The air-fuel ratio of a Diesel engine is 29:1. If the compression ratio is 16:1 and the temperature at the end of compression is 900 K, find at what cylinder volume the combustion is complete? Express this volume as a percentage of stroke. Assume that the combustion begins at the top dead centre and takes place at constant pressure. Take calorific value of the fuel as 42000 kJ/kg, R-0.287 kJ/kg K and Cv0.709+0.000028 TkJ/kg K.
The correct answer and explanation is:
To solve this problem, we need to find the cylinder volume at which the combustion is complete, expressed as a percentage of the stroke. Here’s a step-by-step explanation and calculation:
Given Data:
- Air-fuel ratio, AFR=29:1AFR = 29:1
- Compression ratio, CR=16:1CR = 16:1
- Temperature at the end of compression, Tc=900 KT_c = 900 \, \text{K}
- Calorific value of the fuel, CV=42000 kJ/kg\text{CV} = 42000 \, \text{kJ/kg}
- Specific gas constant, R=0.287 kJ/kg\cdotpKR = 0.287 \, \text{kJ/kg·K}
- Specific heat at constant volume, Cv=0.709+0.000028T kJ/kg\cdotpKC_v = 0.709 + 0.000028T \, \text{kJ/kg·K}
Assumptions:
- Combustion occurs at constant pressure.
- Combustion begins at Top Dead Center (TDC).
Step-by-Step Solution:
- Volume at Top Dead Center (TDC): At TDC, the cylinder is at its minimum volume, VTDCV_{TDC}, which corresponds to the clearance volume, VcV_c. CR=VBDCVTDC=16⇒VTDC=VBDC16CR = \frac{V_{BDC}}{V_{TDC}} = 16 \quad \Rightarrow \quad V_{TDC} = \frac{V_{BDC}}{16}
- Heat Released by Combustion: The heat released by the combustion, QQ, is: Q=CVAFR=4200029=1448.28 kJ/kgQ = \frac{\text{CV}}{AFR} = \frac{42000}{29} = 1448.28 \, \text{kJ/kg}
- Temperature Rise Due to Combustion: Using Q=m⋅Cv⋅ΔTQ = m \cdot C_v \cdot \Delta T and assuming CvC_v varies with temperature, the mean value of CvC_v over the temperature range can be used. For simplicity, we approximate CvC_v at the end of compression: Cv=0.709+0.000028⋅900=0.7342 kJ/kg\cdotpKC_v = 0.709 + 0.000028 \cdot 900 = 0.7342 \, \text{kJ/kg·K} Assuming unit mass of air, the temperature rise is: ΔT=QCv=1448.280.7342≈1973.23 K\Delta T = \frac{Q}{C_v} = \frac{1448.28}{0.7342} \approx 1973.23 \, \text{K} The final temperature after combustion: Tf=Tc+ΔT=900+1973.23≈2873.23 KT_f = T_c + \Delta T = 900 + 1973.23 \approx 2873.23 \, \text{K}
- Volume at End of Combustion: Since the combustion occurs at constant pressure, the relation between temperature and volume is: TcVTDC=TfVf⇒Vf=VTDC⋅TfTc\frac{T_c}{V_{TDC}} = \frac{T_f}{V_f} \quad \Rightarrow \quad V_f = V_{TDC} \cdot \frac{T_f}{T_c} Substituting values: Vf=VBDC16⋅2873.23900=VBDC16⋅3.192V_f = \frac{V_{BDC}}{16} \cdot \frac{2873.23}{900} = \frac{V_{BDC}}{16} \cdot 3.192 Vf=3.19216⋅VBDC=0.1995⋅VBDCV_f = \frac{3.192}{16} \cdot V_{BDC} = 0.1995 \cdot V_{BDC} The volume as a percentage of the stroke is: %Vf=19.95%\%V_f = 19.95\%
Final Answer:
The combustion is complete at 19.95% of the stroke.
Explanation:
- At TDC, the engine cylinder is at minimum volume, corresponding to the clearance volume.
- During combustion at constant pressure, the temperature rises significantly, causing the gas to expand.
- Using thermodynamic relationships and given data, the final volume is determined based on the temperature rise and compression ratio.
- Expressing the final volume as a percentage of the total stroke allows us to understand how much expansion has occurred relative to the engine’s full range of motion.