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<p>Figure 18-46 shows a hydraulic lift operated by pumping fluid into the hydrau

ID: 1702527 • Letter: #

Question

<p>Figure 18-46 shows a hydraulic lift operated by pumping fluid into the hydraulic system. The large cylinder is <span>50</span> cm in diameter, while the small tube leaving the pump is 1.7 cm in diameter. A total load of <span>2500</span> kg is raised <span>2.2</span> m at a constant rate. Neglect pressure variations with height, and neglect also the weight of the fluid raised.</p>
<div class="figure"><img src="http://www.webassign.net/wp3/18-46.gif" border="0" alt="" /><br />Figure 18-46.</div>
<p>(a) What volume of fluid passes through the pump?<br /><span class="qTextField"><span class="box_num" title="Points: /1">1</span></span> m<sup>3</sup><br />(b) What is the pressure at the pump outlet?<br /><span class="qTextField"><span class="box_num" title="Points: /1">2</span></span> atm<br />(c) How much work does the pump do?<br /><span class="qTextField"><span class="box_num" title="Points: /1">3</span></span> kJ<br />(d) If the lifting takes 40 s, what is the pump power?<br /><span class="qTextField"><span class="box_num" title="Points: /1">4</span></span> kW</p>

Explanation / Answer

The volume of liquid that is pushed into the tube is

V = A . h

Here A is the area , h is the height of the rise.

V = ( d^2 / 4 ) h

Given d = 50cm = 0.5m and h = 2.2m

= [ (0.5)^2 / 4 ) 2.2

= 0.4316m^3

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b) The pressure in the both tubes is constant

        P = F /A

          = m g / A

           = 2500kg (9.8) / [ (0.5)^2 / 4 ]

          = 1.2477x10^5 Pa

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c) Work done by the pump is given as

     W = m g h

        = (2500)(9.8)(2.2)

         = 5.39x10^4 J

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d) Power P = W /t

                 = 5.39x10^4 / 40S

                 = 1347.5 W

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