2.3

A water pump delivers 12,5 litres of water per stroke if the effective force on the piston is 27 kN. The diameter of the piston is 250 mm.
2.3.1
Calculate the stroke length of the piston
2.3.2
Calculate the work done per stroke.

1 answer

To calculate the stroke length of the piston and the work done per stroke, we'll use the provided information.

2.3.1 Calculate the Stroke Length of the Piston

  1. Determine the cross-sectional area of the piston: The area \( A \) of a circle is given by the formula:

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

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

    Here, the diameter \( d \) is 250 mm, which we need to convert to meters:

    \[ d = 250 , \text{mm} = 0.250 , \text{m} \]

    Now, we can calculate the area:

    \[ A = \pi \left( \frac{0.250}{2} \right)^2 = \pi \left( 0.125 \right)^2 \approx \pi (0.015625) \approx 0.049087 , \text{m}^2 \]

  2. Determine the stroke length \( L \): We know the volume of water delivered per stroke is 12.5 liters, which we need to convert to cubic meters:

    \[ 12.5 , \text{liters} = 0.0125 , \text{m}^3 \]

    The volume \( V \) produced by the piston during each stroke is also given by the formula for the volume of a cylinder:

    \[ V = A \cdot L \]

    Therefore, we can rearrange this to solve for \( L \):

    \[ L = \frac{V}{A} \]

    Substituting in the values:

    \[ L = \frac{0.0125 , \text{m}^3}{0.049087 , \text{m}^2} \approx 0.2546 , \text{m} = 254.6 , \text{mm} \]

So, the stroke length of the piston is approximately 254.6 mm.

2.3.2 Calculate the Work Done Per Stroke

The work done \( W \) is given by the formula:

\[ W = F \cdot d \]

where \( F \) is the effective force and \( d \) is the stroke length. We have:

  • Effective force \( F = 27 , \text{kN} = 27000 , \text{N} \)
  • Stroke length \( d = 0.2546 , \text{m} \)

Now substituting the values into the work equation:

\[ W = 27000 , \text{N} \cdot 0.2546 , \text{m} \approx 6864.2 , \text{J} \]

So, the work done per stroke is approximately 6864.2 Joules.