Let's solve the **air-standard Diesel cycle** problem step by step. --- ### **Given Data** - Initial pressure, \( p_1 = 95 \) kPa - Initial temperature, \( T_1 = 300 \) K - Maximum temperature, \( T_3 = 195 \) K - Mass of air, \( m = 12 \) g \( = .012 \) kg - Compression ratio, \( r = \frac{V_1}{V_2} = 15 \) - Working fluid: Air (\( R = .287 \) kJ/kg·K, \( c_v = .718 \) kJ/kg·K, \( c_p = 1.005 \) kJ/kg·K, \( \gamma = 1.4 \)) --- ## **Step 1: Diesel Cycle Process Overview** The Diesel cycle has four processes: 1. **1-2:** Isentropic compression 2. **2-3:** Constant pressure heat addition 3. **3-4:** Isentropic expansion 4. **4-1:** Constant volume heat rejection --- ## **Step 2: Find State Points** ### **State 1:** - \( p_1 = 95 \) kPa - \( T_1 = 300 \) K ### **State 2: Isentropic Compression** - \( r = 15 \) - \( \frac{T_2}{T_1} = r^{\gamma-1} \) \[ T_2 = T_1 \times r^{\gamma-1} \] \[ T_2 = 300 \times 15^{1.4-1} = 300 \times 15^{.4} \] \[ 15^{.4} \approx 2.953 \] \[ T_2 = 300 \times 2.953 \approx 886 \text{ K} \] ### **State 3: Constant Pressure Heat Addition** - \( T_3 = 195 \) K ### **Cut-off Ratio (\( r_c \)):** \[ r_c = \frac{V_3}{V_2} = \frac{T_3}{T_2} \] \[ r_c = \frac{195}{886} = 2.20 \] ### **State 4: Isentropic Expansion** \[ \frac{T_4}{T_3} = \left(\frac{V_3}{V_4}\right)^{\gamma-1} \] But \( \frac{V_4}{V_3} = \frac{V_1}{V_2} \div \frac{V_3}{V_2} = \frac{r}{r_c} \) So, \[ T_4 = T_3 \left(\frac{r_c}{r}\right)^{\gamma-1} \] \[ \left(\frac{r_c}{r}\right)^{\gamma-1} = \left(\frac{2.20}{15}\right)^{.4} = (.147)^{.4} \approx .468 \] \[ T_4 = 195 \times .468 \approx 912 \text{ K} \] --- ## **Step 3: Net Work (in kJ)** ### **Heat Added (\( q_{in} \))** \[ q_{in} = c_p (T_3 - T_2) \] \[ q_{in} = 1.005 \times (195 - 886) = 1.005 \times 1064 = 1,069 \text{ kJ/kg} \] ### **Heat Rejected (\( q_{out} \))** \[ q_{out} = c_v (T_4 - T_1) \] \[ q_{out} = .718 \times (912 - 300) = .718 \times 612 = 439.4 \text{ kJ/kg} \] ### **Net Work per kg (\( w_{net} \)):** \[ w_{net} = q_{in} - q_{out} = 1,069 - 439.4 = 629.6 \text{ kJ/kg} \] ### **Net Work for .012 kg:** \[ W_{net} = w_{net} \times m = 629.6 \times .012 = 7.56 \text{ kJ} \] --- ## **Step 4: Thermal Efficiency** \[ \eta = \frac{w_{net}}{q_{in}} \times 100\% \] \[ \eta = \frac{629.6}{1,069} \times 100\% = 58.9\% \] --- ## **Step 5: Mean Effective Pressure (MEP)** \[ \text{MEP} = \frac{W_{net}}{\text{Displacement Volume}} \] ### **Find \( V_1 \) and \( V_2 \):** \[ p_1 V_1 = m R T_1 \implies V_1 = \frac{m R T_1}{p_1} \] \[ V_1 = \frac{.012 \times .287 \times 300}{95} \] \[ = \frac{1.032}{95} = .01087 \text{ m}^3 \] \[ V_2 = \frac{V_1}{r} = \frac{.01087}{15} = .000725 \text{ m}^3 \] \[ \text{Displacement Volume} = V_1 - V_2 = .01087 - .000725 = .01015 \text{ m}^3 \] ### **Calculate MEP:** \[ \text{MEP} = \frac{7.56}{.01015} = 745 \text{ kPa} \] --- ## **Summary Table** | Quantity | Value | |-------------------------|--------------------------| | Net Work (\(W_{net}\)) | \(7.56\) kJ | | Thermal Efficiency | \(58.9\%\) | | Mean Effective Pressure | \(745\) kPa | --- ### **Final Answers** **(a) Net work developed:** \[ \boxed{7.56 \text{ kJ}} \] **(b) Percent thermal efficiency:** \[ \boxed{58.9\%} \] **(c) Mean effective pressure:** \[ \boxed{745 \text{ kPa}} \]