Question

You are designing a hydraulic ram. The ram is used to crush large boulders for processing at a treatment plant for construction materials. The hydraulic cylinder used in such as device has a cylinder diameter of 4m. The system must drive that cylinder at a speed of 2m/s during operation, but the space budget only accounts for a small pump capable of a dispacement of 0.0005m3/rev and 1000 rpm.

Your team managed to increase the space budge. You now have a larger pump. The pump isn't large enough to run the cylinder directly, but is big enough where an accumulator can be replaced with a regenerative circuit.

Using the geometry of the cylinder what is the speed of the cylinder if you don't use regeneration?

Answer 1

sol . Given Cylinder diameter um - Cylinder Speed :amisec Small pump Capable displacement :0.0005 m Speeda) = 1000rpm Stroke length :0.5m Velocity : 2m 1sec flow rate require obtain the am 1sec of is given by . Required to calor ram Speed y Arequired ? 25.13 m3 / see Qpump = 0. soʻlmin (tu Qpump - 0,00834ml see accumulator : @required pump = 25.13 -0.00834 = 25.12 m2 (sec Paccamulator - 25.17 MB/sec i in order to size the accumulato) we need time to which this flow rate is to be delived. time & Stroke length velocity to know I time 20.25 Sece,

> Volume of accumulator (minimum) Time taken to fill the - Volume flow • 628 0.00834 25.12 X0-25 = 6.28m3, accumulator of accumulator rate pump 3952.995ec time require of to fill the accumulator 152.99 See 12.5umin ch