Let's analyze and solve each part step-by-step. --- ## Given Data: - Maximum temperature, \( T_{max} =200\,K \) - pressure, \( p_1 = 95kPa \- Initial temperature \( T_1 = 300K \) Mass of air, m = 12\,g = .012\,kg \) - Compression ratio, \( r = 20 \) --- ## Assumptions: - Air behaves as an ideal gas. - The cycle is adiabatic and reversible (ideal Diesel cycle). --- ### (a) Net work of the cycle, in kJ **Step 1:** Find \( p_2 \) and \( T_2 \) after compression: Using adiabatic relations: \[ T_2 = T_1 \times r^{\frac{\gamma - 1}{\gamma}} \] \[ p_2 = p_1 \times r^{\gamma} \] where \( \gamma = 1.4 \) for air. Calculate \( T_2 \): \[ T_2 = 300 \times 20^{\frac{.4}{1.4}} \approx 300 \times 20^{.2857} \] \[ T_2 \approx 300 \times 2.97 \approx 891\,K \] Calculate \( p_2 \): \[ p_2 = 95 \times 20^{1.4} \approx 95 \times 20^{1.4} \] \[ 20^{1.4} \approx e^{1.4 \times \ln 20} \approx e^{1.4 \times 2.9957} \approx e^{4.193} \approx 66 \] \[ p_2 \approx 95 \times 66 \approx 627\,kPa \] --- **Step 2:** Heat addition during constant pressure process (from \( T_2 \) to \( T_3 \)): Since maximum temperature is \( T_3 = 200\,K \): \[ Q_{in} = m C_p (T_3 - T_2) \] Where: \[ C_p = \frac{\gamma R}{\gamma - 1} \approx 1004\,J/(kg \cdot K) \] Calculate \( Q_{in} \): \[ Q_{in} = .012 \times 1004 \times (200 - 891) \approx .012 \times 1004 \times 1109 \] \[ Q_{in} \approx .012 \times 1,113,436 \approx 13,361\,J \] --- **Step 3:** Work done during the cycle: - Work during compression: \[ W_{comp} = \frac{p_1 V_1 - p_2 V_2}{\text{(assuming work done on air)}} \] Alternatively, for adiabatic processes: \[ W_{compression} = \frac{p_2 V_2 - p_1 V_1}{\text{but easier to use energy difference from heat interactions}} \] - Work during expansion (power stroke): \[ W_{exp} = Q_{in} \quad \text{(since it's a heat addition at constant pressure, approximate)} \] The net work is approximately: \[ W_{net} = Q_{in} - Q_{out} \] But for an ideal Diesel cycle, the net work per cycle is: \[ W_{net} = Q_{in} - Q_{out} \] Where \( Q_{out} \) is heat rejected during the power stroke. --- ### **Using the Diesel cycle efficiency:** \[ \eta_{Diesel} = 1 - \frac{1}{r^{\gamma - 1}} \times \frac{\rho^{\gamma} - 1}{\rho - 1} \] with \( \rho = \frac{T_3}{T_2} = \frac{200}{891} \approx 2.245 \). Calculate efficiency: \[ \eta_{Diesel} \approx 1 - \frac{1}{20^{.4}} \times \frac{2.245^{1.4} - 1}{2.245 - 1} \] Calculate \( 20^{.4} \): \[ 20^{.4} \approx e^{.4 \times 2.9957} \approx e^{1.198} \approx 3.314 \] Calculate \( 2.245^{1.4} \): \[ 2.245^{1.4} \approx e^{1.4 \times \ln 2.245} \approx e^{1.4 \times .809} \approx e^{1.134} \approx 3.11 \] Now: \[ \eta_{Diesel} \approx 1 - \frac{1}{3.314} \times \frac{3.11 - 1}{2.245 - 1} \] \[ = 1 - .301 \times \frac{2.11}{1.245} \approx 1 - .301 \times 1.695 \approx 1 - .51 \approx .49 \] **Net work:** \[ W_{net} = \eta_{Diesel} \times Q_{in} \approx .49 \times 13,361\,J \approx 6,543\,J \approx 6.54\,kJ \] --- ## **Answer for (a):** **Net work of the cycle ≈ 6.54 kJ** --- ### (b) Percent thermal efficiency \[ \boxed{ \eta_{thermal} \approx 49\% } \] --- ### (c) Mean effective pressure (MEP): \[ \text{MEP} = \frac{W_{net}}{V_{displaced}} \] Using ideal gas law: \[ V_1 = \frac{m R T_1}{p_1} \] where \( R = 287\,J/(kg \cdot K) \). Calculate \( V_1 \): \[ V_1 = \frac{.012 \times 287 \times 300}{95,000} \approx \frac{1032.6}{95000} \approx .01086\,m^3 \] Since the net work per cycle: \[ W_{net} \approx 6.54\,kJ \] \[ \text{MEP} = \frac{654\,J}{.01086\,m^3} \approx 602,000\,Pa = 602\,kPa \] **Answer:** ## **(c) Mean effective pressure ≈ 602 kPa** --- **Summary:** | Part | Result | |--------|---------| | (a) Net work | **≈ 6.54 kJ** | | (b) Thermal efficiency | **≈ 49%** | | (c) Mean effective pressure | **≈ 602 kPa** |