Calculate external gear pump geometric displacement, Hagen-Poiseuille laminar slip backflow across radial and axial clearances, true delivered flow, volumetric efficiency ηv, and required electric motor drive torque per Hydraulic Institute HI 3.1-3.5 standards.
1. Gear Tooth Profile & Clearances
2. Fluid Properties & System Pressure
3. Performance & Hydraulic Results
→ [ Meshing Teeth Expel Fluid into High-Pressure Discharge ] ↔ [ Relief Grooves Prevent Trapped Volume Spikes ]
← [ Laminar Slip Backflow Across Radial Tip Clearances (δr) & Axial Wear Plates (δa) ] ←
Mathematical Foundations & Hydraulic Institute Slip Equations
Positive displacement gear pump rating balances geometric chamber volume against laminar slot leakage per Hagen-Poiseuille:
$$V_d approx 2 pi cdot b cdot m^2 cdot z cdot left(1 + rac{pi^2 cos^2 alpha}{12 z^2} ight) quad [ ext{cm}^3/ ext{rev}]$$ $$Q_{th} = rac{V_d cdot N}{1000} quad [ ext{L}/ ext{min}]$$
$$Q_{slip} = left( C_r cdot rac{b cdot delta_r^3}{L_r} + C_a cdot rac{d_p cdot delta_a^3}{L_a} ight) cdot rac{Delta P}{mu} quad [ ext{L}/ ext{min}]$$ Slip scales with the **cube of clearance gap** ($delta^3$).
$$Q_{act} = Q_{th} - Q_{slip}, quad eta_v = rac{Q_{act}}{Q_{th}} imes 100%$$ $$eta_{total} = eta_v cdot eta_{mh}$$
$$ au = rac{V_d cdot Delta P}{20 pi cdot eta_{mh}} quad [ ext{N}cdot ext{m}]$$ $$P_{motor} = rac{Q_{th} cdot Delta P}{600 cdot eta_{mh}} = rac{ au cdot 2pi N}{60,000} quad [ ext{kW}]$$
5 Fatal Traps in Positive Displacement Gear Pump Systems
When two standard involute teeth mesh, the contact points seal off a small volume of liquid in the tooth root valley. As meshing advances, this trapped volume physically contracts. Because hydraulic oils have high bulk modulus, squeezing trapped liquid creates instantaneous pressure peaks exceeding 400 to 600 bar. This shock force hammers against gear shafts, causes high-frequency 2,000 Hz piercing screams, and shears needle bearing cages within 200 hours. CNC-machined relief decompression slots on the side wear plates are strictly mandatory.
Testing or running an industrial gear pump engineered for ISO VG 46 hydraulic oil with diesel, kerosene, or water (viscosity < 2 cSt) causes internal slip to skyrocket because slip scales inversely with viscosity ($Q_{slip} propto 1/mu$). At 100 bar discharge pressure, back-leakage through standard 45-micron clearances equals 100% of theoretical displacement: the pump consumes full motor horsepower while delivering **zero net flow**, instantly boiling the fluid inside the casing and causing catastrophic pump seizure.
Pumping heavy lubricating oil or polymer melts on a cold winter morning when viscosity reaches 3,000 to 10,000 cSt creates massive pipe friction in the suction line. The atmospheric inlet pressure is insufficient to push the viscous liquid into the unmeshing tooth cavities fast enough. The cavities pull a deep vacuum (NPSHa deficit), boiling dissolved air and volatile fractions. As the vapor bubbles enter the high-pressure discharge port, they implode with immense local shock waves that erode tooth flanks and pit side plates. High-viscosity service demands reducing pump RPM to 300 to 500 RPM.
Fine abrasive particles in recycled oil or slurry score the bronze or aluminum axial side plates. Because Hagen-Poiseuille laminar slip scales with the **cube of the clearance gap** ($delta^3$), doubling the axial clearance from 30 $mu$m to 60 $mu$m does not double leakage—**it increases slip flow by $(2)^3 = 800%$!** Delivered flow rate plummets, system cylinder cycle times crawl to a halt, and operators mistakenly blame the drive motor. Continuous 10-micron beta-rated suction/return filtration is non-negotiable.
In hot bitumen, asphalt, or heat transfer fluid service at 180°C to 250°C, alloy steel gears expand thermally at a different rate than cast iron or stainless steel pump casings. If cold manufacturing axial clearances are set to standard room-temperature tolerances (e.g. 30 $mu$m), differential axial thermal growth closes the clearance gap to zero when heated. The spinning gear faces micro-weld (gall) against the stationary end covers, locking the pump shaft instantaneously and shearing motor drive keys.
Step-by-Step Worked Engineering Example
Application: Industrial Hydraulic Power Unit (HPU) External Gear Pump.
- Gear Dimensions: Module $m = 3.5 ext{ mm}$, Tooth count $z = 12$, Face width $b = 32 ext{ mm} = 0.032 ext{ m}$, Pressure angle $alpha = 20^circ$.
- Clearances: Radial tip clearance $delta_r = 45;mu ext{m}$, Axial side plate clearance $delta_a = 35;mu ext{m}$.
- Operating Parameters: Speed $N = 1,450 ext{ RPM}$, Differential pressure $Delta P = 65 ext{ bar} = 6.5 imes 10^6 ext{ Pa}$.
- Fluid Properties: ISO VG 46 hydraulic oil at operating temperature ($ u = 46 ext{ cSt}$, $ ho = 875 ext{ kg/m}^3$, $mu = 0.04025 ext{ Pa}cdot ext{s}$).
- Efficiency: Mechanical-hydraulic efficiency $eta_{mh} = 88%$.
Step 1: Geometric Displacement per Revolution ($V_d$):
$$V_d approx 2 pi cdot (0.35 ext{ cm})^2 cdot 12 cdot 3.2 ext{ cm} cdot left(1 + rac{pi^2 cos^2 20^circ}{12 imes 12^2} ight)$$ $$V_d = 2 pi imes 0.1225 imes 12 imes 3.2 imes 1.005 = 6.283 imes 4.704 imes 1.005 = 29.70 ext{ cm}^3/ ext{rev} quad (1.812 ext{ in}^3/ ext{rev})$$ $$ ext{Theoretical Flow Rate: } Q_{th} = rac{29.70 ext{ cm}^3/ ext{rev} imes 1,450 ext{ RPM}}{1000} = 43.07 ext{ L/min} quad (11.38 ext{ GPM})$$Step 2: Hagen-Poiseuille Internal Slip Flow ($Q_{slip}$):
$$Delta P = 6.5 imes 10^6 ext{ Pa}, quad mu = 0.04025 ext{ Pa}cdot ext{s}$$ $$ ext{Total internal slip leakage calculated across radial tips and axial side plates: } Q_{slip} = 3.65 ext{ L/min}$$ $$Q_{act} = Q_{th} - Q_{slip} = 43.07 - 3.65 = 39.42 ext{ L/min} quad (10.41 ext{ GPM})$$Step 3: Volumetric & Total Pump Efficiency:
$$eta_v = rac{Q_{act}}{Q_{th}} imes 100% = rac{39.42}{43.07} imes 100% = 91.52% implies mathbf{ ext{Healthy High Volumetric Efficiency}}$$ $$eta_{total} = eta_v imes eta_{mh} = 0.9152 imes 0.88 = 0.8054 = 80.54%$$Step 4: Input Drive Torque & Electric Motor Power:
$$ au = rac{V_d cdot Delta P}{20 pi cdot eta_{mh}} = rac{29.70 imes 65}{20 pi imes 0.88} = rac{1,930.5}{55.29} = 34.92 ext{ N}cdot ext{m} quad (25.75 ext{ ft}cdot ext{lb})$$ $$P_{motor} = rac{Q_{th} cdot Delta P}{600 cdot eta_{mh}} = rac{43.07 ext{ L/min} imes 65 ext{ bar}}{600 imes 0.88} = rac{2,799.55}{528} = 5.30 ext{ kW} quad (7.11 ext{ HP})$$ $$mathbf{ ext{Select Standard } 5.5 ext{ kW (7.5 HP) 4-Pole Electric Motor}}.$$