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Figure shows a self-acting water lifting device called a hydraulic ram . Its working is based on the phenomenon of hydraulic impact - when a liquid flowing along a tube is suddenly stopped , then there is a sharp increase in the pressure of the liquid . Its flow can be suddenly stopped , for example , by the shutting of a valve that discharges the water from the tube . A tube with a length of l= 2 m and a diameter of d = 20 cm is lowered into a stream with a current velocity of v = 4 m/s . First let valve V_(2) be open and valveV_(1) be shut . A sharp increase in pressure will cause value V_1 to open (valve V_(2) will close at the same time) and the water will flow up into vessel A . The pressure drops , value V_(1) shuts and V_(2) opens . The water in the tube assumes its course and the phenomenon is repeated in the previous sequence . Find the amount of water raised by the ram in one hour to a height of h = 30 m if each valve opens thirty times a minute. |
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Answer» Solution :Thus , on the basis of the LAW of conservation of energy , we can write `(Mv^(2))/(2) = mgh` where M is the mass of the water in the tube stopped by valve `V_(2)` and m is the mass of the water raised to the height h . Therefore , `(rho l pi d^(2))/(4) xx (v^(2))/(2) = rho V_(0) GH` where `V_(0)` is the volume of the water having mass m . The average volume raised in two seconds is `V_(0)= ( l pi d^(2) v^(2))/(8 gh) = 1.7 xx 10^(-3) m^(3)` One HOUR of ram operation will RAISE `V = 1.7 xx 10^(-3) xx 30 xx 60 ~~ 3 m^(3)` |
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