Question 14.4: A differential accumulator has fixed ram diameter 150 mm and......

Given: d=150  mm =0.15  m ; D=165  mm =0.165  m ; L=1.5  m ; p=12  MPa =12 \times 10^6  N / m ^2

Load on the ram                                 = p × Annular area

\begin{aligned}&=\frac{p \times \pi\left(D^2-d^2\right)}{4}=\frac{12 \times 10^6 \times \pi\left(0.165^2-0.15^2\right)}{4}\\&=44520  N=44.52  kN\end{aligned}

Capacity of the ram                                = Load × Stroke length

\begin{aligned}&=44.52 \times 10^3 \times 1.5=66798  Nm \\&=\frac{66198}{1000 \times 60 \times 600}=0.0185  kWh .\end{aligned}