The relationship between pump inlet/outlet pressure and head

Core Formula (The Most Critical One)

Head (H) = (Pump Outlet Pressure - Pump Inlet Pressure) / (Liquid Density × Gravitational Acceleration)

Represented with symbols:

H = (P₂ - P₁) / (ρ × g)

Where:

H: The Head generated by the pump, measured in meters (m).

P₂: The absolute pressure at the pump outlet flange, typically in Pascals (Pa).

P₁: The absolute pressure at the pump inlet flange, typically in Pascals (Pa).

ρ: The density of the pumped liquid, in kilograms per cubic meter (kg/m³). For water at room temperature, ρ ≈ 1000 kg/m³.

g: The gravitational acceleration, approximately 9.81 m/s².

Explanation of Key Concepts

What is Head?

Not Height: Head is not simply the physical lift height. It is an energy concept, representing the total mechanical energy imparted to a unit weight of liquid by the pump. Its unit is meters (m), which can be understood as the "theoretical height to which the pump can elevate the liquid."

Independent of Medium: Head is a performance parameter of the pump itself. The same pump, at the same speed, will generate the same head (H) value whether it is pumping water, oil, or another liquid. However, the power consumption and the resulting pressure will differ.

What is Pressure?

Pressure is force per unit area. The outlet gauge pressure generated by the pump intuitively reflects the magnitude of its "thrust."

Closely Related to Medium: According to the formula P = ρ × g × H, the pressure (P) generated by the pump directly depends on the liquid's density (ρ). Pumping a denser liquid (like oil) will produce greater pressure at the same head.

Core Difference and Connection

Head is the "Cause," Pressure is the "Effect." The pump's characteristics determine the head it can provide. This head, acting on a liquid of specific density, ultimately manifests as the pressure difference between inlet and outlet.

Think of head as the pump's "capability rating," and pressure as the "effect" produced when that capability acts on a specific object (a certain liquid).

Application Example (Using water, simplified with ρ ≈ 1000 kg/m³, g ≈ 10 m/s²)

Assume a pump has a head of 100 meters.

Calculate the pressure difference it generates:

ΔP = ρ × g × H = 1000 kg/m³ × 10 m/s² × 100 m = 1,000,000 Pa = 1 MPa ≈ 10 bar

This means if the inlet pressure is atmospheric (0 bar gauge), its outlet gauge pressure would be about 10 bar.

Estimating Head On-Site:

If you measure on-site that the pump outlet gauge reads 0.8 MPa (8 bar), and the inlet gauge reads 0.1 MPa (1 bar).

Then the pressure difference ΔP = 0.8 - 0.1 = 0.7 MPa = 700,000 Pa.

Calculate the head: H = ΔP / (ρ × g) = 700,000 / (1000 × 10) = 70 meters.

This 70 meters is the effective head the pump is actually providing under the current operating conditions.

Important Notes

Absolute Pressure Must Be Used for Calculation: The P₁ and P₂ in the formula are, in theory, absolute pressures. However, in practical engineering, when both inlet and outlet pressures are measured with gauges using the same reference (typically local atmospheric pressure), using the gauge pressure difference yields a perfectly correct result. That is, H = (Outlet Gauge Pressure - Inlet Gauge Pressure) / (ρ × g).

Inlet Pressure Must Exceed NPSH Requirement: If the inlet pressure (P₁) is too low, the liquid will vaporize inside the pump, causing cavitation, which severely damages the pump. The Required Net Positive Suction Head (NPSHr) on the pump performance curve is the key parameter set to ensure P₁ is sufficiently high.

System Resistance Determines Operating Point: The pump's actual outlet pressure in a pipeline system is determined by the intersection point of the pump's head-flow curve and the pipeline system's resistance curve. The pump adjusts its output until the head it generates exactly equals the resistance required by the system at that flow rate (including elevation lift, pipe friction, valve resistance, etc.).