A 2.5-cm-thick sheet of plastic initially at 21°C is placed between two heated steel plates that are maintained at 138°C. The plastic is to be heated just long enough for its midplane temperature to reach 132°C. If the thermal conductivity of the plastic is 1.1 × 10–3 W/(m K), the thermal diffusivity is 2.7 × 10–6 m2/s, and the thermal resistance at the interface between the plates and the plastic is negligible, calculate: (a) the required heating time, (b the temperature at a plane 0.6 cm from the steel plate at the moment the heating is discontinued, and (c) the time required for the plastic to reach a temperature of 132°C 0.6 cm from the steel plate.

Question

GIVEN

• A sheet of plastic is placed between two heated steel plates

• Sheet thickness (2L) = 2.5 cm = 0.025 m

• Initial temperature (To) = 21°C

• Temperature of steel plates (Ts) = 138°C

• Heat until mid-plane temperature of sheet (Tc) = 132°C

• The thermal conductivity of the plastic (k) = 1.1 × 10–3 W/(m K)

• The thermal diffusivity (?) = 2.7 × 10–6 m2/s

• The thermal resistance at the interface between the plates and the plastic is negligible

FIND

(a) The required heating time (b) The temperature at a plane 0.6 cm from the steel plate at the moment the heating is discontinued (c) The time required for the plastic to reach a temperature of 132°C 0.6 cm from the steel.

ASSUMPTIONS

• The initial temperature of the sheet is uniform

• The temperature of the steel plates is constant

• The thermal conductivity of the sheet is constant

SKETCH

Answer 1

The significance of the internal resistance is determined from the Biot number



Therefore, the internal resistance is significant and the approximate solution will be used.

At center of rubber x=0, we have





for infinite slab