Jet Fuel Temperature Density Petroleum Volume Correction ASTM D1250 Gravity

Overview

Engineering reference data for Jet Fuel Temperature Density Petroleum Volume Correction ASTM D1250 Gravity in thermodynamics.

Key Formulas

First Law

$\Delta U = Q - W$

Energy is conserved — heat added minus work done.

Ideal Gas Law

$PV = nRT$

Relates pressure, volume, and temperature of an ideal gas.

Heat Transfer

$Q = mc\Delta T$

Sensible heat transfer.

Carnot Efficiency

$\eta = 1 - T_C/T_H$

Maximum efficiency between two temperatures.

Variables

Symbol	Description	Unit
$U$	Internal energy	J
$Q$	Heat	J
$W$	Work	J
$P$	Pressure	Pa
$V$	Volume	m³
$T$	Temperature	K

Volume Correction Factors

Volume correction factors (VCF) adjust for the change in petroleum product volume as temperature changes. The relationship follows a key principle: volume increases as temperature increases. These factors allow conversion between observed volume at a given temperature and the equivalent volume at a standard base temperature (typically 15°C or 59°F).

Core Formula

The fundamental relationship between observed volume, corrected volume, and the volume correction factor is:

$V_{base} = V_{observed} \times VCF_{T \rightarrow base}$

Conversely, to find the volume at a new temperature:

$V_{new} = V_{base} \times VCF_{base \rightarrow T}$

Practical Examples

The following examples illustrate how to apply the volume correction factors shown in the reference figures.

Example 1: Correcting to Base Temperature

• Given: 100 L of jet fuel with density 757 kg/m³ at 100°C.
• Find: Volume at 15°C (base temperature).
• Procedure:
  1. Locate the fuel's density line (yellow) on the Density vs. Temperature figure.
  2. Use the same line on the Volume Correction Figure (Observed to 15°C).
  3. Read the correction factor at 100°C: 0.923.
• Calculation: $V_{15°C} = 100 \, \text{L} \times 0.923 = 92.3 \, \text{L}$
• Check: The corrected volume (92.3 L) is lower than the observed volume (100 L), consistent with cooling the fuel.

Example 2: Calculating Volume at a New Temperature

• Given: 1000 m³ of jet fuel with density 800 kg/m³ at 15°C (base volume).
• Find: Volume if heated to 250°C.
• Procedure:
  1. Locate the fuel's density line (orange) on the Density vs. Temperature figure.
  2. Use the same line on the Volume Correction Figure (15°C to Observed T).
  3. Read the correction factor at 250°C: 1.292.
• Calculation: $V_{250°C} = 1000 \, \text{m³} \times 1.292 = 1292 \, \text{m³}$
• Check: The final volume (1292 m³) is larger than the base volume (1000 m³), consistent with heating the fuel.

References

• Original Source