Let's go step-by-step to solve the problem: --- ### **Given Data:** - **Fuel:** Ethanol (\( C_2H_5OH \)) - **Equivalence Ratio (\( \phi \)):** 0.75 (lean mixture) - **Compression Ratio (\( r \)):** 9.0 - **Displacement (\( V_d \)):** 0.4 L = 0.0004 m³ - **Intake Manifold Pressure (IMP):** 80 kPa - **Intake Manifold Temperature (IMT):** 350 K - **Exhaust Manifold Pressure (EMP):** 100 kPa - **Lower Heating Value (LHV) of Ethanol:** 26,950 kJ/kg - **Specific Heat Ratio (\( k \)):** 1.35 - **Combustion Efficiency (\( \eta_c \)):** 1 - **Air-Fuel Ratio at \(\phi=1\):** 9.0 --- ## (a) **Find \( \Delta T \) using the chemical equation for combustion** ### **Step 1: Write the Balanced Chemical Equation** For ethanol: \[ C_2H_5OH + O_2 + N_2 \rightarrow CO_2 + H_2O + N_2 \] Balance for stoichiometric combustion (\( \phi = 1 \)): \[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \] For air, add nitrogen (since air is 21% O₂ by volume, 79% N₂): \[ C_2H_5OH + 3O_2 + 3 \times 3.76 N_2 \rightarrow 2CO_2 + 3H_2O + 3 \times 3.76 N_2 \] \[ C_2H_5OH + 3O_2 + 11.28 N_2 \rightarrow 2CO_2 + 3H_2O + 11.28 N_2 \] ### **Step 2: Find Amount of Fuel and Air Used** #### **Stoichiometric Air-Fuel Ratio (\( AF_{st} \)):** \[ AF_{st} = 9 \text{ (given)} \] #### **Actual Air-Fuel Ratio (\( AF \)) at \( \phi = 0.75 \):** \[ AF = \frac{AF_{st}}{\phi} = \frac{9}{0.75} = 12 \] #### **Moles of Fuel and Air:** Let’s assume 1 kmol of fuel. - Mass of 1 kmol ethanol (\( C_2H_5OH \)): Molar mass = \( 2 \times 12 + 6 \times 1 + 16 = 46 \) kg/kmol - Air supplied per 1 kmol fuel: \[ AF = 12 \implies 1 \text{ kmol fuel} \rightarrow 12 \text{ kmol air} \] ### **Step 3: Heat Released per kmol Fuel (\( Q_{in} \)):** - Mass of 1 kmol ethanol = 46 kg - LHV = 26,950 kJ/kg \[ Q_{in} = 46 \times 26,950 = 1,239,700 \text{ kJ per kmol ethanol} \] ### **Step 4: Use First Law of Thermodynamics for Heat Addition** Assume air-standard (ideal gas, constant \( C_v \)): \[ Q_{in} = m_{tot} C_v (T_3 - T_2) \] But we want \( \Delta T = T_3 - T_2 \): - \( m_{tot} \) = total mass in cylinder (air + fuel, per cycle) - \( C_v \) = specific heat at constant volume (air, approx. 0.718 kJ/kg·K) - \( Q_{in} \) = energy released per cycle #### **Find number of cycles (for 0.4 L displacement):** For per cycle (1 cylinder, 1 cycle): - Total mass in cylinder per cycle: The mass in the cylinder is for the air + fuel for one engine cycle: - Volume at intake: \( V = 0.4 \) L = 0.0004 m³ - Pressure at intake: \( P = 80 \) kPa - Temperature at intake: \( T = 350 \) K Use ideal gas law: \[ n_{tot} = \frac{PV}{RT} \] Where: - \( R = 0.287 \) kJ/kg·K (for air) \[ m_{tot} = \frac{PV}{RT} \] \[ m_{tot} = \frac{80 \times 0.0004}{0.287 \times 350} = \frac{0.032}{100.45} \approx 0.0003185 \text{ kg} \] #### **Fuel mass per cycle:** \[ AF = \frac{m_{air}}{m_{fuel}} = 12 \implies m_{fuel} = \frac{m_{air}}{12} \] \[ m_{air} + m_{fuel} \approx m_{tot} \] But since fuel is much less than air, \( m_{air} \approx m_{tot} \): \[ m_{fuel} \approx \frac{0.0003185}{12} \approx 2.65 \times 10^{-5} \text{ kg} \] #### **Heat released per cycle:** \[ Q_{in,cycle} = m_{fuel} \times LHV = 2.65 \times 10^{-5} \times 26,950 \approx 0.714 \text{ kJ} \] #### **Total mass in cylinder (for heat absorption):** Use \( m_{tot} \approx 0.0003185 \) kg #### **Find \( \Delta T \):** \[ Q_{in,cycle} = m_{tot} C_v \Delta T \] \[ \Delta T = \frac{Q_{in,cycle}}{m_{tot} C_v} = \frac{0.714}{0.0003185 \times 0.718} = \frac{0.714}{0.0002287} \approx 3124 \text{ K} \] --- **Final Answer (a):** \[ \boxed{\Delta T \approx 3124 \ \text{K}} \] --- ## (b) **Find the Thermal Efficiency of the Engine (\( \eta_{th} \))** For an **Otto Cycle** (ideal SI engine), the thermal efficiency is: \[ \eta_{th} = 1 - \frac{1}{r^{k-1}} \] Where: - \( r \) = compression ratio = 9.0 - \( k \) = 1.35 \[ \eta_{th} = 1 - \frac{1}{9^{1.35-1}} = 1 - \frac{1}{9^{0.35}} \] Calculate \( 9^{0.35} \): \[ \ln(9) = 2.1972 \] \[ 0.35 \times 2.1972 = 0.769 \] \[ e^{0.769} \approx 2.157 \] \[ \eta_{th} = 1 - \frac{1}{2.157} \approx 1 - 0.463 = 0.537 \] --- **Final Answer (b):** \[ \boxed{\eta_{th} \approx 53.7\%} \] --- ## (c) **Draw the P-V Diagram for the Engine** ### **Steps to Draw:** - **Label 4 points:** 1 (intake), 2 (end of compression), 3 (end of combustion), 4 (end of expansion) - **Process 1-2:** Compression (adiabatic, up and left) - **Process 2-3:** Heat addition at constant volume (vertical up) - **Process 3-4:** Expansion (adiabatic, down and right) - **Process 4-1:** Heat rejection at constant volume (vertical down) [You would draw a standard **Otto cycle** \( P \)-\( V \) diagram] --- ### **Summary Table** | Part | Formula Used | Final Answer | |------|--------------|--------------------| | (a) | \( \Delta T = \frac{Q_{in}}{m_{tot} C_v} \) | \( \Delta T \approx 3124 \) K | | (b) | \( \eta_{th} = 1 - \frac{1}{r^{k-1}} \) | \( \eta_{th} \approx 53.7\% \) | | (c) | Otto cycle P-V diagram | (see above description) | --- If you need a hand-drawn P-V diagram, let me know!

**Final Answers:** **(a) Temperature Rise (\( \Delta T \)) during combustion:** \(\boxed{ \Delta T \approx 3124\, \text{K} }\) **(b) Thermal efficiency of the engine:** \(\boxed{ \eta_{th} \approx 53.7\% }\) **(c) P-V diagram:** A standard Otto cycle diagram with four processes: - Compression (adiabatic, upward curve) - Heat addition at constant volume (vertical line upward) - Expansion (adiabatic, downward curve) - Heat rejection at constant volume (vertical line downward) *(Please refer to a typical Otto cycle P-V diagram for visualization.)*