Pressure

Pressure is force applied perpendicular to a surface per unit area i.e.

\begin{align} p=\frac{F}{A} \end{align}

Using the SI system this will lead to [N/m²] as the unit used for force (F) is Newton [N] and area (A) is square meters [m²]. The unit in the SI system for pressure is Pascal [Pa]. Pascal is equal to \( [Pa] = \frac{[N]}{[m]^2} \) hence using the equation above will lead to Pascal. Several units are used for pressure which can lead to confusion and mistakes. Pascal is quite often the unit used in various thermodynamic or fluid dynamic equations as it matches the rest of the SI system. It do however have one drawback in my opinion and recently I saw a video on Youtube where Elon Musk had the same complain regarding Pascal. It tend to be a rather large values that can be hard to relate to. Other notably units used for pressure is bar, psi, mLc (meter liquid column), mWc (meter water column), inHg (inches of mercury), mmHg (millimeter of mercury) and many others.

Pressure per square inch [psi] is the main unit used for pressure in the USA whereas the rest of the world mainly use pressure units related to SI system or derivatives hereof. Especially bar and kPa and MPa is used a lot as well as meter liquid column [mLc] or meter water column [mWc].

It is very important to be aware of the unit used as input when calculating calculating the pressure and when using the pressure as input in calculations. One way to ensure that that the units match is by using software like Mathcad or the free Smath Studio where it is possible to add units to the equation hence reduce the risk of unit errors.

Pressure convertion

We have created a pressure convertion tool to the site to allow conversion to and from the most common pressure units including atm, bar, cmH₂O, hPa, inH₂O, inHg, kg/cm², kPa, ksi, mH₂O (mWc), mmH2O, mmHg, MPa (N/mm²), Mpsi, Pa (N/m²), psf, psi (lbf/in²) and Torr.

The table below can also be used however it is much easier just to use the pressure convertion tool.

Pressure conversion between the most common units
	atm	bar	cmH2O	hPa	inH2O	inHG	kg/cm2	kPa	ksi	mH2O	mmH2O	mmHg	MPa	Mpsi	Pa	psf	psi	Torr
1 atm =	1	1.013	1033.227	1013.25	407.189	29.921	1.033	101.325	1.470E-02	10.332	1.033E+04	760	0.101	1.470E-05	1.013E+05	2116.215	14.696	760
1 bar =	0.987	1	1019.716	1000	401.865	29.53	1.02	100	1.450E-02	10.197	1.020E+04	750.062	0.1	1.450E-05	1.000E+05	2088.542	14.504	750.062
1 cmH2O =	9.678E-04	9.807E-04	1	0.981	0.394	2.896E-02	1.000E-03	9.807E-02	1.422E-05	1.000E-02	10	0.736	9.807E-05	1.422E-08	98.067	2.048	1.422E-02	0.736
1 hPa =	9.869E-04	1.000E-03	1.02	1	0.402	2.953E-02	1.020E-03	0.1	1.450E-05	1.020E-02	10.197	0.75	1.000E-04	1.450E-08	100	2.089	1.450E-02	0.75
1 inH2O =	2.456E-03	2.488E-03	2.537	2.488	1	7.348E-02	2.537E-03	0.249	3.609E-05	2.537E-02	25.375	1.866	2.488E-04	3.609E-08	248.84	5.197	3.609E-02	1.866
1 inHG =	3.342E-02	3.386E-02	34.532	33.864	0.000E+00	1	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00	0.000E+00
1 kg/cm2 =	0.968	0.981	1000	980.665	394.095	28.959	1	98.067	1.422E-02	10	1.000E+04	735.559	9.807E-02	1.422E-05	9.807E+04	2048.16	14.223	735.559
1 kPa =	9.869E-03	1.000E-02	10.197	10	4.019	0.295	1.020E-02	1	1.450E-04	0.102	101.972	7.501	1.000E-03	1.450E-07	1000	20.885	0.145	7.501
1 ksi =	68.046	68.948	7.031E+04	6.895E+04	2.771E+04	2036.02	70.307	6894.757	1	703.07	7.031E+05	5.171E+04	6.895	1.000E-03	6.895E+06	1.440E+05	1000	5.171E+04
1 mH2O =	9.678E-02	9.807E-02	100	98.067	39.409	2.896	0.1	9.807	1.422E-03	1	1000	73.556	9.807E-03	1.422E-06	9806.65	204.816	1.422	73.556
1 mmH2O =	9.678E-05	9.807E-05	0.1	9.807E-02	3.941E-02	2.896E-03	1.000E-04	9.807E-03	1.422E-06	1.000E-03	1	7.356E-02	9.807E-06	1.422E-09	9.807	0.205	1.422E-03	7.356E-02
1 mmHg =	1.316E-03	1.333E-03	1.36	1.333	0.536	3.937E-02	1.360E-03	0.133	1.934E-05	1.360E-02	13.595	1	1.333E-04	1.934E-08	133.322	2.784	1.934E-02	1
1 MPa =	9.869	10	1.020E+04	1.000E+04	4018.647	295.3	10.197	1000	0.145	101.972	1.020E+05	7500.615	1	1.450E-04	1.000E+06	2.089E+04	145.038	7500.617
1 Mpsi =	6.805E+04	6.895E+04	7.031E+07	6.895E+07	2.771E+07	2.036E+06	7.031E+04	6.895E+06	1000	7.031E+05	7.031E+08	5.171E+07	6894.757	1	6.895E+09	1.440E+08	1.000E+06	5.171E+07
1 Pa =	9.869E-06	1.000E-05	1.020E-02	1.000E-02	4.019E-03	2.953E-04	1.020E-05	1.000E-03	1.450E-07	1.020E-04	0.102	7.501E-03	1.000E-06	1.450E-10	1	2.089E-02	1.450E-04	7.501E-03
1 psf =	4.725E-04	4.788E-04	0.488	0.479	0.192	1.414E-02	4.882E-04	4.788E-02	6.944E-06	4.882E-03	4.882	0.359	4.788E-05	6.944E-09	47.88	1	6.944E-03	0.359
1 psi =	6.805E-02	6.895E-02	70.307	68.948	27.708	2.036	7.031E-02	6.895	1.000E-03	0.703	703.07	51.715	6.895E-03	1.000E-06	6894.757	144	1	51.715
1 Torr =	1.316E-03	1.333E-03	1.36	1.333	0.536	3.937E-02	1.360E-03	0.133	1.934E-05	1.360E-02	13.595	1	1.333E-04	1.934E-08	133.322	2.784	1.934E-02	1

Standard atmospheric pressure

The standard atmospheric pressure is a international agreed approximation of the average air pressure at mean sea level. The pressure is equal to:

\begin{align} \text{Standard atmospheric pressure} &= 1 atm = 101325Pa = 1013.25mbar\\ &=1.01325bar=29.921inHg=760mmHg \end{align}

Pressure change as function of altitude

The atmospheric pressure decreases with altitude and this can be calculated using the barometric formula. The below equation is only valid for the lower part of the atmosphere (the troposphere) up to approximately 11km above average sea level.

The temperature lapse rate is taken as 0.00976K/m. What this means is that the temperature decreases 0.00976K per meter height increase or 9.76K/km. This approximation in only usable and valid in the lowest part of the atmosphere.

The pressure decrease is approximately 1.2kPa for every 100m of altitude increase in the lower part of the atmosphere.

\begin{align} p=p_0\cdot \left( 1-\frac{Lh}{T_0} \right) ^{\frac{gM}{R_0L}} \end{align}

where:

Table 1 - Constants for calculating the pressure within the troposphere using the barometric formula
Constant	Value	Description
p₀	101325Pa	Sea level standard atmospheric pressure
L	approx. 0.00976K/m	Temperature lapse rate = g/c_p
T₀	288.16K	Sea level standard temperature
g	9.80665m/s²	Gravitational acceleration sea level
M	0.02896968kg/mol	Dry air molar mass
R₀	8.314462618 J/(mol×K)	Sea level standard atmospheric pressure

Pressure change in the lower part of the atmosphere

Hydrostatic Pressure

Next topic in the this article is how to calculate the pressure at a given water or other liquid height above the point of interest. The pressure at a given liquid height is a function of density and water level height.

\begin{align} p = \rho gh \end{align}

where:

\begin{align} p &= \text{Pressure}\\ g &= \text{Gravitational acceleration } 9.81m/s^2\\ \rho &= \text{Density} \end{align}

As seen the pressure at the bottom only depends upon water column height (h) and not the area. Below sketch show two water basins. The water level in each are identical but the large basin is considerably larger than the small one and can contain more water hence the mass of the water is bigger in large basin compared to small basin.

The pressure at the bottom is however identical.

Going back to pressure definition i.e. \( p=\frac{F}{A} \) it can be demonstrated why. Lets assume that the ground plane of the basins are square with a side length = L. The Force applied perpendicular to the basin bottom can be calculated using Newtons Second Law

\begin{align} F=m\cdot g \end{align}

Where m is the mass of water i.e.

\begin{align} m=A\cdot h \cdot \rho \end{align}

with A being the basin bottom area. This will as expected lead to a higher mass of water in the large basin as the area is bigger.

Thus the force is acting perpendicular to the basin bottom:

\begin{align} F= A \cdot h \cdot \rho \cdot g \end{align}

The total force on the large basin bottom will be larger than on the small basin bottom because of the larger area.

The pressure on the basin bottom can now be calculated

\begin{align} p = \frac{F}{A} = \frac{A \cdot h \cdot \rho \cdot g}{A} =\rho gh \end{align}

As seen the area is canceled hence the pressure only relies on density and water column height as shown in the first equation in this section.

Gauge vs absolute pressure

The terms gauge and absolute pressure is often used. It is typically shown as Bar(A), PSIA, bara etc. for absolute pressure and Bar(g), PSIG, barg etc. for gauge pressure.

Gauge pressure is the pressure relative to the atmospheric pressure. This is typically the pressure shown when looking at a pressure gauge. An example is tire inflation. The pressure written in the instruction manual of the car is gauge pressure i.e. pressure above atmospheric pressure (pressure relative to the atmospheric pressure). The one see on the pressure gauge when using the compressor to pump the tire is also gauge pressure i.e. pressure above ambient pressure. It is easy to check that the compressor indeed show gauge pressure. The pressure gauge will show 0 pressure if turned off and pressure is equalized with the atmosphere. This do not mean that perfect vacuum has formed in the compressor but instead show that the ambient pressure outside the compressor is equal to the pressure inside the compressor tank.

Absolute pressure is the sum of gauge pressure and atmospheric pressure. It is almost always this pressure that shall be used when making thermodynamic property look-ups in tables (including on this website) and typically also the pressure needed in some thermodynamic equaition.

\begin{align} p_{Absolute} = p_{Gauge}+\text{Atmospheric pressure} \end{align}

Example - Gauge vs Absolute Pressure

The water basin example illustrated by the sketch below has a pressure gauge mounted at it's bottom. The water level is 10m above the pressure gauge and the density of the water is 1000kg/m³.

Calculating the pressure p₁ using the equation in Hydrostatic pressure section:

\begin{align} &p_1 = \rho gh=1000\cdot 9.81\cdot 10 = 98100Pa(G) = 9.81Bar(g)\\ &\text{or in absolute pressure}\\ &p_1 = 9.81Bar(g) + 1.01325Bar = 10.82Bar(a) \end{align}

What this means is that the needle on the regular pressure gauge will be showing 9.8Bar(g) and that the absolute pressure is 10.8Bar(a)

Stagnation pressure or pitot pressure

The stagnation pressure is sometimes also called pitot pressure. The reason being that the stagnation pressure often is measured with a pitot tube. It is used to determine airspeed in for example on an airplane where the stagnation pressure is measured using the pitot tube and static pressure measured though static air ports. It is with these two measurements possible to determine the dynamic pressure and consequently the air speed as:

Stagnation pressure = static pressure + dynamic pressure

\begin{align} p_{stagnation}=p_{static}+\frac{1}{2}\rho u^2 \end{align}

where:

\begin{align} p_{stagnation} &= \text{Stagnation pressure}\\ \rho &= \text{Density}\\ w &= \text{Fluid speed} \end{align}

The equation above is however only valid for incompressible flow as the density depends upon the pressure for compressible flow.