Hydraulic power

The third master relation, and the one that ties the first two together. Pressure times flow is power — and it comes out exactly equal to force times speed.

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

Before you start

What you need first

Force from pressure — $F=pA$ (Topic 2).
Speed from flow — $v=\tfrac{Q}{A}$ (Topic 3).

What you'll be able to do

Use $N=pQ$ to find hydraulic power.
Show why $N=pQ=Fv$ — the same power, two ways.
See the "effort × flow" pattern across all systems.
Work in watts and kilowatts with the right unit conversions.

Start here · the relation

Power = pressure × flow

Power is how fast work is done — and recall from Topic 2 that work means a force moved through a distance (force × distance). In a hydraulic line that work is done by oil pressure moving oil, so the power is the pressure times the flow rate:

$$N = p\,Q$$

where:
Symbol	Meaning	SI unit
N	hydraulic power	W
p	pressure	Pa
Q	flow rate	m³/s

The same power, two ways

The hydraulic power going into a cylinder equals the mechanical power coming out (for an ideal, loss-free cylinder):

$$N = p\,Q = F\,v$$

Why are they equal? Substitute the two relations you already know, $F=pA$ and $Q=Av$ (so $v=Q/A$):

$$F\,v = (p\,A)\!\left(\dfrac{Q}{A}\right) = p\,Q$$

The area $A$ cancels. That is the whole point of hydraulics: it carries power as pressure×flow, then hands it back as force×speed — losing nothing but the area.

"Effort × flow" — the same idea everywhere

Every power system multiplies an effort by a flow. Hydraulics is just one row of the same table:

System	Effort	Flow	Power
Mechanical (linear)	Force $F$ (N)	Speed $v$ (m/s)	$N=Fv$
Mechanical (rotary)	Torque $T$ (N·m)	Speed $\omega$ (rad/s)	$N=T\omega$
Electrical (DC)	Voltage $e$ (V)	Current $i$ (A)	$N=ei$
Hydraulic	Pressure $p$ (Pa)	Flow $Q$ (m³/s)	$N=pQ$

After Table 1.2 — the same idea wearing different clothes.

Units, and a word on real losses

With $p$ in pascals and $Q$ in m³/s, $N$ comes out in watts. Machines are usually quoted in kilowatts ($1~\text{kW}=1000~\text{W}$).

Convert before multiplying: bar → Pa (×10⁵) and L/min → m³/s (÷60,000). A quick shortcut for hydraulics: $N~[\text{kW}] \approx \dfrac{p~[\text{bar}]\times Q~[\text{L/min}]}{600}$.

🔭 Looking ahead: $N=pQ$ is the ideal power. Real pumps and motors lose some to leakage and friction, so the input is larger than the output. We handle those efficiencies in the pumps theme.

✏️ Try it yourself — no numbers needed

A hydraulic system runs at some pressure and flow. You then double the pressure but halve the flow. What happens to the hydraulic power — and what real change would do that?

Power is unchanged. $N=pQ$; doubling $p$ and halving $Q$ gives $(2p)(Q/2)=pQ$ — the same power. What does that: a heavier load raises the pressure, and to keep the same power the actuator simply moves slower (less flow). Same power, traded from "fast and gentle" to "slow and strong" — exactly like $N=Fv$.

Common mistakes to avoid

Mistake	Fix
Multiplying bar by L/min directly	Convert to Pa and m³/s first — or use the $\div 600$ shortcut for kW.
Thinking "more pressure = more power"	Power is $p\times Q$. High pressure at tiny flow can be low power.
Confusing ideal power with input power	$N=pQ$ is ideal; the pump's input is larger by its efficiency (later topic).

Recap — the three master relations

$$F = p\,A\qquad v=\dfrac{Q}{A} \qquad N = p\,Q = F\,v$$

Idea	What you own now
Pressure → force	$F=pA$ (Topic 2)
Flow → speed	$v=Q/A$ (Topic 3)
Pressure × flow → power	$N=pQ=Fv$ (this topic)
The big picture	Hydraulics carries power as $pQ$, returns it as $Fv$

Next topic

The basic hydraulic circuit & ISO symbols

You now own the three relations that govern every hydraulic machine. Next we meet the real hardware — the pump, valves, cylinder and tank — and learn to read the standard circuit that connects them.

→ The basic hydraulic circuit & ISO symbols

System	Effort	Flow	Power
Mechanical (linear)	Force \(F\) (N)	Speed \(v\) (m/s)	\(N=Fv\)
Mechanical (rotary)	Torque \(T\) (N·m)	Speed \(\omega\) (rad/s)	\(N=T\omega\)
Electrical (DC)	Voltage \(e\) (V)	Current \(i\) (A)	\(N=ei\)
Hydraulic	Pressure \(p\) (Pa)	Flow \(Q\) (m³/s)	\(N=pQ\)

Idea	What you own now
Pressure → force	\(F=pA\) (Topic 2)
Flow → speed	\(v=Q/A\) (Topic 3)
Pressure × flow → power	\(N=pQ=Fv\) (this topic)
The big picture	Hydraulics carries power as \(pQ\), returns it as \(Fv\)