Question

The interior of a refrigerator having inside dimensions of 0.5 m × 0.5 m

base area and 1 m height, is to be maintained at 6°C. The walls of the refrigerator are constructed

of two mild steel sheets 3 mm thick (k = 46.5 W/m°C) with 50 mm of glass wool insulation (k =

0.046 W/m°C) between them. If the average heat transfer coefficients at the inner and outer

surfaces are 11.6 W/m2°C and 14.5 W/m2°C respectively, calculate :

(i) The rate at which heat must be removed from the interior to maintain the specified

temperature in the kitchen at 25°C, and

(ii) The temperature on the outer surface of the metal sheet.

Answer 1

The variables that we have are:

Thickness of each mild steel refrigerator wall, L_a = L_c = 3 mm = 0.003 m

Thickness of glass wool insulation wall, L_b = 50 mm = 0.050 m

Temperature of hot fluid (outside), $t_{o} = 25°C$

Temperature of cold fluid (inside of refrigerator), $t_{ref} = 6°C$

Thermal conductivity of mild steel, $k_a=k_c= 46.5 \cfrac{W}{m°C}$

Thermal conductivity of glass wool insulation, $k_b= 0.046 \cfrac{W}{m°C}$

Heat transfer coefficients: Hot fluid (outside), $h_{o} = 11.6 \cfrac{W}{m^2°C}$

Cold fluid (inside of refrigerator), $h_{ref} = 14.5\cfrac{W}{m^2°C}$

First, we use the relation to find the rate at which heat must be removed from the interior to maintain the specified temperature in the kitchen at 25°C as:

$q = UA(t_{o} – t_{i})= \cfrac{A(t_{o} – t_{i})}{\cfrac{1}{h_{o} }+\cfrac{L_a}{k_a}+\cfrac{L_b}{k_b}+\cfrac{L_c}{k_c}+\cfrac{1}{h_{i} }}$ 

Before we proceed, we have to find the area of the refrigerator as

 $A=2\times(0.5\,m)(0.5\,m)+4\times(1\,m)(0.5\,m)=2.5\,m^2$

We substitute in the formula for q and find:

$q = \cfrac{(2.5\,m^2)(25 – 6)°C}{\cfrac{1}{11.6 \frac{W}{m^2°C} }+\cfrac{0.003\,m}{46.5 \frac{W}{m°C}}+\cfrac{0.050\,m}{0.046 \frac{W}{m°C}}+\cfrac{0.003\,m}{46.5 \frac{W}{m°C}}+\cfrac{1}{ 14.5\frac{W}{m^2°C} }}$

$\\ \implies q= 38.24 W$

After that, we can also establish another equation for the heat loss on the outside to find the temperature of the shell:

$q = h_{o}A (t_{o} – t_{1}) \implies t_{1} =t_{o} - \cfrac{q}{h_{o}A}$

We substitute and we can confirm the temperature of the outside surface of the metal sheet:

$t_{1}=25°C-\cfrac{38.24\,W}{(11.6 \frac{W}{m^2°C})(2.5\,m^2)} =23.68 °C$

In conclusion, (i) the rate of heat loss of the refrigerator at which heat must be removed from the interior to maintain the specified temperature in the kitchen at 25°C is 38.24 W, and (ii) the temperature of the outer surface of the metal sheet is 23.68 °C.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.