*2.5.1 Determination of the yield (material balance) of the furnace*

The role of the furnace is to heat the oil to a temperature in the range of 330–385°C in order to achieve a vapor-liquid balance before the distillation column [1, 7, 25, 29]. The most popular furnace in the Russian Federation is the tube furnace, as shown in **Figure 11**. The tube furnace is a unit designed to heat and force chemical processes inside the chamber, which are achieved by using the heat generated when fuel is burned. To calculate the heat balance of the atmospheric column, it is necessary to determine the heat of the furnace by the vapor phase and the liquid phase, therefore determine the material balance of the furnace. For example, the furnace material balance we used Bonga crude and the following data [25]:


We use the method of gradual approximation to calculate the percentage of the top product's distillate, with the Eq. (35).

$$\text{Fn} \left( \text{emol} \right) = \sum\_{i} \frac{\text{Cmol}\_{i}}{1 + \text{emol} \times (\text{K}\_{i} - 1)} \tag{35}$$

**Figure 11.** *Cross-sectional with heat exchange chamber of a tube furnace.*

Where, Cmol-molar composition of the upper product (distillate); emol- the mole fraction of the distillate; Ki-phase equilibrium constant of components *i* under conditions. The graph or **Figure 12** shows the proportion of distillation through the formula (33) of Bonga crude – 138158.9 kg/h.

From the graph (**Figure 12**) we determine that the molar fraction of distillation is emol = 0.855. The molar composition of the liquid is also found by the formula (36).

$$Y\_i = K\_i \times Y\_i,\tag{36}$$

Molecular masses are calculated according to the rule of additivity (38):

**Components kg/kmol kg/h % mass Mole fraction Steam** C2H6 30 22.11 0.02 0.00 1028.85 C3H8 33 388.23 0.28 0.01 404.90 Iso-C4H10 58 563.69 0.41 0.01 135.04 N-C4H10 58 886.98 0.64 0.02 126.43 Iso-C5H12 72 124.34 0.09 0.00 85.12 N-C5H12 72 31.78 0.02 0.00 84.19 –105 95.1 4863.19 3.52 0.08 103.15 –125 107.72 3550.68 2.57 0.05 25.25 –145 118.72 4738.85 3.43 0.06 19.53 –165 130.52 9035.59 6.54 0.10 15.72 –185 143.12 2528.31 1.83 0.03 12.51 –205 156.52 5526.36 4.00 0.05 9.84 –225 170.72 5567.80 4.03 0.05 7.66 –245 185.72 7046.10 5.10 0.06 5.88 –265 201.52 4324.37 3.13 0.03 4.47 –285 218.12 6769.79 4.90 0.05 3.35 –305 235.52 7046.10 5.10 0.04 2.47 –325 253.72 8469.14 6.13 0.05 1.80 –345 272.72 15017.87 10.87 0.08 1.30 –375 297.6 21456.08 15.53 0.11 0.92 >375 443.91 30132.46 21.81 0.10 0.58 H2O 18 69.08 0.05 0.01 0.05

The amount of steam and liquid mG phases is determined by the formulas

Accordingly, MsrР = 180.54; MsrC = 206.77. emas = 0.747.

(39), (40):

**129**

**Figure 12.**

*Graph of the vapor-liquid state isotherm.*

**Table 4.**

*Data of the molar and mass composition of desalted crude.*

*Crude Distillation Unit (CDU)*

*DOI: http://dx.doi.org/10.5772/intechopen.90394*

MsrP <sup>¼</sup> <sup>X</sup>*Mi* � *Yi* (38)

The mass fraction of distillation is calculated by the formula (37):

$$\text{emas} = \text{emol} \times \frac{\text{MsrP}}{\text{MsrC}},\tag{37}$$

where MsrP is average molecular weight of the vapor phase; MsgC is average molecular weight of feedstocks.
