In a parallel-flow, liquid-to-liquid heat exchanger, the inlet and outlet temperatures of the hot fluid are \(150^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) while that of the cold fluid are \(30^{\circ} \mathrm{C}\) and \(70^{\circ} \mathrm{C}\), respectively. For the same overall heat transfer coefficient, the percentage decrease in the surface area of the heat exchanger if counter-flow arrangement is used is (a) \(3.9 \%\) (b) \(9.7 \%\) (c) \(14.5 \%\) (d) \(19.7 \%\) (e) \(24.6 \%\)

Let's denote the mass flow rate of the hot fluid as \(m_h\), the mass flow rate of the cold fluid as \(m_c\), the specific heat capacities of hot and cold fluids as \(C_{p_h}\) and \(C_{p_c}\), respectively. The heat capacities of the fluids are calculated as: \(C_h = m_h C_{p_h}\) and \(C_c = m_c C_{p_c}\)

The heat exchanger effectiveness (\(\epsilon\)) is defined as the ratio of the actual heat transfer rate to the maximum possible heat transfer rate in the heat exchanger. For parallel flow, we have: \(\epsilon_{PF} = \frac{T_{h, in} - T_{h, out}}{T_{h, in} - T_{c, in}} \) Plug in the given temperatures and get: \(\epsilon_{PF} = \frac{150 - 90}{150 - 30} = \frac{60}{120} = 0.5\)

The heat capacities ratio (\(C_r\)) and the Number of Transfer Units (NTU) for the parallel-flow heat exchanger can be calculated using the effectiveness (\(\epsilon\)), assuming that there are no phase changes occurring in the fluids and the overall heat transfer coefficient remains constant: \(C_r = \frac{C_c}{C_h}\) and \(NTU_{PF} = \frac{\epsilon_{PF} (1 - C_r)}{1 - C_r \cdot e^{-NTU_{PF} (1 - C_r)}}\) As we don't know the mass flow rates or specific heat capacities of the fluids, we can't directly calculate \(C_r\) and \(NTU_{PF}\). However, we can use another formula to express the effectiveness of counter-flow heat exchanger in terms of NTU and \(C_r\):

For the counter-flow heat exchanger, effectiveness \(\epsilon_{CF}\) can be calculated with the same NTU and \(C_r\) as in the parallel flow: \(\epsilon_{CF} = \frac{1 - e^{-NTU_{CF} (1 + C_r)}}{1 + C_r}\) We know that the overall heat transfer coefficient is the same for both parallel-flow and counter-flow heat exchangers. Therefore, the NTU for the two arrangements is the same: \(NTU_{CF} = NTU_{PF}\) Substitute \(\epsilon_{PF}\), \(C_r\), and \(NTU_{CF}\) into the \(\epsilon_{CF}\) equation: \(\epsilon_{CF} = \frac{1 - e^{-NTU_{PF} (1 + C_r)}}{1 + C_r}\)

The area ratio can be calculated by using the relationship between NTU and area of the heat exchanger. The NTU is defined as: \(NTU = \frac{UA}{C_\text{min}}\) Where \(U\) is the overall heat transfer coefficient, \(A\) is the heat transfer area, and \(C_\text{min}\) is the minimum heat capacity of the fluids. Since \(U\) and \(C_\text{min}\)are constant for both heat exchangers, the area ratio is equal to the NTU ratio: \(\frac{A_{PF}}{A_{CF}} = \frac{NTU_{PF}}{NTU_{CF}}\) As \(NTU_{PF} = NTU_{CF}\), we get: \(\frac{A_{PF}}{A_{CF}} = 1\)

To find the percentage decrease in surface area when switching from a parallel-flow to a counter-flow heat exchanger, we can use the area ratio calculated in Step 5: Percentage decrease in surface area \(= \frac{A_{PF} - A_{CF}}{A_{PF}} \times 100\%\) Since \(\frac{A_{PF}}{A_{CF}} = 1\), the percentage decrease in surface area is: Percentage decrease in surface area \(= \frac{1 - 1}{1} \times 100\% = 0\%\) The correct answer is none of the given options (a), (b), (c), (d), or (e) because the surface area doesn't decrease when switching from a parallel-flow to a counter-flow heat exchanger when the overall heat transfer coefficient is constant. Note: The problem statement appears to be incorrect or missing important information because there is no decrease in the heat exchanger area with the given data. The answer should be revised with appropriate premises, or the problem statement needs to be more precise.