Axial Piston Pump Calculation

Axial Piston Pump Calculation: Displacement, Flow, Pressure, Power & Torque

A Practical Engineering Guide to Sizing an Axial Piston Pump

How do you determine whether an axial piston pump is correctly sized for a hydraulic system?

The answer starts with a few fundamental parameters:

Displacement, speed, pressure, flow and efficiency.

But these parameters cannot be evaluated independently.

Pump displacement determines how much fluid is theoretically moved during each revolution. Pump speed determines how frequently that displacement occurs. Volumetric efficiency determines how much of the theoretical flow actually reaches the hydraulic circuit, while pressure determines the hydraulic power being transmitted.

Once these values are combined, we can calculate the pump’s actual flow, hydraulic output power, required shaft power and input torque.

This article works through a complete numerical example using an axial piston pump and explains what each calculation means from an engineering perspective.

Important: This is a preliminary engineering calculation intended to explain pump sizing principles. A production hydraulic system requires additional analysis of the specific pump model, inlet conditions, pressure limits, speed ratings, temperature, fluid viscosity, control system, duty cycle and manufacturer specifications.

⸻

1. The Pump We Are Going to Calculate

Let’s consider a hypothetical axial piston pump with the following operating conditions:

Parameter Value
Pump displacement 45 cm³/rev
Shaft speed 1,800 rpm
Differential pressure 280 bar
Volumetric efficiency 95%
Mechanical efficiency 90%

The pressure used in the calculations should technically be the differential pressure across the pump:

[
\Delta p=p_{out}-p_{in}
]

For simplicity, we will use:

[
\boxed{\Delta p=280\ bar}
]

The overall efficiency is:

[
\eta_t=\eta_v\times\eta_m
]

Therefore:

[
\eta_t=0.95\times0.90
]

[
\boxed{\eta_t=0.855}
]

or:

[
\boxed{\eta_t=85.5%}
]

Danfoss defines overall efficiency in the same way, as volumetric efficiency multiplied by mechanical efficiency. (assets.danfoss.com)

⸻

2. What Does 45 cm³/rev Actually Mean?

The first number we need to understand is pump displacement.

A displacement of:

[
45\ cm^3/rev
]

means that, theoretically, the pump moves 45 cm³ of hydraulic fluid for every shaft revolution.

This does not mean that the pump will necessarily deliver 45 cm³/rev to the hydraulic circuit.

Internal leakage and other losses mean that the actual delivered volume is lower.

This is where volumetric efficiency becomes important.

⸻

3. Theoretical Pump Flow

The theoretical flow of a fixed-displacement pump can be calculated from:

[
Q_{th}=\frac{V_g\times n}{1000}
]

where:

* (Q_{th}) = theoretical flow in L/min
* (V_g) = displacement in cm³/rev
* (n) = rotational speed in rpm

For our pump:

[
Q_{th}=\frac{45\times1800}{1000}
]

[
\boxed{Q_{th}=81\ L/min}
]

So, if the pump had 100% volumetric efficiency, it would theoretically deliver:

81 L/min

at 1,800 rpm.

This is the value you can obtain directly from displacement and speed.

But real hydraulic pumps do not operate at 100% volumetric efficiency.

⸻

4. Actual Pump Flow

Now we introduce volumetric efficiency:

[
Q=Q_{th}\times\eta_v
]

Therefore:

[
Q=81\times0.95
]

[
\boxed{Q=76.95\ L/min}
]

So our pump produces approximately:

76.95 L/min actual flow

under the assumed operating conditions.

This is a critical distinction when reading hydraulic pump specifications.

A pump described as 45 cm³/rev does not automatically mean that the machine will receive 81 L/min at 1,800 rpm.

The actual flow depends on operating conditions and pump efficiency.

Parker and Danfoss both use the same basic relationship for pump output flow:

[
Q=\frac{V_g\times n\times\eta_v}{1000}
]

(Parker Hannifin Corporation)

⸻

5. What Happens When Pump Speed Changes?

Because displacement is expressed in volume per revolution, pump speed has a direct effect on theoretical flow.

Let’s increase the pump speed from:

1,800 rpm → 2,100 rpm

The new theoretical flow becomes:

[
Q_{th}=\frac{45\times2100}{1000}
]

[
\boxed{Q_{th}=94.5\ L/min}
]

At the same assumed 95% volumetric efficiency:

[
Q=94.5\times0.95
]

[
\boxed{Q\approx89.78\ L/min}
]

So increasing speed by approximately 16.7% increases theoretical flow by approximately the same proportion.

However, this does not mean that a pump can simply be operated at any higher speed.

The manufacturer’s maximum speed, inlet conditions, fluid viscosity, temperature and pump design must all be respected.

For example, current Danfoss axial piston pump specifications show that maximum allowable speed varies significantly with displacement and pump series. (PowerSource)

⸻

6. Hydraulic Power

Now we move from flow to power.

Hydraulic power represents the rate at which energy is being transferred to the hydraulic fluid.

For metric units:

[
P_h=\frac{Q\times\Delta p}{600}
]

where:

* (P_h) = hydraulic power in kW
* (Q) = flow in L/min
* (\Delta p) = pressure difference in bar

Using our actual flow:

[
P_h=\frac{76.95\times280}{600}
]

[
\boxed{P_h=35.91\ kW}
]

Therefore, under these assumptions, the pump delivers approximately:

35.9 kW of hydraulic power

to the hydraulic system.

This is the power actually represented by the combination of flow and pressure.

⸻

7. Why Isn’t the Required Motor Power 35.9 kW?

This is one of the most important questions in hydraulic pump sizing.

If the hydraulic output is 35.9 kW, it would be incorrect to simply select a 35.9 kW prime mover and assume everything is fine.

The pump itself has losses.

We calculated the overall efficiency as:

[
\eta_t=85.5%
]

Therefore:

[
P_{in}=\frac{P_h}{\eta_t}
]

[
P_{in}=\frac{35.91}{0.855}
]

[
\boxed{P_{in}\approx42.0\ kW}
]

The pump therefore requires approximately:

42.0 kW shaft input power

to produce approximately:

35.9 kW hydraulic output power

under the assumed conditions.

Danfoss gives the same relationship in its pump sizing documentation:

[
P_e=\frac{Q_e\times\Delta p}{600\times\eta_t}
]

(assets.danfoss.com)

⸻

8. Where Does the Lost Power Go?

The difference is:

[
42.0-35.9
]

[
\boxed{\approx6.1\ kW}
]

This approximately 6.1 kW is not simply “disappearing.”

It represents energy converted into losses within the hydraulic system and pump.

These losses can include:

* Internal leakage
* Friction
* Bearing losses
* Piston/slipper losses
* Valve plate losses
* Mechanical friction
* Fluid shear
* Other internal losses

Ultimately, much of this energy appears as heat.

This is why efficiency is not merely a specification to compare in a catalog.

It directly affects:

required engine/motor power + heat generation + fuel/energy consumption.

⸻

9. Pump Shaft Torque

We can now calculate the torque required at the pump shaft.

The metric relationship is:

[
T=\frac{V_g\times\Delta p}
{20\pi\times\eta_m}
]

For our pump:

[
T=
\frac{45\times280}
{20\pi\times0.90}
]

[
\boxed{T\approx222.8\ Nm}
]

Therefore, approximately:

223 Nm shaft torque

is required under the assumed operating conditions.

This is an important result because the pump must be mechanically compatible with its prime mover.

The engine, electric motor or gearbox driving the pump must be capable of supplying the required torque at the required speed.

Parker and Danfoss publish the same fundamental pump input torque relationship. (Parker Hannifin Corporation)

⸻

10. Check the Power Calculation Another Way

We can verify our result using shaft torque and rotational speed.

Mechanical power is:

[
P=\frac{T\times n\times2\pi}{60000}
]

Using:

[
T=222.8\ Nm
]

and:

[
n=1800\ rpm
]

we obtain:

[
P=\frac{222.8\times1800\times2\pi}{60000}
]

[
\boxed{P\approx42.0\ kW}
]

The two methods produce essentially the same result.

This type of cross-check is useful in engineering calculations because it helps identify unit or formula errors before they reach the design stage.

⸻

11. What Happens if Pressure Increases?

Now let’s keep the same pump and speed but increase the differential pressure.

Suppose:

[
\Delta p=320\ bar
]

The actual flow may remain approximately similar for this simplified example:

[
Q=76.95\ L/min
]

Hydraulic power becomes:

[
P_h=\frac{76.95\times320}{600}
]

[
\boxed{P_h\approx41.04\ kW}
]

At the same assumed overall efficiency:

[
P_{in}=\frac{41.04}{0.855}
]

[
\boxed{P_{in}\approx48.0\ kW}
]

So increasing pressure from 280 bar to 320 bar increases the required input power substantially.

This demonstrates an important hydraulic principle:

Flow and pressure together determine hydraulic power.

Increasing either one increases the power requirement.

⸻

12. Pressure Does Not Simply “Come From the Pump”

A common oversimplification is to say that a hydraulic pump “creates pressure.”

More precisely, the pump supplies flow, while system resistance creates the pressure required to move that flow through the hydraulic circuit.

For example, if the pump supplies 77 L/min into a circuit with very little resistance, pressure may remain relatively low.

If the same pump supplies the same flow against a high load, system pressure rises.

This distinction becomes fundamental when analyzing:

* Hydraulic cylinders
* Hydraulic motors
* Control valves
* Relief valves
* Load-sensing systems
* Mobile hydraulic circuits

The pump is therefore best understood as a flow source whose required pressure is determined by the system load and resistance.

⸻

13. Connecting Pump Flow to a Hydraulic Cylinder

Now let’s connect our pump calculation to an actual machine function.

Assume the pump supplies:

[
Q=76.95\ L/min
]

to a hydraulic cylinder with a piston diameter of:

[
D=100\ mm
]

The piston area is:

[
A=\frac{\pi D^2}{4}
]

[
A=\frac{\pi\times100^2}{4}
]

[
\boxed{A=7854\ mm^2}
]

or:

[
A=0.007854\ m^2
]

The cylinder speed is approximately:

[
v=\frac{Q}{A}
]

After converting the flow to cubic metres per second:

[
Q=76.95\ L/min
]

[
Q=0.0012825\ m^3/s
]

Therefore:

[
v=\frac{0.0012825}{0.007854}
]

[
\boxed{v\approx0.163\ m/s}
]

So our 45 cm³/rev pump operating at 1,800 rpm and delivering approximately 76.95 L/min would move a 100 mm bore cylinder at roughly:

163 mm/s

under these simplified conditions.

Now the pump calculation has become a machine-performance calculation.

⸻

14. Cylinder Force at 280 bar

The theoretical extension force of the cylinder is:

[
F=P\times A
]

With:

[
P=280\ bar=28,000,000\ Pa
]

and:

[
A=0.007854\ m^2
]

we obtain:

[
F=28,000,000\times0.007854
]

[
\boxed{F\approx219,900\ N}
]

or approximately:

[
\boxed{F\approx220\ kN}
]

This is the theoretical hydraulic force before considering cylinder efficiency, friction and other system losses.

We have therefore connected four different engineering quantities:

[
\boxed{\text{Pump displacement}}
]

↓

[
\boxed{\text{Flow}}
]

↓

[
\boxed{\text{Cylinder speed}}
]

and:

[
\boxed{\text{Pressure}}
]

↓

[
\boxed{\text{Cylinder force}}
]

This is why pump sizing cannot be separated from the machine function it is expected to perform.

⸻

15. Complete Example Summary

For our simplified example:

Parameter Result
Pump displacement 45 cm³/rev
Speed 1,800 rpm
Differential pressure 280 bar
Volumetric efficiency 95%
Mechanical efficiency 90%
Overall efficiency 85.5%
Theoretical flow 81.0 L/min
Actual flow 76.95 L/min
Hydraulic power 35.91 kW
Required shaft power ≈42.0 kW
Shaft torque ≈222.8 Nm
100 mm cylinder speed ≈0.163 m/s
100 mm cylinder theoretical force ≈220 kN

These values are based on the simplified assumptions stated in the example and should not be interpreted as the rated performance of a particular commercial pump.

⸻

16. What This Calculation Does Not Tell Us

A numerical calculation is powerful, but it is not a substitute for the pump manufacturer’s technical data.

Before selecting a real axial piston pump, the following must also be checked:

Pump limitations

* Maximum continuous pressure
* Maximum intermittent pressure
* Maximum and minimum speed
* Maximum displacement
* Inlet pressure
* Case pressure
* Case drain requirements

Hydraulic conditions

* Oil viscosity
* Oil temperature
* Filtration
* Contamination level
* Reservoir conditions
* Suction line sizing
* Aeration
* Cavitation risk

Control system

* Fixed or variable displacement
* Pressure compensation
* Load sensing
* Electronic control
* Swash plate control
* Displacement control range

Machine requirements

* Required actuator speed
* Required actuator force
* Duty cycle
* Peak load
* Continuous load
* Available engine/motor power

For example, a current Danfoss H1 axial piston pump range includes different displacement classes, pressure ratings and maximum speeds, demonstrating why the calculated operating point must always be checked against the specific pump model’s limits. (PowerSource)

⸻

17. The Engineering Lesson

The most important lesson is not the final number.

It is the relationship between the variables.

Displacement determines volume per revolution.

[
D\rightarrow V/rev
]

Speed determines how many revolutions occur per minute.

[
D+n\rightarrow Q
]

Volumetric efficiency determines how much of the theoretical flow is actually delivered.

[
Q_{th}\times\eta_v\rightarrow Q
]

Pressure and flow determine hydraulic power.

[
Q+\Delta p\rightarrow P_h
]

Efficiency determines the required input power.

[
P_h\div\eta_t\rightarrow P_{in}
]

Pressure and displacement determine shaft torque.

[
D+\Delta p+\eta_m\rightarrow T
]

And finally:

[
\boxed{\text{Pump}\rightarrow\text{Flow}\rightarrow\text{Actuator Speed}}
]

[
\boxed{\text{Pressure}\rightarrow\text{Force}}
]

[
\boxed{\text{Flow}\times\text{Pressure}\rightarrow\text{Hydraulic Power}}
]

This is the engineering chain that connects a hydraulic pump specification to actual machine performance.

⸻

The Key Takeaway

An axial piston pump is more than a displacement number printed on a nameplate.

A 45 cm³/rev pump at 1,800 rpm theoretically represents 81 L/min. Once volumetric efficiency is considered, the actual flow in our example becomes approximately 76.95 L/min. At 280 bar, that flow represents approximately 35.9 kW of hydraulic power, while the pump requires approximately 42 kW of shaft input power under the assumed efficiencies.

The important point is that these numbers are connected.

Displacement determines flow potential.
Speed determines how quickly that displacement is delivered.
Pressure determines the work that can be performed.
Efficiency determines the power required to achieve it.

For construction machinery and industrial hydraulic systems, these parameters must be evaluated together.

At Shop & Supply, we approach hydraulic components from the same engineering perspective. Selecting the correct pump is not simply about matching a part number, displacement or pressure rating. It means understanding the application’s required flow, pressure, speed, torque, efficiency and duty cycle — and then determining whether the component is technically suited to the job.

The right part is more than the right number — it is the right engineering solution for the application.

Shop & Supply — Right Part. Right Knowledge. Stronger Performance.

⸻

Technical references

The fundamental pump sizing equations used in this article are consistent with published technical documentation from major hydraulic manufacturers, including Parker Hannifin and Danfoss.

* Parker — Axial Piston Pumps: Conversions & Fluid Power Formulas
* Danfoss — H1 Axial Piston Pumps: Basic Information & Sizing Equations
* Danfoss — MP1 Axial Piston Pumps: System Design Parameters