# Calculate the density of petroleum products

In order to determine with the help of this table, the density of the oil product at a given temperature, you should:

Table of average temperature corrections of the density of petroleum products
Density at 20 °С Temperature correction of 1 °С Density at 20 °С Temperature correction of 1 °С
0.6500–0.6590 0.000962 0.8300–0.8399 0.000725
0.6600–0.6690 0.000949 0.8400–0.8499 0.000712
0.6700–0.6790 0.000936 0.8500–0.8599 0.000699
0.6800–0.6890 0.000925 0.8600–0.8699 0.000686
0.6900–0.6999 0.000910 0.8700–0.8799 0.000673
0.7000–0.7099 0.000897 0.8800–0.8899 0.000660
0.7100–0.7199 0.000884 0.8900–0.8999 0.000647
0.7200–0.7299 0.000870 0.9000–0.9099 0.000633
0.7300–0.7399 0.000857 0.9100–0.9199 0.000620
0.7400–0.7499 0.000844 0.9200–0.9299 0.000607
0.7500–0.7599 0.000831 0.9300–0.9399 0.000594
0.7600–0.7699 0.000818 0.9400–0.9499 0.000581
0.7700–0.7799 0.000805 0.9500–0.9599 0.000567
0.7800–0.7899 0.000792 0.9600–0.9699 0.000554
0.7900–0.7999 0.000778 0.9700–0.9799 0.000541
0.8000–0.8099 0.000765 0.9800–0.9899 0.000528
0.8100–0.8199 0.000752 0.9900–1.0000 0.000515
0.8200–0.8299 0.000738
1. found on the passport of the density of petroleum at +20 °С;
2. measure the average temperature in the cargo tank;
3. determine the difference between +20 °C and an average temperature of the cargo;
4. on a graph of the temperature correction to find the correction of 1 °C, corresponding to the density of the product at +20 °C;
5. multiply the temperature correction to the density of the temperature difference;
6. obtained in Section "d" work to subtract from the density at +20 °C, if the average temperature of the oil product in the tank is above +20 °C, or add this product if the product temperature below +20 °C.

## Example №1

The density of petroleum product at +20 °C, according to the passport 0.8240. The temperature of mineral oil in the tank +23 °C. Determine the density of mineral oil on the table at this temperature.

### We find:

1. temperature difference of 23 °C − 20 °C = 3 °C;
2. temperature correction of 1 °C on the table for the density of 0.8240, representing 0.000738;
3. temperature correction of 3 °C: 0.000738 × 3 = 0.002214 or 0.0022 is rounded;
4. required density of mineral oil at a temperature of +23 °C (amendment must be subtracted, as the temperature in the cargo tank above +20 °C), equal to 0.8240 − 0.0022 = 0.8218 or 0.8220 is rounded.

## Example №2

The density of petroleum at +20 °C, according to a passport, 0.7520. The temperature in the cargo tank −12 °C. Determine the density of petroleum product at this temperature.

### We find:

1. the temperature difference between +20 °C − (−12 °C) = 32 °C;
2. temperature correction of 1 °C on the table for the density of 0.7520, representing 0.000831;
3. temperature correction to 32 °C, equal to 0.000831 × 32 = 0.026592 or 0.0266 is rounded;
4. required density of mineral oil at a temperature of −12 °C (the amendment should be added, as the temperature of the load in the tank below +20 °C), equal to 0.7520 + 0.0266 = 0.7786 or 0.7785 is rounded.