Question

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting of 1.25 cm thick plywood (k = 0.104 W/m-K) on two sides with 5 cm thick Styrofoam (k = 0.04 W/m-K) in the middle. On a cold day when the exterior temperature is 0 C what is the heat loss from the wall per unit area? What are the interface temperatures between the plywood and Styrofoam?

Answer 1

`(\dot Q)/(A) =20.129\ W.m^(-2)`

`T_1=27.58\ ^(\circ)C` &
`T_2=2.41875\ ^(\circ)C`

Step-by-step explanation:

Given:

• interior temperature of box,
  `T_i=30^(\circ)C`

• height of the walls of box,
  `h=3\ m`

• thickness of each layer of bi-layered plywood,
  `x_p=1.25\ cm=0.0125\ m`

• thermal conductivity of plywood,
  `k_p=0.104\ W.m^(-1).K^(-1)`

• thickness of sandwiched Styrofoam,
  `x_s=5\ cm=0.05\ m`

• thermal conductivity of Styrofoam,
  `k_s=0.04\ W.m^(-1).K^(-1)`

• exterior temperature,
  `T_o=0^(\circ)C`

From the Fourier's law of conduction:

`\dot Q=(dT)/(((x)/(kA)) )`

`\dot Q=(dT)/(R_(th) )` ....................................(1)

Now calculating the equivalent thermal resistance for conductivity using electrical analogy:

`R_(th)=R_p+R_s+R_p`

`R_(th)=(x_p)/(k_p.A)+(x_s)/(k_s.A)+(x_p)/(k_p.A)`

`R_(th)=(1)/(A) ((x_p)/(k_p)+(x_s)/(k_s)+(x_p)/(k_p))`

`R_(th)=(1)/(A) ((0.0125)/(0.104)+(0.05)/(0.04)+(0.0125)/(0.104))`

`R_(th)=(1.4904)/(A)` .....................(2)

Putting the value from (2) into (1):

`\dot Q=(30-0)/((1.4904)/(A) )`

`\dot Q=(30\ A)/(1.4904)`

`(\dot Q)/(A) =20.129\ W.m^(-2)` is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

For plywood-Styrofoam interface from inside:

`(\dot Q)/(A) =k_p.(T_i-T_1)/(x_p)`

`20.129=0.104* (30-T_1)/(0.0125)`

`T_1=27.58\ ^(\circ)C`

&For Styrofoam-plywood interface from inside:

`(\dot Q)/(A) =k_s.(T_1-T_2)/(x_s)`

`20.129=0.04* (27.58-T_2)/(0.05)`

`T_2=2.41875\ ^(\circ)C`