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Atmospheric Pressure given Total or Absolute Pressure Formula

GoAtmospheric Pressure given Gauge Pressure Formula

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Atmospheric Pressure is the force per unit area exerted against a surface by the weight of air above that surface in the Earth’s atmosphere. Check FAQs

P atm = P abs - (ρ \cdot [g] \cdot H \cdot \cosh (2 \cdot π \cdot \frac{D Z+d}{λ})) \cdot \frac{\cos (θ)}{2 \cdot \cosh (2 \cdot π \cdot \frac{d}{λ})} + (ρ \cdot [g] \cdot Z)

P_atm - Atmospheric Pressure?P_abs - Absolute Pressure?ρ - Mass Density?H - Wave Height?D_Z+d - Distance above the Bottom?λ - Wavelength?θ - Phase Angle?d - Water Depth?Z - Seabed Elevation?[g] - Gravitational acceleration on Earth?[g] - Gravitational acceleration on Earth?π - Archimedes' constant?

Atmospheric Pressure given Total or Absolute Pressure Example

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Here is how the Atmospheric Pressure given Total or Absolute Pressure equation looks like with Values.

Here is how the Atmospheric Pressure given Total or Absolute Pressure equation looks like with Units.

Here is how the Atmospheric Pressure given Total or Absolute Pressure equation looks like.

100964.782 = 100000 - (997 \cdot 9.8066 \cdot 3 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{2}{26.8})) \cdot \frac{\cos (60)}{2 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{1.05}{26.8})} + (997 \cdot 9.8066 \cdot 0.908)

100964.782Pa = 100000Pa - (997kg/m³ \cdot 9.8066m/s² \cdot 3m \cdot \cosh (2 \cdot 3.1416 \cdot \frac{2m}{26.8m})) \cdot \frac{\cos (60°)}{2 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{1.05m}{26.8m})} + (997kg/m³ \cdot 9.8066m/s² \cdot 0.908)

100964.782 = 100000 - (997 \cdot 9.8066 \cdot 3 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{2}{26.8})) \cdot \frac{\cos (60)}{2 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{1.05}{26.8})} + (997 \cdot 9.8066 \cdot 0.908)

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Atmospheric Pressure given Total or Absolute Pressure Solution

Follow our step by step solution on how to calculate Atmospheric Pressure given Total or Absolute Pressure?

FIRST Step Consider the formula

P atm = P abs - (ρ \cdot [g] \cdot H \cdot \cosh (2 \cdot π \cdot \frac{D Z+d}{λ})) \cdot \frac{\cos (θ)}{2 \cdot \cosh (2 \cdot π \cdot \frac{d}{λ})} + (ρ \cdot [g] \cdot Z)

Next Step Substitute values of Variables

∴

P atm = 100000Pa - (997kg/m³ \cdot [g] \cdot 3m \cdot \cosh (2 \cdot π \cdot \frac{2m}{26.8m})) \cdot \frac{\cos (60°)}{2 \cdot \cosh (2 \cdot π \cdot \frac{1.05m}{26.8m})} + (997kg/m³ \cdot [g] \cdot 0.908)

Next Step Substitute values of Constants

∴

P atm = 100000Pa - (997kg/m³ \cdot 9.8066m/s² \cdot 3m \cdot \cosh (2 \cdot 3.1416 \cdot \frac{2m}{26.8m})) \cdot \frac{\cos (60°)}{2 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{1.05m}{26.8m})} + (997kg/m³ \cdot 9.8066m/s² \cdot 0.908)

Next Step Convert Units

∴

P atm = 100000Pa - (997kg/m³ \cdot 9.8066m/s² \cdot 3m \cdot \cosh (2 \cdot 3.1416 \cdot \frac{2m}{26.8m})) \cdot \frac{\cos (1.0472rad)}{2 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{1.05m}{26.8m})} + (997kg/m³ \cdot 9.8066m/s² \cdot 0.908)

Next Step Prepare to Evaluate

∴

P atm = 100000 - (997 \cdot 9.8066 \cdot 3 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{2}{26.8})) \cdot \frac{\cos (1.0472)}{2 \cdot \cosh (2 \cdot 3.1416 \cdot \frac{1.05}{26.8})} + (997 \cdot 9.8066 \cdot 0.908)

Next Step Evaluate

∴

P atm = 100964.781979786Pa

LAST Step Rounding Answer

∴

P atm = 100964.782Pa

Atmospheric Pressure given Total or Absolute Pressure Formula Elements

Variables

Constants

Functions

Atmospheric Pressure

Atmospheric Pressure is the force per unit area exerted against a surface by the weight of air above that surface in the Earth’s atmosphere.

Symbol: P_atm

Measurement: PressureUnit: Pa

Note: Value should be greater than 0.

Formulas to find Atmospheric Pressure

Absolute Pressure

Absolute Pressure is the total pressure measured with respect to absolute zero, which is a perfect vacuum. It is the sum of the gauge pressure and the atmospheric pressure.

Symbol: P_abs

Measurement: PressureUnit: Pa

Note: Value should be greater than 0.

Formulas to find Absolute Pressure

Mass Density

Mass Density is crucial for understanding the distribution of pressures exerted by overlying soil or water layers on underground structures like foundations, tunnels, or pipelines.

Symbol: ρ

Measurement: Mass ConcentrationUnit: kg/m³

Note: Value should be greater than 0.

Formulas to find Mass Density

Wave Height

Wave Height is the vertical distance between the crest and the trough of a wave. Higher wave heights correspond to greater wave forces, which leads to increased structural loading.

Symbol: H

Measurement: LengthUnit: m