3.3.1. Burner

(a) Reactant (natural gas) exergy *Er* is calculated as follows: [43]

$$E\_r = 0.95 Q\_{\% \mu, \upsilon, ad} \tag{26}$$

(b) *Eair* is the exergy of air:

$$E\_{a\dot{r}} = 0\tag{27}$$

(c) Exergy loss of irreversible combustion in the burner calculated by reactant exergy *Er* and resultant exergy *Ers* [43]:

$$\begin{cases} \begin{array}{c} E\_{b,irr} = E\_r - E\_{rs} \\ E\_{rs} = (Q\_{ar,net} - Q\_{ar,net} \varepsilon) \left( 1 - \frac{T\_0}{T\_b - T\_0} \ln \frac{T\_b}{T\_0} \right) \end{array} \tag{28}$$

(d) Exergy loss of incomplete combustion is energy loss of incomplete combustion according to the definition of exergy:

$$E\_{b,inc} = Q\_{b,inc} \tag{29}$$

(e) Exergy loss of heat leakage is calculated by heat leakage energy and heat leakage temperature, and the calculation of burner, pipe, well and soil is similar. The tube surface temperature *Tb*,*w*,*o*, *Tp*,*w*,*o*, *Tw*,*w*,*<sup>o</sup>* are regarded as leakage temperature calculated in Appendix B (c). The heat leakage temperature of soil is the soil temperature itself:

$$E\_{b,l} = Q\_{b,l}(1 - \frac{T\_0}{T\_{b,w,o}}) \tag{30}$$

$$E\_{p,l} = Q\_{p,l}(1 - \frac{T\_0}{T\_{p,w,\rho}}) \tag{31}$$

$$E\_{w,l} = Q\_{w,l}(1 - \frac{T\_0}{T\_{w,\mu\nu\rho}}) \tag{32}$$

$$E\_{s,l} = Q\_{s,l}(1 - \frac{T\_0}{T\_s})\tag{33}$$

(f) Exergy loss of flow is energy loss of flow on account of the definition of exergy and the pipe as well as well is calculated in the same way as the burner:

$$E\_{b,f} = Q\_{b,f} \tag{34}$$

$$E\_{p,f} = Q\_{p,f} \tag{35}$$

$$E\_{w,f} = Q\_{w,f} \tag{36}$$

(g) Exergy out of the burner *Eb,to,p*, is solved by the conservation of exergy equation. The calculation of the exergy out of the pipe, well and soil is similar, so the equation will not be repeated below:

$$E\_{b,to,p} = E\_r + E\_{air} - E\_{b,irr} - E\_{b,juc} - E\_{b,l} - E\_{b,f} \tag{37}$$
