Jet Fuel Head Calculator
Inches of Fuel: PSI:
Specific Gravity (S.G.):  
Density (lb/gallon):  
Density (Lb/ft³):  
Result (PSI): Result (Inches of Fuel):  
Inches of Water (inH₂O):  
Kilopascals (kPa):  
Inches of Mercury (inHg):  

*Fuels are mixtures and have variation, see other calculators*

*S.G. based on water @ 70°F = 998kg/m³*

*Click here to use density of 999kg/m³*

Specific Gravity is a Contentious Unit

Jet Fuel Head Calculator with Temperature Compensation
Inches of Fuel: PSI:
Degrees (°F)
Degrees (°C)
Density (kg/m³):  
Density (lb/in³):  
Density (lb/gal):  
Density (lb/ft³):  
Specific Gravity*:  
Specific Gravity**:  
Specific Gravity***:  
°API****:  
Inches of Water (inH₂O):  
Result (PSI): Result (Inches of Fuel):  

*Fuels are mixtures and have variation, see other calculators*

* SG based on static density of water of 1000 (kg/m³) *

** SG based on static density of water of 998 (kg/m³) **

*** SG based on temperature dependent density of water ***

**** °API is typically referenced at 60°F ****

Specific Gravity is a Contentious Unit

Also See:  

Jet Fuel Nominal Density

Also See:  

Min/Max Jet Fuel Nominal Density

Also See:  

Universal Fluid Head Pressure Calculator

Also See:  

Water Density Compensated Universal Fluid Head Calculator
Reference Guide: Fuel Head Pressure
01

What Is Hydrostatic Pressure?

A column of liquid presses down on everything below it. The pressure at any depth depends on only two things: the height of liquid above that point and the liquid's density. The shape and width of the tank do not matter, so a thin standpipe and a wide tank filled to the same height give the same pressure at the bottom.

That is why this pressure is called head and is often quoted as a height: inches of fuel or inches of water.

Hydrostatic pressure\[ P = \rho\,g\,h \]

\(\rho\) is the fluid density, \(g\) the acceleration of gravity and \(h\) the height of fluid above the point. This is gauge pressure. Add the atmospheric pressure (14.696 psia at sea level) for absolute pressure.

A column of fuel open to the air. Pressure arrows on the wall grow longer with depth, and a gauge at the bottom reads P = rho g h. A graph beside it shows pressure rising in a straight line with depth.
Pressure Grows With Depth
Pressure rises in a straight line from zero at the surface. Twice the height gives twice the pressure.
02

Every Unit Cancels

Written out with every unit, the head conversion shows each unit cancelling until only pounds per square inch are left:

Inches of fluid → psi, unit by unit\[ P\ \frac{\text{lb}}{\text{in}^2} = h\ \text{in} \times \text{SG} \times \rho_{\text{water}}\ \cancel{\frac{\text{kg}}{\text{m}^3}} \times \frac{1\ \text{lb}/\text{in}^3}{27{,}679.90471\ \cancel{\frac{\text{kg}}{\text{m}^3}}} \]

The remaining units are \(\text{in} \times \text{lb}/\text{in}^3 = \text{lb}/\text{in}^2\), which is psi.

psi → inches of fluid, unit by unit\[ h\ \text{in} = P\ \frac{\cancel{\text{lb}}}{\text{in}^2} \times \frac{27{,}679.90471\ \cancel{\frac{\text{kg}}{\text{m}^3}}}{1\ \cancel{\text{lb}}/\text{in}^3} \times \frac{1}{\text{SG}} \times \frac{1}{\rho_{\text{water}}\ \cancel{\frac{\text{kg}}{\text{m}^3}}} \]
Why the Numbers Are Exact

None of the constants is a rounded measurement. The pound and the inch are defined exactly in SI units (1 lb = 0.45359237 kg and 1 in = 0.0254 m), the US gallon is defined as exactly 231 in³, and an hour is 60 minutes. That makes 27,679.90471 kg/m³ exactly one pound per cubic inch:

One pound per cubic inch\[ 1\ \frac{\text{lb}}{\text{in}^3} = \frac{0.45359237\ \text{kg}}{(0.0254\ \text{m})^3} = 27{,}679.90471\ \frac{\text{kg}}{\text{m}^3} \]

The only inputs that carry real uncertainty are the specific gravity and the water density it is referenced to.

03

Inches of Fluid ↔ PSI

In US units the \(g\) drops out. A density in lb/in³ times a height in inches gives psi directly, because one pound of mass weighs one pound of force under standard gravity. With the fluid's density written as SG times the density of water in kg/m³:

Inches of fluid → psi\[ P_{\text{psi}} = h_{\text{in}} \times \text{SG} \times \frac{\rho_{\text{water}}}{27{,}679.9} \]
psi → inches of fluid\[ h_{\text{in}} = \frac{P_{\text{psi}} \times 27{,}679.9}{\text{SG} \times \rho_{\text{water}}} \]

27,679.9 kg/m³ is exactly 1 lb/in³, so dividing by it converts the density to lb/in³.

Example

100 inches of jet fuel with SG = 0.80, referenced to water at 60 °F (999.0 kg/m³):

\[ P = 100 \times 0.80 \times \frac{999.0}{27{,}679.9} = 2.887\ \text{psi} \]
Fluid SGpsi per inchpsi per foot
0.750.02710.325
0.80 (typical jet fuel)0.02890.346
0.850.03070.368
1.00 (water)0.03610.433

Water at 999.0 kg/m³ (60 °F).

04

Density & Specific Gravity Conversions

Specific gravity (SG) is the fluid's density divided by the density of water. It has no units, which makes it easy to quote, but it is only exact once you know which water it is compared to.

Specific gravity\[ \text{SG} = \frac{\rho_{\text{fluid}}}{\rho_{\text{water}}} \qquad \rho_{\text{fluid}} = \text{SG} \times \rho_{\text{water}} \]

Water is 999.0 kg/m³ at 60 °F and 1,000.0 kg/m³ at 4 °C. Fuel specifications usually quote SG at 60 °F. The water-density-compensated calculators compare against water at the fluid's own temperature instead. See why specific gravity is a contentious unit.

kg/m³ → lb/gal and lb/in³\[ \begin{gathered} \rho_{\text{lb/gal}} = \rho_{\text{kg/m}^3} \times \frac{231}{27{,}679.9} = \frac{\rho_{\text{kg/m}^3}}{119.83} \\ \rho_{\text{lb/in}^3} = \frac{\rho_{\text{kg/m}^3}}{27{,}679.9} \end{gathered} \]
UnitEquals
1 lb/in³27,679.9 kg/m³
1 lb/gal (US)119.83 kg/m³
1 lb/ft³16.018 kg/m³
1 US gallon231 in³ = 3.7854 L

Cite This Work