Modular pump drives
A modular pump drive approach provides thousands of choices for gear ratios, pump adapters, and mounting options. With so many combinations, most machine manufacturers can find a pump drive that meets the configuration and reliability needs of its applications.
For example, John Deere’s Funk brand of pump drives come in 5000 variations when you consider the number of models, each one of which can have five or six different spline configurations and use an SAE A, B, C, or D output pad. Also, each of these configurations can be driven through a planetary gear drive, by a flange or shaft power takeoff (PTO) input, or it can have a clutch in front of it. These configurations are just for the pump drives and do not even take pump features into account.
However, the modular drive does not necessarily have to power a hydraulic pump. An alternative approach is to use one or more of the drive’s outputs as a mechanical PTO. Some equipment builders use one pump drive to power another using the drive shaft of the first pump drive. Some of Deere’s pump drives have PTO outputs available to power another pump drive, or that PTO could power other equipment directly, such as an air compressor or generator.
The flexibility of a modular pump drive allows equipment manufacturers to match pump drive speed to the optimum operating speed of an engine to achieve the maximum performance and fuel economy.
The flexibility of a modular pump drive also lends itself well to hybrid drivetrain systems because modular pump drives are efficient at routing and using all of an engine’s power.
Selection by torque
The most reliable technique for selecting a pump drive is the torque method. Suggested here is a step-bystep procedure:
1. Determine the net peak torque (lb-ft or N-m) transmitted from the engine. This will serve as the maximum input torque to the pump drive. Remember to deduct any continuous parasitic losses.
2. Determine the maximum input torque (again, in lb-ft or N-m) required to drive each and every hydraulic pump attached to the pump drive. If a pump will be driven at a speed other than the engine’s flywheel speed, be sure to consider the drive ratio in your calculations. A drive ratio other than 1:1 may be chosen so the engine can run at a speed for optimum fuel economy — say, 1800 to 2200 rpm — while the pumps may run most effectively at, say 2200 to 2700 rpm.
3. Using the lesser of the torque values calculated above, select a pump drive series with a maximum input torque capacity that exceeds the torque requirements of the pumps.
Sometimes you can take a shortcut. If the engine’s peak torque is less than the pump drive’s peak input torque, you don’t need to go any further because the engine cannot overpower the pump drive. A smaller pump drive could be used if the engine’s peak torque rating isn’t oversized to increase reliability. Additionally, the torque rating might be oversized because the engine is also driving some other equipment. If that’s the case, you can proceed to step two.
Step two helps determine the sum of the load of all the hydraulic pumps that will be mounted to the pump drive, whether it’s one pump or nine pumps. If the sum of the load required from all the pumps is lower than the engine peak load, then that’s the torque needed to size the pump drive.
For example, say an engine’s net peak torque is 825 lb-ft and has the following requirements:
• torque required at Pump A is 225 lb-ft,
• torque required at Pump B is 250 lb-ft,
• torque required at Pump C is 143 lb-ft, and
• the engine will run at 2000 rpm while the pumps will be driven at 2390 rpm.
This means the sum of all pump torques is 618 lb-ft, and the pump drive ratio is 0.836:1 (speed increase). Dividing 618 by 0.836 indicates that the effective input shaft torque required is 739.2 lb-ft, meaning a pump drive with a torque capacity above 740 lb-ft is needed.