Question

An inexpensive food and beverage container is fabricated from 25-mni-thick polystyrene (k = 0.023 W/m K) and has interior dimensions of 1.2 m × 1.4 m × 1.2 m. Under conditions for which an inner surface temperature of approximately 2ºC is maintained by an ice-water mixture and an outer surface temperature of 30ºC is maintained by the ambient, what is the heat flux through the container wall? Assuming negligible heat gain through the 1.4 m × 1.2 m base of the cooler, what is the total heat load for the prescribed conditions?

Answer 1

Answer:

The total heat load is 230.81 W

Explanation:

The heat flux is given as

                                     $q^*=k\frac{T_2-T_1}{L}$

Here

• q* is the heat flux
• k is the thermal conductivity of the polystyrene
• T2 is the outer temperature which is 30 C
• T1 is the inner temperature which is 2 C
• L is the thickness which is given 25 mm

                                        $q^*=k\frac{T_2-T_1}{L}\\q^*=0.023 \frac{30-2}{0.025}\\q^*=25.76 W/m^2$

Overall heat is given as

$q=q^* \times A\\q=25.76 \times [H(2W_1+2W_2)+W_1 \times W_2]\\q=25.76 \times [1.4(2 \times 1.2 +2 \times 1.4)+1.4\times 1.2]\\q=25.76 \times 8.96\\q=230. 81 W$

So the total heat load is 230.81 W