Let's solve the problem step by step. --- ## **Given Data** - Inlet Pressure, \( P_1 = 3\, \text{MPa} = 300\, \text{kPa} \) - Inlet Temperature, \( T_1 = 500\, \text{K} \) - Mass flow rate, \( \dot{m} = 2\, \text{kg/s} \) - Exit Temperature, \( T_2 = 450\, \text{K} \) - Working fluid: **CO₂** - Behave as **ideal gas** for (a) and (c); **real gas** for (b). --- ## **(a) Density of CO₂ as an Ideal Gas at Inlet** The **ideal gas equation** is: \[ PV = nRT \] or \[ P = \rho R_{\text{specific}} T \] where - \( \rho \) = density (kg/m³) - \( R_{\text{specific}} = \frac{R_u}{M} \) - \( R_u \) = Universal gas constant = \( 8.314\, \text{kJ}/(\text{kmol}\cdot\text{K}) \) - \( M \) = Molar mass of CO₂ = **44.01 kg/kmol** So, \[ R_{\text{specific}} = \frac{8.314}{44.01} = .189 \, \text{kJ}/(\text{kg}\cdot\text{K}) = 189\, \text{J}/(\text{kg}\cdot\text{K}) \] Now, rearrange for \(\rho\): \[ \rho = \frac{P}{R_{\text{specific}} T} \] Plug in the values (convert P to Pa): - \( P = 3\, \text{MPa} = 3 \times 10^6\, \text{Pa} \) - \( R = 189\, \text{J}/(\text{kg}\cdot\text{K}) \) - \( T = 500\, \text{K} \) \[ \rho = \frac{3 \times 10^6}{189 \times 500} \] \[ \rho = \frac{3 \times 10^6}{94,500} \] \[ \rho = 31.75\, \text{kg/m}^3 \] --- ### **Final Answer (a):** \[ \boxed{ \rho_{ideal,\,inlet} = 31.75\, \text{kg/m}^3 } \] --- ## **(b) Density of CO₂ as a Real Gas (Using Compressibility Factor \(Z\))** The real gas equation is: \[ \rho = \frac{P}{Z R_{\text{specific}} T} \] We need to find the compressibility factor \( Z \) using the **generalized compressibility chart**. #### **Step 1: Find Reduced Properties** - \( P_{cr} = 7.38\, \text{MPa} \) - \( T_{cr} = 304.2\, \text{K} \) \[ P_r = \frac{P}{P_{cr}} = \frac{3}{7.38} = .406 \] \[ T_r = \frac{T}{T_{cr}} = \frac{500}{304.2} = 1.643 \] #### **Step 2: Find Z from Chart** At \( P_r = .406 \), \( T_r = 1.643 \): - From the generalized compressibility chart, \( Z \approx .85 \) (value may slightly vary based on chart). #### **Step 3: Calculate Density** \[ \rho = \frac{P}{Z R T} \] \[ \rho = \frac{3 \times 10^6}{.85 \times 189 \times 500} \] \[ \rho = \frac{3 \times 10^6}{80,325} \] \[ \rho = 37.36\, \text{kg/m}^3 \] --- ### **Final Answer (b):** \[ \boxed{ \rho_{real,\,inlet} = 37.36\, \text{kg/m}^3 } \] --- ## **(c) Volumetric Flow Rate of Ideal Gas CO₂ at Exit** \[ \dot{V} = \frac{\dot{m}}{\rho} \] First, calculate the **exit density** as ideal gas: - \( P = 3 \times 10^6\, \text{Pa} \) - \( R = 189\, \text{J}/(\text{kg}\cdot\text{K}) \) - \( T_2 = 450\, \text{K} \) \[ \rho_{exit} = \frac{3 \times 10^6}{189 \times 450} \] \[ \rho_{exit} = \frac{3 \times 10^6}{85,050} \] \[ \rho_{exit} = 35.28\, \text{kg/m}^3 \] Now, \[ \dot{V}_{exit} = \frac{2}{35.28} = .0567\, \text{m}^3/\text{s} \] --- ### **Final Answer (c):** \[ \boxed{ \dot{V}_{exit,\,ideal} = .0567\, \text{m}^3/\text{s} } \] --- ## **Summary Table** | Part | Parameter | Value | |------|-----------|-------| | (a) | Inlet density (ideal) | \( 31.75\, \text{kg/m}^3 \) | | (b) | Inlet density (real, \(Z \approx .85\)) | \( 37.36\, \text{kg/m}^3 \) | | (c) | Volumetric flow (ideal, exit) | \( .0567\, \text{m}^3/\text{s} \) | --- If you need more explanation on any step, **let me know!**