ME3311 · Hydraulic & Pneumatic

Theme 4 · Pumps

Positive displacement & pump flow

The pump is the heart of the system — but it does just one job: it makes a flow of oil. How much it delivers comes down to its size and its speed.

Source: Rabie, Fluid Power Engineering, Ch. 4.

Before you start

What you need first

Flow = area × speed, $Q=Av$ — flow is a volume per second (Topic 3).
The pump's place in the circuit — driven by a motor, it feeds the pressure line (Topic 5).

What you'll be able to do

Explain why a pump makes flow, not pressure.
State the displacement $V_g$ and find the theoretical flow $Q_t = V_g\,n$.
Correct the ideal flow for leakage with the volumetric efficiency $\eta_v$.

Start here · the one big idea

A pump makes flow — not pressure

This is the idea the whole theme turns on. A pump pushes a flow of oil through the system. By itself it does not make pressure.

Pump → flow · Load → pressure

So where does the pressure come from? Pressure builds whenever something resists the flow — the load on a cylinder, a closed valve, a narrow passage. No resistance → almost no pressure. Big resistance → high pressure.

The pump sets how fast the oil moves (the flow); the load sets how hard it has to push (the pressure). Keep these two jobs separate and the rest of hydraulics falls into place.

The principle

The positive-displacement principle

A hydraulic pump traps a fixed volume of oil and pushes it out — over and over, once per chamber per revolution. Two check valves keep the oil going one way:

Suction stroke — the piston draws back, the inlet valve opens, oil is sucked from the tank.
Delivery stroke — the piston pushes in, the outlet valve opens, oil is forced to the system.

Because each turn moves a set amount of oil, the flow is fixed by the pump's size and its speed — nothing else.

A single-piston pump: it sucks oil from the tank, then pushes it to the system (after Rabie Fig. 4.1).

Hydraulics uses positive-displacement pumps (which trap and push a set volume) because they hold a steady flow even against very high pressure. The other family — rotodynamic pumps, which just spin the oil faster (like a water pump) — give big flow but little pressure, so they are not used to drive hydraulic loads.

Displacement — the oil moved per turn

The displacement $V_g$ is the volume of oil the pump delivers in one revolution of its shaft. It is purely geometric — set by the chamber size and how many chambers there are — so it is also called the geometric volume. A bigger pump has a bigger $V_g$.

$$V_g = (V_{\max} - V_{\min})\,z\,i$$

where:
Symbol	Meaning	SI unit
$V_g$	displacement — oil delivered per revolution	m³/rev
$V_{\max}-V_{\min}$	swept volume of one chamber (full minus empty)	m³
$z$	number of pumping chambers	—
$i$	pumping strokes per revolution	—

You will almost always be given $V_g$ on a pump's data sheet (in cm³/rev). The formula above just shows what it is built from.

The headline relation

Theoretical flow rate

Move $V_g$ of oil every revolution and spin the shaft $n$ times a second, and the ideal flow is simply the two multiplied:

$$Q_t = V_g\,n$$

where:
Symbol	Meaning	SI unit
$Q_t$	theoretical (ideal) flow rate	m³/s
$V_g$	displacement	m³/rev
$n$	pump speed	rev/s

Watch the units. Catalogues quote $V_g$ in cm³/rev and speed in rev/min. Convert first: $1~\text{cm}^3 = 10^{-6}~\text{m}^3$, and $n[\text{rev/s}] = n[\text{rev/min}]\div 60$. Then $Q_t$ comes out in m³/s ($\times 60{,}000$ for L/min).

The real flow is a little less — leakage

A real pump delivers less than $Q_t$. The main reason is internal leakage: some oil slips back through the tiny clearances inside the pump, from the high-pressure side to the low-pressure side, and never reaches the system.

Leakage grows with pressure (the harder you squeeze, the more slips back), with thinner oil (low viscosity), and steeply with wear (it rises with the cube of the clearance). So the actual flow drops as the pressure climbs.

Actual flow falls as pressure rises; the gap is the leakage (after Rabie Fig. 4.5).

We capture this with the volumetric efficiency $\eta_v$ — the fraction of the ideal flow that actually survives to the outlet:

$$\eta_v = \dfrac{Q}{Q_t} \qquad\Longrightarrow\qquad Q = V_g\,n\,\eta_v$$

where:
Symbol	Meaning	SI unit
$\eta_v$	volumetric efficiency — how much flow survives the leakage	— (0–1)
$Q$	actual delivered flow	m³/s
$Q_t$	theoretical flow, $V_g\,n$	m³/s

Pump type	Typical $\eta_v$
Piston pumps	high — about 0.95–0.99
Gear & vane pumps	lower — about 0.8–0.9

Displacement pumps overall run $\eta_v \approx 0.8\!-\!0.99$. The tighter the seal (piston pumps), the less leaks back, the higher the efficiency.

✏️ Try it yourself

A pump has $V_g = 16~\text{cm}^3/\text{rev}$ and runs at $1500~\text{rev/min}$.

Find the theoretical flow $Q_t$ in L/min.
If its volumetric efficiency is $\eta_v = 0.90$, what is the actual flow?
Does that lost flow depend on the load pressure? Why?

1. $n = 1500/60 = 25~\text{rev/s}$; $Q_t = 16\times10^{-6}\times25 = 4\times10^{-4}~\text{m}^3/\text{s} = \mathbf{24~\text{L/min}}$. 2. $Q = Q_t\,\eta_v = 24\times0.90 = \mathbf{21.6~\text{L/min}}$. 3. Yes — the lost flow is leakage, which rises with pressure. At higher load pressure more oil slips back, so $\eta_v$ (and the delivered flow) falls.

Common mistakes to avoid

Mistake	Fix
Leaving speed in rev/min	Convert: $n[\text{rev/s}] = n[\text{rev/min}]\div 60$.
Leaving $V_g$ in cm³/rev	Use m³/rev: $1~\text{cm}^3 = 10^{-6}~\text{m}^3$.
Treating $Q_t$ as the delivered flow	Real flow is smaller: $Q = Q_t\,\eta_v$.
Saying "the pump makes the pressure"	The pump makes flow; the load makes the pressure.

Recap — the whole topic on one screen

$$Q_t = V_g\,n \qquad \eta_v = \dfrac{Q}{Q_t} \qquad Q = V_g\,n\,\eta_v$$

Idea	What you own now
The pump's job	It makes flow; the load makes the pressure
Displacement	$V_g$ = oil moved per revolution (geometric volume)
Ideal flow	$Q_t = V_g\,n$ — size × speed
Real flow	$Q = V_g\,n\,\eta_v$; leakage rises with pressure

Next topic

Pump efficiency, torque & power

We have the flow. Now: how much torque must the motor twist to push that oil against pressure, and how much drive power does the pump need once all the losses are counted?

→ Pump efficiency, torque & power

Symbol	Meaning	SI unit
\(V_g\)	displacement — oil delivered per revolution	m³/rev
\(V_{\max}-V_{\min}\)	swept volume of one chamber (full minus empty)	m³
\(z\)	number of pumping chambers	—
\(i\)	pumping strokes per revolution	—

Symbol	Meaning	SI unit
\(Q_t\)	theoretical (ideal) flow rate	m³/s
\(V_g\)	displacement	m³/rev
\(n\)	pump speed	rev/s

Symbol	Meaning	SI unit
\(\eta_v\)	volumetric efficiency — how much flow survives the leakage	— (0–1)
\(Q\)	actual delivered flow	m³/s
\(Q_t\)	theoretical flow, \(V_g\,n\)	m³/s

Mistake	Fix
Leaving speed in rev/min	Convert: \(n[\text{rev/s}] = n[\text{rev/min}]\div 60\).
Leaving \(V_g\) in cm³/rev	Use m³/rev: \(1~\text{cm}^3 = 10^{-6}~\text{m}^3\).
Treating \(Q_t\) as the delivered flow	Real flow is smaller: \(Q = Q_t\,\eta_v\).
Saying "the pump makes the pressure"	The pump makes flow; the load makes the pressure.