A 3.2mm stainless steel wire, 30cm long has a voltage of 10 volt impressed on it. The outer surface temperature of the wire is maintained at $93^0C$. Calculate the center temperature of the wire. Take resistivity ($\rho$) of the wire as $70×10^{-8}$ ohm-m and the thermal conductivity as 22.5W/m-K.

Given,

d = 3.2mm = 0.0032m (Diameter of wire)

l = 30cm = 0.3m (Length of wire)

V = 10V (Voltage)

ts = 93℃ = 366K (Surface temperature of wire)

$ρ = 70×10^{-8}Ωm$ (Specific resistance of wire)

K = 22.5W/mK (Thermal conductivity of wire material)

Find: (1) tc (center temperature of wire)

Solution:

Electrical resistance

Re = ρ.lAc = ρ.lπ4 × d2 = 70 × 10 - 8 × 0.3π4 × 0.00322

Re = 0.0261Ω

Heat transfer rate

Q = I2Re = (VRe)2Re = V2Re

Q = 1020.0261

Q = 3831.417 W

Heat transfer rate/Volume

qg = QVolume = Qπ4 × d2 × l = 3831.417π4 × 0.00322 × 0.3

qg = 1587.992 × 106 W/m3

Temperature distribution equation

tc - ts = qgR24K

tc - 366 = 1587.992 × 106 × 0.001624 × 22.5

tc = 411.169K = 138.169℃