Calculate the thermal resistance for each layer.

The overall coefficient of heat transfer $U$ is the reciprocal of the total thermal resistance $R_{total}$. The total thermal resistance is the sum of individual thermal resistances for each layer, given by $R_i = \frac{x_i}{K_i}$, where $x_i$ is the thickness and $K_i$ is the thermal conductivity.

First, convert all thicknesses from mm to m:

• Air film: $x_1 = 1.5 \text{ mm} = 0.0015 \text{ m}$
• Condensate film: $x_2 = 1.0 \text{ mm} = 0.0010 \text{ m}$
• Scale (i): $x_3 = 2.0 \text{ mm} = 0.0020 \text{ m}$
• Copper: $x_4 = 10 \text{ mm} = 0.010 \text{ m}$
• Scale (ii): $x_5 = 1.5 \text{ mm} = 0.0015 \text{ m}$
• Water film: $x_6 = 2.0 \text{ mm} = 0.0020 \text{ m}$

Now, calculate the thermal resistance for each layer:

$$R_1 = \frac{0.0015 \text{ m}}{0.03 \text{ W/mK}} = 0.05 \text{ m}^2\text{K/W}$$ $$R_2 = \frac{0.0010 \text{ m}}{0.60 \text{ W/mK}} \approx 0.001667 \text{ m}^2\text{K/W}$$ $$R_3 = \frac{0.0020 \text{ m}}{0.40 \text{ W/mK}} = 0.005 \text{ m}^2\text{K/W}$$ $$R_4 = \frac{0.010 \text{ m}}{379 \text{ W/mK}} \approx 0.000026 \text{ m}^2\text{K/W}$$ $$R_5 = \frac{0.0015 \text{ m}}{0.40 \text{ W/mK}} = 0.00375 \text{ m}^2\text{K/W}$$ $$R_6 = \frac{0.0020 \text{ m}}{0.60 \text{ W/mK}} \approx 0.003333 \text{ m}^2\text{K/W}$$

$$R_{total} = R_1 + R_2 + R_3 + R_4 + R_5 + R_6$$ $$R_{total} = 0.05 + 0.001667 + 0.005 + 0.000026 + 0.00375 + 0.003333$$ $$R_{total} \approx 0.063776 \text{ m}^2\text{K/W}$$

$$U = \frac{1}{R_{total}} = \frac{1}{0.063776 \text{ m}^2\text{K/W}}$$ $$U \approx 15.68 \text{ W/m}^2\text{K}$$

The overall coefficient of heat transfer is $\boxed{\text{15.68 W/m}^2\text{K}}$.

b) What happens to the value of the overall coefficient of heat transfer if:

i) Copper is to be changed with Steel (lower 'K' value)?
If copper is replaced by steel, which has a lower thermal conductivity ($K$), the thermal resistance of that layer ($R = x/K$) will increase. Since the overall heat transfer coefficient $U = 1/R_{total}$, an increase in $R_{total}$ will lead to a decrease in the overall heat transfer coefficient $U$.

ii) The values of first and second scales were to be reduced to half?
If the thicknesses of Scale (i) and Scale (ii) are reduced to half, their individual thermal resistances ($R = x/K$) will decrease. This reduction in individual resistances will decrease the total thermal resistance $R_{total}$. Consequently, the overall heat transfer coefficient $U = 1/R_{total}$ will increase.