Absolutely, let's walk through each part of this heat exchanger problem step by step, just like we'd do in a tutoring session: --- ### (a) Rate of Heat Transfer in the Heat Exchanger First, let's find out how much heat is being transferred. Since heat loss to the surroundings is negligible, the heat lost by the hot fluid is gained by the cold fluid. You can use the formula: \[ Q = \dot{m} \cdot c_p \cdot (T_{\text{in}} - T_{\text{out}}) \] Let's use the hot fluid since both should give the same answer. - Mass flow rate, \(\dot{m}_h = 2.5 \, \text{kg/s}\) - Specific heat, \(c_{p,h} = 2.8 \, \text{kJ/kg} \cdot \text{K}\) - Inlet temperature, \(T_{h,in} = 160^\circ \text{C}\) - Outlet temperature, \(T_{h,out} = 90^\circ \text{C}\) Plugging in the numbers: \[ Q = 2.5 \times 2.8 \times (160 - 90) \] \[ Q = 2.5 \times 2.8 \times 70 = 490 \, \text{kJ/s} = 490,000 \, \text{W} \] --- ### (b) Outlet Temperature of the Cooling Water Now, let's figure out how hot the cooling water gets by the time it leaves the exchanger. Use the same heat transfer rate but for the cold fluid: \[ Q = \dot{m}_c \cdot c_{p,c} \cdot (T_{c,out} - T_{c,in}) \] Rearranging: \[ T_{c,out} = \frac{Q}{\dot{m}_c \cdot c_{p,c}} + T_{c,in} \] Plug in the numbers: - \(\dot{m}_c = 3. \, \text{kg/s}\) - \(c_{p,c} = 4.18 \, \text{k/kg} \cdot \text{K}\) - \(T_{c,in} = 30^\circ \text{C}\) \[ T_{c,out} = \frac{490}{3. \times 4.18} + 30 \] \[ T_{c,out} = \frac{490}{12.54} + 30 \approx 39.08 + 30 = 69.08^\circ \text{C} \] --- ### (c) Logarithmic Mean Temperature Difference (LMTD) LMTD is used because the temperature difference between the fluids changes along the length of the exchanger. For counter-current flow: \[ \Delta T_1 = T_{h,in} - T_{c,out} \] \[ \Delta T_2 = T_{h,out} - T_{c,in} \] So, \[ \Delta T_1 = 160 - 69.08 = 90.92^\circ \text{C} \] \[ \Delta T_2 = 90 - 30 = 60^\circ \text{C} \] LMTD formula: \[ \Delta T_{m} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)} \] Plug in the values: \[ \Delta T_{m} = \frac{90.92 - 60}{\ln(90.92/60)} \] \[ \Delta T_{m} = \frac{30.92}{\ln(1.515)} \] \[ \ln(1.515) \approx .415 \] \[ \Delta T_{m} = \frac{30.92}{.415} \approx 74.52^\circ \text{C} \] --- ### (d) Required Heat Transfer Area Now, let's find out how much area is needed to transfer that much heat. The basic heat exchanger equation is: \[ Q = U \cdot A \cdot \Delta T_{m} \] Rearrange for \(A\): \[ A = \frac{Q}{U \cdot \Delta T_{m}} \] Plug in the numbers: - \(Q = 490,000 \, \text{W}\) - \(U = 600 \, \text{W/m}^2\cdot\text{K}\) - \(\Delta T_{m} = 74.52^\circ \text{C}\) \[ A = \frac{490,000}{600 \times 74.52} \] \[ A = \frac{490,000}{44,712} \approx 10.96 \, \text{m}^2 \] --- ### (e) Effectiveness of the Heat Exchanger Effectiveness (\(\epsilon\)) tells us how good the heat exchanger is at transferring the maximum possible heat. It's defined as: \[ \epsilon = \frac{Q}{Q_{\text{max}}} \] \(Q_{\text{max}}\) is the maximum possible heat transfer, which occurs if the fluid with the minimum heat capacity rate (\(C_{\text{min}}\)) is brought up to the temperature difference between the two inlets. Calculate heat capacity rates: \[ C_h = \dot{m}_h \cdot c_{p,h} = 2.5 \times 2.8 = 7. \, \text{kW/K} \] \[ C_c = \dot{m}_c \cdot c_{p,c} = 3. \times 4.18 = 12.54 \, \text{kW/K} \] So \(C_{\text{min}} = 7. \, \text{kW/K}\). Now, \[ Q_{\text{max}} = C_{\text{min}} \times (T_{h,in} - T_{c,in}) = 7. \times (160 - 30) = 7. \times 130 = 910 \, \text{kW} \] So the effectiveness: \[ \epsilon = \frac{490}{910} \approx .538 \] --- ### (f) If Hot-Liquid Flow Rate is Increased by 20%, Find New Outlet Temperature of Hot Liquid If the hot-liquid flow rate increases by 20%, the new mass flow rate is: \[ \dot{m}_{h,\text{new}} = 2.5 \times 1.2 = 3. \, \text{kg/s} \] Now recalculate the heat capacity rates: - \(C_{h,\text{new}} = 3. \times 2.8 = 8.4 \, \text{kW/K}\) - \(C_c = 12.54 \, \text{kW/K}\) - \(C_{\text{min,new}} = 8.4 \, \text{kW/K}\) For the same area, U, and LMTD, the heat transfer rate remains the same as before (\(Q = 490 \, \text{kW}\)), but in reality, increasing flow changes the temperature profiles. However, **if the area is fixed and effectiveness remains the same**, the new exit temperature of hot fluid can be found by: \[ Q = \dot{m}_{h,\text{new}} \cdot c_{p,h} \cdot (T_{h,in} - T_{h,out,new}) \] \[ 490 = 3. \times 2.8 \times (160 - T_{h,out,new}) \] \[ 490 = 8.4 \times (160 - T_{h,out,new}) \] Now, solve for \(T_{h,out,new}\): \[ 160 - T_{h,out,new} = \frac{490}{8.4} = 58.33 \] \[ T_{h,out,new} = 160 - 58.33 = 101.67^\circ \text{C} \] So, the new outlet temperature of the hot liquid would be about **101.7°C**. --- Let me know if you need more explanation on any step!