ME3311 · Hydraulic & Pneumatic

Theme 4 · Pumps

Pump efficiency, torque & power

To push oil against pressure, the motor must twist the pump shaft. Here we find that torque, the three efficiencies that nibble at the pump, and the drive power you must install.

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

Before you start

What you need first

Displacement & flow — $V_g$, $Q_t = V_g\,n$, volumetric efficiency $\eta_v$ (Topic 19).
Hydraulic power — $N = p\,Q$ (Topic 4).

What you'll be able to do

Find the driving torque $T = \tfrac{V_g\,\Delta P}{2\pi}$.
Use the three efficiencies $\eta_v,\eta_m,\eta_h$ and the overall $\eta_T$.
Size the drive power $N_{in} = \tfrac{pQ}{\eta_T}$, and avoid cavitation & pulsation.

Start here · recall

The power the pump delivers

The useful (hydraulic) power leaving the pump is the same pressure × flow you met in Topic 4 — here the pressure is the rise the pump produces and the flow is what it actually delivers:

$$N_h = p\,Q$$

That is the output. The motor must supply more than this, because torque and power are lost on the way through. The rest of this topic accounts for those losses.

The torque to drive the pump

In one revolution the shaft does work $2\pi T$, and that work goes into raising the pressure of a volume $V_g$ of oil by $\Delta P$ (work $= V_g\,\Delta P$). Equating the two gives the ideal driving torque:

$$T_t = \dfrac{V_g\,\Delta P}{2\pi}$$

It says exactly what intuition expects: to push oil against a higher pressure you must twist harder, and a bigger pump (larger $V_g$) needs more torque for the same pressure.

where:
Symbol	Meaning	SI unit
$T_t$	theoretical (ideal) driving torque	N·m
$V_g$	displacement	m³/rev
$\Delta P$	pressure rise across the pump, $P - P_i$	Pa

The inlet pressure $P_i$ is usually tiny next to the delivery pressure, so in practice $\Delta P \approx P$.

Three losses → three efficiencies

A real pump loses a little of each thing it handles. Each loss has its own efficiency:

Efficiency	What it measures	Definition
Volumetric $\eta_v$	lost flow (leakage)	$\eta_v = Q/Q_t$
Mechanical $\eta_m$	lost torque (friction)	$\eta_m = (T-T_F)/T$
Hydraulic $\eta_h$	lost pressure inside the pump	$\eta_h = P/P_C$

Here $T_F$ is the torque eaten by friction, and $P_C$ is the pressure built inside the pumping chamber (a little above the exit pressure $P$). The hydraulic loss is negligible at normal speeds (below ~50 rev/s, oil speed below ~5 m/s), so $\eta_h\approx 1$ for most problems. Multiply the three to get the overall efficiency:

$$\eta_T = \eta_v\,\eta_m\,\eta_h$$

A good displacement pump reaches $\eta_T \approx 0.85\!-\!0.90$ overall. Because each $\eta$ is below 1, the product is always smaller than any one of them.

Real driving torque and drive power

Friction and internal pressure loss make the real torque larger than the ideal one:

$$T = \dfrac{V_g\,\Delta P}{2\pi\,\eta_m\,\eta_h} \;\approx\; \dfrac{V_g\,P}{2\pi\,\eta_m\,\eta_h}$$

And the drive (input mechanical) power the motor must supply is the useful hydraulic power divided by the overall efficiency:

$$N_{in} = \dfrac{N_h}{\eta_T} = \dfrac{p\,Q}{\eta_T}$$

where:
Symbol	Meaning	SI unit
$N_{in}$	drive power the motor must supply (input)	W
$N_h = pQ$	useful hydraulic power out of the pump	W
$\eta_T$	overall pump efficiency	— (0–1)

Dividing by $\eta_T$ (a number below 1) makes $N_{in}$ larger than the output $pQ$. The difference is power turned into heat inside the pump — which is why hydraulic systems need cooling.

Watch out: cavitation

If the pump inlet pressure falls below the oil's vapour pressure, vapour bubbles form and then collapse violently inside the pump (the cavitation you met in Topic 12). The result: a loud sharp noise, falling flow (volumetric efficiency drops ~1%), and pitting damage to the pump's surfaces from impact pressures that can reach thousands of bar.

Keep the inlet pressure above the vapour pressure by a small margin — the cavitation reserve, about 0.3–0.4 bar. You do this by:

a short, wide suction line, and no (or a clean) inlet filter — less inlet loss;
mounting the pump below the tank (flooded suction);
using a booster pump or a pressurised tank.

Watch out: flow pulsation

The chambers deliver one after another, so the flow is not perfectly steady — it ripples around its mean value, causing small pressure ripples and vibration. We measure the ripple with the flow-pulsation coefficient:

$$\sigma_Q = \dfrac{Q_{\max}-Q_{\min}}{Q_m}\times100\%$$

Pumps with an odd number of pistons (3, 5, 7…) pulsate the least — a small but real reason odd piston counts are common.

Flow ripples around the mean (after Rabie Fig. 4.9).

✏️ Try it yourself

A pump delivers $Q = 30~\text{L/min}$ at $p = 120~\text{bar}$ with overall efficiency $\eta_T = 0.85$.

Find the useful hydraulic power $N_h$.
Find the drive power $N_{in}$ the motor must supply.
Where does the difference go?

1. $Q = 30/60{,}000 = 5\times10^{-4}~\text{m}^3/\text{s}$, $p = 1.2\times10^{7}~\text{Pa}$; $N_h = pQ = \mathbf{6000~\text{W}}$. 2. $N_{in} = N_h/\eta_T = 6000/0.85 \approx \mathbf{7.1~\text{kW}}$. 3. The extra ~1.1 kW is lost to leakage, friction and internal pressure loss — it becomes heat in the oil.

Common mistakes to avoid

Mistake	Fix
Dividing $pQ$ by $\eta_T$ wrong way	Drive power is bigger: $N_{in}=pQ/\eta_T$ (input > output).
Adding the efficiencies	Multiply them: $\eta_T = \eta_v\eta_m\eta_h$.
Using $P$ when $\Delta P$ is meant	Torque uses the pressure rise $\Delta P = P-P_i$ (≈ $P$ if inlet is small).
Pressure in bar / flow in L/min	Convert to Pa and m³/s before $N=pQ$.

Recap — the whole topic on one screen

$$T = \dfrac{V_g\,\Delta P}{2\pi\,\eta_m\,\eta_h} \qquad \eta_T = \eta_v\,\eta_m\,\eta_h \qquad N_{in} = \dfrac{pQ}{\eta_T}$$

Idea	What you own now
Driving torque	$T_t = V_g\Delta P/2\pi$; higher pressure or bigger pump → more torque
Three efficiencies	$\eta_v$ flow, $\eta_m$ torque, $\eta_h$ pressure; $\eta_T = \eta_v\eta_m\eta_h$
Drive power	$N_{in} = pQ/\eta_T$ — always bigger than the output $pQ$
Watch out	Cavitation (low inlet) and pulsation (uneven delivery)

Next topic

Pump types

We have treated the pump as a black box that makes flow and needs power. Next we open it up: gear, vane and piston pumps — how each traps the oil, and what each is good and bad at.

→ Pump types (gear, vane, piston)

Mistake	Fix
Dividing \(pQ\) by \(\eta_T\) wrong way	Drive power is bigger: \(N_{in}=pQ/\eta_T\) (input > output).
Adding the efficiencies	Multiply them: \(\eta_T = \eta_v\eta_m\eta_h\).
Using \(P\) when \(\Delta P\) is meant	Torque uses the pressure rise \(\Delta P = P-P_i\) (≈ \(P\) if inlet is small).
Pressure in bar / flow in L/min	Convert to Pa and m³/s before \(N=pQ\).

Symbol	Meaning	SI unit
\(T_t\)	theoretical (ideal) driving torque	N·m
\(V_g\)	displacement	m³/rev
\(\Delta P\)	pressure rise across the pump, \(P - P_i\)	Pa

Efficiency	What it measures	Definition
Volumetric \(\eta_v\)	lost flow (leakage)	\(\eta_v = Q/Q_t\)
Mechanical \(\eta_m\)	lost torque (friction)	\(\eta_m = (T-T_F)/T\)
Hydraulic \(\eta_h\)	lost pressure inside the pump	\(\eta_h = P/P_C\)

Symbol	Meaning	SI unit
\(N_{in}\)	drive power the motor must supply (input)	W
\(N_h = pQ\)	useful hydraulic power out of the pump	W
\(\eta_T\)	overall pump efficiency	— (0–1)

Idea	What you own now
Driving torque	\(T_t = V_g\Delta P/2\pi\); higher pressure or bigger pump → more torque
Three efficiencies	\(\eta_v\) flow, \(\eta_m\) torque, \(\eta_h\) pressure; \(\eta_T = \eta_v\eta_m\eta_h\)
Drive power	\(N_{in} = pQ/\eta_T\) — always bigger than the output \(pQ\)
Watch out	Cavitation (low inlet) and pulsation (uneven delivery)