Q. I know that hydraulic pumps vary in efficiency, but how do I figure out the horsepower required to do a certain job?
A. Work is defined as the amount of force applied over a certain distance:
Work = Force X Distance
Power is the rate that work is done:
|Power =||Force X Distance|
In the hydraulic system, we represent speed and force by flow and pressure, typically measured in Gallons Per Minute (GPM) and Pounds per Square Inch (PSI) respectively. Thus hydraulic horsepower is the flow multiplied by the pressure and may be determined by the formula:
HP = GPM X PSI X .000583
where the constant .000583 resolves the units of measure. In Figure 1, we are using a fixed displacement 30 GPM pump which will deliver a constant flow assuming it is always turned at the same speed by the drive motor..
The pressure in the system will be determined by the resistance to this flow. Currently, the only resistance to the flow is the load that is being moved, which requires 1300 PSI to overcome. But a different size load would require a different amount of pressure to move, so the horsepower requirements should be determined at the maximum pressure and flow. The system relief valve should be set 200 PSI above the pressure required to move the biggest load so that the relief valve setting will determine the maximum pressure that will ever be reached in the system. In Figure 2, the handvalve is closed leaving no path for fluid to travel except through the relief valve.
The hydraulic horsepower developed by the system can be determined using the horsepower formula:
HP = GPM X PSI X .000583
HP = 30 GPM X 1500 PSI X .000583
HP = 45,000 X .000583
HP = 26.2
If the pump was 100% efficient, then the formula could also be used to determine the required electric motor size. But because of internal pump bypassing and friction of the bearings and pump elements, the pump is somewhat less than 100% efficient. Therefore, slightly more electrical horsepower will be called for to drive the pump. While most pumps are 85-90% efficient, piston pumps are more efficient than vane or gear pumps. The efficiency is established by the pump manufacturer when the pump is tested. In our example, we assume an 87% pump efficiency and alter the constant to reflect it.
|87% (or .87)|
HP = GPM X PSI X .00067
HP = 30 GPM X 1500 PSI X .00067
HP = 45,000 X .00067
HP = 30
As a rule of thumb, it takes about 1 HP to pump 1 GPM at 1500 PSI. In the example, since our pump will deliver a constant 30 GPM a 30 HP electric motor may be used. Note that our fixed displacement pump will deliver its full flow regardless of the resistance tha must be overcome. If the flow is blocked and the only path to tank is through the relief valve, there will be a 1500 PSI pressure drop across the valve. Since energy can neither be created nor destroyed, the full 26.2 hydraulic horsepower will be converted to heat. Whenever the flow is not being used to move a load, it must be either reduced to 0 GPM (i.e. turned off) or diverted back to tank at 0 PSI. One method of returning flow back to tank at low pressure is illustrated in figure 3:
By installing another handvalve which is left open to dump the oil back to tank at low pressure, the system will require the electric motor to pull only enough electrical current to drive the hydraulic system and will thus be very low. About 5% of the system horsepower is required to pump the fluid back to tank at 0 PSI. 5% of 30 HP is 1.5 HP, so only 1.5 HP is converted to heat.