Question 14.6: The plunger of a hydraulic press has a diameter of 15 mm and......
The plunger of a hydraulic press has a diameter of 15 mm and stroke of 200 mm. It is required to raise a load of 30 kN on the ram which is 125 mm in diameter. Find the force required to be applied on the plunger to raise the load. If the load is to be raised by 500 mm, find the required number of strokes of plunger. If the time taken to lift the load is 10.5 minutes, what will be the power required to lift the plunger? Neglect the effect of friction.
Given: d=15 mm =0.015 m ; L=200 mm =0.2 m , W=30 kN =30 \times 10^3 N; D = 125 mm = 0.125 m; h = 500 mm = 0.5 m; t = 10.5 min. = 630 s
Intensity of pressure on fluid
p=\frac{W}{A}=\frac{3 \times 10^3 \times 4}{\pi \times 0.125^2}=2444.6 N / m ^2
Force to be applied on plunger
F=p \times a=2444.6 \times \frac{\pi}{4} \times 0.015^2=432 N
Number of strokes to lift load by 0.5 m
=\frac{\text { Volume of liquid to be displaced }}{\text { Volume displaced in one stroke }}
=\frac{\pi \times 0.125^2 \times 0.5 / 4}{\pi \times 0.15^2 \times 0.2 / 4}=174 \text { strokes }
Work done by the press in 10.5 minutes
=W \times L=30 \times 10^3 \times 0.5=15000 Nm
Work done per second (power required to lift the plunger)
=\frac{15000}{630}=23.81 Nm / s