Question

Fresh food product is held in cold storage at 278K. It is packed in a container in the shape of a slab with all facesinsulated except for the top flat surface, which is exposed to airat 278oK. The slab is 154.2 mm thick and the exposedsurface area is 0.18m2. The density of the foodstuff is 641 kg/m3. The heat of respiration is 0.070 kJ/kg-hr.and the thermal conductivity is 0.346 W/mK.

Answer 1

To calculate : 1) temperature in the food product at steady state and 2) total heat given off in W

SOLUTION:

Given that heat of respiration = heat of generation = q^. = 0.070 KJ/ Kg.hr = 0.070 (KJ/ Kg.hr) x 641 (kg/m3) = 44.87 KJ/ m³ hr = 44.87 X 10³ /3600 J/ m³ sec = 12.464 J/ m³ sec

Length of slab , L = 152.4 mm = 0.1524 m

Thermal conductivity , k = 0.346 W/m K

For the problems of conduction with heat generation, at steady state we have

T_o,max = T_w + (q^. L²/2k) = 278 + (12.464 X (0.1524)² / (2 x 0.346) = 278.42 K

Total heat given off, q^._total = q^. x Volume of slab = 12.464 J/ m³ sec x (0.186 m² x 0.1524 m)

= 0.3533 W