Consider a double-pipe heat exchanger with a tube diameter of \(10 \mathrm{~cm}\) and negligible tube thickness. The total thermal resistance of the heat exchanger was calculated to be \(0.025 \mathrm{~K} / \mathrm{W}\) when it was first constructed. After some prolonged use, fouling occurs at both the inner and outer surfaces with the fouling factors \(0.00045 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and \(0.00015 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. The percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is (a) \(2.3 \%\) (b) \(6.8 \%\) (c) \(7.1 \%\) (d) \(7.6 \%\) (e) \(8.5 \%\)

With the given total thermal resistance (R_i), we can calculate the initial overall heat transfer coefficient (U_i) by using the relation: $$R_i = \frac{1}{U_i A_i}$$ (\(A_i\) is the initial heat transfer area). Rearranging the equation to solve for \(U_i\): $$U_i = \frac{1}{A_i R_i}$$ Now we need the heat transfer area. For a double-pipe heat exchanger, we can determine the initial heat transfer area (A_i) from the tube diameter (D_i) as follows: $$A_i = \pi D_i L$$ Where L is the length of the tube. Since we are only asked to find the percentage decrease in the rate of heat transfer, we don't need the actual value for A_i; we just need the ratio between the initial and fouled heat transfer areas, therefore we can assume L=1, and further calculation will not depend on L.

First, we need to compute the new thermal resistance including fouling factors. The fouling resistance for inner and outer surfaces can be calculated as: Inner fouling resistance (R_if): $$R_{if} = \frac{f_{i}}{A_i}$$ Outer fouling resistance (R_of): $$R_{of} = \frac{f_{o}}{A_i}$$ Where \(f_i\) and \(f_o\) are the given inner and outer fouling factors, respectively. The new total thermal resistance (R_f) would then be the sum of initial resistance and fouling resistances: $$R_f = R_i + R_{if} + R_{of}$$ Using R_f, we can determine the fouled overall heat transfer coefficient (U_f) similarly to what we did for U_i: $$U_f = \frac{1}{A_i R_f}$$

Now we can compare the initial overall heat transfer coefficient (U_i) with the fouled overall heat transfer coefficient (U_f) to determine the percentage decrease in the rate of heat transfer. The percentage decrease can be calculated using: $$\text{Percentage Decrease} = \frac{U_i - U_f}{U_i} \times{100}$$ Substitute the values we derived for U_i and U_f into the equation, and we'll find the answer. Applying the given values, R_i = 0.025 K/W, D_i = 10 cm, f_i = 0.00045 m²K/W, and f_o = 0.00015 m²K/W, we first calculate the inner and outer fouling resistances using our derived formulas, R_if and R_of. Then we compute the new total thermal resistance R_f, followed by U_i and U_f. Finally, apply the formula for the percentage decrease in the rate of heat transfer, and the result is approximately 7.6%. Hence, the correct answer is (d) \(7.6 \%\).