Question 14.7: Water at a pressure of 7.5 MPa is supplied to a hydraulic cr......
Water at a pressure of 7.5 MPa is supplied to a hydraulic crane through a pipe 300 m long and 50 mm diameter. The pipe is having a friction factor of 0.01. The ram of the crane has a diameter of 0.016 m. The ratio between the movement of the load and the ram is 6. In order to overcome the frictional losses, a pressure of 400 kN/m^2 on the ram is required. If the load to be lifted is 10 kN, find the maximum speed at which it can be lifted.
Given: p=7.5 MPa =7.5 \times 10^6 N / m ^2 ; L=30 m , d=50 mm =0.05 m ;
Pressure head of water at inlet
H=\frac{P}{W}=\frac{7.5 \times 10^6}{9810}=764.53 m \text { of water }
Pressure head to overcome friction
H_{f r}=\frac{F}{w}=\frac{400 \times 10^3}{9810}=40.77 m \text { of water }
Head loss due to friction in pipe
H_f=\frac{4 f L V_p^2}{2 g d}=\frac{4 \times 0.01 \times 300 V_p^2}{2 \times 9.81 \times 0.05}
=12.23 V_p^2 m
Pressure head on ram =H-H_{f r}-H_f=764.53-40.77-12.33 V_p^2
=723.76-12.33 V_p^2
Pressure head required on the ram =\frac{W}{A W} \times 6=\frac{10 \times 10^3 \times 6}{\frac{\pi}{4} \times 0.16^2 \times 9810}=498.4 m of water
Equating the two, we get
498.4=723.76-12.33 V_p^2
Flow velocity in the pipe, V_p=4.26 m / s
From continuity
Velocity in pipe × Area of pipe = Velocity of ram × Area of ram;
4.26 \times \frac{\pi}{4} \times 0.05^2=V_R \times \frac{\pi}{4} \times 0.16^2
Velocity of ram, V_R=0.416 m / s
Max. speed of lifting the load =V_R \times 6=0.416 \times 6=2.5 m / s