Calculate bearing oil flow from bearing length, steady load, journal diameter, clearance factor, and rotational speed in gpm or lpm.
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Bearing Oil Flow Formula
The bearing oil flow calculator uses an empirical oil flow correlation. Inputs are converted to base units first: bearing length and journal diameter in inches, steady load in pounds-force, and rotational speed in RPM. The base result is US gallons per minute.
Q_{gpm} = 29.3 \times 10^{-6} \times (L + 0.0043 \times W/D) \times C \times D^2 \times NIf you choose liters per minute, the calculator converts the gallon-per-minute result:
Q_{L/min} = Q_{gpm} \times 3.78541If rotational speed is entered in radians per minute, it is converted to RPM before the oil flow formula is applied:
N_{rpm} = N_{rad/min} \times 0.159155- Qgpm = oil flow in US gallons per minute
- QL/min = oil flow in liters per minute
- L = bearing length, in inches
- W = steady load supported by the bearing, in pounds-force
- D = journal diameter, in inches
- C = bearing clearance factor, unitless
- N = rotational speed, in RPM
The calculator converts your selected input units into the base units required by the formula, applies the empirical oil flow equation, then converts the result to either gallons per minute or liters per minute. The clearance factor is entered directly because it is unitless.
Unit Conversions Used in the Calculation
| Quantity | Entered Unit | Base Unit Conversion |
|---|---|---|
| Bearing length or journal diameter | in | 1 in = 1 in |
| Bearing length or journal diameter | ft | 1 ft = 12 in |
| Bearing length or journal diameter | cm | 1 cm = 0.393701 in |
| Bearing length or journal diameter | m | 1 m = 39.3701 in |
| Steady load | kgf | 1 kgf = 2.20462 lbf |
| Rotational speed | rad/min | 1 rad/min = 0.159155 RPM |
How Input Changes Affect Oil Flow
| Input | Effect on Calculated Oil Flow |
|---|---|
| Bearing length | Higher bearing length increases the flow term directly. |
| Steady load | Higher load increases the load correction term, depending on journal diameter. |
| Journal diameter | Diameter has a strong effect because it appears as D², while also affecting the load correction term. |
| Clearance factor | Oil flow changes in direct proportion to the clearance factor. |
| Rotational speed | Oil flow changes in direct proportion to RPM. |
Example Calculations
Example 1: Oil flow in gallons per minute
Suppose you enter these values:
- Bearing length = 4 in
- Steady load = 1000 lb
- Journal diameter = 2 in
- Clearance factor = 1.5
- Rotational speed = 1800 RPM
Using the formula:
Q_{gpm} = 29.3 \times 10^{-6} \times (4 + 0.0043 \times 1000/2) \times 1.5 \times 2^2 \times 1800The calculated oil flow is:
Q = 1.9461 gallons/min
Example 2: Oil flow with metric inputs
Suppose you enter these values:
- Bearing length = 10 cm
- Steady load = 500 kgf
- Journal diameter = 5 cm
- Clearance factor = 1.2
- Rotational speed = 1000 rad/min
The calculator first converts the inputs to the base units used by the formula:
- 10 cm = 3.9370 in
- 500 kgf = 1102.31 lbf
- 5 cm = 1.9685 in
- 1000 rad/min = 159.155 RPM
The calculated result is approximately:
Q = 0.1376 gallons/min, or 0.5208 liters/min
FAQ
What unit should the clearance factor use?
The clearance factor is unitless. Enter it as a plain number. For example, use 1.2, not 1.2 inches or 1.2 millimeters.
Does the calculator use kg as mass or force?
The kgf option is treated as kilogram-force, not mass in kilograms. The calculator converts kgf to pounds-force using 1 kgf = 2.20462 lbf.
Why does journal diameter affect the result so much?
Journal diameter appears as D² in the formula, so increasing diameter can increase the calculated oil flow significantly. It also appears in the load correction term, where the steady load is divided by diameter.
