Question 14.3: The sliding ram of an accumulator which moves through a dist......
The sliding ram of an accumulator which moves through a distance of 6 m in 2.5 minutes is 0.35 m in diameter. The weight on the ram including its self weight is 250 kN and the package friction is equivalent to 5% of the load. If the pump supplies water at a rate of 0.48 m^3/minute, find the intensity of pressure and the power delivered by the accumulator. Also find the power required to drive the pump if its efficiency is 80%.
Given: L=6 m ; t=2.5 min =150 s ; D=0.35 m ; W=250 \times 10^3 N ; F=5 \% \text { of W; } Q_1=0.48 m ^3 / min =8 \times 10^{-3 } m ^3 / s ; \eta_p=0.8
Net load on ram = Load – Frictional load
=250 \times 10^3(1-0.05)=237.5 \times 10^3 N
Intensity of pressure p=\frac{\text { Load }}{\text { Area }}=\frac{237.5 \times 10^3 \times \pi}{4 \times 0.35^2}=2.4685 \times 10^6 N / m ^2
Pressure head, H=\frac{2.4685 \times 10^6}{9810}=251.63 m \text { of water }
Discharge from the accumulator, Q_2
=\frac{\text { Volume }}{\text { Time }}=\frac{\pi}{4} D^2 \frac{L}{t}=\frac{\pi}{4} \times 0.35^2 \times \frac{6}{150}=3.85 \times 10^{-3} m ^3 / s
Total flow delivered by accumulator =Q_1+Q_2=(8+3.85) \times 10^{-3}=0.01185 m ^3 / s
\begin{aligned}P &=w Q H=9810 \times 0.01185 \times 251.63 \\&=29252 W=29.252 kW\end{aligned}
Power required to drive the pump =w Q_1 H / \eta_p
=9810 \times 8 \times 10^{-3} \times \frac{251.63}{0.8}=24685 W =24.684 kW.