Sure! Let's break down the **problem** and solve it step by step. --- ## **Problem Statement (Rewritten for Clarity)** - **Air compressor**: Compresses air from **100 kPa, 17°C** to **1 MPa** at a **rate of 100 L/min**. - **Exit temperature**: **80°C** - **Find**: 1. **Power input** to the compressor (**kW**). 2. **Isentropic efficiency** of the compressor. --- ## **Step 1: System Diagram and Given Data** ### **Given:** - **Inlet pressure (\(P_1\))**: 100 kPa - **Inlet temp (\(T_1\))**: 17°C = 290 K - **Exit pressure (\(P_2\))**: 1 MPa = 100 kPa - **Exit temp (\(T_2\))**: 80°C = 353 K - **Volumetric flow (\(\dot{V}_1\))**: 100 L/min = 1 m³/min = 1/60 m³/s ≈ .01667 m³/s --- ## **Step 2: Calculate Mass Flow Rate (\(\dot{m}\))** Use the **ideal gas law** at the inlet: \[ \dot{m} = \frac{P_1 \dot{V}_1}{R T_1} \] Where: - \(R\) for air = .287 kJ/kg·K = 287 J/kg·K Plug in values (convert \(P_1\) to Pa): \[ P_1 = 100\,kPa = 100,000\,Pa \] \[ T_1 = 17^\circ C = 290\,K \] \[ \dot{V}_1 = .01667\,m^3/s \] \[ \dot{m} = \frac{100{,}000 \times .01667}{287 \times 290} \] \[ \dot{m} = \frac{1667}{83230} \] \[ \dot{m} \approx .020\,kg/s \] --- ## **Step 3: Power Input Calculation** Compressor power input (steady flow energy eq., neglecting kinetic and potential energy): \[ \dot{W}_{\text{input}} = \dot{m} (h_2 - h_1) \] For air, use \(h = c_p T\) (approx. for ideal gas): - \(c_p\) for air ≈ 1.005 kJ/kg·K \[ h_1 = c_p T_1 = 1.005 \times 290 = 291.5\,kJ/kg \] \[ h_2 = c_p T_2 = 1.005 \times 353 = 354.8\,kJ/kg \] \[ \Delta h = h_2 - h_1 = 354.8 - 291.5 = 63.3\,kJ/kg \] \[ \dot{W}_{\text{input}} = .02 \times 63.3 = 1.266\,kW \] --- ## **Step 4: Isentropic Efficiency Calculation** Isentropic efficiency (\(\eta_c\)) of a compressor: \[ \eta_c = \frac{h_{2s} - h_1}{h_2 - h_1} \] Where \(h_{2s}\) is the enthalpy at the exit pressure if the process were isentropic. ### **Calculate \(T_{2s}\):** For an isentropic process in an ideal gas: \[ \frac{T_{2s}}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \] - For air, \(k = 1.4\) \[ \frac{T_{2s}}{290} = \left( \frac{100}{100} \right)^{\frac{.4}{1.4}} \] \[ = (10)^{.286} \] \[ = 1.93 \] \[ T_{2s} = 1.93 \times 290 = 560 \,K \] Wait! This is much higher than the actual exit temperature. Let's double-check this: \[ n = \frac{k-1}{k} = \frac{.4}{1.4} \approx .286 \] \[ T_{2s} = 290 \times 10^{.286} \approx 290 \times 1.93 = 560\,K \] That is correct. \[ h_{2s} = c_p T_{2s} = 1.005 \times 560 = 563\,kJ/kg \] Now use the efficiency formula: \[ \eta_c = \frac{h_{2s} - h_1}{h_2 - h_1} = \frac{563 - 291.5}{354.8 - 291.5} \] \[ = \frac{271.5}{63.3} = 4.29 \] This is much greater than 1, which is **not physically possible**. Let's check the steps for the isentropic temperature: **The isentropic exit temperature should be lower than the actual, but here it's much higher.** Wait, let's check the equations: \[ \frac{T_{2s}}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} \] \[ \frac{T_{2s}}{290} = (10)^{.286} = 1.93 \] \[ T_{2s} = 1.93 \times 290 = 560\,K \] Given that the actual exit temperature is 353 K (80°C), the process is highly **irreversible** (real compressors usually have actual temperature higher than isentropic, not lower). So, \(\eta_c\) is usually defined as: \[ \eta_c = \frac{\text{Isentropic work}}{\text{Actual work}} \] In this case, \[ \eta_c = \frac{h_{2s} - h_1}{h_2 - h_1} \] With our numbers: \[ h_1 = 291.5\,kJ/kg \] \[ h_2 = 354.8\,kJ/kg \] \[ h_{2s} = 563\,kJ/kg \] \[ \eta_c = \frac{563 - 291.5}{354.8 - 291.5} = \frac{271.5}{63.3} = 4.29 \] Again, this is not possible. The actual temperature should be higher than the isentropic, so something is off in the physicality of the problem's numbers. But, following the method, **the calculation steps are correct**. The issue must be with the provided values (the exit temperature is usually *higher* than the isentropic, not lower). --- ## **Final Answers (with Explanation)** ### **1. The Power Input** \[ \boxed{1.27 \text{ kW}} \] *(rounded from 1.266 kW)* ### **2. Isentropic Efficiency** \[ \boxed{4.29} \] *(which is not physical; normally, efficiency should be less than 1. This might be a typo in the problem statement. Double-check exit temperature!)* --- ### **Summary Table** | Quantity | Value | |---------------------------|--------------------------| | Mass flow rate (\(\dot{m}\)) | .02 kg/s | | Power input (\(\dot{W}\)) | 1.27 kW | | Isentropic efficiency | 4.29 (not physically possible) | --- ### **Key Steps:** 1. **Find mass flow rate using ideal gas law**. 2. **Calculate power input using enthalpy change**. 3. **Compute isentropic temperature and corresponding enthalpy**. 4. **Calculate isentropic efficiency**. --- **Let me know if you want to discuss the result or clarify the steps further!**