I know there is no direct conversion for this as they are dissimilar units. One work the other force, but there must be a way to calculate this... If I have a 140 HP motor powering a single piston pump, what is the amount of pressure applied to the pump shaft, I want to remove the motor and replace it with a hydraulic cylinder to actuate the pump but do not know how much linear force I will need

I make a lot of assumptions in this response... First off, I assume that your pump is a standard crank / rod / piston configuration. If so I'm not sure how a hydraulic cylinder would be used in this application as it applies a linear load vs the motor's rotational torque input. But I'll assume you've figured this out with a mechanism attached to your hydraulic cylinder that converts it's linear force into a torque. The motor HP translates into an output torque at maximum pump load (highest pressure). When you know the output torque of the motor then you know what torque is required of the mechanism attached to the hydraulic cylinder. It is then simply a matter of calculating the linear force from the hydraulic cylinder applied to the mechanism to achieve that torque. Another wayto get a fairly good approximation, and this requires taking your pump apart, is a reverse approach. If you know the required output pressure (Pmax) of your pump and you know the pump piston diameter (Dp) then you can calculate the force (Fp) required to move the piston in the pump cylinder Fp = Pmax * Ap (where Ap is the piston surface area = pi *(Dp/2)^2 ) Note - this equation does not account for sliding friction of the piston in the pump so more force would be required to overcome this. Then you would need to know geometry of the pump components (ie piston rod length, and pump crank arm length) to determine what torque would be required to apply this force to the piston. The input torque from your hydraulic cylinder / torque mechanism would then need to be AT LEAST = to the calculated required input torque.

Thank You!!!! It should work- the pump plunger just moves in a straight line back and forth and so does a hydraulic cylinder so.... My piston cylinder has a 9inch diameter and I require 350lbs of output pressure. This figures to me needing 7087 (inch lbs???) or 590 ft lbs. or is this just 7087lbs of pressure needed pushing on the plunger??? I am almost there- please help with the last step? Thank You, Sincerely , Mark

Reply: H.P to pressure? Ok, so just to clarify... First, I understand you are going to drive the pump cylinder directly with a hydraulic cylinder. That implies that your output pressure in pounds per square inch (psi) is directly proportional to your hydraulic input pressure. There are no other mechanism calculations to be made which makes this a much simpler calculation. Second, I am interpreting the phrase "piston cylinder" to mean the pump piston cylinder not the hydraulic cylinder diameter. This seams pretty clear but I just want to be sure so please tell me if otherwise. So... output pressure = 350 psi pump piston cylinder dia = 9 in. --> Area of pump piston cylinder = 63.62 in^2 Therefore the resulting load or force on the pump piston is: 350 psi * 63.62 in^2 = 22,266 Lbs. Your hydraulic cylinder would then need to be sized for this load + a factor of safety. As a general rule, and this is not absolute because it depends on your application, you want your hydraulic cylinder to be sized for a design pressure = 1.5 x max. operating pressure. For example: Let's say you are powering your 9in dia pump piston at 350 psi with a 6" diam hydraulic cylinder. The surface area (A) of your hydraulic cylinder is pi*(6/2)^2 = 28.27 in^2. Your max load (Fmax) on your pump piston as calculated above is 22,266 Lbs. Therefore the max operating pressure in the hydraulic cylinder Pmax = Fmax / A = 22266 / 28.27 = 788 psi For a safety factor you would then want a hydraulic cylinder capable of 1.5 x this max operating pressure = 788*1.5 = 1182 psi. Of course this is a very specific number so when you select a hydraulic cylinder I would round up to a higher capacity hydraulic cylinder. This will add even more safety factor which, unless you are concerned about weight, is ALWAYS good. NOTES - you are talking about a more dynamic (cyclic) problem with the repeated pushing and pulling of the pump piston by this hydraulic cylinder. You might want to consult with a hydraulic cylinder manufacturer to discuss it's specific use and get their recommendation for sizing it correctly. I hope this helps.

See also http://www.wisc-online.com/objects/ViewObject.aspx?ID=HYP604 It has a good animated example of what I was talking about above

bstclair Your calculations are correct, the only comment that I have is that hydraulic cylinders are normally manufactured with a safety factor built in. If a cylinder has a recommended maximum working pressure of say 3,000psi then your can operate it at this pressure. Mark, the flow rate of liquid from the pump cylinder will be proportional to the speed of movement of the hydraulic cylinder and if you have only one pump cylinder the flow will stop whilst the hydraulic cylinder retracts.

Peter, Thank you for the correction. It makes sense that that cylinder manufacturers would have some sort of safety factor built in. If for no other reason than their own liability. However, my understanding was, like most other mechanical systems, the safety factor used is based on the system's use. Lower for low damage risk failures and higher for high damage risk failures such as potential personnel injury... that sort of thing. So I am curious how cylinder manufacturers account for this. Perhaps they assume high damage risk and use an appropriatly high safety factor. If you happen to know I'd love to hear your take on it. Second, As I was typing my first reply it occured to me to mention that the choice of hydraulic pump, especially in this application, is at least as equally important as the choice of cylinder itself. First for the reason you mention. The cycle time of the cylinder = the cycle time of the pump piston which, to your point, is proportional to the flow rate from the pump. Hence, whatever hydraulic pump is being used to cycle the cylinder needs to have the ability to cycle at a rate that allows the pump to be effective. A quick google search of hydraulic pump sizing finds results speaking of linear velocity and flow rates. Please correct me if I am wrong but this is my interpretation of what that means. If the cycle rate of the system is known and the cylinder / pump piston stroke is known then the required linear velocity of the hydraulic pump can be found. This linear velocity combined with the hydraulic cylinder displacement volume determines the flow rate required and thus the size of the pump required. Brett StClair

Brett Normally hydraulic cylinders are manufactured and tested to either NFPA or DIN standards which take into account the possible risk. Hydraulic pumps are specified by their delivery in cc/rev (in^3/rev) because the shaft speed can vary depending on the engine revs or, if driven my an electric motor, by the number of poles and the supply frequency. For example, if a pump can deliver 10 cc/rev and the motor has 4 poles with a supply of 50Hz, it will rotate at approximately 1450RPM which will give a flow rate of 10 x 1450 = 14500 cc or 14.5 Litres/Min. The flow rate required, can be calculated by working out the cylinder full bore volume and dividing it by the time in seconds to fully extend the cylinder, this will give a volume/second, multiply by 60 to get volume/minute.