# Problem Overview An adiabatic compressor compresses saturated steam from initial conditions to a higher pressure, delivering shaft power. We are asked to find the outlet temperature in two cases: - (a) Outlet velocity is negligible - (b) Outlet velocity is **50 m/s** --- # Known Data | Parameter | Value | |--------------|---------| | Mass flow rate (\(\dot{m}\)) | 5 kg/s | | Inlet temperature (\(T_1\)) | 100°C | | Initial pressure (\(P_1\)) | Saturated steam at 100°C \(\approx .1013\, MPa\) (atmospheric pressure) | | Final pressure (\(P_2\)) | 3. MPa | | Shaft power (\(W_{shaft}\)) | 4.875 MW = 4875 kW | | Inlet velocity (\(v_1\)) | Negligible | --- # Assumptions - The process is adiabatic (no heat transfer). - The steam behaves as a real substance, and we use steam tables. - Kinetic energy change is significant only in case (b). --- # Step 1: Find inlet specific enthalpy \(h_1\) From steam tables at \(T_1 = 100^\circ C\): - Saturation pressure at 100°C: \(P_{sat} \approx .1013\, MPa\) - Enthalpy of saturated vapor \(h_{g}\) at 100°C: approximately **2676.2 kJ/kg** Since the inlet is saturated steam at 100°C: \[ h_1 = h_{g} \approx 2676.2\, \text{kJ/kg} \] --- # Step 2: Determine the inlet properties - Inlet enthalpy \(h_1 \approx 2676.2\, \text{kJ/kg}\) - Inlet temperature \(T_1 = 100^\circ C\) --- # Step 3: Apply steady-flow energy equation **General form:** \[ W_{shaft} = \dot{m} (h_2 + KE_2 - h_1 - KE_1) \] Since inlet velocity is negligible (\(v_1 \approx \)), and in case (a), outlet velocity is negligible: \[ W_{shaft} = \dot{m}(h_2 - h_1) \] ### For case (a): \[ h_2 = h_1 + \frac{W_{shaft}}{\dot{m}} \] Calculate: \[ h_2 = 2676.2 + \frac{4875\, \text{kW}}{5\, \text{kg/s}} = 2676.2 + 975\, \text{kJ/kg} = 3651.2\, \text{kJ/kg} \] --- # Step 4: Find the outlet temperature \(T_2\) in case (a) - From steam tables, locate \(h = 3651.2\, \text{kJ/kg}\) at \(P_2 = 3\, MPa\): At \(P = 3\, MPa\), the saturated vapor enthalpy \(h_g \approx 2778\, \text{kJ/kg}\), which is less than 3651.2 kJ/kg. **Interpretation:** - The enthalpy exceeds saturated vapor at 3 MPa, indicating superheated steam. - To find \(T_2\), we look at superheated steam tables at \(P=3\, MPa\): | Temperature (°C) | Enthalpy \(h\) (kJ/kg) | |------------------|-------------------------| | 300 | 2862. | | 350 | 2984. | | 400 | 3104. | | 450 | 3224. | | 500 | 3342. | | 550 | 346. | | 600 | 3578. | Our \(h_2 = 3651.2\, \text{kJ/kg}\): - Slightly above 600°C data point, so approximate \(T_2 \approx 620^\circ C\). **Final answer for (a):** \[ \boxed{ T_2 \approx 620^\circ C } \] --- # Step 5: Case (b) - Outlet velocity is 50 m/s - Now include kinetic energy change: \[ W_{shaft} = \dot{m}(h_2 - h_1) + \frac{1}{2}\dot{m}(v_2^2 - v_1^2) \] Since \(v_1 \approx \): \[ W_{shaft} = \dot{m}(h_2 - h_1) + \frac{1}{2} \times 5 \times (50)^2 \] Calculate kinetic energy term: \[ KE_{out} = \frac{1}{2} \times 5\, \text{kg/s} \times (50)^2\, \text{m}^2/\text{s}^2 = 2.5 \times 250 = 625\, \text{J/s} = 6.25\, \text{kW} \] **Note:** Since power is in kW, convert to kJ: \[ 6.25\, \text{kW} = 6.25\, \text{kJ/s} \] But the power supplied is 4875 kW, so the energy balance: \[ W_{shaft} = \dot{m}(h_2 - h_1) + KE_{out} \] Rearranged: \[ h_2 = h_1 + \frac{W_{shaft} - KE_{out}}{\dot{m}} = 2676.2 + \frac{4875 - 6.25}{5} = 2676.2 + \frac{4868.75}{5} = 2676.2 + 973.75 = 365\, \text{kJ/kg} \] This is very similar to the case (a) result, with negligible difference due to \(KE_{out}\). --- # Step 6: Find outlet temperature \(T_2\) in case (b) Again, \(h_2 \approx 365\, \text{kJ/kg}\): - As before, this exceeds saturated vapor enthalpy at 3 MPa, indicating superheated steam. - Using superheated steam tables at 3 MPa: \[ h \approx 365\, \text{kJ/kg} \] Interpolating: - Slightly above 600°C data point, so: \[ T_2 \approx 620^\circ C \] **Final answer for (b):** \[ \boxed{ T_2 \approx 620^\circ C } \] --- # **Summary of Results** | Case | Outlet Temperature \(T_2\) | Remarks | |---------|------------------------------|---------| | (a) | **~620°C** | No outlet velocity | | (b) | **~620°C** | Outlet velocity = 50 m/s | --- # **Note:** - The slight difference in kinetic energy affects the enthalpy calculation marginally. - Actual values depend on precise steam table data, but the approximations provide a good estimate. --- If you have any further questions, feel free to ask!