TITLE: How A Gear Pump WorksAUTHOR: Martin L. CulpepperCOURSE: 2YEAR: G

MAIN FUNCTIONAL REQUIREMENT: Convert mechanical power into fluid power.

DESIGN PARAMETER: Pump (a gear pump is one type of pump which can satisfy this functionalrequirement)

GEOMETRY/STRUCTURE AND PARTS:

EXPLANATION OF HOW IT WORKS:

1. One shaft is driven by a motor or some other means2. The gear mounted to this shaft (driving gear) engages the other gear (driven gear)3. Fluid on the inlet side flows into and is trapped between the rotating gear teeth and the housing4. The fluid is carried around the outside of the gears to the outlet side of the pump5. As the fluid can not seep back along the path it came, nor between the engaged gear teeth (they

create a seal,) it must exit the outlet port.

In some gear pumps, there are side plates usually made of brass which can be replaced or re-ground whenthe gap between the face of the gear and the end housing becomes too large

DOMINANT PHYSICS:

Variable Description Metric Units English UnitsPin Power input to shaft Watts HorsepowerPout Power output to fluid system Watts Horsepower

PlossPower loss (i.e. to coloumb friction and viscousdissipation) Watts Horsepower

w Shaft rotational speed rad/s RPMDp Pressure increase between inlet and outlet Pascals psiQ Flow rate through the pump liters3/s in3/shm Mechanical efficiency --- ---

The pump takes power from a rotating shaft:

Pin = T x w

A portion of this power is dissipated in the pump through coloumb friction and viscous dissipation. Thisis not easily quantified theoretically and is often determined experimentally. This power will be denotedat Ploss.

Ploss = f(friction, viscous effects......)

Some fluid will seep through the gap between the sides of the gears and the endplates (see figure below.) This gap must be small in order to maintain the pressure increase across the pump. Increasing the gapdiminishes the pumps ability to hold a pressure difference between the inlet and outlet. The gap istypically around 0.0005 inches.

The power which can then be derived from the fluid which comes out of the pump is:

Pout = (Dp x Q) = Pin - Ploss = T x w - Ploss

This can also be expressed using the efficiency:

Pout = hm x Pin

LIMITING PHYSICS

The performance/use of the pump is limited by its:

Efficiency

hm=of a pump is Pout/Pin. This is a function of the fluid viscosity, clearance between internal components,friction between mating components, and other variables.

Typically, gear pumps have efficiencies around 85%.

Bearings

Many external gear pumps use journal bearings to support the rotating shafts. In order for these bearingsto work, a minimum speed is required (depends upon pressure of the pump.) In addition to imposinglimits on the operational speed, in many cases, the bearings determine the maximum pressure the pumpcan operate at. Should the pressure drop across the pump be too large, the journal bearings will not beable to support the loads on the shafts (which come mainly from the pressure difference.)

IF NEEDED, PLOTS/GRAPHS/TABLES:

None to include here

WHERE YOU CAN FIND GEAR PUMPS:

These pumps have few moving parts, making them inexpensive. These pumps are typically used wherelow to medium pressure (about 2500 - 4000 psi) is needed and mechanical efficiency is not extremelyimportant (typical efficiency is about 85%.)

You can find gear pumps on the following machines:

1. Usually your car's oil pump2. Hydraulically driven lawn care equipment3. Some hydraulically driven log splitters4. Hydraulic power units on trucks and construction equipment5. Metering applications (gear pumps are good at controlling volume flow rate)

REFERENCES/MORE INFORMATION:

Viking Pump's Web Page on External Gear Pumps

Don't forget to add book had on hydraulic component design

Others....ADMINISTRATIVE: Home | General Info | Syllabus

STUDENT RESOURCES: Gear Pump Resources | Workshops | Labs | Field Trips | Final Reports | Design Tools

MISCELLANEOUS: Students | Prizes | Machine Gallery | Class Gallery | Comments | Links | Sponsors