Hydraulic Pump Volume Capacity

Overview

Engineering reference data for Hydraulic Pump Volume Capacity in hydraulics pneumatics.

Key Formulas

Pascal's Law

$P = \frac{F}{A}$

Pressure is equal in all directions.

Hydraulic Force

$F_2 = F_1 \frac{A_2}{A_1}$

Force amplification by area ratio.

Variables

Unit Conversions

Here are common unit conversions relevant to hydraulic pump capacity calculations:

• 1 gpm (US) = 6.30888×10⁻⁵ m³/s = 0.227 m³/h = 0.06309 dm³(litre)/s = 2.228×10⁻³ ft³/s = 0.1337 ft³/min = 0.8327 gpm (UK)
• 1 inch (in) = 25.4 mm = 2.54 cm = 0.0254 m = 0.08333 ft
• 1 in² = 6.452 cm² = 6.452×10⁻⁴ m² = 6.944×10⁻³ ft²

Calculation Example

Problem: A 3" hydraulic cylinder with a cylinder area of 7.065 in² makes a 30-inch stroke in 3 seconds. Calculate the required pump capacity.

Solution: Using the Imperial units formula:

$q = \frac{0.26 \times A \times s}{t}$

Where:

• $A = 7.065 \text{ in}^2$
• $s = 30 \text{ in}$
• $t = 3 \text{ s}$

Calculation: $q = \frac{0.26 \times 7.065 \times 30}{3} = 18.4 \text{ gpm}$

Result: The required pump output capacity is 18.4 gallons per minute (gpm).

References

• Original Source