**Figure 8.**

*The descriptive diagram of the desalting process.*

status "Expense" is similar to the desalting- dewatering process. 140 t of emulsion � 1.4 t of water in emulsion = 138.6 t of oil. The status "Expense" is similar to the desalting-dewatering process. Loss is equal: ð Þ <sup>138</sup>*:*6�0*:*<sup>3</sup> <sup>100</sup> = 0.4191 t. The amount of dewatered oil or desalted oil will be 138.6 t – 0.4158 t = 138.0897 t. According to this task and Russian GOST R 51828-2002, the water content in the desalted oil shall not exceed 0.5% of the mass. Thus, the water contained in the desalted oil will be:

$$138.0897\text{ t} - 99.95\% \text{ mass.} \tag{4}$$

$$\mathbf{X} - \mathbf{0}.\mathbf{0}5\% \text{ mass.}\tag{5}$$

The amount of water X is 0.0691 t.

The water content-X is 0.0691 t in the desalted oil. The total water content phase in the "incoming" status is 14 t + 0.14 t = 14.14 t.

Based on the previous results and by the possibility that we can really have more losses. We arbitrarily decide to multiply our losses by 2; the total loss is: 0.4191 t + 0.4191 t = 0.8383 t. Consequently, the summary material balance of the installation is made. The total amount of drainage water is equal to the drainage water by oil treatment stage: 0.9332 t + 14.07 t = 15.0032 t. The material balance of the electrostatic desalter installation is presented in **Table 2**.
