# Diesel Cycle Problem Solution## **Given Data** - **Initial pressure ($P_1$):** 100 kPa - **Initial temperature ($T_1$):** $50^\circ$C = 323.15 K - **Heat added per kg ($q_{in}$):** 140 kJ/kg - **Compression ratio ($r$):** Varies from 16 to 25 - **Air properties (assume ideal gas, constant specific heats):** - $c_p = 1.005$ kJ/kg·K - $c_v = .718$ kJ/kg·K - $\gamma = \frac{c_p}{c_v} \approx 1.4$ ## **Diesel Cycle Process Overview** 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 1: Isentropic Compression (1 → 2)** - Compression ratio: $r = \frac{V_1}{V_2}$ **Isentropic relations:** \[ \frac{T_2}{T_1} = r^{\gamma - 1} \] \[ \frac{P_2}{P_1} = r^{\gamma} \] --- ## **Step 2: Constant Pressure Heat Addition (2 → 3)** - $P_2 = P_3$ - Heat added: $q_{in} = c_p (T_3 - T_2)$ \[ T_3 = T_2 + \frac{q_{in}}{c_p} \] --- ## **Step 3: Isentropic Expansion (3 → 4)** - Expansion ratio: $\rho = \frac{V_3}{V_2}$ Note: For Diesel cycle, **cut-off ratio** $\rho$ is defined as $V_3/V_2$. \[ \frac{T_4}{T_3} = \left(\frac{V_3}{V_4}\right)^{1-\gamma} \] but $V_4 = V_1$ and $V_3 = \rho V_2$. --- ## **Step 4: Find Cut-Off Ratio ($\rho$)** From the heat addition relation: \[ q_{in} = c_p (T_3 - T_2) \] Also, \[ \frac{V_3}{V_2} = \rho \] At constant pressure: \[ \frac{T_3}{T_2} = \frac{V_3}{V_2} = \rho \] So, \[ T_3 = \rho T_2 \] So, \[ q_{in} = c_p (T_3 - T_2) = c_p T_2 (\rho - 1) \] \[ \Rightarrow \rho = 1 + \frac{q_{in}}{c_p T_2} \] --- ## **Step 5: Step-by-Step Calculations** Let's use $r = 16$ and $r = 25$ as examples. ### **A. For $r = 16$** #### 1. **Find $T_2$** \[ T_2 = T_1 \cdot r^{\gamma - 1} \] \[ T_2 = 323.15 \cdot 16^{.4} \] Calculate $16^{.4}$: \[ \ln(16) = 2.7726 \] \[ .4 \cdot 2.7726 = 1.109 \] \[ e^{1.109} = 3.031 \] \[ T_2 = 323.15 \cdot 3.031 \approx 980.1\ \text{K} \] #### 2. **Find Cut-off Ratio $\rho$** \[ \rho = 1 + \frac{q_{in}}{c_p T_2} \] \[ \rho = 1 + \frac{140}{1.005 \times 980.1} \] \[ 1.005 \times 980.1 = 985. \] \[ \frac{140}{985.} = 1.422 \] \[ \rho = 1 + 1.422 = 2.422 \] #### 3. **Find $T_3$ and Maximum Temperature** \[ T_3 = \rho T_2 = 2.422 \times 980.1 = 2,374.6\ \text{K} \] - **Maximum temperature: $T_3 = 2,375$ K (rounded)** #### 4. **Find Maximum Pressure $P_3$** Isobaric process: $P_3 = P_2$ First, find $P_2$: \[ P_2 = P_1 \cdot r^{\gamma} \] \[ P_2 = 100 \cdot 16^{1.4} \] Calculate $16^{1.4}$: \[ 1.4 \times 2.7726 = 3.8817,\quad e^{3.8817} = 48.6 \] \[ P_2 = 100 \times 48.6 = 4,860\ \text{kPa} \] - **Maximum pressure: $P_3 = 4,860$ kPa** #### 5. **Thermal Efficiency** Diesel cycle efficiency: \[ \eta = 1 - \frac{1}{r^{\gamma-1}} \cdot \frac{\rho^{\gamma} - 1}{\gamma (\rho - 1)} \] Calculate for $r=16$, $\rho=2.422$, $\gamma=1.4$: - $r^{\gamma-1} = 16^{.4} = 3.031$ - $\frac{1}{3.031} = .330$ - $\rho^{\gamma} = 2.422^{1.4}$ - $\ln(2.422) = .8847$ - $1.4 \times .8847 = 1.2386$ - $e^{1.2386} = 3.450$ - $\rho^{\gamma} - 1 = 2.450$ - $\gamma(\rho - 1) = 1.4 \times 1.422 = 1.991$ - $\frac{2.450}{1.991} = 1.231$ Now substitute: \[ \eta = 1 - .330 \times 1.231 = 1 - .406 = .594 \] - **Thermal efficiency: $\eta = 59.4\%$** #### 6. **Mean Effective Pressure (MEP)** \[ \text{MEP} = \frac{W_{net}}{v_{max} - v_{min}} \] $W_{net} = \eta \cdot q_{in}$ $v_{max} - v_{min} = v_1 - v_2 = v_1 (1 - \frac{1}{r})$ Specific volume at state 1: \[ v_1 = \frac{R T_1}{P_1} \] $R = .287$ kJ/kg·K = 287 J/kg·K \[ v_1 = \frac{.287 \times 323.15}{100} = \frac{92.74}{100} = .927\ \text{m}^3/\text{kg} \] \[ v_{max} - v_{min} = .927 \left(1 - \frac{1}{16}\right) = .927 \times .9375 = .868\ \text{m}^3/\text{kg} \] \[ W_{net} = .594 \times 140 = 831.6\ \text{kJ}/\text{kg} \] \[ \text{MEP} = \frac{831.6}{.868} = 958.3\ \text{kPa} \] --- ### **B. For $r = 25$** #### 1. **$T_2$** \[ T_2 = 323.15 \times 25^{.4} \] \[ \ln(25) = 3.2189 \] \[ .4 \times 3.2189 = 1.2876,\quad e^{1.2876} = 3.624 \] \[ T_2 = 323.15 \times 3.624 = 1,171.7\ \text{K} \] #### 2. **$\rho$** \[ \rho = 1 + \frac{140}{1.005 \times 1,171.7} = 1 + \frac{140}{1,177.6} = 1 + 1.189 = 2.189 \] #### 3. **$T_3$** \[ T_3 = 2.189 \times 1,171.7 = 2,564.9\ \text{K} \] - **Maximum temperature: $T_3 = 2,565$ K** #### 4. **$P_2$ and $P_3$** \[ P_2 = 100 \times 25^{1.4} \] \[ 1.4 \times 3.2189 = 4.5065 \] \[ e^{4.5065} = 90.5 \] \[ P_2 = 100 \times 90.5 = 9,050\ \text{kPa} \] - **Maximum pressure: $P_3 = 9,050$ kPa** #### 5. **Thermal Efficiency** - $r^{\gamma-1} = 25^{.4} = 3.624$ - $\frac{1}{3.624} = .276$ - $\rho^{\gamma} = 2.189^{1.4}$ - $\ln(2.189) = .783$ - $1.4 \times .783 = 1.096$ - $e^{1.096} = 2.993$ - $\rho^{\gamma} - 1 = 1.993$ - $\gamma(\rho - 1) = 1.4 \times 1.189 = 1.665$ - $\frac{1.993}{1.665} = 1.197$ \[ \eta = 1 - .276 \times 1.197 = 1 - .330 = .670 \] - **Thermal efficiency: $\eta = 67.\%$** #### 6. **MEP** \[ v_1 = \frac{.287 \times 323.15}{100} = .927\ \text{m}^3/\text{kg} \] \[ v_{max} - v_{min} = .927 \left(1 - \frac{1}{25}\right) = .927 \times .96 = .890\ \text{m}^3/\text{kg} \] \[ W_{net} = .67 \times 140 = 938.\ \text{kJ}/\text{kg} \] \[ \text{MEP} = \frac{938.}{.890} = 1,054.\ \text{kPa} \] --- ## **Summary Table** | $r$ (Compression Ratio) | $\rho$ (Cutoff Ratio) | $T_3$ (K) | $P_3$ (kPa) | $\eta$ (Efficiency) | MEP (kPa) | |:----------------------:|:--------------------:|:---------:|:-----------:|:-------------------:|:---------:| | 16 | 2.422 | 2,375 | 4,860 | 59.4% | 958 | | 25 | 2.189 | 2,565 | 9,050 | 67.% | 1,054 | --- ## **Final Answers** ### For $r = 16$: - **Maximum temperature:** $2,375$ K - **Maximum pressure:** $4,860$ kPa - **Thermal efficiency:** $59.4\%$ - **Mean effective pressure:** $958$ kPa ### For $r = 25$: - **Maximum temperature:** $2,565$ K - **Maximum pressure:** $9,050$ kPa - **Thermal efficiency:** $67.\%$ - **Mean effective pressure:** $1,054$ kPa --- ## **Key Formulas Used** - $T_2 = T_1 \cdot r^{\gamma - 1}$ - $\rho = 1 + \frac{q_{in}}{c_p T_2}$ - $T_3 = \rho T_2$ - $P_2 = P_1 \cdot r^{\gamma}$, $P_3 =