How to Select The Proper Hydraulic Pump and Electric Motor for Your System 
The matching of hydraulic pumps and motors is extremely critical in hydraulic systems. Excellent hydraulic system depends on efficient and stable hydraulic pump and motor combination.
When properly configuring hydraulic pumps and motors, a number of factors need to be carefully considered.
Main Formula
The selection process involves determining the hydraulic pump specifications and calculating the motor power based on the working pressure and flow rate,
following the formula
P = (p × Q) / 60,
where P represents the motor power (kW), p is the working pressure of the hydraulic pump (MPa), and Q is the output flow rate of the hydraulic pump (L/min). This process is demonstrated in the following example.
Example Analysis:
It is known that the hydraulic pump displacement of 136 ml/rev, the motor speed of 970 r / min, the maximum working pressure of the system is 12 MPa. We calculate the output flow of the hydraulic pump and the required motor power.
- Calculate the output flow rate of the hydraulic pump:Pump output flow rate Q = Displacement q × Speed n = 136 ml/rev × 970 r/min = 131920 ml/min = 131.92 L/min.
- Calculate the power required by the system: First, calculate the theoretical power N = (p × Q) / 60 = (12 MPa × 131.92 L/min) / 60 ≈ 26.38 kW based on the maximum working pressure of the system and the output flow rate of the pump.
- In practice, considering the operating efficiency of the hydraulic pump (usually between 70% and 85%), the motor should be selected slightly larger to ensure stable operation of the system at full load. Assuming a safety factor of 1.15, the required motor power ND = 26.38 kW × 1.15 ≈ 30.33 kW.
- Selection of motor: According to the calculation, a motor close to 30.33 kW is selected, and a 30 kW motor is a more appropriate choice. For actual selection, it is also necessary to refer to the motor manual to ensure that the rated power, speed, voltage and other parameters of the motor meet the system requirements.
Points of Attention in Calculation
- In the calculation, it should be based on the actual work of the hydraulic system flow and pressure, to ensure that the motor selection close to the actual working conditions, in order to avoid too large a waste of energy or too small to affect the performance of the system.
- Another formula, P = Q × p / 612, is also valid, but note that the pressure unit in this formula is kgf/cm². When converting, 1 MPa ≈ 10.197 kgf/cm², make sure the units are the same before calculating, the result will be similar to the previous formula, but need to be adjusted to match the different pressure units.
How to calculate hydraulic cylinder pressure: thrust and tension?
Main Formula
The force formula: F = PS (P: pressure; S: pressurized area)
From the above formula, it can be seen that the force generated is different due to the different pressurized area of the cylinder when making push and pull, i.e.:
Pushing force F1 = P×π(D/2)² = P×π/4*D²
Pulling force F2 = P×π[(D/2)²-(d/2)²] = P×π/4* (D²-d²)
(φD: cylinder bore; d: piston rod diameter)
In practical applications, it is also necessary to add a loading rate β. Because the force generated by the cylinder will not be used 100% for pushing or pulling, β is often selected 0.8, so the formula becomes:
Pushing force F1 = 0.8 x P x π/4 x D²
Pulling force F2 = 0.8×P×π/4×(D²-d²)
From the above formula, we can see that as long as we know the inner diameter of the cylinder φD and piston diameter φd and the pressure P (generally a constant), we can calculate the force that can be generated by this type of cylinder.
Example analysis:
Commonly used standard column hydraulic cylinder P value can withstand pressure to 140kgf/cm2.
Assumptions: cylinder bore D = 100mm live rod diameter d = 56mm. note that the diameter of the unit calculation needs to be reduced to cm.
Then:
Thrust F1 = P×πD²/4×0.8 = 140×π×10²/4×0.8 ≈ 8796(kgf);
Tension F2 = P×π(D²-d²)/4×0.8 = 140×π(10²-5.6²)×0.8 ≈ 6037(kgf)
Data tables for quick reference
The following table shows operation pressures according to the diameter of the cylinder
