Hydrostatic pressure represents the force per unit area exerted by a column of fluid at a particular depth due to the action of gravity. Hydrostatic calculations apply to a fluid at rest, meaning other pressure changes associated with flow are not accounted for, which simplifies the calculation tremendously and is usually sufficiently accurate for basic analysis.

To calculate the hydrostatic pressure in a well, engineers use an equation like the following:

\[\require{textmacros}\textit{Hydrostatic pressure (HSP)} = {\textit{Density}} \times {\textit{Gravity}} \times \textit{Height}\]

### U.S. Petroleum Industry Uses Oilfield Units

In the U.S. petroleum industry, calculations are done in “oilfield units”, and pressure is measured in units of pounds force (lb_{f}) per square inch, abbreviated as psi, so the units for *HSP* will be psi. In oilfield units, fluid density is measured in pounds mass (lb_{m}) per gallon, written as lb_{m}/gal or abbreviated as ppg. Freshwater has a fluid density of 8.34 ppg. Comparing these units to the SI system, lb_{m} is comparable to kilograms (kg), and lb_{f} is comparable to Newtons (N).

### Engineers Regularly Have to Convert Units

The Gravity term in the equation above refers to the gravitational constant, \(g\), divided by a unit conversion factor, \(g_c\), or \(\text{Gravity} = \frac{g}{g_c}\). In oilfield units, \(g\) = 32.2 ft/s^{2}, and \(g_c\) = 32.2 (lb_{m}-ft/s^{2})/lb_{f}. This sounds complicated, but if you multiply the gravity term out, the result is pretty simple.

\[\textit{Gravity} =\frac{g}{g_{c}} =\frac{32.2\frac{\text{ft} }{\text{s}^{2} } }{\frac{32.2\ \text{lb}_{\text{m} } \cdot \frac{\text{ft} }{\text{s}^{2} } }{\text{lb}_{\text{f} } } } =\frac{1\ \text{lb}_{\text{f} } }{1\ \text{lb}_{\text{m} } } \]

Or in other words, the force caused by gravity, in oilfield units, is 1 pound force (lb_{f}) for every 1 pound mass (lb_{m}). Another way to say this is 1 lb_{m} “weighs” 1 lb_{f}. Using this conversion information and given that water has a density (mass/volume) of 8.34 lb_{m}/gal, we can conclude that water weighs 8.34 lb_{f}/gal. For drilling operations, this normalized fluid weight (force/volume) is called the mud weight. Given that in Oilfield Units the mass and weight have the same numerical value (because 1 lb_{m} weights 1 lb_{f}), the two terms are used synonymously, and has led to the practice where the density of the mud in drilling is called the mud weight.

### An Aside Concerning SI Units

A comparable representation of Gravity in SI units may look more familiar to you. The gravitational constant is \(g\) = 9.8 m/s^{2}. The unit conversion factor is \(g_c\) = 1 (kg-m/s^{2})/N. Multiplying as before,

\[\textit{Gravity} =\frac{g}{g_{c}} =\frac{9.8\frac{\text{m} }{\text{s}^{2} } }{\frac{1\ \text{kg} \cdot \frac{\text{m} }{\text{s}^{2} } }{\text{N} } } =\frac{9.8\ \text{N} }{1\ \text{kg} } \]

In other words, this equation says the force caused by gravity exerts 9.8 N for every kg, or a kg weighs 9.8 N. So unlike the Oilfield Unit system, the mass and weight of an object will have different numeric values, and fluid density and normalized weight cannot be used synonymously. In SI units, water has a density of 1,000 kg/m^{3}, but since a kg weighs 9.8 N, water “weighs” 9,800 N/m^{3}.

The Height in this equation refers to the fluid column and is described as the “true vertical depth (*TVD*)” expressed in feet for oilfield units. We use true vertical depth because sometimes a well deviates slightly from vertical, making the “measured depth”, or the distance traveled by the drill bit, greater than or equal to the *TVD*. But gravity acts vertically toward the center of the earth, so *TVD* is the value that is important.

### Example Calculation in Oilfield Units

An example calculation can help illustrate how hydrostatic calculations are made, and can guide you through the unit conversions required. Assume a vertical wellbore is 1,000 ft deep (*TVD*), and it is filled with fresh water, which has a density of 8.34 ppg. What is the hydrostatic pressure at the bottom of this well?

\[\textit{HSP} =8.34\frac{\text{lb}_{\text{m} } }{\text{gal} } \times 1\frac{\text{lb}_{\text{f} } }{\text{lb}_{\text{m} } } \times 1000\ \text{ft} =8340\frac{\text{lb}_{\text{f} } \cdot \text{ft} }{\text{gal} } \]

Now you can see the units for this result seem odd, but let’s analyze it. Lb_{f} is a unit of force. Gallons (gal) is a unit of volume, which is length^{3}. “ft” is a unit of length. Multiplying the units gives force × length/length^{3} or force/length^{2} which is force/area, which is the definition of pressure. But in oilfield units, we express pressure in psi, not lb_{f}-ft/gal, so we need to convert “ft” and “gal” into “in” and “in^{3}”, respectively, as follows:

\[\textit{HSP} =8340\frac{\text{lb}_{\text{f} } \cdot \text{ft} }{\text{gal} } \times 12\frac{\text{in} }{\text{ft} } \times 1\frac{\text{gal} }{231\ \text{in}^{3} } =433\ \frac{\text{lb}_{\text{f} } }{\text{in}^{2} } =\ 433\ \text{psi} \]

### Conversions Result in a Handy Formula Format

So the fluid pressure at the bottom of a 1,000 ft tall column of fresh water in our well exerts 433 psi. Multiplying out this entire string of conversions every time you want to know the pressure in your well can be time consuming, so engineers develop shortcuts for common calculations to make things more efficient. Our answer above says that gravity exerts 433 psi of pressure for every 1,000 ft of freshwater column height (433 psi per 1,000 ft), which we can normalize to 0.433 psi per ft or 0.433 psi/ft. We call this value of 0.433 psi/ft the freshwater Pressure Gradient (*PG*), which is the rate at which pressure changes with increasing freshwater column height. Knowing this, we can modify our hydrostatic pressure equation to a more “handy” format, assuming freshwater, as

\[\textit{HSP} =0.433\frac{\text{psi} }{\text{ft} } \times \text{height} \]

What if we have something other than freshwater, or more specifically, something with a *density *different from freshwater? Let’s rewrite our *HSP* equation to allow for a generic Pressure Gradient (*PG*), which is to relax the assumption that we have freshwater with a *PG* = 0.433 psi/ft.

\[\textit{HSP} =\textit{PG} \times \textit{height} \]

Comparing to our original *HSP* equation, we can see that the following must be true:

\[\textit{PG}=\textit{density} \times \textit{gravity} \]

These two equations show that *PG* and consequently *HSP* are directly proportional to the density of the fluid. Being “directly proportional” means if the density changes by a certain ratio, the pressure or the pressure gradient will change by that exact same ratio. The ratio of the density of one substance compared to some other substance considered a standard is called the Specific Gravity (*SG*), and for fluids it is convenient to use freshwater as the standard. Drilling mud is usually more dense than freshwater, being made of saline water with some solids added (typically clay).

##### Imagine you made a drilling fluid that had a density of 11 ppg. What is its *SG*?

## 0.76

#### Incorrect.

## 1.00

#### Incorrect.

## 1.32

#### Correct.

11 ppg is greater than 8.34 ppg by 1.32 times, which is the specific gravity of the drilling mud relative to water.

## 11.00

#### Incorrect.

##### Given the specific gravity of the drilling mud, what would be the Pressure Gradient (*PG*) in a wellbore filled with that fluid?

## 8.34 ppg

#### Incorrect.

## 0.433 psi/ft

#### Incorrect.

## 0.572 psi/ft

#### Correct.

(*PG* of drilling mud) = (*SG* of drilling mud) × (*PG* of freshwater) = 1.32 × (0.433 psi/ft) = 0.572 psi/ft

## 572 psi

#### Incorrect.

##### What would be the hydrostatic pressure if our 1,000 ft deep wellbore were filled with this mud?

## 433 psi

#### Incorrect.

## 572 psi

#### Correct.

*HSP* = (0.572 psi/ft) × (1,000 ft) = 572 psi

Increasing the *SG* of the drilling mud increases the hydrostatic pressure in the wellbore proportionally.

## 11,000 psi

#### Incorrect.

## None of the above

#### Incorrect.

This example calculation, examining the pressure of a fluid with a higher *SG* than water, demonstrates the predominant mechanism by which drillers exert primary well control: they increase the pressure at the bottom of the well by increasing the density of the wellbore fluid, or in driller’s language, they “increase the mud weight.”