Engineering guide
The VSP module calculates the leakage flow through a narrow annular gap of the kind that occurs as a sealing gap at shaft passages, pistons, throttle bushings or canned seals of pumps and valves. From the diameter and length of the sealing gap, the gap width and the applied pressure difference, the gap flow — i.e. the leakage volume flow — is determined.
The properties of the fluid in the gap enter via density and kinematic viscosity. As an intermediate result, the calculation provides the friction factor of the gap flow, from which — together with the pressure difference — the flow velocity in the gap and hence the leakage flow are obtained. In practice, this calculation is needed to estimate leakage rates of throttle gaps, to size barrier fluid quantities, or to account for gap losses in hydraulic systems.
Typical workflow
Define the gap geometry
The diameter of the sealing gap, the length of the sealing gap and the gap width describe the annulus. Since the gap width is small compared with the diameter, the annulus can be treated as an unrolled plane gap; the hydraulic diameter equals twice the gap width.
Enter operating data and fluid properties
The pressure difference across the gap as the driving head, together with the density and kinematic viscosity of the fluid in the gap, are specified.
Determine the flow regime and friction factor
From the velocity in the gap and the hydraulic diameter, the Reynolds number is formed; depending on whether the flow is laminar or turbulent, the friction factor of the gap flow follows. Since the velocity is initially unknown, the pressure-loss equation is solved for the velocity, iteratively if necessary.
Calculate the gap flow
From the flow velocity and the annular gap area (circumference times gap width), the gap flow follows as the leakage volume flow.
Frequently asked questions
Why does the gap width have such a strong influence on the leakage?
For laminar gap flow, the leakage flow grows with the third power of the gap width: doubling the gap means eight times the leakage. That is why manufacturing tolerances, wear and thermal widening of the gap dominate the leakage behavior far more than any other input quantity — the gap width should always be considered together with its tolerance limits.
Does the calculation also apply to eccentric gaps?
The basic calculation assumes a concentric annular gap. If the shaft is eccentric, the leakage flow increases for laminar flow — in the limiting case of full eccentricity to up to 2.5 times the concentric value. This allowance should be taken into account when assessing real sealing gaps.
When is the gap flow laminar and when turbulent?
The Reynolds number, formed with the gap velocity and the hydraulic diameter (twice the gap width), is decisive. Below about Re ≈ 2,300 the flow is laminar and the friction factor then follows the law λ ∝ 1/Re. Narrow sealing gaps with viscous media are almost always in the laminar range; with water and large pressure differences, however, the flow can become turbulent, in which case the leakage scales less strongly with the gap width.
Does the calculation account for shaft rotation?
The pure pressure-driven gap flow is considered axially. Rotation of the shaft superimposes a drag flow in the circumferential direction; it hardly changes the axial leakage flow for laminar conditions, but it can promote transition to turbulence and significantly increase the friction power dissipated in the gap. For high-speed shafts this must be assessed separately.
Worked example
Estimate the leakage flow of water (20 °C) through the throttle gap of a shaft passage — a worked example of an annular gap leakage calculation. The gap is concentric; the flow is first assumed laminar and then checked via the Reynolds number.
Given
| Diameter of the sealing gap d | 50 mm |
|---|---|
| Length of the sealing gap L | 30 mm |
| Gap width s | 0.1 mm |
| Pressure difference Δp | 1.0 bar |
| Density ρ (water, 20 °C) | 998 kg/m³ |
| Kinematic viscosity ν | 1.0 · 10⁻⁶ m²/s |
Solution
Mean gap velocity (laminar gap flow)
For the plane laminar gap, Δp = 12 · μ · L · w / s² with μ = ρ · ν = 998 · 1.0 · 10⁻⁶ = 9.98 · 10⁻⁴ Pa·s. Solved for the mean velocity:
w = s² · Δp / (12 · μ · L) = (1.0 · 10⁻⁴)² · 10⁵ / (12 · 9.98 · 10⁻⁴ · 0.03) = 2.78 m/s
Check of the flow regime
Hydraulic diameter dh = 2 · s = 0.2 mm.
Re = w · dh / ν = 2.78 · 2.0 · 10⁻⁴ / 1.0 · 10⁻⁶ ≈ 557 < 2,300 → laminar, assumption confirmed.
Friction factor: λ = 96 / Re = 96 / 557 ≈ 0.172
Gap flow
Annular gap area A = π · d · s = π · 0.05 · 1.0 · 10⁻⁴ = 1.571 · 10⁻⁵ m².
Q = w · A = 2.78 · 1.571 · 10⁻⁵ = 4.37 · 10⁻⁵ m³/s ≈ 157 l/h
Result
| Flow velocity in the gap | 2.78 m/s |
|---|---|
| Reynolds number | 557 (laminar) |
| Friction factor λ | 0.172 |
| Gap flow (leakage) | 157 l/h |
Technical content source: LV-Solved-Problems-Content 628d313247e7