A hydraulic accumulator must work under a pressure of 2 MPa.

Calculate the mass needed for it to act under the pressure of 2 MPa if the ram has a mass of 750 kg and a diameter of 450 mm

1 answer

To determine the mass needed for a hydraulic accumulator to act under a pressure of 2 MPa, we can use the relationship between pressure, force, and area.

Step 1: Calculate the area of the ram

The area \( A \) of the ram can be calculated using the formula for the area of a circle:

\[ A = \pi \left(\frac{d}{2}\right)^2 \]

where \( d \) is the diameter of the ram.

Given:

  • Diameter \( d = 450 , \text{mm} = 0.450 , \text{m} \)

Calculating the radius:

\[ r = \frac{d}{2} = \frac{0.450}{2} = 0.225 , \text{m} \]

Now, calculate the area:

\[ A = \pi (0.225)^2 = \pi (0.050625) \approx 0.159 , \text{m}^2 \]

Step 2: Calculate the force required using the pressure

The force \( F \) exerted by the hydraulic accumulator can be calculated using the formula:

\[ F = P \times A \]

where \( P \) is the pressure in pascals and \( A \) is the area in square meters.

Given:

  • Pressure \( P = 2 , \text{MPa} = 2 \times 10^6 , \text{Pa} \)

Now, calculate the force:

\[ F = 2 \times 10^6 , \text{Pa} \times 0.159 , \text{m}^2 \approx 318,000 , \text{N} \]

Step 3: Calculate the additional mass needed

The net force exerted by the hydraulic system must balance the weight of the ram plus the additional weight (mass) contributing to the pressure.

The weight \( W \) of the ram and the additional mass \( m_{\text{additional}} \) can be represented as:

\[ W = (m_{\text{ram}} + m_{\text{additional}}) \times g \]

where \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)).

To solve for the additional mass needed, we set \( F = W \):

\[ 318,000 , \text{N} = (750, \text{kg} + m_{\text{additional}}) \times 9.81 , \text{m/s}^2 \]

Step 4: Solve for \( m_{\text{additional}} \)

Rearranging the equation:

\[ 750 + m_{\text{additional}} = \frac{318,000}{9.81} \]

Calculating the right side:

\[ \frac{318,000}{9.81} \approx 32,400.82 , \text{kg} \]

Now, solve for \( m_{\text{additional}} \):

\[ m_{\text{additional}} \approx 32,400.82 - 750 \approx 31,650.82 , \text{kg} \]

Conclusion

The additional mass needed for the hydraulic accumulator to act under a pressure of 2 MPa, considering the 750 kg mass of the ram, is approximately 31,650.82 kg.