Sizing the line

Choosing a pipe bore is one short calculation: pick a speed from the recommended band, and the flow fixes the diameter.

Source: Rabie, Fluid Power Engineering, Ch. 3.

Before you start

What you need first

Flow = area × speed, $Q=Av$ (Topic 3).
Recommended line speeds (Topic 14).

What you'll be able to do

Find oil speed from flow and bore, $v=\tfrac{4Q}{\pi D^2}$.
Size a bore from a target speed, $D=\sqrt{\tfrac{4Q}{\pi v}}$.
Check a line against the recommended band.

Start here · the relation

Speed from flow and bore

The pipe's cross-section is a circle, $A=\tfrac{\pi}{4}D^2$, so $v=Q/A$ becomes:

$$v = \dfrac{Q}{A} = \dfrac{4Q}{\pi D^2}$$

where:
Symbol	Meaning	SI unit
v	oil velocity in the pipe	m/s
Q	flow rate	m³/s
D	pipe inside diameter (bore)	m

Because $D$ is squared, velocity is very sensitive to bore: $v\propto 1/D^2$.

Pick a diameter from a target speed

To design a line, choose a speed from the recommended band and rearrange for the bore:

$$D = \sqrt{\dfrac{4Q}{\pi v}}$$

Then round up to the next standard pipe size (rounding up keeps the actual speed at or below your target).

Same $Q$: halve the bore and the speed rises four-fold ($v\propto 1/D^2$).

Which speed to choose

Take the speed from the recommended band for that line (Topic 14), then size the bore:

Line	Design speed	Note
Pressure	2 – 6 m/s	Higher → the thinnest pipe
Suction	0.6 – 1.6 m/s	Kept lowest → the fattest pipe (avoids cavitation)
Return / low-pressure	0.6 – 1.6 m/s	Same low band as suction (Rabie)

For the same flow, the low-pressure lines (suction & return) are the fattest because they run far slower than the pressure line — and the suction, designed at the lowest speed, is the largest of all. A frequent exam point.

✏️ Try it yourself — no numbers needed

A line carries a fixed flow at 2 m/s. To save space, someone swaps it for a pipe of half the bore. What happens to the oil speed, and why does that matter for losses?

Speed ×4 → 8 m/s. Since $v=4Q/\pi D^2$, halving $D$ makes the area $4\times$ smaller, so the oil runs four times faster. Why it matters: friction loss grows with $v^2$ (and the laminar loss with $1/D^4$ — next topics), so a "small" bore saving causes a large jump in lost pressure and heat. 8 m/s is also well above the recommended band.

Common mistakes to avoid

Mistake	Fix
Leaving $Q$ in L/min	Convert: ÷60,000 to get m³/s.
Putting bore in mm into the formula	Use metres: $25~\text{mm}=0.025~\text{m}$.
Rounding the bore down to a standard size	Round up — a bigger bore keeps the speed at/below target.

Recap — the whole topic on one screen

$$v=\dfrac{4Q}{\pi D^2}\qquad D=\sqrt{\dfrac{4Q}{\pi v}}$$

Idea	What you own now
Speed from bore	$v=4Q/\pi D^2$; $v\propto 1/D^2$
Size a bore	Choose $v$ from the band, then $D=\sqrt{4Q/\pi v}$, round up
Suction is fattest	Lowest design speed → biggest bore

Next topic

Laminar, turbulent & the Reynolds number

Now that we know the speed, we can ask: is the flow smooth or chaotic? That decides how much pressure friction will cost — and it's settled by the Reynolds number.

→ Laminar, turbulent & Reynolds number

Mistake	Fix
Leaving \(Q\) in L/min	Convert: ÷60,000 to get m³/s.
Putting bore in mm into the formula	Use metres: \(25~\text{mm}=0.025~\text{m}\).
Rounding the bore down to a standard size	Round up — a bigger bore keeps the speed at/below target.

Idea	What you own now
Speed from bore	\(v=4Q/\pi D^2\); \(v\propto 1/D^2\)
Size a bore	Choose \(v\) from the band, then \(D=\sqrt{4Q/\pi v}\), round up
Suction is fattest	Lowest design speed → biggest bore