Question 14.2: An accumulator has a ram of diameter 0.25 m and stroke 5 m. ......
An accumulator has a ram of diameter 0.25 m and stroke 5 m. The ram takes 100 seconds to fall through its full stroke. Discharge from the pump is 0.45 m^3/min . If the load on the accumulator is 380 kN, find the power delivered to the hydraulic machine. Consider the frictional resistance as 5%.
Given: D=0.25 m ; L=5 m ; t=100 s ; Q=0.456 m ^3 / min =0.0075 m ^3 / s; W = 380 kN
Distance travelled by ram in 1 second =\frac{L}{t}=\frac{5}{100}
=0.05 m / s
Net load on accumulator, considering friction
=380 \times 0.95=361 kN
Power delivered by accumulator = Load × Distance travelled per second
=361 \times 10^3 \times 0.05=18050 W
Pressure head to be developed by pump to raise the ram,
\begin{aligned}H &=\frac{p}{W}=\frac{W}{A \times W}=\frac{W \times 4}{\pi D^4 \times W}\\&=\frac{361 \times 10^3 \times 4}{\pi \times 0.25^2 \times 9810}=750 m\end{aligned}
Power delivered by pump = wQH
=9810 \times 0.0075 \times 750=55156 W
Total power supplied to hydraulic machine
\begin{aligned}&=18050+55156 \\&=73206 W=73.206 kW .\end{aligned}