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.
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
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
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