Question

In a hypothetical experiment, electrical current of 15 AMPs driven by an applied potential of 100 volts passes through a cylindrical rod (3 mm in diameter and 7 cm long). Recall, the power is calculated by power(Watts) = potential(VOLTS) x current(AMPS). The electrical power heats the cylindrical rod to a high temperature. The rod is inside of a ceramic tube (ID = 12 mm). The wall temperature of the ceramic tube is 600 C (maintained at this stead temperature by air cooling on the outside). Assuming that convection and conduction between the resistor and the ceramic tube are negligible compared to thermal radiation, estimate the surface temperature of the rod. The emissivity of the ceramic tube is 0.9 and the emissivity of the resistor is 0.8.

Answer 1

In this problem two cylinders are placed concentrically ( or one in side the other) and heat exchange takes places between outter surface of inner cylinder and inner surface of outter cylinder by radiation. Heat transfer by conduction and convection are absent or negligible.

Net heat transfer between two cylindrical surfaces as discribed above can be expressed mathematically as

Q - Net heat transfer [Current (I) X Potential (V)]

Q = I * V = 15A * 100V = 1500 W { I=15A, V= 100V }

- Stefan–Boltzmann constant= 5.67 X 10^-8 w/m² k⁴

T₁ - surface temperature of inner cylinder(Rod) in Kelvin

T₂- Surface temperature of outter cylinder(ceramic tube) = 600^oC = 873 K

D₁ - Outter dia of rod = 3mm = 0.003m

D₂ - Inner dia of ceramic tube = 12mm =0.012m

L - length of cylinder = 7cm =0.07m

A₁ - Outter surface area of inner cylinder = D₁L = 3.14 * 0.003 * 0.07 = 0.6594 X 10^-3 m²

A₂ - Inner surface area of outter cylinder = D₂L = 3.14 * 0.012 * 0.07 = 2.6376 X 10^-3 m²

- ^{Emissivity of inner cylinder(rod) = 0.8}

^{Emissivity of outter cylinder(cerami) = 0.9}

Substituting all the values in above equation and solving for T₁

T₁ = 2654.03 K = 2381.03 ^oC [ surface temperature of rod]