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Question:

Hydraulic Power and speed?

The boom cylinders (2) of a 350D LC John Deere excavator have a bore diameter of 145 mm and a stroke of 1520mm. The hydraulic pressure of the system can be boosted up to 36.3 MPa when breaking hard ground or lifting wet sand or gravel. What is the maximum extending speed of both cylinders. Assume all the engine power of 202 kW is converted into hydraulic power.I got 0.17 m/s but wanted to make sure if thats right. If you can show the step by step method of solving this problem. I would greatly appreciate it.thank you.

Answer:

The basic formula for hydraulic power is proportional to the force and the flow. We can calculate from the power to find time and so speed. This is: power_watts = (force_newtons x displacement_meters) / time_seconds We need the time so: time = (force * displacement) / power The pressure is due to the resistance to flow. Determine the force from the pressure and the cross sectional area of the cylinder, where one pascal is one newton per square meter. area = pi * r^2 = pi * 0.0725m^2 = 0.016513m^2 The pressure is 36.3MPa. The force is: Force = pressure * area = 599422N. The displacement is the stroke of 1.52m, so from the above formula for time: time = (599422N * 1.52m) / 202000W = 4.51s The extension speed (velocity) is: velocity = distance / time = 1.52m /4.51s = 0.337m/s Edited later: I missed the point there are 2 cylinders. As there are 2 cylinders for the same flow, the speed is half of one cylinder: 0.337 / 2 = 0.1685m/s <== Answer (as for your answer)
In order to calculate the Hydraulic Motor Speed n (rpm), you must enter the following figures:- 1) Pump Flow Q (lpm) and is the flow input to the motor from eg hydraulic power pack 2) Motor Efficiency, for hydraulic motors this is in the range 0.85-0.95. 3) Displacement Vg,, sizes for hydraulic motors from 5cm3to 250cm3 Useful conversions factors: 1 cm3 = 0.061 cu.inch 1 Litre = 0.22 UK Gallons = 0.26 US Gallon

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