ME3311 · Hydraulic & Pneumatic
Theme 5 · Valves

Flow control & the orifice equation

A cylinder's speed follows its flow — so to set the speed we squeeze the flow through an adjustable hole. How much gets through is the orifice equation, the calc heart of the valves theme.

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

Before you start

What you need first

  • Speed from flow, \(v = Q/A\) — control the flow, control the speed (Topic 3).
  • Oil density \(\rho\) (about 870 kg/m³) (Topic 8).
  • A pressure drop drives flow through a restriction (Topic 17).

What you'll be able to do

  • Read the throttle (adjustable orifice) symbol.
  • Use the orifice equation \(Q = C_d A\sqrt{2\,\Delta P/\rho}\) to find the flow through a hole.
  • Explain why a plain throttle's speed wanders with the load, and why throttling makes heat.
  • Read a throttle-check valve (control one way, free the other).

Start here · the one big idea

Flow sets the speed

From the flow theme: a cylinder's speed is its flow divided by its area.

$$v = \dfrac{Q}{A}$$piston speed
So if we can control the flow into an actuator, we control its speed. A flow-control valve does exactly that — by forcing the oil through a small, adjustable hole.

The throttle valve — an adjustable hole

A throttle valve is just an adjustable orifice — a small hole whose size you can change. Make the hole smaller → less flow gets through → the actuator moves slower.

On the symbol, the arrow through it means the restriction is adjustable (a knob sets the hole size). A fixed throttle has a set hole; an adjustable one you can tune.

The arrow marks it adjustable (after Rabie Fig. 5.57).

The headline relation

How much flows through a hole?

The flow through any sharp-edged hole is given by the orifice equation:

$$Q = C_d\,A\,\sqrt{\dfrac{2\,\Delta P}{\rho}}$$
where:
SymbolMeaningSI unit
\(Q\)flow rate through the holem³/s
\(C_d\)discharge coefficient — how "lossy" the hole is (≈ 0.6 for a sharp orifice)—
\(A\)orifice (hole) aream²
\(\Delta P\)pressure drop across the holePa
\(\rho\)oil densitykg/m³

Read off what controls the flow:

  • a bigger hole (\(A\)) → more flow (this is the knob you turn);
  • a bigger pressure drop (\(\Delta P\)) → more flow;
  • but flow grows only with \(\sqrt{\Delta P}\), not \(\Delta P\) itself — double the pressure drop and the flow rises by just \(\sqrt2 \approx 1.41\).
Watch the units. Convert before substituting: \(1~\text{mm}^2 = 10^{-6}~\text{m}^2\), \(1~\text{bar} = 10^{5}~\text{Pa}\). Then \(Q\) comes out in m³/s (\(\times 60{,}000\) for L/min).

Why a plain throttle is not enough

Look again at the equation: the flow depends on \(\Delta P\), the pressure drop across the hole. But \(\Delta P\) is set by the load — and the load changes. A heavier load leaves less pressure to spare across the throttle, so \(\Delta P\) falls, the flow falls, and the speed wanders.

We usually want a steady speed whatever the load does. A plain throttle cannot give that — its flow drifts with \(\Delta P\). The cure is a pressure-compensated flow-control valve (Topic 28), which holds \(\Delta P\) across the throttle constant.

A throttle also wastes energy: the oil squeezed through the hole loses the pressure \(\Delta P\), and that lost pressure becomes heat. The wasted power is \(N = \Delta P \times Q\) across the throttle — heavy throttling means hot oil.

Throttle-check — control one way, free the other

Put a check valve (Topic 26) in parallel with the throttle and you get a throttle-check (one-way flow-control) valve. It throttles the flow in one direction, and lets it run free in the other (the check bypasses the hole).

Typical use: a slow, controlled extend stroke, then a fast free retract.

Throttle with a bypass check (after Rabie Fig. 5.57b).

✏️ Try it yourself

A throttle has \(C_d = 0.6\), \(A = 6~\text{mm}^2\), \(\Delta P = 25~\text{bar}\), \(\rho = 870~\text{kg/m}^3\).

  1. Find the flow \(Q\) in L/min.
  2. If the load grows so \(\Delta P\) falls to \(15~\text{bar}\) (same hole), does the speed rise or fall, and roughly by what factor does \(Q\) change?
  3. What single valve would keep the flow steady through this change?
1. \(\sqrt{2\times2.5\times10^{6}/870} = \sqrt{5.75\times10^{3}} = 75.8~\text{m/s}\);\; \(Q = 0.6\times6\times10^{-6}\times75.8 = 2.73\times10^{-4}~\text{m}^3/\text{s} = \mathbf{16.4~\text{L/min}}\). 2. The flow (and speed) falls. Since \(Q\propto\sqrt{\Delta P}\), the factor is \(\sqrt{15/25} = \sqrt{0.6} \approx \mathbf{0.77}\) — about a 23% drop. 3. A pressure-compensated flow-control valve (it holds \(\Delta P\) across the throttle constant) — Topic 28.

Common mistakes to avoid

MistakeFix
Using \(Q \propto \Delta P\)It is \(Q \propto \sqrt{\Delta P}\): quadruple \(\Delta P\) only doubles the flow.
Leaving \(A\) in mm² or \(\Delta P\) in barConvert: \(1~\text{mm}^2=10^{-6}~\text{m}^2\), \(1~\text{bar}=10^{5}~\text{Pa}\).
Forgetting \(C_d\) (using \(C_d=1\))A real sharp orifice passes only ≈ 60% of the ideal — use \(C_d \approx 0.6\).
Expecting a plain throttle to hold speedIts flow drifts with the load's \(\Delta P\); you need a compensated FCV.

Recap — the whole topic on one screen

$$Q = C_d\,A\,\sqrt{\dfrac{2\,\Delta P}{\rho}} \qquad (C_d \approx 0.6)$$
IdeaWhat you own now
The jobSet the actuator speed by setting the flow (\(v=Q/A\))
The throttleAn adjustable orifice — smaller hole, slower
The equation\(Q = C_d A\sqrt{2\Delta P/\rho}\); flow \(\propto \sqrt{\Delta P}\)
The catchPlain throttle drifts with load; throttling wastes heat (\(N=\Delta P\,Q\))
Throttle-checkControl one way, free flow the other

Next topic

Speed-control circuits

Now we place the throttle in a real circuit: meter-in, meter-out and bleed-off — and the pressure-compensated valve that finally holds the speed steady.

→ Speed-control circuits