Let's solve the problem step by step: ### 1. Calculate Heat Duty (Q) First, convert all units to SI (Joules, seconds, Kelvin). Given: - Hot fluid: \(\dot{m}_h = 1.8\, \text{kg/s}\), \(C_{p,h} = 3.\, \text{kJ/kg·K} = 300\, \text{J/kg·K}\), \(T_{h,in} = 160^\circ\text{C}\) - Cold fluid: \(\dot{m}_c = 2.5\, \text{kg/s}\), \(C_{p,c} = 4.2\, \text{kJ/kg·K} = 420\, \text{J/kg·K}\), \(T_{c,in} = 30^\circ\text{C}\) The maximum possible heat transfer is limited by the fluid with the smaller heat capacity rate: \[ C_h = \dot{m}_h \cdot C_{p,h} = 1.8 \times 300 = 540\, \text{W/K} \] \[ C_c = \dot{m}_c \cdot C_{p,c} = 2.5 \times 420 = 10500\, \text{W/K} \] Since \(C_h < C_c\), the hot fluid determines the maximum possible heat transfer. \[ Q = C_h \cdot (T_{h,in} - T_{h,out}) \] But \(T_{h,out}\) is unknown. Let's use the energy balance: \[ Q = C_h (T_{h,in} - T_{h,out}) = C_c (T_{c,out} - T_{c,in}) \] We need to find \(T_{h,out}\) and \(T_{c,out}\). For a counter-current heat exchanger: \[ \frac{T_{h,in} - T_{c,out}}{T_{h,out} - T_{c,in}} = \exp\left(-\frac{UA}{C_h}\left(1 - \frac{C_h}{C_c}\right)\right) \] But without the area, use the effectiveness-NTU method for counter-current exchangers: Effectiveness \(\epsilon = \frac{Q}{Q_{max}}\), where \(Q_{max} = C_{min}(T_{h,in} - T_{c,in})\). Assume maximum possible transfer (\(T_{h,out} = T_{c,in}\)), then \[ Q_{max} = C_{min}(T_{h,in} - T_{c,in}) = 540 \times (160 - 30) = 540 \times 130 = 702,000\, \text{W} = 702\, \text{kW} \] But actual \(Q\) depends on area; since area is unknown, assume maximum for now: **Heat duty \(Q = 702\, \text{kW}\).** ### 2. Outlet Temperature of Cold Fluid (\(T_{c,out}\)) \[ Q = C_c (T_{c,out} - T_{c,in}) \] \[ 702,000 = 10,500 (T_{c,out} - 30) \] \[ T_{c,out} = \frac{702,000}{10,500} + 30 = 66.86 + 30 = 96.86^\circ\text{C} \] **Outlet temperature of cold fluid is \(96.9^\circ\text{C}\) (rounded to one decimal).** ### 3. Log Mean Temperature Difference (LMTD) For counter-current: \[ \Delta T_1 = T_{h,in} - T_{c,out} = 160 - 96.86 = 63.14^\circ\text{C} \] \[ \Delta T_2 = T_{h,out} - T_{c,in} = 30 - 30 = ^\circ\text{C} \] But \(T_{h,out}\) isn't 30°C exactly (as the fluids don't exchange all their heat in practice), so let's check energy balance: \[ Q = C_h (T_{h,in} - T_{h,out}) \] \[ 702,000 = 5,400 (160 - T_{h,out}) \] \[ 160 - T_{h,out} = \frac{702,000}{5,400} = 130 \] \[ T_{h,out} = 160 - 130 = 30^\circ\text{C} \] So, calculated as above. \[ LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln\left( \frac{\Delta T_1}{\Delta T_2} \right)} \] But \(\Delta T_2 = \), so LMTD approaches zero, which is not physical. In real cases, the outlet temperature of hot fluid can't reach the inlet temperature of the cold fluid. Let's backtrack: The maximum possible \(T_{c,out}\) is less than \(T_{h,in}\), so let's use heat capacity rates: \[ T_{c,out} = T_{c,in} + \frac{C_h}{C_c}(T_{h,in} - T_{c,in}) \] \[ T_{c,out} = 30 + \frac{540}{10500}(160 - 30) = 30 + .514 \times 130 = 30 + 66.86 = 96.86^\circ \text{C} \] So values are consistent. **LMTD calculation:** \[ \Delta T_1 = 160 - 96.86 = 63.14^\circ\text{C} \] \[ \Delta T_2 = 30 - 30 = ^\circ\text{C} \] But again, LMTD is not defined for \(\Delta T_2 = \). In practice, outlets don't match exactly, so let's assume \(T_{h,out}\) is slightly higher, say 31°C, so: \[ \Delta T_2 = 31 - 30 = 1^\circ\text{C} \] Now, \[ LMTD = \frac{63.14 - 1}{\ln(63.14/1)} = \frac{62.14}{4.146} = 14.99^\circ\text{C} \] **LMTD is approximately \(15^\circ\text{C}\).** ### 4. Required Heat Transfer Area (A) Given \( U = 700\, \text{W/m}^2\cdot\text{K} \): \[ Q = UA \cdot LMTD \] \[ 702,000 = 700 \times A \times 15 \] \[ A = \frac{702,000}{700 \times 15} = \frac{702,000}{10,500} = 66.86\, \text{m}^2 \] **Required heat transfer area is \(66.9\, \text{m}^2\) (rounded).** ### 5. Effect if Exchanger is Co-current Instead For co-current flow, the outlet temperature of the cold fluid will be lower than with counter-current, and the LMTD will be lower for the same area, making the heat exchanger less efficient. The maximum temperature that the cold fluid can reach in co-current flow is always less than the outlet temperature in counter-current flow. Therefore, for the same area and operating conditions, the heat recovery is lower in a co-current exchanger. --- **Final Answers Summary:** 1. Heat duty \(Q = 702\, \text{kW}\) 2. Outlet temperature of cold fluid \(T_{c,out} = 96.9^\circ\text{C}\) 3. LMTD \(\approx 15^\circ\text{C}\) 4. Required heat transfer area \(A = 66.9\, \text{m}^2\) 5. If the exchanger is co-current instead, the outlet temperature of the cold fluid will be lower and the exchanger will be less efficient.