Questions on Basic Knowledge of Oil Measurement

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I. Fill in the blanks: 1. When measuring light oil, one should measure the (actual level). Measure twice in succession; the reading error should not exceed 1 millimeter. The result of the first measurement shall be used as the oil level, and if it exceeds this value, re-measurement is required. 2. For vertical metal tanks, after the heavy viscous oil has been collected, the liquid level must remain stable for (3) hours; after discharge, it must remain stable for (1) hour before measurements can be taken ； After light oil is received, the liquid level must remain stable for (2) hours; after the oil is delivered, it must stay stable for (30) minutes before level measurement can be taken. 3. When taking temperature readings, the minimum immersion time required for the cup-and-box thermometer to be placed in the area being measured is: for light oils and other oils with a dynamic viscosity of 20 mm2/s or less at 40°C, the minimum immersion time is (5) minutes ； For crude oil, lubricating oils, and other oils with a dynamic viscosity greater than 20 mm2/s at 40°C and a dynamic viscosity less than 36 mm2/s at 100°C, the minimum immersion time is (15) minutes ； For heavy lubricating oils (cylinder oils, gear oils, residue oils, and other oils with a dynamic viscosity at 100°C equal to or greater than 36 mm2/s), the minimum immersion time is (30) minutes. 4. When measuring heavy oil, the (empty gauge) should be checked. The distance should be measured continuously (2) times, and the reading error must not exceed (2) millimeters. If the error between the two readings is within (1) millimeter, the value from the first measurement is taken; if it exceeds (1) millimeter, the average of the two measurements is used. II. Essay Questions: 1. How is the temperature measurement location specified when testing oil temperatures? Answer: If the oil level is below 3 meters, measure the oil temperature at the middle part of the oil column ； The oil level is 3 to 5 meters high; two measurements are taken at a point 1 meter below the oil surface and at a point 1 meter above the oil surface, and the arithmetic average of these values is used as the temperature of the oil ； The oil level is more than 5 meters high; three measurements are taken at one meter below the oil surface, in the middle of the oil layer, and one meter above the oil surface, and the arithmetic average of these values is used as the temperature of the oil. If the temperature at any one of these points differs from the average temperature by more than 1°C, an additional measurement point must be taken between the upper and middle measurement points, and another point between the middle and lower measurement points; the arithmetic average of these five points is then used as the temperature of the oil. III. Calculation problem: A gasoline tank (without insulation) is currently under observation. The floating height of the floating roof is 1.681 meters, the weight of the floating roof is 5.412 tons, the average temperature of the oil inside is 37.1°C, the oil level is 11.713 meters, the water level is 0.026 meters, and the air temperature is 19.7°C. According to the certificate of conformity, the standard density of this gasoline at 20°C is 767.6 kg/m3. Determine the weight of the oil in this tank. Solution: (1) To calculate the volume correction factor VCF20 at 20°C, use the standard density at 20°C, which is 767.6 kg/m3. Refer to Table 60B using a oil temperature of 37.1°C; since no interpolation is required for temperature values, the value closest to 37.1°C, namely 37.00°C, is used. The standard density at 37.00°C is 766.0 kg/m3, and the volume correction factor for this value is 0.9801 ； At 37.00℃ and a standard density of 768.0 kg/m3, the volume correction factor is 0.9805. Calculate the volume correction factor VCF20 using interpolation:
VCF20 = 0.9801 + (0.9801 – 0.9805) / (766.0 – 768.0) × (767.6 – 766.0) = 0.9801 + 0.00032 = 0.98042

(2) Calculate the volume at 37.1°C:
Refer to the tank capacity table (units: liters; all values in the table are also in liters). The volume at 11.7 meters is 12,537,618 liters. Using the capacity table for the range of 10.799 meters to 12.597 meters, the volume at 0.01 meters is 10,753 liters, and at 0.003 meters it is 3,226 liters.
Thus, V_table = 12,537,618 + 10,753 + 3,226 = 12,551,597 liters.
Refering to the tank bottom area table, the volume at 0.02 meters is 3,738 liters, and at 0.03 meters it is 8,619 liters. Using interpolation:
V_water = 3,738 + (8,619 – 3,738) × 0.6 = 6,667 liters.
Using the static pressure correction table, the volume at 11.7 meters is 8,592 liters, and at 11.8 meters it is 8,758 liters. Using interpolation:
ΔVp = … × 0.7676 × 0.98042 = 6,482 liters.
Therefore, the volume of the diesel at the measurement temperature of 37.1°C is:
Vt = (V_table + ΔVp – V_water) × {1 + 0.000024 × } = (12,551,597 + 6,482 – 6,667) × {1 + 0.000024 × } = 12,553,942 liters.

(3) Calculate the weight:
From the table, the lifting height of the tank’s floating roof is 1.681 meters, and the weight of the floating roof is 5.412 tons.
Weight = V20 × (ρ20 – 0.0011) – Weight of floating roof
= Vt × VCF20 × (ρ20 – 0.0011) – 5,412
= 12,553,942 × 0.98042 × (0.7676 – 0.0011) – 5,412
= 9,428,774 kg = 9,428.774 tons.
Answer: The total weight of the oil in the tank is 9,428.774 tons.