Let's solve the plate-fin (crossflow) heat exchanger **condenser design** step by step--- ## **Given Data- **Reant:** R134a (condensing at 40°C) - **Condensation rate:** \( \dot{m}_{\text{refr}} = .015\, \text{kg/s} \) - **Air inlet temperature:** \( T_{a,i} = 30^\circ C \) - **Air velocity at inlet:** \( V_{a,i} = 2\, \text{m/s} \) - **Overall heat transfer coefficient:** \( U = 50\, \text{W}/\text{m}^2\cdot\text{K} \) - **Convective h (refrigerant side):** \( h_{\text{refr}} = 500\, \text{W}/\text{m}^2\cdot\text{K} \) - **Frontal area of condenser:** \( A_{fr} = .25\, \text{m}^2 \) - **Two-tube passes** - **Find:** (a)-(f) --- ## **Step 1: Calculate Heat Duty (\( Q \))** Condensing vapor means all vapor is condensed at constant temperature (\( T_{\text{sat}} = 40^\circ C \)), so \[ Q = \dot{m}_{\text{refr}} \cdot h_{fg} \] From **R-134a tables** at 40°C: - \( h_{fg} \) (Enthalpy of vaporization) ≈ **150 kJ/kg** = **150,000 J/kg** \[ Q = .015\, \text{kg/s} \times 150,000\, \text{J/kg} = \boxed{225\, \text{W}} \] --- ## **Step 2: Calculate Log Mean Temperature Difference (LMTD)** For a **crossflow condenser (both fluids unmixed):** \[ \Delta T_1 = T_{\text{refr,sat}} - T_{a,i} \] \[ \Delta T_2 = T_{\text{refr,sat}} - T_{a,o} \] But \( T_{a,o} \) (air outlet temperature) is not given directly. We'll use energy balance: \[ Q = \dot{m}_a c_{p,a} (T_{a,o} - T_{a,i}) \] Let’s **assume properties for air:** - \( c_{p,a} = 1005\, \text{J/kg.K} \) - Air density (\( \rho_a \)) ~ 1.2 kg/m³ **Air mass flow rate:** \[ \dot{m}_a = \rho_a \cdot V_{a,i} \cdot A_{fr} \] \[ \dot{m}_a = 1.2\, \text{kg/m}^3 \cdot 2\, \text{m/s} \cdot .25\, \text{m}^2 = .6\, \text{kg/s} \] Now, \[ Q = \dot{m}_a c_{p,a} (T_{a,o} - T_{a,i}) \] \[ 225 = .6 \cdot 1005 \cdot (T_{a,o} - 30) \] \[ 225 = 603 (T_{a,o} - 30) \] \[ T_{a,o} - 30 = \frac{225}{603} \approx 3.73 \] \[ T_{a,o} \approx 33.73^\circ C \] #### **Now calculate LMTD:** \[ \Delta T_1 = 40 - 30 = 10^\circ C \] \[ \Delta T_2 = 40 - 33.73 = 6.27^\circ C \] \[ \Delta T_m = \frac{10 - 6.27}{\ln(10 / 6.27)} = \frac{3.73}{\ln(1.595)} = \frac{3.73}{.467} \approx 8.^\circ C \] --- ## **Step 3: Required Heat Transfer Area** \[ Q = U \cdot A_{hx} \cdot \Delta T_m \] \[ A_{hx} = \frac{Q}{U \Delta T_m} = \frac{225}{50 \times 8} = \frac{225}{400} = 5.63\, \text{m}^2 \] --- ## **Step 4: Tube Design Calculation** ### **(a) Tube Length and Diameter** Assume typical **tube diameter \( d \) = 1 cm = .01 m** (you can adjust if needed). Let’s find the **total tube length required**: - For two-pass, total active surface area: \[ A_{hx} = N_{tubes} \times \pi d L_{tube} \] \[ 5.63 = N_{tubes} \times \pi \times .01 \times L_{tube} \] \[ N_{tubes} \times L_{tube} = \frac{5.63}{\pi \times .01} \approx \frac{5.63}{.0314} \approx 179.3\, \text{m} \] Assume **length per tube** = 1 m (typical for compact heat exchangers): \[ N_{tubes} = \frac{179.3}{1} \approx 179 \text{ tubes} \] **Or** if you want to use fewer tubes, increase length accordingly. ### **(b) Number of Tubes:** \[ \boxed{179 \text{ tubes (if length = 1 m each, d = 1 cm)}} \] ### **(c) Number of Fins (Plates)** Assume fin spacing = 2 mm = .002 m. \[ \text{Number of fins} = \frac{L_{tube}}{\text{fin spacing}} = \frac{1}{.002} = 500 \text{ fins per tube} \] --- ### **(d) Refrigerant Velocity** Each tube area: \[ A_{tube} = \frac{\pi}{4} d^2 = \frac{\pi}{4} (.01)^2 = 7.85 \times 10^{-5}\, \text{m}^2 \] Total cross-sectional area (for 179 tubes): \[ A_{total} = 179 \times 7.85 \times 10^{-5} = .01405\, \text{m}^2 \] \[ V_{\text{refr}} = \frac{\dot{m}_{\text{refr}}}{\rho_{\text{refr}} \cdot A_{total}} \] At 40°C, liquid R-134a density ≈ 111 kg/m³ \[ V_{\text{refr}} = \frac{.015}{111 \times .01405} = \frac{.015}{15.5955} \approx .00096\, \text{m/s} \] --- ### **(e) Heat Transfer Area** Already found: \[ A_{hx} = \boxed{5.63\, \text{m}^2} \] --- ### **(f) Air Outlet Temperature** Already found: \[ T_{a,o} = \boxed{33.73^\circ C} \] --- ## **Summary Table** | Parameter | Value | |-----------------------------------|---------------------------| | (a) Tube length, diameter | 1 m, 1 cm | | (b) Number of tubes | 179 | | (c) Number of fins (per tube) | 500 | | (d) Refrigerant velocity | .00096 m/s | | (e) Heat transfer area | 5.63 m² | | (f) Air outlet temperature | 33.73°C | --- ### **If you want to change assumptions (e.g., tube diameter or length), recalculate as shown above!** Let me know if you need any explanation or want to see how things change with different tube dimensions!