A hydraulic press has a ram of 150 mm diameter and plunger of 30 mm. The stroke of the plunger is 250 mm and weight lifted is 600 N. If the distance moved by the weight is 1.20 m in 20 minutes, determine : (a) the force applied on the plunger, (b) power required to drive the plunger, and (c) number of strokes performed by the plunger.

Hydraulic Press Performance Analysis

Problem Statement

Given Data & Constants

Diameter of Ram, $D = 150 \, \text{mm} = 0.15 \, \text{m}$
Diameter of Plunger, $d = 30 \, \text{mm} = 0.03 \, \text{m}$
Stroke of Plunger, $L = 250 \, \text{mm} = 0.25 \, \text{m}$
Weight Lifted, $W = 600 \, \text{N}$
Distance moved by weight, $h = 1.20 \, \text{m}$
Time taken, $t = 20 \, \text{minutes} = 1200 \, \text{s}$

Solution

(a) Force Applied on the Plunger (F)

Using Pascal's Law, the pressure on the plunger and ram are equal. This allows us to find the force using the ratio of the areas.

$$ \text{Area of Ram, } A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.15)^2 \approx 0.01767 \, \text{m}^2 $$ $$ \text{Area of Plunger, } a = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.03)^2 \approx 0.0007068 \, \text{m}^2 $$ $$ \frac{F}{a} = \frac{W}{A} \implies F = W \times \frac{a}{A} $$ $$ F = 600 \times \frac{0.0007068}{0.01767} = 600 \times \frac{1}{25} $$ $$ F = 24 \, \text{N} $$

(c) Number of Strokes Performed

The total volume of fluid displaced to lift the ram must equal the volume of fluid pushed by the plunger over a number of strokes.

$$ \text{Volume displaced by ram} = A \times h = 0.01767 \, \text{m}^2 \times 1.20 \, \text{m} \approx 0.0212 \, \text{m}^3 $$ $$ \text{Volume per plunger stroke} = a \times L = 0.0007068 \, \text{m}^2 \times 0.25 \, \text{m} \approx 0.0001767 \, \text{m}^3 $$ $$ \text{Number of strokes} = \frac{\text{Volume displaced by ram}}{\text{Volume per plunger stroke}} $$ $$ \text{Number of strokes} = \frac{0.0212}{0.0001767} \approx 120 $$

(b) Power Required to Drive the Plunger

Power is the total work done by the plunger divided by the time taken.

$$ \text{Total distance moved by plunger} = \text{Number of strokes} \times L = 120 \times 0.25 \, \text{m} = 30 \, \text{m} $$ $$ \text{Work Done} = F \times (\text{Total distance}) = 24 \, \text{N} \times 30 \, \text{m} = 720 \, \text{J} $$ $$ \text{Power} = \frac{\text{Work Done}}{t} = \frac{720 \, \text{J}}{1200 \, \text{s}} = 0.6 \, \text{W} $$

Final Results:

(a) Force applied on the plunger: $ 24 \, \text{N} $

(b) Power required to drive the plunger: $ 0.6 \, \text{W} $

Explanation of the Method

(a) Force Calculation: This is a direct application of Pascal's Law. The ratio of the forces is equal to the ratio of the areas. Since the ram's area is 25 times larger than the plunger's area, the force required on the plunger is 1/25th of the weight being lifted.

(c) Number of Strokes: This calculation is based on the conservation of volume. The large volume of fluid needed to lift the ram by 1.2 meters is supplied by many small-volume strokes of the plunger. By dividing the total required volume by the volume of a single stroke, we find how many strokes are necessary.

(b) Power Calculation: Power is the rate at which work is done. We first calculate the total work performed by the plunger, which is the force applied (24 N) multiplied by the total distance the plunger traveled (120 strokes x 0.25 m/stroke = 30 m). This total work is then divided by the total time (1200 seconds) to find the average power.