Figure 18-46 shows a hydraulic lift operated by pumping fluid into the hydraulic
ID: 1302802 • Letter: F
Question
Figure 18-46 shows a hydraulic lift operated by pumping fluid into the hydraulic system. The large cylinder is 20 cm in diameter, while the small tube leaving the pump is 1.7 cm in diameter. A total load of 2700 kg is raised 2.3 m at a constant rate. Neglect pressure variations with height, and neglect also the weight of the fluid raised.
(a) What volume of fluid passes through the pump?
m3
(b) What is the pressure at the pump outlet?
atm
(c) How much work does the pump do?
kJ
(d) If the lifting takes 40 s, what is the pump power?
kW
Explanation / Answer
The volume of liquid that is pushed into the tube V = A h
V = ( pi d2 / 4 ) h
d = 20 cm = 0.2 m, h = 2.3 m
= [ pi (0.2)2 / 4 ) 2.3
= 0.0722 m3
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b) The pressure in the both tubes is constant P = F /A = m g / A
= 2700kg (9.8) / [ ? (0.2)2 / 4 ]
= 8.42 * 105 Pa
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c) Work done by the pump is given as W = m g h
= (2700) (9.8) (2.3)
= 6.09 * 104 J
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d) Power P = W / t = 6.09 * 104 / 40S
= 1521.45 W
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