The steady state solution is that the rate of heat flow will be -kAdT/dx or -kA(T(.5 m)-T(0m))/.5 m
A = π(.1m)2/4 Plug in and find the rate of heat flow in J/s
Now divide 1000 J by the rate to get the time in seconds.
The H.
asked 12/17/21If one end of a solid copper pipe with a diameter of 10 cm is kept at 100 degrees centigrade, and the other side is kept at 0 degrees centigrade, how long will it take for 1000 Joules of energy to travel along the 50 cm long pipe? The thermal conductivity is 385 W/mK.
The steady state solution is that the rate of heat flow will be -kAdT/dx or -kA(T(.5 m)-T(0m))/.5 m
A = π(.1m)2/4 Plug in and find the rate of heat flow in J/s
Now divide 1000 J by the rate to get the time in seconds.
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