Question

A domestic freezer has an internal capacity of 1m * 10.5m * 0.5m. The temperature inside is maintained at -5°c in ambient temperature of up to 35°c. Calculate the maximum rate of heat transfer to the freezer if the wall and door consists of:
5mm plastic. K = 1W/mK
15mm insulation K = 0.05W/mK
2mm steel. K = 40W/mK
and the convective heat transfer coefficient is 8W/m²K on both sides of the freezer

Answer 1

Question # 79002

A domestic freezer has an internal capacity of $1\mathrm{m} \times 10.5\mathrm{m} \times 0.5\mathrm{m}$. The temperature inside is maintained at $-5^{\circ}\mathrm{C}$ in ambient temperature of up to $35^{\circ}\mathrm{C}$. Calculate the maximum rate of heat transfer to the freezer if the wall and door consists of:

- 5 mm plastic. K = 1 W/mK

- 15 mm insulation. K = 0.05 W/mK

- 2 mm steel. K = 40 W/mK

and the convective heat transfer coefficient is $8\mathrm{W/m^2K}$ on both sides of the freezer

Answer:

The maximum rate of the heat transfer $Q$ to the freezer is given by:

$$Q = U A (T_{out} - T_{in}),$$

where

$$U = \left(\frac{1}{h_{in}} + \frac{t_1}{K_1} + \frac{t_2}{K_2} + \frac{t_3}{K_3} + \frac{1}{h_{out}}\right)^{-1},$$

is the overall heat transfer coefficient,

$h_{in} = h_{out} = 8\mathrm{W/m^2K}$ — the convective heat transfer coefficients at the inner and outer surfaces,

$t_1 = 0.005\mathrm{m}, t_2 = 0.015\mathrm{m}, t_3 = 0.02\mathrm{m}$ — the thickness of the plastic, the insulation and the steel layers, respectively,

$K_1 = 1\mathrm{W/m^2K}, K_2 = 0.05\mathrm{W/m^2K}, K_3 = 40\mathrm{W/m^2K}$ — the thermal conductivity coefficient of the plastic, the insulation and the steel, respectively.

$$A = 2 (ab + bc + ac),$$

the area of the freezer walls,

$a = 1\mathrm{m}, b = 10.5\mathrm{m}, c = 0.5\mathrm{m}$ — the dimensions of the freezer,

$T_{in} = -5^{\circ}\mathrm{C}, T_{out} = 35^{\circ}\mathrm{C}$ — the temperature inside and outside the freezer, respectively.

Substitute into (3), (2) and (1):

$$\begin{aligned} A &= 2 (1 \cdot 10.5 + 10.5 \cdot 0.5 + 1 \cdot 0.5) = 32.5 \mathrm{m^2}, \\ U &= \left(\frac{1}{8} + \frac{0.005}{1} + \frac{0.015}{0.05} + \frac{0.002}{40} + \frac{1}{8}\right)^{-1} = 1.802 \mathrm{W/m^2K}, \\ Q &= 1.802 \cdot 32.5 \cdot (35 + 5) = 2342 \mathrm{W}. \end{aligned}$$