How much horsepower is required to drive a pump in a hydraulic cylinder system with a return stroke force of 6000 lbs at a speed of 350 in/min?

To determine the horsepower required to drive a hydraulic pump based on the provided parameters, it is essential to understand the relationship between force, speed, and horsepower in a hydraulic system.

The formula used to calculate the hydraulic horsepower is:

[ \text{Horsepower} = \frac{\text{Force (lbs)} \times \text{Speed (in/min)}}{33,000} ]

In this case, the return stroke force is 6000 lbs, and the speed is 350 in/min. Plugging these values into the formula gives:

[ \text{Horsepower} = \frac{6000 \text{ lbs} \times 350 \text{ in/min}}{33,000} ]

Calculating this:

1. Multiply the force and speed:

( 6000 \times 350 = 2,100,000 )

2. Divide by 33,000 to convert to horsepower:

( \frac{2,100,000}{33,000} \approx 63.64 )

Since we are looking for the hydraulic horsepower which generally also includes consideration for system inefficiencies, we must also account for the efficiency factor of the hydraulic system in the overall calculation. Given that the