Hydraulic Cylinder

Calculate thrust, velocity, and power for double-acting hydraulic cylinders

Inputs

D

Cylinder bore D

d

Rod diameter d

p_1

Piston-side pressure p₁

MPa

p_2

Rod-side pressure p₂

MPa

Back pressure on the opposite side

Q

Supply flow rate Q

L/min

\lambda

Load coefficient λ

Friction coefficient, typically 0.95–0.98

Formula Interpretation

Thrust Formula

F_1 = \lambda\!\left(A_1 p_1 - A_2 p_2\right) = \lambda\dfrac{\pi}{4}\!\left[D^2 p_1 - (D^2\!-\!d^2)p_2\right]

Theoretical thrust $F$ is the net pressure force. Actual thrust multiplies by load coefficient $\lambda$ to account for friction losses.

Velocity & Power

v = \lambda\dfrac{Q}{A}\qquad P = F' \cdot v

Velocity $v$ equals flow rate divided by the active piston area. Power $P$ is the product of actual thrust and velocity.

Key Concepts

Double-Acting Cylinder

Hydraulic fluid acts on both sides of the piston, enabling controlled force in both advance and retract directions.

Differential Areas

The advance stroke uses full bore area A₁; the retract stroke uses annular rod-side area A₂ = A₁ − Arod, giving different forces and speeds.

Load Coefficient λ

λ accounts for seal friction and flow losses, typically 0.95–0.98. It reduces both actual thrust and effective velocity equally.

Worked Example

A double-acting cylinder has bore $D=45\,\text{mm}$ , rod diameter $d=16\,\text{mm}$ . Supply pressure $p_1=3.8\,\text{MPa}$ , back pressure $p_2=0.1\,\text{MPa}$ , flow rate $Q=10\,\text{L/min}$ , λ = 0.97. Find actual thrust, velocity, and power for the advance stroke.

D = 45 mm, d = 16 mm, p₁ = 3.8 MPa, p₂ = 0.1 MPa, Q = 10 L/min, λ = 0.97

Step 1 — Calculate piston areas

A_1 = \dfrac{\pi}{4}D^2 = \dfrac{3.14}{4}\!\times\!\left(\!\dfrac{45}{1000}\!\right)^{\!2} \approx 1.590\times10^{-3}\,\text{m}^2

A_2 = \dfrac{\pi}{4}(D^2-d^2) = \dfrac{3.14}{4}\!\left[\!\left(\!\dfrac{45}{1000}\!\right)^{\!2}\!-\!\left(\!\dfrac{16}{1000}\!\right)^{\!2}\right] \approx 1.388\times10^{-3}\,\text{m}^2

Step 2 — Theoretical thrust

F_1 = A_1 p_1 - A_2 p_2 = 1.590\times10^{-3}\times3.8\times10^6 - 1.388\times10^{-3}\times0.1\times10^6 \approx 5903\,\text{N}

Step 3 — Actual thrust, velocity, and power

F_1' = \lambda F_1 = 0.97\times5903 \approx 5726\,\text{N}\;(5.72\,\text{kN})

v_1 = \lambda\dfrac{Q_1}{A_1} = 0.97\times\dfrac{10}{1000\times60}\div1.590\times10^{-3} \approx 0.102\,\text{m/s}

P = F_1'\times v_1 = 5726\times0.102 \approx 584\,\text{W}

Actual thrust F₁′ ≈ 5720 N (5.72 kN), velocity v₁ ≈ 0.102 m/s, power P ≈ 583 W

Extended Knowledge

•Hydraulic circuits also include pipelines, motors, oil reservoirs, and control valves for adjusting pressure, flow, and direction.
•Oil viscosity decreases at elevated temperatures, which can cause air entrainment and cavitation — adequate cooling is essential.
•Hydraulic pumps used in high-pressure systems are typically axial-piston or radial-piston types with variable displacement.
•Cylinder speed can be increased by using a regenerative circuit that directs rod-side return oil back to the advance side.