A 2-cm-diameter bare aluminum electric power transmission line with an emissivity of 0.07 carries 500 A at 400 kV. The wire has an electrical resistivity of 1.72 micro-ohms cm2/cm at 20°C and is suspended horizontally between two towers separated by 1 km. Determine the surface temperature of the transmission line if the air temperature is 20°C. What fraction of the dissipated power is due to radiation heat transfer?

Question

GIVEN

• An aluminum electric power transmission line suspended horizontally

• Emissivity (?) in air = 0.3

• Line diameter (D) = 2 cm = 0.02 m

• Current (I) = 500 amp

• Voltage (V) = 400 kV

• Electrical resistivity (?e) = 1.72 ? cm2/cm at 20°C

• Space between towers (L) = 1 km

• Air temperature (T?) = 20°C = 293 K

FIND

(a) Surface temperature of wire (Tw)

(b) Fraction of dissipated power due to radiation

ASSUMPTIONS

• Steady state

• The wire radiates to the surroundings which behave as a black body enclosure at T? SKETCH



PROPERTIES AND CONSTANTS

the Stephan-Boltzmann constant (?) = 5.67 × 10–8 W/m2 K4.

Answer 1

The power dissipation is given by Ohm’s Law



This must equal the rate of heat transfer by convection and radiation per meter



Since hc varies with Tw, an iterative procedure must be used.

For the first iteration, let Tw = 60°C.

for dry air at the mean temperature of 40°C

Thermal expansion coefficient (?) = 0.00319 1/K

Thermal conductivity (k) = 0.0265 W/(m K)

Kinematic viscosity (?) = 17.6 × 10–6 m2/s

Prandtl number (Pr) = 0.71

The Grashof number based on the wire diameter is



The Nusselt number for this geometry and Grashof number is



Checking the units then eliminating them for clarity



By trial and error Tw = 317 K = 44°C

Performing further iterations



The equilibrium surface temperature is 46°C.

(b) The rate of heat transfer by convection is



The rate of heat transfer by radiation is



As a check on the results



The fraction of the power dissipation by radiation is