Question 14.1: The diameter of the ram of an accumulator is 300 mm and its ......

Given: D=0.3  m ; L=8  m ; p=50  MPa =5 \times 10^6  N / m ^2

Load on the ram, W=p \times A=p \times \frac{\pi D^2}{4}

=5 \times 10^6 \times \pi \times \frac{0.3^2}{4}=353429  N

Capacity of accumulator = Load × Stroke

\begin{aligned}&=353429 \times 8=2827433  Nm=2827.433  kNm \\&=\frac{2827433}{1000 \times 60 \times 60}=0.7854  kWh .\end{aligned}