Effects of using ethyl acetate as a surprising additive in SI engine pertaining to an environmental perspective

Abstract

Blending oxygenated fuels with gasoline improve exhaust emissions. In this study, tests were performed in a single-cylinder spark-ignition engine at a constant speed of 1600 rpm for different loads (8, 10, 12, and 14 kg) and compression ratios (8:1, 9:1, and 10:1), and carbon monoxide, hydrocarbon, nitrogen monoxide, and carbon dioxide emissions were determined. Ethyl acetate was mixed with gasoline at 4%, 8%, and 12% by volume and used as alternative fuels in the experiments. The performance of ethyl acetate blends compared to gasoline was investigated by energy, exergy, and exergoeconomic analyses. Carbon monoxide, hydrocarbon, and nitrogen monoxide occurring in the engine were lower for ethyl acetate blends than gasoline. When the load is 14 kg and the compression ratio is 10:1, hydrocarbon emissions in 12% ethyl acetate blend and gasoline are 128.66 ppm and 161 ppm, respectively. According to the data obtained from energy analysis, the difference between the thermal efficiency of 12% ethyl acetate blend and gasoline fuels is a maximum of 5%. This difference decreases even more at low engine loads. When the load is 14 kg and the compression ratio is 10:1 in 12% ethyl acetate blend and gasoline, the exergy efficiency is 34.27% and 37.17%, respectively. According to the exergoeconomic analysis, the engine power cost varies according to different loads and compression ratios and is higher by 20–27% in 12% ethyl acetate blend than gasoline. In case pump prices of ethyl acetate are reduced, fuel blends with gasoline can be used as an alternative fuel.

Appendix

The uncertainty values of the measured characteristics are as follows: load (∆m) and engine speed (∆N) are considered as ± 0.1 kg and 1 rpm, respectively. For the fuel samples, the uncertainty of the volume (∆f) is taken as ± 0.1 cc and the uncertainty of the time (∆t) is accepted as ± 0.2 s, respectively. A sample calculation of the uncertainty analysis at the load of 14 kg operating condition is presented underneath (Table 11).

Table 11 The engine parameters and their values

$$\frac{\Delta N}{N} = \frac{1}{1600} = 0.0625 \%$$

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$$\frac{\Delta m}{m} = \frac{0.1}{{14}} = 0.7143 \%$$

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• 1. Uncertainty in brake power (BP)

  $$BP = \frac{2\pi NT}{{60. 1000}} = \frac{2\pi Nmgl}{{60000}} = \frac{2 x 3.14 x 1600 x 14 x 9.81 x 0.230 }{{60000}} = 5.2927 kW$$

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  $$\frac{\partial BP}{{\partial N}} = \frac{2\pi mgl}{{60000}} = \frac{2 x 3.14 x 14 x 9.81 x 0.230}{{60000}} = 0.00330791$$

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  $$\frac{\partial BP}{{\partial m}} = \frac{2\pi Ngl}{{60000}} = \frac{2 x 3.14 x 1600 x 9.81 x 0.230}{{60000}} = 0.3780467$$

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  $$\Delta BP = \sqrt {\left( {\Delta N\frac{\partial BP}{{\partial N}}} \right)^{2} + \left( {\Delta m\frac{\partial BP}{{\partial m}}} \right)^{2} }$$

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  $$\Delta BP = \sqrt {\left( {1 x 0.00330791} \right)^{2} + \left( {0.1 x 0.3780467} \right)^{2} }$$

  (34)
  $$\Delta BP = 0.037949115 kW$$

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  $$\frac{\Delta BP}{{BP}} = \frac{0.037949115}{{5.2927}} = 0.717\%$$

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2.Uncertainty in mass of the fuel consumption (MFC)

$$MFC = \frac{f x 3600 x \rho }{{t x 1000}} = \frac{25.00 x 3600 x 0.715}{{60 x 1000}} = 1.0725 \frac{kg}{h}$$

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$$\frac{\partial MFC}{{\partial t}} = - \frac{f x 3600 x \rho }{{\left( t \right)^{2} x 1000}} = - \frac{25.00 x 3600 x 0.715}{{60 x 60 x 1000}} = - 0.017875$$

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$$\frac{\partial MFC}{{\partial f}} = \frac{3600 x \rho }{{t x 1000}} = \frac{3600 x 0.715}{{60 x 1000}} = 0.04290$$

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$$\Delta MFC = \sqrt {\left( {\Delta t\frac{\partial MFC}{{\partial t}}} \right)^{2} + \left( {\Delta f\frac{\partial MFC}{{\partial f}}} \right)^{2} }$$

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$$\Delta MFC = \sqrt {\left( {0.2 x - 0.017875} \right)^{2} + \left( {0.1 x 0.04290 } \right)^{2} }$$

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$$\Delta MFC = 0.005584329 \frac{kg}{{kWh}}$$

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$$\frac{\Delta MFC}{{MFC}} = \frac{0.005584329}{{1.0725}} = 0.521\%$$

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3.Uncertainty in brake specific fuel consumption (BSFC)

$$BSFC = \frac{MFC}{{BP}} = \frac{1.0725 }{{5.2927}} = 0.202637595 \frac{kg}{{kWh}}$$

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$$\frac{\partial BSFC}{{\partial BP}} = - \frac{MFC}{{\left( {BP} \right)^{2} }} = - \frac{1.0725}{{\left( {5.2927} \right)^{2} }} = - 0.038286242$$

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$$\frac{\partial BSFC}{{\partial MFC}} = \frac{1}{BP} = \frac{1}{5.2927} = 0.188939482$$

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$$\Delta BSFC = \sqrt {\left( {\Delta BP\frac{\partial BSFC}{{\partial BP}}} \right)^{2} + \left( {\Delta MFC\frac{\partial BSFC}{{\partial MFC}}} \right)^{2} }$$

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$$\Delta BSFC = \sqrt {\left( {0.037949115 x - 0.038286242} \right)^{2} + \left( {0.005584329 x - 0.188939482} \right)^{2} }$$

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$$\Delta BSFC = 0.001795617 \frac{kg}{{kWh}}$$

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$$\frac{\Delta BSFC}{{BSFC}} = \frac{0.001795617}{{0.202637595}} = 0.886\%$$

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4.Uncertainty in brake thermal efficiency (BTE)

$$BTE = \frac{BP x 3600 x 100}{{MFC x LHV}} = \frac{5.2927 x 3600 x 100}{{1.0725 x 44250}} = 40.14849\%$$

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$$\frac{\partial BTE}{{\partial BP}} = \frac{3600 x 100}{{MFC x LHV}} = \frac{3600 x 100}{{1.0725 x 44250}} = 7.585635$$

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$$\frac{\partial BTE}{{\partial MFC}} = - \frac{BP x 3600 x 100}{{\left( {MFC} \right)^{2} x LHV}} = - \frac{5.2927 x 3600 x 100}{{\left( {1.0725} \right)^{2} x 44250}} = - 37.43449$$

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$$\Delta BTE = \sqrt {\left( {\Delta BP\frac{\partial BTE}{{\partial BP}}} \right)^{2} + \left( {\Delta MFC\frac{\partial BTE}{{\partial MFC}}} \right)^{2} }$$

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$$\Delta BTE = \sqrt {\left( {0.037949115 x 7.585635 } \right)^{2} + \left( {0.005584329 x - 37.43449} \right)^{2} }$$

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$$\Delta BTE = 0.355764677\%$$

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$$\frac{\Delta BTE}{{BTE}} = \frac{0.355764677}{{40.14849}} = 0.886\%$$

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