Flow rate & piston speed

The second master relation. Pressure made the force; flow makes the motion. Pour oil into a cylinder faster and the piston moves faster.

Source: Rabie, Fluid Power Engineering, Ch. 1.

Before you start

What you need first

What fluid power is (Topic 1).
Area of a circle — $A=\tfrac{\pi}{4}D^2$.

What you'll be able to do

Say what flow rate $Q$ is.
Use $Q=Av$ and $v=\tfrac{Q}{A}$.
Explain why a smaller area moves faster for the same flow.
Convert L/min ↔ m³/s safely.

Start here · the idea

What is flow rate?

Flow rate $Q$ is how much oil volume passes per second. The pump's job is to make flow; pour more oil into a cylinder each second and the piston moves faster.

Pressure decides force. Flow rate decides speed. They are two separate questions — never mix them up.

Flow rate = area × speed

The oil filling behind the piston has nowhere else to go, so the volume it adds per second equals the piston's face area times how fast the piston moves:

$$Q = A\,v$$

where:
Symbol	Meaning	SI unit
Q	volume flow rate	m³/s
A	piston area	m²
v	piston speed	m/s

Oil enters at $Q$; the piston of area $A$ rises at $v$.

Rearranged — speed from flow

$$v = \dfrac{Q}{A}$$

Want a slower actuator? Reduce the flow. That is exactly what a flow-control valve does — we meet it later in the valves theme.

Same flow, smaller area → faster

Oil is almost incompressible, so whatever volume goes in per second must show up as piston sweep per second. For a fixed flow $Q$:

$$v = \dfrac{Q}{A}\quad\Longrightarrow\quad \text{smaller } A \Rightarrow \text{larger } v$$

🔭 Looking ahead: this is why a cylinder retracts faster than it extends — the rod side has the smaller (ring) area. We use this directly in Cylinder speed later in the course.

A note on flow units

The SI unit is m³/s, but pumps are rated in litres per minute (L/min):

$$1~\text{L/min} = \dfrac{10^{-3}}{60}~\text{m}^3/\text{s} = 1.667\times10^{-5}~\text{m}^3/\text{s}$$

Always convert L/min → m³/s before using a formula. A handy check: divide L/min by 60,000 to get m³/s.

✏️ Try it yourself — no numbers needed

The same pump (same flow $Q$) feeds two cylinders. Cylinder B's piston area is half of cylinder A's. Which piston moves faster, and by how much?

Cylinder B — twice as fast. Speed is $v=Q/A$. With the same flow, halving the area doubles the speed. Why: the same oil per second has to fill a thinner column, so the piston must travel further each second. Smaller area, faster motion.

Common mistakes to avoid

Mistake	Fix
Leaving flow in L/min inside $v=Q/A$	Convert first: divide L/min by 60,000 to get m³/s.
Using the diameter where the formula wants area	Find $A=\tfrac{\pi}{4}D^2$ first, then divide.
Thinking more flow gives more force	Flow sets speed; pressure sets force. Different jobs.

Recap — the whole topic on one screen

$$Q = A\,v\qquad v=\dfrac{Q}{A} \qquad 1~\text{L/min}=1.667\times10^{-5}~\text{m}^3/\text{s}$$

Idea	What you own now
Flow rate	Volume of oil per second; made by the pump
Flow → speed	$v=Q/A$ — flow sets how fast, not how hard
Area effect	Same flow, smaller area → faster motion
Units	Convert L/min → m³/s (÷60,000) before any formula

Next topic

Hydraulic power

We now have force (from pressure) and speed (from flow). Multiply the right pair and you get power — and a surprisingly neat identity, $N=pQ=Fv$.

→ Hydraulic power

Mistake	Fix
Leaving flow in L/min inside \(v=Q/A\)	Convert first: divide L/min by 60,000 to get m³/s.
Using the diameter where the formula wants area	Find \(A=\tfrac{\pi}{4}D^2\) first, then divide.
Thinking more flow gives more force	Flow sets speed; pressure sets force. Different jobs.