Answer:
steady state temperature =88.7deg C
t=time within 1 deg C of it steady state is 8.31s
Explanation:
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu
The diameter of the wire is known to be=1mm
properties=
The density of the wire is 8,000 kg/m3,
heat capacity is 500 J/kgK
themal conductivity is 20W/m.K
electrical resistance per unit length of 0.01 Ω/m
from lump capavity method
[tex]B_{i} =\frac{hr/2}{k}[/tex]
500*(2.5*10^-4)/20
0.006<0.1
we know also, to find steady state temperature
[tex]\pi[/tex]Dh(T-Tinf)=[tex]I^{2} R_{e}[/tex]
make T the subject of the equation , we have
T=25+[tex]\frac{100^2*0.01}{\pi*0.001*500 }[/tex]
T=88.7 degC
rate of chnage in temperature
dT/dt=[tex]\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)[/tex]
at t=o and integrating both sides[tex]\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}[/tex]
we have
[tex]\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}[/tex]
t=8.31s
steady state temperature =88.7deg C
t=time within 1 degC of it steady stae is 8.31s
