Pressure & Head Calculations

Pressure and head are two ways of describing the energy associated with water in a system. Operators encounter them when reading pressure gauges, evaluating storage tanks, checking distribution system pressures, analyzing pump operation, and comparing elevations.

The key relationship is simple: a vertical column of water approximately 2.31 feet high produces 1 pound per square inch of pressure at its base. Understanding this relationship makes it possible to convert between pressure in psi and pressure head in feet of water.

Pressure and Head

Pressure is force applied over an area. In water system calculations, pressure is commonly expressed in pounds per square inch, or psi.

Pressure head expresses pressure as the equivalent vertical height of a water column. It is normally expressed in feet of water.

For water under typical operator-calculation conditions:

2.31 ft of water ≈ 1 psi

This gives two basic formulas:

Pressure, psi = Head, ft / 2.31

Head, ft = Pressure, psi × 2.31

These formulas are among the most useful hydraulic relationships for drinking water and wastewater operators.

Converting Head to Pressure

If the height of the water column is known, divide by 2.31 to determine pressure.

Formula: Pressure, psi = Head, ft / 2.31

Suppose the vertical distance from the water surface in an elevated tank to a pressure gauge is 115.5 feet.

Pressure = 115.5 ft / 2.31

Pressure = 50 psi

The theoretical static pressure at the gauge elevation is approximately 50 psi.

Converting Pressure to Head

If pressure is known, multiply by 2.31 to determine the equivalent pressure head.

Formula: Head, ft = Pressure, psi × 2.31

If a pressure gauge reads 72 psi:

Head = 72 psi × 2.31 ft/psi

Head = 166.32 ft

The pressure corresponds to approximately 166 feet of water head.

Pressure Depends on Vertical Height

Static water pressure depends on the vertical height of the water column above the point where pressure is measured. It does not depend directly on the total number of gallons in the tank.

Consider two open tanks that both have a water depth of 100 feet. One tank is narrow and the other is very wide. The wider tank contains much more water, but the pressure at the bottom of both tanks is approximately the same because the water depth is the same.

Pressure at the bottom is:

100 ft / 2.31 = 43.3 psi

This is an important exam concept. Tank volume affects how much water is stored, but water-column height determines hydrostatic pressure.

Static Pressure from an Elevated Tank

Elevated storage can provide pressure without a pump operating continuously. The elevation difference between the water surface and the point of use creates static head.

Suppose the water surface in a storage tank is at elevation 720 feet and a hydrant is at elevation 570 feet.

Step 1: Find the elevation difference.

Head = 720 ft - 570 ft = 150 ft

Step 2: Convert head to pressure.

Pressure = 150 ft / 2.31

Pressure = 64.9 psi

Ignoring friction loss and other hydraulic effects, the theoretical static pressure at the hydrant is approximately 65 psi.

Elevation Changes and Pressure

In a static water system, pressure generally decreases as elevation increases and increases as elevation decreases.

A vertical elevation change of 2.31 feet corresponds to approximately 1 psi.

For example, if two points in the same static system differ in elevation by 46.2 feet:

Pressure difference = 46.2 ft / 2.31

Pressure difference = 20 psi

The lower point would have approximately 20 psi more static pressure than the higher point, assuming both points are connected to the same hydraulic source and other conditions are equal.

Example: Pressure at Two Elevations

A reservoir water surface is at elevation 800 feet. Location A is at elevation 650 feet and Location B is at elevation 700 feet.

For Location A:

Head = 800 - 650 = 150 ft

Pressure = 150 / 2.31 = 64.9 psi

For Location B:

Head = 800 - 700 = 100 ft

Pressure = 100 / 2.31 = 43.3 psi

Location A has greater pressure because it is at the lower elevation.

The pressure difference is:

64.9 - 43.3 = 21.6 psi

This also can be checked directly from the 50-foot elevation difference:

50 ft / 2.31 = 21.6 psi

Static Head

Static head refers to the elevation difference that must be overcome or that produces pressure when water is not moving.

In a simple pumping situation, static head may be determined by comparing the elevation of the source water surface with the elevation of the discharge water surface.

Suppose water is pumped from a clearwell with a water-surface elevation of 420 feet to a storage tank with a water-surface elevation of 565 feet.

Static head = 565 ft - 420 ft

Static head = 145 ft

The pump must overcome at least this elevation difference. When water is actually flowing, additional energy is normally required to overcome friction and other losses. Those additional hydraulic and pump relationships are covered in later articles.

Suction and Discharge Head

Pump problems may describe pressure or elevation conditions on the suction and discharge sides of a pump.

A pressure gauge reading can be converted to feet of head so that pressure information can be combined with other hydraulic terms expressed in feet.

For example, a pump discharge gauge reads 60 psi.

Discharge pressure head = 60 × 2.31

Discharge pressure head = 138.6 ft

If a suction gauge indicates a positive pressure of 10 psi:

Suction pressure head = 10 × 2.31

Suction pressure head = 23.1 ft

These converted values can later be used in pump head and total dynamic head calculations.

Pressure Head, Elevation Head, and Velocity Head

Hydraulic systems contain several forms of head.

• Pressure head represents energy associated with water pressure.
• Elevation head represents energy associated with elevation above a reference point.
• Velocity head represents energy associated with water velocity.

Water moving through a real system also loses energy because of friction in pipes, fittings, valves, and other components.

For basic operator calculations, the important point is that different hydraulic energy terms can all be expressed as feet of head. This allows pressure, elevation, velocity, and losses to be compared using a common unit.

Pressure While Water Is Flowing

The simple elevation-to-pressure relationship is most direct under static conditions. When water is flowing, actual gauge pressure can be lower than the theoretical static pressure because energy is lost to friction and other hydraulic effects.

For example, an elevation difference might theoretically provide 70 psi of static pressure. During high flow, a downstream gauge might read less than 70 psi because water is losing head as it moves through mains, valves, fittings, meters, and other restrictions.

This distinction helps operators understand why a distribution system can have acceptable static pressure but significantly lower pressure during periods of high demand.

Pressure Gauges and Elevation

The elevation of the pressure gauge matters. A gauge installed at a lower elevation will normally show a higher pressure than a gauge connected to the same static hydraulic source at a higher elevation.

Suppose a gauge is moved 23.1 feet lower in the same static system.

Pressure increase = 23.1 / 2.31

Pressure increase = 10 psi

Similarly, moving 23.1 feet higher would reduce static pressure by approximately 10 psi.

Finding an Unknown Elevation

The pressure-head relationship can also be used to estimate an elevation difference.

Suppose a pressure gauge connected to an elevated reservoir reads 52 psi under static conditions.

Head = 52 × 2.31

Head = 120.12 ft

If the reservoir water surface elevation is 680 feet:

Gauge elevation = 680 - 120.12

Gauge elevation = 559.88 ft

The gauge is therefore at an elevation of approximately 560 feet, assuming the reservoir water surface is the controlling hydraulic level and hydraulic losses are not involved.

Changing Water Level in a Storage Tank

As the water level in an elevated tank rises or falls, the available static head changes.

Suppose the water surface rises by 15 feet.

Pressure increase = 15 / 2.31

Pressure increase = 6.49 psi

At locations hydraulically connected to the tank and at unchanged elevations, theoretical static pressure would increase by approximately 6.5 psi.

This is one reason distribution pressure can vary as storage tank levels change.

Gauge Pressure Versus Absolute Pressure

Most routine operator pressure readings are gauge pressures. A standard pressure gauge normally indicates pressure relative to atmospheric pressure.

Absolute pressure includes atmospheric pressure in addition to gauge pressure. Absolute pressure becomes especially important in subjects such as pump suction conditions, vacuum measurements, and cavitation.

For ordinary pressure-head conversion questions, use the gauge pressure given in the problem unless the problem specifically asks for absolute pressure.

Common Pressure and Head Mistakes

• Multiplying by 2.31 when converting head to psi instead of dividing.
• Dividing by 2.31 when converting psi to feet of head instead of multiplying.
• Using tank volume instead of vertical water depth to determine hydrostatic pressure.
• Using horizontal distance instead of vertical elevation difference.
• Ignoring whether a location is above or below the reference water surface.
• Assuming flowing pressure must equal theoretical static pressure.
• Combining values in psi with values in feet without first converting them to common units.
• Confusing elevation head with tank volume or pipe length.

A Reliable Method for Pressure and Head Problems

1. Identify whether the problem asks for pressure in psi or head in feet.
2. Determine the relevant vertical elevation or water-depth difference.
3. Use 2.31 ft of water per psi.
4. For head to pressure, divide feet by 2.31.
5. For pressure to head, multiply psi by 2.31.
6. Keep pressure head and elevation head in feet when they must be combined.
7. Determine whether the problem describes static or flowing conditions.
8. Check whether the answer is reasonable for the elevation difference.

Quick Reasonableness Checks

Memorizing a few approximate relationships can help identify calculator or decimal errors:

• 23.1 ft ≈ 10 psi
• 46.2 ft ≈ 20 psi
• 69.3 ft ≈ 30 psi
• 115.5 ft ≈ 50 psi
• 231 ft ≈ 100 psi

If a calculation says that 23 feet of water produces 100 psi, the conversion has clearly been performed incorrectly.

What to Remember for the Exam

• Pressure is commonly expressed in psi.
• Pressure head is commonly expressed in feet of water.
• 2.31 ft of water is approximately equal to 1 psi.
• Pressure, psi = Head, ft / 2.31.
• Head, ft = Pressure, psi × 2.31.
• Hydrostatic pressure depends on vertical water-column height, not total tank volume.
• Lower elevations generally have greater static pressure when supplied from the same hydraulic level.
• A 23.1-foot elevation difference corresponds to approximately 10 psi.
• Static head is based on elevation difference.
• Pressure head, elevation head, and velocity head are all forms of hydraulic energy that can be expressed in feet.
• Actual pressure during flow can be lower than static pressure because of friction and other head losses.
• Convert pressure and head to common units before combining them in hydraulic calculations.