Question
The following information refers to a single-acting hydraulic press:
Diameter of ram = 465 mm
Force applied to plunger = 360 kN
Diameter of the plunger = 120 mm
Plunger stroke = G,02 m
If the mechanical efficiency of the hydraulic systemas following:The distance that the ram
piston will move after 25 pumping
strokes in mm
Diameter of ram = 465 mm
Force applied to plunger = 360 kN
Diameter of the plunger = 120 mm
Plunger stroke = G,02 m
If the mechanical efficiency of the hydraulic systemas following:The distance that the ram
piston will move after 25 pumping
strokes in mm
Answers
GPT-4o mini
To calculate the distance that the ram piston will move after 25 pumping strokes in a single-acting hydraulic press, we can use the concept of hydraulic fluids and the principle of Pascal's Law.
### Step 1: Calculate the area of the plunger and ram
1. **Area of the plunger**:
\[
A_{plunger} = \pi \left(\frac{D_{plunger}}{2}\right)^2
\]
Where \(D_{plunger} = 120 \, \text{mm} = 0.120 \, \text{m}\).
\[
A_{plunger} = \pi \left(\frac{0.120}{2}\right)^2 \approx \pi \times 0.006^2 \approx 0.0001131 \, \text{m}^2
\]
2. **Area of the ram**:
\[
A_{ram} = \pi \left(\frac{D_{ram}}{2}\right)^2
\]
Where \(D_{ram} = 465 \, \text{mm} = 0.465 \, \text{m}\).
\[
A_{ram} = \pi \left(\frac{0.465}{2}\right)^2 \approx \pi \times 0.058 \approx 0.1826 \, \text{m}^2
\]
### Step 2: Calculate the volume of fluid moved by the plunger per stroke
The volume of fluid moved by the plunger in one stroke can be calculated from its area and stroke length:
Given that the plunger stroke length is \(0.02 \, \text{m}\) (20 mm):
\[
V_{plunger} = A_{plunger} \times \text{stroke length}
\]
\[
V_{plunger} = 0.0001131 \, \text{m}^2 \times 0.02 \, \text{m} \approx 0.000002262 \, \text{m}^3
\]
### Step 3: Calculate the distance moved by the ram after 25 strokes
The volume of hydraulic fluid that goes into the ram must equal the volume displaced by the ram. The volume displaced by the ram after 25 strokes is:
\[
V_{ram} = A_{ram} \times \text{distance moved by ram}
\]
Let \(d_{ram}\) be the distance moved by the ram. We can relate the volume moved by the plunger to the volume displaced by the ram:
\[
25 \times V_{plunger} = A_{ram} \times d_{ram}
\]
Substituting the values:
\[
25 \times 0.000002262 \, \text{m}^3 = 0.1826 \, \text{m}^2 \times d_{ram}
\]
\[
0.00005655 = 0.1826 \, \text{m}^2 \times d_{ram}
\]
Now, solve for \(d_{ram}\):
\[
d_{ram} = \frac{0.00005655}{0.1826} \approx 0.000309 \, \text{m}
\]
### Step 4: Convert to millimeters
To convert the distance \(d_{ram}\) to millimeters:
\[
d_{ram} \approx 0.000309 \, \text{m} \times 1000 = 0.309 \, \text{mm}
\]
### Conclusion
The ram piston will move approximately **0.309 mm** after **25 pumping strokes**.
### Step 1: Calculate the area of the plunger and ram
1. **Area of the plunger**:
\[
A_{plunger} = \pi \left(\frac{D_{plunger}}{2}\right)^2
\]
Where \(D_{plunger} = 120 \, \text{mm} = 0.120 \, \text{m}\).
\[
A_{plunger} = \pi \left(\frac{0.120}{2}\right)^2 \approx \pi \times 0.006^2 \approx 0.0001131 \, \text{m}^2
\]
2. **Area of the ram**:
\[
A_{ram} = \pi \left(\frac{D_{ram}}{2}\right)^2
\]
Where \(D_{ram} = 465 \, \text{mm} = 0.465 \, \text{m}\).
\[
A_{ram} = \pi \left(\frac{0.465}{2}\right)^2 \approx \pi \times 0.058 \approx 0.1826 \, \text{m}^2
\]
### Step 2: Calculate the volume of fluid moved by the plunger per stroke
The volume of fluid moved by the plunger in one stroke can be calculated from its area and stroke length:
Given that the plunger stroke length is \(0.02 \, \text{m}\) (20 mm):
\[
V_{plunger} = A_{plunger} \times \text{stroke length}
\]
\[
V_{plunger} = 0.0001131 \, \text{m}^2 \times 0.02 \, \text{m} \approx 0.000002262 \, \text{m}^3
\]
### Step 3: Calculate the distance moved by the ram after 25 strokes
The volume of hydraulic fluid that goes into the ram must equal the volume displaced by the ram. The volume displaced by the ram after 25 strokes is:
\[
V_{ram} = A_{ram} \times \text{distance moved by ram}
\]
Let \(d_{ram}\) be the distance moved by the ram. We can relate the volume moved by the plunger to the volume displaced by the ram:
\[
25 \times V_{plunger} = A_{ram} \times d_{ram}
\]
Substituting the values:
\[
25 \times 0.000002262 \, \text{m}^3 = 0.1826 \, \text{m}^2 \times d_{ram}
\]
\[
0.00005655 = 0.1826 \, \text{m}^2 \times d_{ram}
\]
Now, solve for \(d_{ram}\):
\[
d_{ram} = \frac{0.00005655}{0.1826} \approx 0.000309 \, \text{m}
\]
### Step 4: Convert to millimeters
To convert the distance \(d_{ram}\) to millimeters:
\[
d_{ram} \approx 0.000309 \, \text{m} \times 1000 = 0.309 \, \text{mm}
\]
### Conclusion
The ram piston will move approximately **0.309 mm** after **25 pumping strokes**.