findin da temp of any jn?(better if gn wid an example)
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2 Answers
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take this one as an example from the test.... if D is the junc.
then at steady state,
\frac{Q_{AD}}{t}+\frac{Q_{BD}}{t}=\frac{Q_{DC}}{t}
\Rightarrow \frac{KA(90-\theta) }{l} +\frac{KA(90-\theta) }{l} =\frac{KA\theta }{l}
θ=60°C
