Flow divider

One pump, two actuators that must move together. A flow divider splits the flow into set shares — and the same device, run backwards, trades flow for a higher pressure.

Source: Rabie, Fluid Power Engineering, Ch. 5.

Before you start

What you need first

Displacement & flow, $Q = V_g\,n$ — flow per turn × speed (Topic 19).
Speed from flow, $v = Q/A$ (Topic 3).
Flow is conserved — what goes in must come out (continuity) (Topic 3).

What you'll be able to do

Say what a flow divider does and why it synchronizes two actuators.
Split a flow in the displacement ratio $Q_1 : Q_2 = V_{g1} : V_{g2}$.
Explain the pressure-intensifier trick (trade flow for pressure).

Start here · the one big idea

Split one flow into two

A flow divider splits the pump's flow into two (or more) parts — often into two equal halves. Its main use is to make two cylinders or motors move together (synchronized).

Send equal flow to two identical cylinders and they extend at the same speed ($v = Q/A$) — even if their loads are unequal. That is something a simple tee-junction cannot do: a tee just sends more oil to whichever side is easier (lower pressure).

Synchronizing two actuators

Two cylinders raising one platform must rise together, or it tilts. A flow divider keeps them in step. The common spool flow divider even self-corrects: if one side starts to lead, the extra flow on that side shifts the internal spool, which throttles the fast side and opens the slow side until the two are equal again.

A flow divider controls the split of flow, so it controls the relative speed of the two actuators — independent of their individual loads.

The headline relation

The displacement flow divider

One neat design uses two hydraulic motors on a common shaft. Because they share the shaft they turn at the same speed $n$; and each motor passes $Q_i = V_{gi}\,n$ (Topic 19). Dividing the two:

$$\dfrac{Q_1}{Q_2} = \dfrac{V_{g1}}{V_{g2}}$$

So the flow splits in the ratio of the two displacements. Make them equal and you get an even 50 : 50 split.

Two motors on one shaft split the flow by displacement (after Rabie Fig. 5.64).

where:
Symbol	Meaning	SI unit
$Q_1, Q_2$	flow to branch 1, branch 2	m³/s
$V_{g1}, V_{g2}$	displacement of motor 1, motor 2	m³/rev
$n$	shared shaft speed	rev/s

Since flow is conserved ($Q = Q_1 + Q_2$), each branch gets a fixed share of the input:

$$Q_1 = Q\,\dfrac{V_{g1}}{V_{g1}+V_{g2}}, \qquad Q_2 = Q\,\dfrac{V_{g2}}{V_{g1}+V_{g2}}$$

Bonus: a pressure intensifier

Run a displacement divider "backwards" and it becomes a pressure intensifier. A large motor (fed the main flow at pressure $P_1$) drives a small pump on the same shaft; since the shaft torque is shared and $T = V_g\,\Delta P/2\pi$ (Topic 20), the smaller output displacement puts out a higher pressure:

$$P_1 V_{g1} = P_2 V_{g2} \;\Longrightarrow\; P_2 = P_1\dfrac{V_{g1}}{V_{g2}}$$

You trade flow for pressure: the small side delivers less flow ($Q_2 = Q_1 V_{g2}/V_{g1}$) but at a higher pressure — a handy way to reach a high local pressure (e.g. a clamp) without raising the whole system's pressure.

✏️ Try it yourself

A pump gives $30~\text{L/min}$ into a displacement divider with $V_{g1}:V_{g2} = 3:2$. Find $Q_1$ and $Q_2$.
An equal divider feeds two identical cylinders of area $A = 15~\text{cm}^2$ from a $20~\text{L/min}$ pump. Find each cylinder's speed.
Why does an equal flow divider keep two cylinders synchronized even when one carries a heavier load?

1. Total $= 3+2 = 5$ parts: $Q_1 = 30\times\tfrac{3}{5} = \mathbf{18~\text{L/min}}$,\; $Q_2 = 30\times\tfrac{2}{5} = \mathbf{12~\text{L/min}}$. 2. Each gets $Q = 10~\text{L/min} = 1.667\times10^{-4}~\text{m}^3/\text{s}$;\; $A = 15~\text{cm}^2 = 1.5\times10^{-3}~\text{m}^2$;\; $v = Q/A = 1.667\times10^{-4}/1.5\times10^{-3} = \mathbf{0.111~\text{m/s}}$ each. 3. It fixes the split of flow, and speed follows flow ($v=Q/A$) — not load. So both get the same flow and the same speed regardless of which side is harder to push.

Common mistakes to avoid

Mistake	Fix
Thinking a tee can synchronize cylinders	A tee sends more oil to the easier side; a flow divider fixes the split regardless of load.
Splitting flow in the area or pressure ratio	A displacement divider splits in the displacement ratio, $Q_1:Q_2 = V_{g1}:V_{g2}$.
Expecting "free" pressure from an intensifier	You trade flow for pressure: higher $P_2$ means proportionally less $Q_2$ (energy is conserved).
Forgetting to convert L/min for speed	Use $Q[\text{m}^3/\text{s}] = Q[\text{L/min}]/60{,}000$ before $v=Q/A$.

Recap — the whole topic on one screen

$$\dfrac{Q_1}{Q_2} = \dfrac{V_{g1}}{V_{g2}} \qquad P_1 V_{g1} = P_2 V_{g2}$$

Idea	What you own now
The job	Split one flow into set shares to synchronize actuators
The split	$Q_1:Q_2 = V_{g1}:V_{g2}$; equal displacements → 50:50
Self-correct	A spool divider re-balances if one side leads
Intensifier	Run it backwards: trade flow for a higher pressure ($P_1V_{g1}=P_2V_{g2}$)

Next — Theme 6

Accessories — the accumulator

That completes the valves theme: source → control is done. Next we add the support parts — starting with the accumulator, an energy store governed by the gas law — coming soon.

← Back to all ME3311 topics

Symbol	Meaning	SI unit
\(Q_1, Q_2\)	flow to branch 1, branch 2	m³/s
\(V_{g1}, V_{g2}\)	displacement of motor 1, motor 2	m³/rev
\(n\)	shared shaft speed	rev/s