Question 10.7.2: The ram and plunger of a hydraulic press are 25 cm and 3 cm,......
The ram and plunger of a hydraulic press are 25 cm and 3 cm, respectively, and the leverage of the handle is 10:1. With a plunger stroke of 25 cm, the press is able to lift a load of 200 kN through 1.25 m in 2 minutes. Determine (i) force applied at the end of the lever (ii) power required to drive the plunger, and (iii) number of stroke to be performed by the plunger in one second. Assume that the packing friction of the plunger as well as the ram is 4% of the load.
Given: d = 0.03 m, D = 0.25 m, L/l = 10, h = 0.25 m, W = 200 kN, H = 1.25/120 m, k = 0.04
See Fig. 10.6.1. From the given data
p × a = F – kF and W = p × A (1 – k)
→
F=\frac{Wa}{A(1-k)^2}=\frac{200 \times 10^3 \times \pi/4\times 0.03^2}{{\pi}/{4} \times 0.25^2\times 0.96^2} =3120 N
Effort required at the end of the lever = 3120/10 = 312 N
Number of strokes =\frac{\mathrm{Total voulume of liquid to supplied}}{\mathrm{Volume of liquied displaced / stroke in plunger}}
=\frac{AH}{ah}=\left\lgroup\frac{0.25}{0.03} \right\rgroup^2 \frac{1.25/120}{0.25}=2.2 =3 \mathrm{strokes}
Work done at the plunger = F × h × n = 3120 × 0.25 × 2.2 = 2262 W or 2.262 kW
