Question

6. "Car crusher" A hydraulic cylinder is to compress a car body down to bale size in 10 sec. The operation requires a S = 10 ft stroke, and a Fload =8000 lb force. If a P = 1000 psi pump has been selected, and assuming a the cylinder is 80% efficient, compute: a) the required piston area, A = Flod/P (in^2) b) the necessary pump flow rate, Q (cfs) = A (ft^2)*s (ft)/t (sec) c) the necessary pump flow rate, Q (gpm), Q(gpm) 448*Q(ft^3/s) d) the hydraulic horsepower delivered to the cylinder, HPin = P (psi)*Q (gpm)/ 1714 d) the output horsepower delivered by the cylinder to the load, HP out = nHP in

Answer 1

Given t = 10 See S = 10 ft (stroke) Fload = 8000 lbf P= 1000 psi 2 - 80% or 008 2 8000 co02 144 @ The Required piston Area A= Fload 8lin²=8 000555 ft Pepsi) And 6 Pump flow Rote Q (ft3)= A (ft?) * S (ft) = 0.0555 x 10 t (see) 0.0555 ft3/s AND . The necessary pump flow Rate: 10 Q[gpm) 448* Q (ft3/5) 448 x 0-0555 [ Q(8pm) 24-89 gpm And 229 9 gpm

@ hydraulic horsepower delivered to the Cylinder : P(psi)* QC3pm) (HP) in 1714 1000 X 24.89 1714 (MP). 14.52 hp And e Output horse power delivered by the Cylinder to the load. : (MP) n (HP) in out 8 X 14.52 (HP) 1 1 6167 hp out And