INTRODUCTION

Thanks to flexible control of the compression ratio, it is possible to influence the parameters of physical processes in the engine that affect fuel consumption and the emission of toxic components: pressure and temperature at the end of the compression stroke; maximum combustion pressure and temperature; degree of expansion and indicator efficiency; combustion chamber volume; exhaust gas temperature.

Despite the fact that adjusting the compression ratio is a difficult task, known two-shaft designs can be used without significant complications [1–4]. Work [5] showed a decrease in engine efficiency with increasing crankshaft rotation speed, which can be corrected by compensating the compression ratio. The article [6] pointed out the prospects of using twin-shaft engines [1–5] with an adjustable compression ratio. Here, according to the data in article [6], the results of calculating the efficiency and fuel consumption for speeds of 2000 and 4000 rpm with a compression ratio of 8.6 and a speed of 4000 rpm with a compression ratio of 13 and combustion characteristics m = 3 and 1 are presented. The indicated characteristics at higher speeds are shown to approach characteristics at lower speeds with increasing compression ratio. Previously [5] engine parameters were calculated under various modes based on Wiebe theory. Using the data from the above calculations, the efficiency and fuel consumption are determined for the corresponding modes and nature of engine operation.

MATHEMATICAL MODEL

The calculation is made under the following conditions:

Specific volume of working fluid

$$\nu = \frac{{{{\nu }_{a}}}}{{{\varepsilon }}}\left\{ {1 + \frac{{{{\varepsilon }} - 1}}{2}\left[ {\left( {1 + \frac{1}{{{\lambda }}}} \right)} \right] - \left( {{{\cos\alpha }} + \frac{1}{{{\lambda }}}\sqrt {1 - {{{{\lambda }}}^{2}}{{{\text{sin}}}^{2}}{{\alpha }}} } \right)} \right\},$$

where λ is the ratio of the crank radius.

Proportion of fuel burned at the site \(i - \left( {i - 1} \right)\)

$${{\Delta }}{{x}_{n}} = {{e}^{{ - 6.908{{{\left( {{{\Delta }}{{{{\varphi }}}_{n}}/{{{{\varphi }}}_{z}}} \right)}}^{{m + 1}}}}}} - {{e}^{{ - 6.908{{{\left( {{{\Delta }}{{{{\varphi }}}_{{n - 1}}}/{{\Delta }}{{{{\varphi }}}_{z}}} \right)}}^{{m + 1}}}}}}.$$

Pressure of the working fluid in the cylinder during the combustion process

$${{P}_{n}} = \frac{{0.0854{{q}_{z}}{{\Delta }}{{x}_{n}} + {{P}_{{n - 1}}}\left( {k{{{v}}_{{n - 1}}} - {{{v}}_{n}}} \right)}}{{k{{{v}}_{n}} - {{{v}}_{{n - 1}}}}},$$

where k is the heat capacity factor; qz is the total heat of combustion used.

Ignition timing is 24°, compression ratios are 8.6 and 13, rotation speed is 2000 and 4000 rpm, combustion character indicators m = 1 and 3. Flame front inclination angle φr = 46° and 92°.

Table 1 shows the data of this calculation. Based on these data, the corresponding efficiency factors and fuel consumption were determined.

Table 1. Pressure distribution depending on the specific volume for combustion types m = 1 and 3 at 2000 and 4000 rpm crankshaft for compression ratios ε = 8.6, ε = 13, φz is the duration of combustion
Full size table

NUMERICAL MODEL

Figure 1 shows a diagram for determining the operation of the cycle. Based on the points on the graph, the state of the cycle is determined, from which the efficiency is calculated.

Fig. 1.
figure 1

Scheme for determining cycle operation: bMm—top dead center of the engine piston position; HMm is the bottom dead center of the engine piston position.

Full size image

The cycle work is composed of the sum of the combustion work during compression

$${{l}_{{yc}}} = \mathop \smallint \limits_y^c {{p}_{y}}{v}d{{{v}}_{y}}\quad {\text{or}}\quad \sum {{\Delta }}{{p}_{y}}{{\Delta }}{{{v}}_{{y~}}}.$$
(1)

The cycle work is composed of the sum of the combustion work during expansion

$${{l}_{{cz}}} = \mathop \smallint \limits_c^z {{p}_{c}}{{{v}}_{c}}d{{{v}}_{c}}\quad {\text{or}}\quad \sum {{\Delta }}{{p}_{c}}\Delta {{{v}}_{{c~}}}.$$
(2)

The work of the cycle is made up of the sum of the work of pure compression

$${{l}_{{ay}}} = \frac{1}{{{{n}_{1}} - 1}}\left( {{{p}_{y}}{{{v}}_{y}} - {{p}_{a}}{{{v}}_{a}}} \right),\quad {{n}_{1}} = 1.35.$$
(3)

The work of the cycle is composed of the sum of the work of pure expansion

$${{l}_{{zb}}} = \frac{1}{{{{n}_{2}} - 1}}\left( {{{p}_{z}}{{{v}}_{z}} - {{p}_{b}}{{{v}}_{b}}} \right),\quad {{n}_{2}} = 1.28,$$
(4)

where \({{p}_{b}} = {{\left( {\frac{{{{{v}}_{z}}}}{{{{{v}}_{b}}}}} \right)}^{{{{n}_{2}}}}}\), \({{p}_{z}} = {{\left( {\frac{{{{{\text{v}}}_{z}}}}{{{{{\text{v}}}_{a}}}}} \right)}^{{{{n}_{2}}}}}{{p}_{z}}\).

Complete cycle work

$$l = {{l}_{{yc}}} + {{l}_{{cz}}} + {{l}_{{ay}}} + {{l}_{{zb}}}.$$
(5)

Efficiency determined by the formula

$${{{{\eta }}}_{i}} = \frac{{l(1 + {{\gamma }})(1 + {{\alpha }}L_{0}^{'})}}{{427Hu}},$$
(6)

where H is the lower calorific value of the fuel (H = 10500 kcal/kg); (1 + γ) is the coefficient of residual gases, (1 + γ) = 1.088; α is the excess air coefficient (α = 0.85); \(L_{0}^{'}\) is the theoretically required amount of air for complete combustion of 1 kg of fuel (\(L_{0}^{'}\) = 14.8 kg g).

Indicated specific fuel consumption \(~g = \frac{{632}}{{{{{{\eta }}}_{i}}Hu}}\).

RESULTS

I. Consider the case \({{\varepsilon }} = 8.6\); m = 3; φr = 46°

$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 1 \right)\,\,{{l}_{{yc}}} = - 0.515 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 2 \right)\;{{l}_{{cz}}} = 2.859 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 3 \right)\,\,{{l}_{{ay}}} = - 2.476 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 4 \right)\;{{l}_{{zb}}} = 10.75 \times {{10}^{4}}\;{\text{kg m/kg,}}$$
$${\text{as}}\;{\text{a}}\;{\text{result}},\;{{l}_{i}} = 10.63 \times {{10}^{4}}\;{\text{kg m/kg}}{\text{.}}$$

According to formula (6)

$${{\eta }} = \frac{{10.63 \times {{{10}}^{4}} \times 1.088\left( {1 + 0.85 \times 14.8} \right)}}{{427 \times 10{\kern 1pt} {\kern 1pt} 500}} = 0.35.$$

Indicated specific fuel consumption

$$g = \frac{{632}}{{{{{{\eta }}}_{1}}{{H}_{0}}}} = \frac{{632}}{{0.35 \times 10{\kern 1pt} {\kern 1pt} 500}} = 172~\,\,{\text{g/}}\left( {{\text{h}}.{\text{p}}\,\,{\text{h}}} \right),$$

where g/(h.p h)—grams per horsepower hour.

II. In case of \({{\varepsilon }} = 8.6;\,\,m = 1;~\,\,{{{{\varphi }}}_{r}} = 46^\circ \)

$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 1 \right)\;{{l}_{{ye}}} = - 0.958 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 2 \right)\;{{l}_{{cz}}} = 2.7979 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 3 \right)\;{{l}_{{ay}}} = - 2.476 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 4 \right)\;{{l}_{{zb}}} = 10.134~\;{\text{kg m/kg,}}$$
$${\text{as}}\;{\text{a}}\;{\text{result}},\;{{l}_{1}} = 9.498 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${{l}_{i}} = 9.4979.$$

According to formula (6)

$${{\eta }} = \frac{{9.4979 \times {{{10}}^{4}} \times 1.088\left( {1 + 0.85 \times 14.8} \right)}}{{427 \times 10{\kern 1pt} {\kern 1pt} 500}} = 0.31.$$

Indicated specific fuel consumption

$$g = \frac{{632}}{{{{{{\eta }}}_{1}}{{H}_{0}}}} = \frac{{632}}{{0.31 \times 10500}} = 194\,\,{\text{g/(h}}.{\text{p}}\,\,{\text{h)}}.$$

III. In case of case \({{\varepsilon }} = 8.6\); m = 3; φr = 92°

$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 1 \right)\;{{l}_{{ye}}} = - 0.4636 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 2 \right)\;{{l}_{{cz}}} = 3.4686 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 3 \right)\;{{l}_{{ay}}} = - 2.4757 \times {{10}^{4}}\;{\text{kg m/kg ;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 4 \right)\;{{l}_{{zb}}} = 4.1108 \times {{10}^{4}}\,\,{\text{kg m/kg,}}$$
$${\text{as}}\;{\text{a}}\;{\text{result,}}\;{{l}_{i}} = 4.632 \times {{10}^{4}}\,\,{\text{kg m/kg}}{\text{.}}$$

According to formula (6)

$${{\eta }} = 0.1526;$$
$$g = \frac{{632}}{{0.15 \times 10500}} = 0.401\,\,{\text{g/}}\left( {{\text{h}}.{\text{p}}\,\,{\text{h}}} \right).$$

IV. In case of \({{\varepsilon }} = 8.6\); m = 1; φr = 92°

$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 1 \right)\;{{l}_{{ye}}} = - 1.354 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 2 \right)\;{{l}_{{cz}}} = 6.317 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 3 \right)\;{{l}_{{ay}}} = - 2.476 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 4 \right)\;{{l}_{{zb}}} = 3.926 \times {{10}^{4}}\,\,{\text{kg m/kg,}}$$
$${\text{as}}\;{\text{a}}\;{\text{result}},\;{{l}_{i}} = 6.413 \times {{10}^{4}}\,\,{\text{kg m/kg}}{\text{.}}$$

According to formula (6)

$${{\eta }} = 0.2116;$$
$$g = \frac{{632}}{{0.21 \times 10500}} = 286\,\,{\text{g/(h}}.{\text{p}}\,\,{\text{h)}}.$$

V. In case of \({{\varepsilon }} = 13\); m = 3; φr = 64°

$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 1 \right)\;{{l}_{{ye}}} = 0.44548 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 2 \right)\;{{l}_{{cz}}} = 5.926 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 3 \right)\;{{l}_{{ay}}} = - 2.998 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 4 \right)\;{{l}_{{zb}}} = 11.28 \times {{10}^{4}}\;{\text{kg m/kg,}}$$
$${\text{as}}\;{\text{a}}\;{\text{ result}},\;{{l}_{i}} = 13.8 \times {{10}^{4}}\;{\text{kg}}\,\,{\text{m/kg}}{\text{.}}$$

According to formula (6)

$${{\eta }} = 0.455;$$
$$g = \frac{{632}}{{0.455 \times 10{\kern 1pt} {\kern 1pt} {\kern 1pt} 500}} = 132~\,\,{\text{g/}}\left( {{\text{h}}.{\text{p}}\,\,{\text{h}}} \right),$$

VI. In case of \({{\varepsilon }} = 13\); m = 1; φr = 64°

$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 1 \right)\;{{l}_{{ye}}} = 0.9949 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 2 \right)\;{{l}_{{cz}}} = 6.5014 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 3 \right)\;{{l}_{{ay}}} = - 2.998 \times {{10}^{4}}\;{\text{kg m/kg;}}$$
$${\text{according}}\;{\text{to}}\;{\text{formula}}\;\left( 4 \right)\;{{l}_{{zb}}} = 9.39 \times {{10}^{4}}\;{\text{kg m/kg,}}$$
$${\text{as}}\;{\text{a}}\;{\text{result}},\;{{l}_{i}} = 13.78 \times {{10}^{4}}\;{\text{kg m/kg}}{\text{.}}$$

According to formula (6)

$${{\eta }} = 0.3923;$$
$$g = \frac{{632}}{{0.39 \times 10{\kern 1pt} {\kern 1pt} 500}} = 154\,\,~{\text{g/}}\left( {{\text{h}}.{\text{p}}\,\,{\text{h}}} \right).$$

CONCLUSIONS

With an increase in crankshaft rotation speed from 2000 to 4000 rpm, the efficiency of the cycle will decrease by almost half and fuel consumption will increase accordingly. By increasing the compression ratio from 8.6 to 13 at a speed of 4000 rpm, the efficiency is restored and fuel consumption decreases accordingly.

As the crankshaft rotation speed increases, the pressure in the cylinder drops and the efficiency decreases. In order to restore pressure and, accordingly, efficiency, compression ratio control can be applied, which must be provided for in the engine design.