Question 14.8: A hydraulic lift is designed to take a load of 80 kN. The se......

Load to be taken by ram = Load on the lift + Self weight of lift + Frictional resistance.

= 80 + 20 + (80 + 20) × 0.05 = 105 kN

Let $V_s$ be the flow velocity in the supply line. The head loss due friction is:

$H_f=\frac{4 f L_s V_s^2}{2 g d_s}=\frac{4 \times 0.009 \times 600 \times V_s^2}{2 \times 9.81 \times 0.075}=14.63 V_s^2$

Pressure of water at the ram inlet $p_i=p-p_f=p-w H_f$

$p_i=7 \times 10^6-9810 \times 14.63=7 \times 10^6-0.144 \times 10^6 V_s^2$

Force on the hydraulic ram $=p_i \times A$

$=\left(7 \times 10^6-0.144 \times 10^6 V_s^2\right) \times \frac{\pi}{4} \times 0.18^2=\left(178 \times 10^3-3.66 \times 10^3 V_s^2\right)$

The force is equal to the load to be taken by ram

$105 \times 10^3=178 \times 10^3-3.66 \times 10^3 V_s^2$

∴            The flow velocity in the supply line is:

$V_s=4.47  m / s$

Discharge through the pipeline, $Q=a_s V_s=\frac{\pi}{4} d_s^2 \times V_s$

$=\frac{\pi \times 0.075^2 \times 4.47}{4}=0.0197  m ^3 / s$

Since the same discharge flows through the ram of the lift, the velocity of the ram is:

$V_{ ram }=\frac{Q}{\frac{\pi}{4} D^2}=\frac{0.0197 \times 4}{\pi \times 0.18^2}=0.7754   m / s$

Since the velocity of the ram is same as velocity of the lift, the velocity of lift is:

$V_{\text {lift }}=0.7754  m / s$