**3.2 Evaluation of the efficiency of the diesel engine working process based on the anergo-exergy method**

According to the theory [8], any heat and enthalpy can be represented as components of exergy and anergy.

$$Q = E\_Q + A\_Q; I = E + A\_\cdot$$

In addition, the anergy balance equations are valid for any ICE unit

$$\sum D\_i = \sum A\_{out} - \sum A\_{in}$$

and balance of exergy

$$\sum D\_i = \sum E\_{in} - \sum E\_{out}$$

where ΣЕin and ΣEout may include work supplied to the assembly or taken away from it.

From these equations, it can be concluded that the exergy losses ΣD*<sup>i</sup>* arising due to the irreversibility of real processes increase the energy leaving the assembly and decrease the exergy entering the assembly. This allows representing the flows of anergy, exergy, and exergy losses in the form of anergo-exergy scheme of the thermodynamic unit. Based on the above equations and fundamental considerations, which are laid above in the basis for constructing an anergo-exergy scheme of a thermodynamic unit, will make it possible to build an anergo-exergy scheme of a diesel engine, shown in **Figure 4**.

The bifurcation of the ICE assemblies made it possible to reveal the corresponding losses of exergy: *D*sc, *Ds*, Σ*Di*, *Dw*, *D*wp, *D*ex, *Dt*, *D*fp, *Do*, *D*op, *D*мн. To determine the listed exergy losses, it is easier to use the following energy balances of the corresponding assemblies. *<sup>D</sup>*sc <sup>=</sup> *<sup>А</sup>*sc – *<sup>А</sup>*0; *Ds* = (*A*″os - *<sup>A</sup>*<sup>0</sup> os) – (*A*sc – *A*s); *A<sup>т</sup>* = *A*exh – *<sup>A</sup>т*; *<sup>A</sup>*<sup>w</sup> <sup>=</sup> *<sup>A</sup>*″*<sup>w</sup>* – *Aw* – *<sup>A</sup>*qw; *Do* <sup>=</sup> *<sup>A</sup>*″*<sup>o</sup>* – *<sup>А</sup>o*; *<sup>D</sup>*op <sup>=</sup> *<sup>А</sup><sup>o</sup>* – *<sup>А</sup>*<sup>0</sup> *<sup>o</sup>*; *D*fp = *A*<sup>0</sup> *<sup>f</sup>* – *Аf*.

For a diesel engine cylinder, you can write

$$E\_{Q\_{dh}} = (Q\_{dh} - Q\_w) - A\_Q,$$

where *AQ* ¼ *Т*<sup>0</sup> Ð *<sup>δ</sup>Qch <sup>T</sup>* � *Т*<sup>0</sup> Ð *<sup>δ</sup>Qw <sup>T</sup>* ¼ *AQch* � *AQw*—heat anergy *Q* ¼ *Qch* � *Qw:* At the same time

$$D\_f = D\_s + D\_B = A\_T - A\_Q - A\_s - A\_f.\tag{12}$$

To find *AQ* and *Df*, we divide the cycle in the engine cylinder into two parts: section (a–e)—compression, combustion, and expansion and section (e–a)—the gas exchange process.

In general, *dA* ¼ *δAQ* þ *δA <sup>f</sup>* þ *δAs* � *δАТ* þ *δDs* þ *δDВ*, so for the process (a—f)

*Research and Innovation to Improve the Efficiency of Modern Diesel Engines DOI: http://dx.doi.org/10.5772/intechopen.102759*

#### **Figure 4.**

*Anergo-exergy scheme of the internal combustion engine. SC—supercharger; S—air cooler; C—cylinders; EX exhaust manifold; T—gas turbine; EM—engine mechanisms; FP—fuel pump; WP—water pump; OP—oil pump; W—water system unit that receives frictional heat transferred to water; O—oil system unit that receives frictional heat transferred to oil;* Ai*,* Е<sup>i</sup> *respectively, are the flows of anergy and exergy of the working fluid; —respectively, with the cooling agent anergy and exergy;* А<sup>0</sup> *os,* a *os,* a<sup>0</sup> <sup>w</sup>*,* A*w,* a <sup>w</sup>*,* A<sup>0</sup> <sup>o</sup>*,* Ao*,* A <sup>o</sup>*,* Af*,* a<sup>0</sup> <sup>f</sup> *and* E<sup>0</sup> *os,* E *os* Е<sup>0</sup> <sup>w</sup>*,* Еw*,* Е <sup>w</sup> Е0 <sup>o</sup>*,* Еo*,* Е <sup>o</sup> Еf*,* Е<sup>0</sup> <sup>f</sup> *respectively are the flows of anergy and exergy with the cooling agent, of water in the liquid cooling system of the engine, of oil in the engine lubrication system, of the fuel in the fuel supply system.*

$$A\_e - A\_a = A\_{Q\_{ch}} - A'\_{Q\_w} + A\_f,$$

For process (e–a)

$$A\_a - A\_\epsilon = -A\_{Q\_w}'' + A\_\sharp - A\_T + D\_f.$$

If the law of heat transfer in the gas exchange section is known, then

$$A\_{Q\_w}^{\prime\prime} = T\_0 \int \frac{\delta Q\_w}{T} = \frac{T\_0}{T\_m^{\prime\prime}} Q\_w^{\prime\prime},$$

where, *Т*<sup>00</sup> *<sup>m</sup>* average process (e–a) temperature. Having found *A*<sup>00</sup> *Qw* , we find the total losses of exergy during gas exchange *Df*. According to Eq. (12)), one can find *AQ* ¼ *AT* � *As* � *A <sup>f</sup>* � *Df :*

If the law of heat transfer is known in the area of compression, combustion, expansion, then

$$A'\_{Q\_w} = T\_0 \int \frac{\delta Q\_w}{T} = \frac{T\_0}{T'\_m} Q'\_w,$$

where *Т*<sup>0</sup> *<sup>m</sup>* average process (a–f) temperature. In this case *AQch* ¼ *A<sup>е</sup>* � *A<sup>а</sup>* � *A <sup>f</sup>* þ *A*<sup>0</sup> *Qw :*

The balance of the exergy flows of the internal combustion engine can be obtained by considering the contour *E*:

$$L\_0 + L\_m + L\_{op} + E\_o' + A\_o' + L\_{fp} + Q\_{ch} + E\_f + A\_w' + E\_w' + L\_{wp} + E\_o' + L\_{cs} = $$

$$L\_t = L\_t + E\_o'' + A\_o'' + L\_p + D\_{fp} + E\_w'' + A\_w'' + E\_{os}'' + D\_{st} + D\_t + A\_Q + D\_w + D\_T + E\_{cth} + L\_{mf}$$

Let us take into account that

$$L\_p = L\_{op} + L\_{wp} + L\_{fp}; L\_m + L\_{sc} = L\_{mi}; D\_{\sum o} = D\_o + D\_{op} = A\_o'' - A\_o';$$

$$D\_{\sum w} + A\_{Qw} = D\_w + D\_{wp} + A\_{Q\_w} = A\_w'' - A\_w'; D\_{f\overline{p}} + D\_s + D\_s = D\_f.$$

With this in mind, we get

$$\begin{aligned} Q\_{ch} &= L\_{\epsilon} + \left[ \left( E\_{o}'' - E\_{o}' \right) + \left( A\_{o}'' - A\_{o}' \right) \right] + \left[ \left( E\_{w}'' - E\_{w}' \right) + \left( A\_{w}'' - A\_{w}' \right) \right] + \left( E\_{os}'' - E\_{os}' \right) + \cdots \\ &+ \left( E\_{exh} - E\_{0} - E\_{f} \right) + D\_{sc} + D\_{s} + D\_{f} + A\_{Q} + D\_{T} \end{aligned}$$

or

$$Q\_{ch} = L\_{\epsilon} + \Delta E\_{o} + \Delta E\_{w} + \Delta E\_{os} + \Delta E\_{exh} + A\_{Q} + D\_{ch} + D\_{s} + D\_{f} + D\_{T} + D\_{\sum w^{2}} + D\_{\sum w^{3}}$$

Note that in this expression

$$
\sum E\_i + A\_{Qch} + \sum D\_i = Q\_2 \quad \text{and} \quad \sum E\_i = \sum Q\_2.
$$

Since *Qch* � *AQch* ¼ *EQch* , then

$$E\_{Q\_{\rm sh}} = L\_{\rm e} + \sum E\_i + \sum D\_i.$$

This dependence is the equation of the energy balance of the engine. It can be seen from it that the exergy *EQch* , is supplied to the working fluid with the heat of the fuel *<sup>Q</sup>*ch, is spent on the efficient operation of the engine *Le* <sup>P</sup> , covering the losses of exergy *Di*. Part of the fuel heat energy is removed to the environment with water Δ*Ew*, with oil Δ*Eo*, with intermediate air cooling Δ*Еos*, and with exhaust gases Δ*Еexh*. According to the well-known theory [9, 10], exergy supplied to the internal combustion engine

$$E\_{\text{sup}} = E\_{Q\_{\text{sh}}} - \sum E\_{i\text{-}}$$

*Research and Innovation to Improve the Efficiency of Modern Diesel Engines DOI: http://dx.doi.org/10.5772/intechopen.102759*

Then

$$E\_{\text{sup}} = L\_e + \sum D\_i \dots$$

The efficiency of converting exergy supplied to the internal combustion engine into useful work can be estimated by the exergy efficiency

$$\eta\_e = \frac{L\_e}{E\_{\text{sup}}} = \mathbf{1} - \frac{\sum D\_i}{E\_{\text{sup}}} \dots$$

Let us give an example of determining exergy losses using the proposed anergoexergy method for a 6ChN12/14 tractor diesel engine with a power of 150 kW in one of its operating modes.

The performed calculation shows that in the diesel cycle, when the heat of the fuel is transferred to the working fluid, 21.3% of the anergy of this heat was formed. In addition, due to the irreversibility of real processes, losses of exergy amounted to 16.75%, that is, 38.05% of inoperable heat was also formed in the cycle. Part of the workable heat (20.95%) is carried away into the environment by heat carriers—oil, water, air, and exhaust gases. The rest of the heat turned into useful work (41.05%).

The largest amount of workable heat is carried away by waste gases (16.84%). The loss of performance in the lubrication and cooling systems is 2.16 and 7.84%, respectively. Attention is drawn to the noticeable total loss of performance during filling and when gases enter the exhaust manifold—1.34%. Noteworthy are DSC = 1.25% and DT = 3.83%—losses of exergy in the supercharger and gas turbine. In reducing the indicated losses of exergy, reserves for increasing the efficiency of a diesel engine are laid. **Figure 5** shows the items of the exergy balance.

In the diesel cycle, the exergy of the chemical heat of the fuel is supplied to the working fluid

$$E\_{Q\_x} = Q\_{dh} - A\_{Q\_{dh}} = \mathbf{3.19kJ/cycle.}$$

**Figure 5**. *Exergy balance of a diesel engine.*

However, exergy took part in the process of converting this heat into work.

$$E\_{\rm sup} = E\_{Q\_{ch}} - \sum E\_i - A\_{Q\_{ch}} = 2.34k \text{J/cycle}.$$

Due to the irreversibility of real processes, 29% of *E*sup turned into loss of exergy, that is, a third of the exergy has irreversibly disappeared (turned into anergy).

The remaining 71% of exergy turned into effective work. The exergy losses in the exergy balance are "significant", since the exergy losses in water are 13.58%, that is, almost half of all exergy losses. Turbine losses are less and amount to about 6%. Both should be dealt with at the same time by the researcher in order to reduce them.

So, the use of anergo-exergy method of analysis makes it possible to identify the mechanisms of the formation of internal and external losses of diesel engines and their systems, and to substantiate the ways to achieve their maximum efficiency.
