Question

A double-acting hydraulic cylinder has bore and rod diameters of 76 mm and 25 mm, respectively. In extension as well as retraction strokes of the cylinder, the oil flow rate to the cylinder from the pump is 40 L min$^{-1}$. The velocity of the piston during retraction stroke in m s$^{-1}$ is _____. \(\textit{[Round off to two decimal places.]}\)

Hint:

In Newton-Raphson method, the next approximation is found by using both the function value and its derivative at the current approximation.

Answer 1

Correct Answer: 0.14

The hydraulic cylinder's velocity during retraction can be calculated using the formula: \[ v = \frac{Q}{A} \] where:
- \( v \) = velocity of the piston (m/s),
- \( Q \) = flow rate (m³/s),
- \( A \) = cross-sectional area of the cylinder (m²).
First, convert the flow rate to m³/s: \[ Q = 40 \, \text{L/min} = \frac{40}{1000 \times 60} \, \text{m}^3/\text{s} = 0.000667 \, \text{m}^3/\text{s}. \] Next, calculate the cross-sectional area \( A \) during the retraction stroke: \[ A = \pi \left( \left( \frac{D_{\text{bore}}}{2} \right)^2 - \left( \frac{D_{\text{rod}}}{2} \right)^2 \right) \] where \( D_{\text{bore}} = 76 \, \text{mm} = 0.076 \, \text{m} \) and \( D_{\text{rod}} = 25 \, \text{mm} = 0.025 \, \text{m} \).
\[ A = \pi \left( \left( \frac{0.076}{2} \right)^2 - \left( \frac{0.025}{2} \right)^2 \right) \] \[ A = \pi \left( (0.038)^2 - (0.0125)^2 \right) = \pi \left( 0.001444 - 0.00015625 \right) = \pi \times 0.00128775 = 0.004050 \, \text{m}^2. \] Now, calculate the velocity: \[ v = \frac{0.000667}{0.004050} = 0.164 \, \text{m/s}. \] Thus, the velocity of the piston during retraction stroke is approximately \( \boxed{0.16} \, \text{m/s} \) (rounded to two decimal places).