Question

Question 1: Your bear cooler is made of 20-mm-thick polystyrene (k = 0.021 W/m.K) and has interior dimensions of 0.5 mx0.3 m X 0.3 m. If you fill the inside with an ice-water mixture (so the temperature is maintained at 4°C) and leave the cooler on the porch on hot summer day (ambient temperature of 30°C ), what is the heat flux through the container wall? Assume negligible heat gain through the base of the cooler. You may have to make additional assumption to answer this question.

Answer 1

The thickness of polystyrene, t = 0.02m

Inner dimensions of the box, l*b*h =0.5m * 0.3m * 0.3m

Outer dimensions of the box = (l+2t) * (b+2t) * (c+2t)

= (0.5+2(0.02))m * (0.3+2(0.02))m * (0.3+2(0.02))m

= 0.54m * 0.34m * 0.34m

Ambient temperature, Ta = 30 C = 303K

Inner wall temperature, Ti = temperature of ice-water = 4 C = 277K

Let outer wall temperature be To

Conductivity of polystyrene, k = 0.021 W/mK

Assuming free convection of air to the exposed surface of the cooler, h ~ 5 to 25 W/m²K

Heat transfer coefficient of air at 30C, h = 10 W/m²K -- assumed, for stagnant air

Area for heat transfer, A = 5 faces of the cooler exposed to air

Assuming that no heat transfer occurs at the bottom face of cooler.

Heat flux through cooler by conduction = Heat flux from cooler to air by convection

h* (Ta )k(To-T;) To

0.021 W/mk 10 W (То - 277K) R * (303K -То) 0.02m

10 * (303K To) 1.05W/m2K (To 277K) m2K

9.524(303K To) (To- 277K

2885.7 9.524T0 To-277K

3162.7 10.5247 T

To 300.53 K

So heat flux through cooler, q:

k(To-T) 0.021W mK (300.53K-277K 0.02m

а3 1.05W/m?К * (23.53К) %3D 24.55W/m