To solve the problem, we need to perform the calculations based on the provided data for the single-acting hydraulic press.
Given Data:
- Area of plunger \( A_p = 0.09 , \text{m}^2 \)
- Stroke of plunger \( s_p = 0.13 , \text{m} \)
- Area of ram \( A_r = 0.45 , \text{m}^2 \)
- Force applied to plunger \( F_p = 300 , \text{N} \)
(a) Volume of Liquid Displaced by the Plunger in 6 Strokes:
The volume of liquid displaced by the plunger for one stroke can be calculated using the formula for the volume of a cylinder:
\[ V = A_p \times s_p \]
Substituting the known values:
\[ V = 0.09 , \text{m}^2 \times 0.13 , \text{m} = 0.0117 , \text{m}^3 \]
For 6 strokes, the total volume displaced will be:
\[ V_{\text{total}} = 6 \times V = 6 \times 0.0117 , \text{m}^3 = 0.0702 , \text{m}^3 \]
(b) Distance Moved by the Ram in mm After One Pumping Stroke of the Plunger:
Using Pascal's principle, we can relate the areas and the distances moved by the plunger and the ram. The volume displaced by the plunger is equal to the volume displaced by the ram.
For one stroke,
\[ A_p \times s_p = A_r \times s_r \]
Where \( s_r \) is the distance moved by the ram. Rearranging gives:
\[ s_r = \frac{A_p \times s_p}{A_r} \]
Substituting the known values:
\[ s_r = \frac{0.09 , \text{m}^2 \times 0.13 , \text{m}}{0.45 , \text{m}^2} \]
Calculating \( s_r \):
\[ s_r = \frac{0.0117 , \text{m}^3}{0.45 , \text{m}^2} = 0.026 , \text{m} \]
Convert \( s_r \) to millimeters:
\[ s_r = 0.026 , \text{m} \times 1000 , \text{mm/m} = 26 , \text{mm} \]
Final Answers:
(a) The volume of liquid displaced by the plunger in 6 strokes is 0.0702 m³. (b) The distance moved by the ram in mm, after one pumping stroke of the plunger is 26 mm.
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