To solve the problems, let's first outline the formulas and calculations required for each part.
Given Data:
- Area of plunger (A₁) = 0.09 m²
- Stroke of plunger (L₁) = 0.13 m
- Force applied to plunger (F₁) = 300 N
- Area of ram (A₂) = 0.45 m²
(a) Volume of Liquid Displaced by the Plunger in 6 Strokes
The volume of liquid displaced by the plunger can be calculated using the formula:
\[ \text{Volume displaced} = A₁ \times L₁ \]
Calculating the volume for one stroke:
\[ \text{Volume displaced (one stroke)} = 0.09 , \text{m}² \times 0.13 , \text{m} = 0.0117 , \text{m}³ \]
Now for 6 strokes:
\[ \text{Volume displaced (6 strokes)} = 0.0117 , \text{m}³ \times 6 = 0.0702 , \text{m}³ \]
(b) Distance Moved by the Ram After One Pumping Stroke of the Plunger
To find the distance moved by the ram, we can use the principle of conservation of volume, which states that the volume displaced by the plunger equals the volume pushed by the ram.
Let the distance moved by the ram (L₂) be what we need to find. We can establish the equation based on volume:
\[ \text{Volume displaced by ram} = A₂ \times L₂ \]
Setting the volumes equal:
\[ A₁ \times L₁ = A₂ \times L₂ \]
Now substituting the known values:
\[ 0.09 , \text{m}² \times 0.13 , \text{m} = 0.45 , \text{m}² \times L₂ \]
Solving for \(L₂\):
\[ L₂ = \frac{0.09 \times 0.13}{0.45} = \frac{0.0117}{0.45} \approx 0.026 \]
Now, to convert \(L₂\) into mm:
\[ L₂ = 0.026 , \text{m} \times 1000 = 26 , \text{mm} \]
(c) Force Exerted by the Ram
The force exerted by the ram (F₂) can be calculated using Pascal's law, where the force exerted is proportional to the area:
\[ \frac{F₁}{A₁} = \frac{F₂}{A₂} \]
Rearranging gives:
\[ F₂ = F₁ \times \frac{A₂}{A₁} \]
Substituting the known values:
\[ F₂ = 300 , \text{N} \times \frac{0.45 , \text{m}²}{0.09 , \text{m}²} = 300 \times 5 = 1500 , \text{N} \]
Summary of Calculations:
(a) Volume of liquid displaced by the plunger in 6 strokes: 0.0702 m³
(b) Distance moved by the ram after one pumping stroke: 26 mm
(c) Force exerted by the ram: 1500 N