**Subject:** Thermodynamics **Topic:** Entropy Generation in a Heat Exchanger (Window Air Conditioner) --- ## Step 1: **Given Data & Simple Conceptual Introduction** ### **Given Data:** - **Air:** - Inlet pressure = 100 kPa - Inlet temperature = 27°C (300.15 K) - Volume flow rate = 6 m³/min = 0.1 m³/s - Leaves at same pressure - **Refrigerant-134a (R-134a):** - Inlet pressure = 120 kPa - Inlet quality (x₁) = 0.3 - Mass flow rate = 2 kg/min = 0.0333 kg/s - Leaves as saturated vapor (x₂ = 1) at 120 kPa - **Assume:** - Insulated system (no heat loss to surroundings) - Air leaves at same pressure - Constant specific heats for air at 300K ### **Conceptual Introduction:** We are to find the **rate of entropy generation** (\(\dot{S}_{gen}\)) in the evaporator section. This requires an entropy balance for the control volume involving both air and refrigerant streams. **Explanation block:** This problem involves applying the entropy balance equation to a well-insulated heat exchanger where air and refrigerant R-134a exchange heat. The change in entropy for both streams, considering their mass flow rates and states, will allow calculation of the entropy generation rate. --- ## Step 2: **Step-by-Step Complete Calculation Part (Air Side)** ### **1. Calculate mass flow rate of air (\(\dot{m}_{air}\)):** \[ \dot{m}_{air} = \frac{\dot{V}}{v} \] Where: - \(\dot{V} = 0.1 \text{ m}^3/\text{s}\) - \(v\) = specific volume of air at 100 kPa, 27°C For air (ideal gas): \[ v = \frac{RT}{P} \] Where: - \(R = 0.287 \text{ kJ/kg·K}\) - \(T = 300.15 \text{ K}\) - \(P = 100 \text{ kPa} = 100,000 \text{ Pa}\) \[ v = \frac{0.287 \times 300.15}{100} = 0.86143 \text{ m}^3/\text{kg} \] \[ \dot{m}_{air} = \frac{0.1}{0.86143} = 0.116 \text{ kg/s} \] ### **2. Entropy Change of Air:** Assuming air leaves at the same pressure, but temperature changes (let \(T_{out,air}\) be the exit temperature): \[ \Delta s_{air} = c_p \ln\left(\frac{T_{out,air}}{T_{in,air}}\right) \] Assume \(c_p \approx 1.005 \text{ kJ/kg·K}\) for air. **But we do not know \(T_{out,air}\) yet.** We need to relate the heat transferred to the refrigerant. **Explanation block:** To calculate the entropy change for the air, we first computed its mass flow rate using the ideal gas equation. The entropy change per kg is then calculated using the relation for ideal gases with constant pressure. --- ## Step 3: **Step-by-Step Complete Calculation Part (Refrigerant Side & Entropy Generation)** ### **1. Change in Entropy for Refrigerant-134a:** \[ \dot{S}_{R-134a} = \dot{m}_{R-134a} \cdot (s_2 - s_1) \] Where: - \(\dot{m}_{R-134a} = 0.0333 \text{ kg/s}\) - At 120 kPa: - \(x_1 = 0.3\) (quality at inlet) - \(x_2 = 1\) (saturated vapor at exit) From R134a tables at P = 120 kPa: - \(s_f = 0.0646 \text{ kJ/kg·K}\) (saturated liquid) - \(s_g = 0.8960 \text{ kJ/kg·K}\) (saturated vapor) \[ s_1 = s_f + x_1(s_g - s_f) = 0.0646 + 0.3 \times (0.8960 - 0.0646) = 0.0646 + 0.3 \times 0.8314 = 0.0646 + 0.2494 = 0.314 \text{ kJ/kg·K} \] \[ s_2 = s_g = 0.8960 \text{ kJ/kg·K} \] \[ \Delta s_{R-134a} = s_2 - s_1 = 0.8960 - 0.314 = 0.582 \text{ kJ/kg·K} \] \[ \dot{S}_{R-134a} = 0.0333 \times 0.582 = 0.0194 \text{ kW/K} \] ### **2. Energy Balance to Find \(T_{out,air}\):** Heat absorbed by refrigerant: \[ \dot{Q} = \dot{m}_{R-134a} \cdot (h_2 - h_1) \] At 120 kPa from R134a tables: - \(h_f = 27.36 \text{ kJ/kg}\) - \(h_g = 237.0 \text{ kJ/kg}\) \[ h_1 = h_f + x_1(h_g - h_f) = 27.36 + 0.3 \times (237.0 - 27.36) = 27.36 + 0.3 \times 209.64 = 27.36 + 62.892 = 90.25 \text{ kJ/kg} \] \[ h_2 = h_g = 237.0 \text{ kJ/kg} \] \[ \Delta h_{R-134a} = h_2 - h_1 = 237.0 - 90.25 = 146.75 \text{ kJ/kg} \] \[ \dot{Q} = 0.0333 \times 146.75 = 4.89 \text{ kW} \] Heat lost by air: \[ \dot{Q} = \dot{m}_{air} \cdot c_p \cdot (T_{in,air} - T_{out,air}) \] \[ 4.89 = 0.116 \times 1.005 \times (300.15 - T_{out,air}) \] \[ 4.89 = 0.11658 \times (300.15 - T_{out,air}) \] \[ \frac{4.89}{0.11658} = 300.15 - T_{out,air} \] \[ 41.94 = 300.15 - T_{out,air} \] \[ T_{out,air} = 300.15 - 41.94 = 258.21 \text{ K} \] ### **3. Entropy Change for Air:** \[ \Delta s_{air} = c_p \ln \left( \frac{T_{out,air}}{T_{in,air}} \right) = 1.005 \ln\left( \frac{258.21}{300.15} \right) \] \[ = 1.005 \times \ln(0.860) \] \[ = 1.005 \times (-0.1507) = -0.1515 \text{ kJ/kg·K} \] \[ \dot{S}_{air} = \dot{m}_{air} \cdot \Delta s_{air} = 0.116 \times (-0.1515) = -0.0176 \text{ kW/K} \] ### **4. Entropy Generation:** \[ \dot{S}_{gen} = (\dot{S}_{R-134a} + \dot{S}_{air}) \] \[ \dot{S}_{gen} = 0.0194 + (-0.0176) = 0.0018 \text{ kW/K} \] **Explanation block:** We calculated the entropy change for R-134a using property tables and its mass flow rate, and for air using its temperature change and mass flow rate. The sum of these entropy changes gives the total entropy generation in the process. --- ## **Final Answer:** **The rate of entropy generation for this process is** \[ \boxed{0.0018 \text{ kW/K}} \] or \[ \boxed{1.8 \text{ W/K}} \]