How to Analyze Electrical Load Data

It will depend on the machine used, the load applied, the height or time in use when you use constant speed to hoist. Your motor will demand more current at full load. Aside of your question, you have also the investment cost, the operation costs, maintenance, insurance .....

You will need a little more information on the rock issue.

Horsepower is defined as 550 pounds of load moving one foot in one second. How fast does the hoist move? How far does it move each lift. How much material (pounds) is moved each load. How many loads will be done each day / week / month? Once you have that data, you can calculate theoretical HP (based on pounds per second)- double that for system friction losses. Then, calculate the power used by multiplying the HP by 0.75 to get kW. For example- 100 million pounds per month, lifted 10 feet at 200 pounds per lift will equal (100,000,000 / ((720 hours per month X 60 minutes per hour X 60 seconds per minute) X 2 to allow for the return empty trip X 2 to allow for filling the basket and dumping it) = 9.65 pounds per second during each lift. Assuming that lifting occurs 25% of the time, the average lift would be 9.65 / 25%, or 38.6 pounds per second for each lift. Therefore, the power used per each load is (38.6 X 10 feet per lift) / 550 Lb-Ft/HP = 0.70 HP per lift. 0.70 X 0.75 = 0.53 kW.

Residentially- There is a wide range of loads dependent on body count and size of house. Assume a load of 3.5 kW per 4 occupants per 1000 SF of house. That will be a reasonable guess for now.

Hope this helps